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Odisha State Board BSE Odisha 10th Class English Solutions Chapter 9 From the Formalin Jar Textbook Exercise Questions and Answers.

BSE Odisha Class 10 English Solutions Chapter 9 From the Formalin Jar

BSE Odisha 10th Class English From the Formalin Jar Text Book Questions and Answers

F. Let’s Understand The Poem:

Question 1.
What is this poem about?
(କବିତାଟି କେଉଁ ବିଷୟରେ ଲିଖ୍ ?)
Answer:
This poem is about the display (ପ୍ରଦର୍ଶନ) of Vicky’s brain, whose life was cut short for careless driving.

Question 2.
Where is Vicky’s brilliant brain?
(ଭିକିର ଚମତ୍କାର ମସ୍ତିଷ୍କ କେଉଁଠାରେ ଅଛି ?)
Answer:
Vicky’s brilliant brain is in a formalin jar.

Question 3.
Why is the brain put inside the formalin jar ?
(ମସ୍ତିଷ୍କକୁ କାହିଁକି ଫର୍ମାଲିନ୍ କାଚପାତ୍ର ବା ନଳୀ ଭିତରେ ରଖାଯାଇଛି ?)
Answer:
The brain is put inside the formalin jar as a display (ଏକ ପ୍ରଦର୍ଶନ ରୂପେ) for students to gain knowledge.

Question 4.
Who does ‘I’ refer to in the poem?
(କବିତାରେ ‘I’ କାହାକୁ ସୂଚିତ କରୁଛି ?)
Answer:
In the poem ‘I’ refers to Vicky’s brain.

Question 5.
How does the brain feel inside the jar ?
(କାଚନଳୀ ବା ପାତ୍ର ଭିତରେ ଥ‌ିବା ଭିକିର ମସ୍ତିଷ୍କ କ’ଣ ଅନୁଭବ କରୁଛି ?)
Answer:
The brain feels itself as an object of disdain or hatred (ଏକ ଘୃଣିତ ବସ୍ତୁ ରୂପେ) inside the jar.

Question 6.
Who is the brain talking about?
(ମସ୍ତିଷ୍କ କାହା ବିଷୟରେ କହୁଛି ?)
Answer:
The brain is talking about a smart and healthy youth ‘Vicky’.

Question 7.
What does the brain say about Vicky ?
(ଭିକି ବିଷୟରେ ମସ୍ତିଷ୍କ କ’ଣ କହୁଛି ?)
Answer:
The brain in the formalin jar says that Vicky was a strong, stout, careful, and brilliant boy.

Question 8.
Do you think that Vicky and the brilliant brain are two persons ? Why ?
କ’ଣ ଭାବୁଛ ଭିକି ଓ ବୁଦ୍ଧିମାନ ମସ୍ତିଷ୍କ ଦୁଇ ବ୍ୟକ୍ତି ଅଟନ୍ତି ? କାହିଁକି ?)
Answer:
No, I don’t think so. Because it is Vicky’s brain that is put separately (ଅଲଗା) in the formalin jar after Vicky’s terrible death in an accident.

Question 9.
Which were Vicky’s favorite subjects?
( ଭିକିର ପ୍ରିୟ ପାଠ୍ୟବିଷୟଗୁଡ଼ିକ କ’ଣ ଥିଲା ?)
Answer:
Vicky’s favourite subjects were computers and mathematics.

Question 10.
What was the motto of his life?
(ତା’ର ଜୀବନର ନୀତିବାକ୍ୟ କ’ଣ ଥିଲା ?)
Answer:
The motto (ନୀତି ବାକ୍ୟ) of Vicky’s life was “No pain, No gain.” (କଷ୍ଟ, ଲାଭ ନାହିଁ )

Question 11.
Was Vicky good at her studies? How do you know this?
(ଭିକି ପାଠପଢ଼ାରେ ଭଲ ଥିଲା କି ? ତୁମେ ଏହା କିପରି ଜାଣୁଛ ?)
Answer:
Yes, Vicky was good at studies. Because he had a brilliant brain. Besides he excelled (ଚମତ୍କାର ପ୍ରଦର୍ଶନ କରୁଥିଲା) in mathematics and computers.

Question 12.
Vicky loved his parents. Which line says so?
(ଭିକି ନିଜର ପିତାମାତାଙ୍କୁ ଭଲ ପାଉଥିଲା । କେଉଁ ଧାଡ଼ି ଏକଥା ଦର୍ଶାଉଛି ?)
Answer:
Vicky loved his parents. The line that says so is “Vicky’s love for his parents was truly insane.”

Question 13.
Which word says that Vicky loved his grandmother very much?
(ଭିକି ତା’ର ଜେଜେମା’କୁ ଖୁବ୍ ଭଲ ପାଉଥିଲା ବୋଲି କେଉଁ ଶବ୍ଦଟି ସୂଚିତ କରୁଛି ?)
Answer:
The word adored’ (ଆଦର କରୁଥିଲା) says that Vicky loved his grandmother very much.

Question 14.
What else, besides studies, did Vicky do during his college days?
(ପାଠପଢ଼ା ଛଡ଼ା ଭିକି ନିଜର କଲେଜ ଦିନମାନଙ୍କରେ କ’ଣ କରୁଥିଲା ?)
Answer:
Besides studies, Vicky in his college days was in love.

Question 15.
Who was Lorraine? Why did he steal a glance at her?
(ଲୋରେନ୍ କିଏ ଥୁଲା ? ସେ (ଭିକି) କାହିଁକି ତାକୁ ଲୁଚିଛପି ଚାହୁଁଥିଲା ?)
Answer:
Lorraine was Vicky’s girl classmate. He stole a glance at her as he loved her deeply.

Question 16.
Why does the poet say that only the brain could explain the cause of Vicky’s heart-beat; thud-thud?
(କବି କାହିଁକି କହିଛନ୍ତି କେବଳ ମସ୍ତିଷ୍କ ହିଁ ଭିକିର ହୃଦୟ ସ୍ପନ୍ଦନର ଭୟମିଶ୍ରିତ ଶବ୍ଦ କହିପାରିବ ?)
Answer:
The poet says that only the brain could explain the cause of Vicky’s heartbeat: thud-thud because it (the brain) is directly related to the heart.

Question 17.
Read stanza 5 and stanza 6 again. What are they about? How are they different from others?
(ପଡ୍‌ ୫ ଓ ପଡ୍‌ ୬କୁ ପୁନଶ୍ଚ ପଢ଼ । ସେଗୁଡ଼ିକ କାହା ସମ୍ପର୍କରେ ?ଏହି ଦୁଇଟି ପଡ୍‌ତ୍ତି କିଭଳି ଅନ୍ୟମାନଙ୍କଠାରୁ ଭିନ୍ନ ?)
Answer:
Stanza 5 and stanza 6 are about Vicky’s death in an accident. They are different from others as (ଯେହେତୁ) they describe how a simple mistake took Vicky’s life causing the family’s tragedy.

Question 18.
Where was Vicky riding? When and why?
(ଭିକି ଗାଡ଼ି ଚଳାଇ କେଉଁଠାକୁ ଯାଉଥିଲା ? କେତେବେଳେ ଓ କାହିଁକି ?)
Answer:
Vicky was riding to his friend’s house in the next lane on a wet rainy day to attend to an ordinary task.

Question 19.
Where was his friend’s house? Was it very far?
(ତାହାର ସାଙ୍ଗର ଘର କେଉଁଠାରେ ଥିଲା ? ଏହା କ’ଣ ବହୁତ ଦୂରରେ ଥିଲା ?)
Answer:
His friend’s house was in the next lane. No, it was not too far.

Question 20.
Why did he go to his friend?
(ସେ କାହିଁକି ତା’ର ସାଙ୍ଗ ପାଖକୁ ଯାଇଥିଲା ?)
Answer:
He (Vicky) went to his friend for some ordinary work.

Question 21.
What caused the accident? Which words in the poem describe the accident?
(କେଉଁଥ‌ିପାଇଁ | କାହିଁକି ଦୁର୍ଘଟଣା ଘଟିଲା ? କବିତାରେ କେଉଁ ଶବ୍ଦସବୁ ଦୁର୍ଘଟଣା ବିଷୟରେ ବର୍ଣ୍ଣନା କରୁଛି ?)
Answer:
Going to his friend’s house without wearing his helmet caused Vicky’s accident and tragic death. The words which describe the accident are “He met with an accident gory and inhumane.”

Question 22.
What meaning does the line – ‘For once, from wearing his helmet he did refrain convey? Was it Vicky’s habit to wear his helmet while riding his bike?
(ଧାଡ଼ି “For once, from wearing his helmet he did refrain’ କେଉଁ ଅର୍ଥ ପ୍ରଦାନ କରୁଛି ? ନିଜର ବାଇକ୍ ଚଳାଇଲାବେଳେ ହେଲମେଟ୍ ପିନ୍ଧିବା ଭିକିର ଏକ ଅଭ୍ୟାସ ଥିଲା କି ?)
Answer:
The line “For once, from wearing his helmet he did refrain” conveys (implies) Vicky’s not wearing his helmet for the first time in his life. Yes, it was Vicky’s habit to wear his helmet while riding his bike.

Question 23.
How did the accident affect Vicky and his family?
(ଦୁର୍ଘଟଣା କିପରି ଭିକି ଓ ତା’ର ପରିବାରକୁ ପ୍ରଭାବିତ କରିଥିଲା ?)
Answer:
Vicky lost his life in the accident and his gruesome (ଶୋଚନୀୟ ) death in the accident plunged (ବୁଡ଼ାଇ ଦେଇଥିଲା) his family into huge shock (ପ୍ରଚଣ୍ଡ ଦୁଃଖରେ ).

Question 24.
What message does the poet have for the young generation through this poem?
(ଏହି କବିତା ମାଧ୍ୟମରେ କବି ଯୁବ ପିଢ଼ିପାଇଁ କେଉଁ ସନ୍ଦେଶ ବା ବାର୍ତ୍ତା ଦେଇଛନ୍ତି ?)
Answer:
In this poem, the poet’s message is clear and serious. She warns (ସତର୍କ କରାଇ ଦେଇଛନ୍ତି ) the younger generation against risky (dangerous) driving.

Question 25.
Can you say why such dreadful accidents occur on road and cause death every moment?
(ତୁମେ କ’ଣ କହିପାରିବ କାହିଁକି ପ୍ରତ୍ୟେକ ମୁହୂର୍ତ୍ତରେ ରାସ୍ତାରେ ଏଭଳି ଭୟଙ୍କର ଦୁର୍ଘଟଣା ଘଟୁଛି ଓ ଜୀବନହାନି ଘଟାଉଛି ? )
Answer:
Such (ଏଭଳି) dreadful (ଭୟଙ୍କର) accidents occur (ଘଟୁଛି) on road and cause (ଘଟାଉଛି ) death every moment due to careless and fast driving and also driving without helmets.

Question 26.
Suggest some ways to reduce Road Traffic Injuries (RTI)?
(ରାସ୍ତା ଗହଳିଜନିତ ମୃତାହତ ସଂଖ୍ୟା କମାଇବାର କେତୋଟି ଉପାୟର ପ୍ରସ୍ତାବ ଦିଅ ।)
Answer:
Rules like (ଭଳି) wearing helmets, safe driving, avoiding (ଦୂରେଇ ରହିବା ) talking on mobiles, refraining (ଦୂରେଇ ରହିବା) from listening to music on mobiles, riding on the right route, etc. can help reduce Road Traffic Injuries.

G. Let’s Appreciate The Poem:

Question 1.
Who donated Vicky’s brain to the Medical College? How is it preserved and used?
(କିଏ ଭିକିର ମସ୍ତିଷ୍କକୁ ମେଡ଼ିକାଲ କଲେଜକୁ ଦାନ ଦେଇଥିଲେ ? ଏହାକୁ କିଭଳି ସଂରକ୍ଷିତ କରି ରଖାଯାଇଛି ଓ ବ୍ୟବହାର କରାଯାଉଛି ?)
Answer:
Vicky’s parents donated Vicky’s brain to the Medical College. It is preserved in a formalin jar on display to promote knowledge gain to the medical students.

Question 2.
What does the line ‘On display to promote knowledge gain’ express?
(ଧାଡ଼ି ‘On display to promote knowledge କ’ଣ ସୂଚିତ କରୁଛି ?)
Answer:
The line “On display to promote knowledge gain” suggests that from the display (ପ୍ରଦର୍ଶନରୁ)of Vicky’s (human) brain the students gain more knowledge about its construction and functions.

Question 3.
What is called a specimen? Why is the brain preserved as a specimen?
(Specimen କ’ଣ ? ମସ୍ତିଷ୍କକୁ ଏକ ନମୁନା ରୂପେ କାହିଁକି ରଖାଯାଇଛି ?)
Answer:
A specimen is a model. The brain is preserved as a specimen to provide more knowledge to the students about this useful organ of the human body.

Question 4.
How was the brilliant brain a part of a living human two years ago?
(କିଭଳିୁ ଇ ବର୍ଷ ପୂର୍ବେ ବୁଦ୍ଧିମାନ ମସ୍ତିଷ୍କ ଏକ ଜୀବନ୍ତ ମାନବର ଗୋଟିଏ ଅଂଶ | ଅଙ୍ଗ ଥିଲା ?)
Answer:
Two years ago the brilliant brain, now kept in the formalin jar, was possessed by a handsome (ସୁନ୍ଦର) and strong youth named Vicky.

Question 5.
What does the phrase – ‘Yet ended up in this jar’ mean to you?
(ବାକ୍ୟାଶ ‘Yet ended up in this jar ‘ର ଅର୍ଥ ତୁମେ କ’ଣ ବୁଝୁଛ ?)
Answer:
The phrase “Yet ended up in this jar” means at last Vicky’s brain rested in the formalin jar though it has no stroke, tumor, or bugs within it.

Question 6.
Which expression suggests that Vicky was hard-working?
(କେଉଁ ଉକ୍ତି ସୂଚିତ କରୁଛି ଯେ ଭିକି ପରିଶ୍ରମୀ ଥିଲା ?)
Answer:
The expression which suggests (ସୂଚିତ କରୁଛି) that Vicky was hard working is “Full of life and vigor, sun, wind or rain”.

Question 7.
Why does the poet say? “Only I could explain”?
(କବି କାହିଁକି କହିଛନ୍ତି ‘କେବଳ ମୁଁ ହିଁ ବୁଝାଇପାରିବି? ?)
Answer:
The poet says. “Only I could explain” because it is Vicky’s brain that can say why his heart got ‘thuds’

Question 8.
Should we consider such demise ‘h chance or by choice’?
(ଭିକିର ଏଭଳି ମୃତ୍ୟୁକୁ ଆମ୍ଭେମାନେ ପରିସ୍ଥିତି ନା ଇଚ୍ଛାସୃଷ୍ଟ ବୋଲି କହିବା ?)
Answer:
We should consider such demise (death of Vicky ‘by chance’.

Question 9.
The poet uses flashback, which Is often used to recount events that had happened before the story started. Which stanzas talk about the past events and which ones the present incident ? (ଗଳ୍ପ ଆରମ୍ଭ ପୂର୍ବରୁ କବି ଅତୀତ ଘଟଣା ଯାହାକି ପୂର୍ବରୁ ଘଟିଛି ତା’ର ସ୍ମୃତିଚାରଣ କରିଛନ୍ତି । କବିତାରେ କେଉଁ ପଙ୍‌କ୍ତିଗୁଡ଼ିକ ଅତୀତ ଘଟଣା ଓ କେଉଁ ପଗୁଡ଼ିକ ବର୍ତ୍ତମାନର ଘଟଣାକୁ ବ୍ୟକ୍ତ କରୁଛି ?)
Answer:
The poet uses flashback, which is often used to recount events that had happened before the story started. The stanzas stating past events are:
“On a wet monsoon day. for a task mundane
Vicky rode his bike to his friend in the next lane;
For once, from wearing his helmet he did refrain
God! He met with an accident gory and inhumane.” (Stanza — 5)
and
“All it look was a stray moment inane
A young life lost, a family crushed with pain;
Dear friends, take care; risky driving can be the bane
DRIVE SAFE — let your precious life not be in vain! (Stanza — 6)
Similarly stanza — 3 and stanza — 4 also talk about past events.
The stanza stating present events is:
“Hi ! ¡ am Vicky’s brilliant brain
Sitting in a tourmaline jar with disdain:
On display to promote knowledge gain
Watching people stare at new again and again. (Stanza – I)
Similarly stanza — 2 also talks about the present incident.

Question 10.
Find the rhyming words in the poem.
(କବିତାରେ ଯତିପାତ ଶବ୍ଦସବୁ (ସମାନ ଉଚ୍ଚାରିତ) ଖୋଜି ବାହାର କର ।)
Answer:
The rhyming words in the poem are (1st stanza) brain — disdain, gain — again. (2nd stanza) specimen — human, within — chagrin. (3rd stanza) sane — rain, domain — gain. 4th stanza insane cane. Lorraine — explain, (5th stanza) mundane — lane. refrain — inhumane. (6th stanza) inane pain, banc — vain.

Question 11.
Why does the poet use capital letters for the first two words in the last line of the poem?
(କବିତାର ଶେଷ ଧାଡ଼ିରେ ପ୍ରଥମ ଦୁଇଟି ଶବ୍ଦ ପାଇଁ କବି କାହିଁକି ବଡ଼ ଅକ୍ଷର ବ୍ୟବହାର କରିଛନ୍ତି ?)
Answer:
The poet uses capital letters for the first two words in the last line of the poem (DRIVE SAFE) to stress upon (ଜୋର ଦେବା ପାଇଁ) ‘safe driving.

Question 12.
You read and understood the poem. Do you like the title of the poem? Why/ Why not? Can you suggest another title for the poem?
(ତୁମେ କବିତାଟି ପଢ଼ିସାରିଛ ଓ ବୁଝିସାରିଛ । କବିତାର ଶୀର୍ଷକ (ନାମ) ତୁମକୁ ଭଲ ଲାଗୁଛି କି ? କାହିଁକି/କାହିଁକି ନୁହେଁ ? ତୁମେ କବିତା ପାଇଁ ଅନ୍ୟ ଏକ ଶୀର୍ଷକ ଦେଇପାରିବ କି ?)
Answer:
I like the title (name) of the poem. Because the formalin jar inside which Vicky’s brain is preserved gives a strong message (ବାର୍ତ୍ତା) how the jar will keep waiting (ଅପେକ୍ଷା କରି ରହିବ) for a careless boy like Vicky who suffered a premature death (ଏକ ଅକାଳ ମୃତ୍ୟୁର ଶିକାର ହୋଇଥିଲେ)without wearing a helmet.

H. Let’s Listen And Speak:

(a) Listen to the following sentences about the poem (Your teacher reads the sentences aloud.) and say whether they are right or wrong. Then correct the sentences in case you find them wrong. (କବିତା ସମ୍ପର୍କରେ ନିମ୍ନଲିଖ୍ ବାକ୍ୟଗୁଡ଼ିକୁ ଶୁଣ (ତୁମ ଶିକ୍ଷକ ବାକ୍ୟଗୁଡ଼ିକୁ ଉଚ୍ଚ ସ୍ୱରରେ ପଢ଼ିବେ) ଏବଂ କୁହ ସେଗୁଡ଼ିକ ଠିକ୍ ନା ଭୁଲ୍ । ତା’ପରେ ଯେଉଁଗୁଡ଼ିକୁ ତୁମେ ଭୁଲ୍ ଦେଖୁଛ ଠିକ୍ କର ।)

(i) Vicky’s brain was sitting in the formalin jar with pleasure and respect.
(ii) The brain didn’t like the people staring at him.
(iii) The brain felt insulted to be a specimen for others.
(iv) Vicky was a strong and stout young man.
(v) He was good at math and computer.
(vi) He believed in taking the pain to succeed in life.
(vii) He liked neither his grandmother nor her wrinkles and stick.
(viii) All his friends in the college disliked him.
(ix) He rode to his friend on a monsoon day for important work.
(x) His friend’s house was very far from his place.
(xi) He usually liked to wear helmets.
(xii) He was killed in a serious accident.
(xiii) He was clever, handsome, and careful.
Answer:
(i) Wrong
Right: Vicky’s brain was sitting in the formalin jar with disdain.
(ii) Right
(iii) Right
(iv) Right
(v) Right.
(vi) Right
(vii) Wrong
Right: He liked both his grandmother and her wrinkles and the stick.
(viii) Wrong
Right: All his friends in the college liked him.
(ix) Wrong
Right: He rode to his friend on a monsoon day for ordinary work.
(x) Wrong
Right: His friend’s house, was very near to his place.
(xi) Right
(xii) Right
(xiii) Right

(b) Pronounce the following words correctly.(ନିମ୍ନଲିଖ୍ ଶବ୍ଦଗୁଡ଼ିକୁ ଠିକ୍ ଭାବରେ ଉଚ୍ଚାରଣ କର ।)
disdain, gain, again, human, sane, rain, insane, cane, explain, mundane, refrain, inhumane, pain, bane, in vain
Answer:
disdain — ଅବମାନନା |
gain — ଗେନ୍
again — ଏଗେନ୍
human — ହ୍ୟୁମାନ୍
sane — ସାନ
rain — କେନ୍
insane — ପାଗଳ |
cane — କେନ୍
explain — ଏକ୍ସପେନ୍
mundane — ସାଂପ୍ରତିକ
refrain — ରିଫ୍ରେନ୍
inhumane — ଇହ୍ୟୁମେନ୍
pain — ସେନ୍
bane —  ବେନ୍
in vain — ଇନ୍ ଭେନ୍

(c) Imagine that Vicky’s friend, Akash visited the Niedleal College, where Vicky’s brain was preserved in a formalin Jar. Given below Is a dialogue between Vicky’s brain and his friend, Akash. Practise the dialogue. (ମନେକର ଭିକିର ସାଙ୍ଗ ଆକାଶ ମେଡ଼ିକାଲ କଲେଜ ପରିଦର୍ଶନରେ ଯାଇଥିଲେ, ଯେଉଁଠାରେ ଭିକିର ମସ୍ତିଷ୍କ ଗୋଟିଏ ଫର୍ମାଲିନ୍ ଜାର୍‌ରେ ସଂରକ୍ଷଣ କରି ରଖାଯାଇଥିଲା । ନିମ୍ନରେ ଭିକିର ମସ୍ତିଷ୍କ ଓ ଆକାଶ ମଧ୍ଯରେ ହୋଇଥ‌ିବା କଥାବାର୍ତ୍ତା ଦିଆଯାଇଛି । ସଂଳାପଗୁଡ଼ିକୁ ଅଭ୍ୟାସ କର ।)
Brain       : Hi Friend ! How’re you?
Akash      : Good. Who’s speaking?
Brain      : I’m Vicky’s brain.
Akash     : Where you’re?
Brain      : ln the formalin jar
Akash     : Who put you here?
Brain      : The doctor.
Akash    : What for?
Brain      : For display! Student’. will gain knowledge.
Akash    : How did the doctor find you?
Brain     : Listen, Vicky was a young college boy. He was strong, stout, careful, and brilliant. One monsoon day he was riding without wearing the helmet to his friend who lived the next lane. On the way, he was killed in an accident. The doctor parted me from his body and put it here for his students to gain knowledge.
Akash   : How sad!
Brain     : Dear Friend, Vicky was a brilliant boy, but on that fateful day he took it casually to wear his helmet as he was to ride to his friend who was staying the next lane. For his carelessness, a precious life was lost. So please be careful and always DRIVE SAFE on road. Don’t lose your life and put your family in sorrow and suffering.
Akash    : Thank you Friend for your advice. I’m very sorry for Vicky, my fast friend, bye, see you!

I. Lets Read And Write:
(a) Given below is a table showing a chart of penalties for the road offences. Read the table and write a sentence for each. (ରାସ୍ତା ନିୟମରେ ଅବମାନନା ନିମନ୍ତେ ଦଣ୍ଡ ବା ଜରିମାନା ପରିମାଣ ଦର୍ଶାଉଥ‌ିବା ଏକ ସାରଣୀ ନିମ୍ନରେ ପ୍ରଦତ୍ତ ହୋଇଛି । ସାରଣୀଟିକୁ ପଢ଼ ଏବଂ ପ୍ରତ୍ୟେକ ପାଇଁ ବାକ୍ୟଟିଏ ଲେଖ ।)

SI. No.	Offence	Penalty (Rs.) (Minimum Amount)
1	General	500
2	Rules of road regulation violation	500
3	Traveling without ticket	500
4	Unauthorized use without license	5,000
5	Driving without licence	5,000
6	Disobedience of orders of authorities	2,000
7	Drink and Drive	10,000
8	Speeding or Racing	5,000
9	Vehicle without permit	10,000
10	Driving without qualification	10,000
11	Without Seat belt	1,000
12	Without helmet	1,000
13	Oversized vehicles	5,000
14	Not providing way for emergency vehicle	10,000
15	Over speeding	1,000/2,000
16	Driving without insurance	2,000
17	Dangerous driving penalty	5,000
18	Offences by juveniles	25,000
19	Overloading of passengers	1,000 / 1 passenger
20	Overloading of two-wheelers	2,000

Answer:
1. As per (ଅନୁସାରେ ) the Motor Vehicle Act, for general violation, you will have to pay (ଦେବାକୁ ପଡ଼ିବ) a fine of Rs. 500/-.
2. As per the Motor Vehicle Act, for the rules of road regulation violation, you will have to pay a fine of Rs. 500/-.
3. As per Motor Vehicle Act, if you travel without a ticket, you will have to pay a fine of Rs.500/-.
4. As per Motor Vehicle Act, for unauthorized use (ଅନଷ୍କୃତ ବ୍ୟବହାର) without a license, you will have to pay a fine of Rs. 5,000/-.
5. As per Motor Vehicle Act, if you are driving without license, you have to pay a fine of Rs. 5,000/-.
6. As per the Motor Vehicle Act, for disobedience (ଉଲ୍ଲଙ୍ଘନ) of orders of authorities, you will have to pay a fine of Rs. 2,000/-.
7. As per Motor Vehicle Act, if you drink and drive on the road, you will have to pay a fine of Rs. 10,000/-.
8. As per Motor Vehicle Act, for speeding or racing, you will have to pay a fine of Rs. 5,000/-.
9. As per Motor Vehicle Act, for driving a vehicle without a permit, you will have to pay a fine of Rs. 10,000/-.
10. As per Motor Vehicle Act, if you drive without qualification, you will have to pay a fine of Rs. 10,000/-.
11. As per Motor Vehicle Act, if you drive without a seat belt, you will have to pay a fine of Rs. 1,000/-.
12. As per Motor Vehicle Act, if you ride without a helmet, you will have to pay a fine of Rs. 1,000/-. –
13. As per Motor Vehicle Act, if you ride an oversized vehicle, you will have to pay a fine of Rs. 5,000/-.
14. As per the Motor Vehicle Act, if you don’t provide a way (ରାସ୍ତା/ବାଟ ନ ଦିଅ) for an emergency vehicle, you will have to pay a fine of Rs. 10,000/-.
15. As per Motor Vehicle Act, for overspeeding, you will have to pay a fine of Rs. 1,000/- or Rs. 2,000/-.
16. As per Motor Vehicle Act, if you drive/ride without insurance, you will have to pay a fine of Rs. 2,000/-.
17. As per Motor Vehicle Act, if you drive dangerously, you will have to pay a fine of Rs. 5,000/-.
18. As per Motor Vehicle Act, if you are a juvenile (ନାବାଳକ) and found riding, you will have to pay a fine of Rs. 25,000/-.
19. As per Motor Vehicle Act, if you are found to have the offence (ଅପରାଧ) of overloading (ଆବଶ୍ୟକତାରୁ ଅଧ‌ିକ ଯାତ୍ରୀ ପରିବହନ), you will have to pay a fine of Rs. 1,000/- for each passenger.
20. As per Motor Vehicle Act, for overloading on two-wheelers, you will have to pay a fine of Rs. 2,000/-.

(b) Read the traffic symbols and write one sentence for each. The first one is done for you.
(ଟ୍ରାଫିକ୍ ସଙ୍କେତ ପଢ଼ ଓ ପ୍ରତ୍ୟେକ ପାଇଁ ଗୋଟିଏ ଲେଖାଏଁ ବାକ୍ୟ ଲେଖ । ପ୍ରଥମଟି ତୁମ ପାଇଁ କରି ଦିଆଯାଇଛି ।)

Answer:
Example :
1. The first symbol indicates that there is a right-hand curve ahead.
2. The second symbol indicates that there is a left-hand curve ahead.
3. The third symbol indicates that there is a right hairpin bend ahead.
4. The fourth symbol indicates that there is a left hairpin bend ahead.
5. The fifth symbol indicates that there is a right reverse bend ahead.
6. The sixth symbol indicates that there is a left reverse bend ahead.
7. The seventh symbol indicates that there is a steep ascent ahead,
8. The eighth symbol indicates that there is a steep descent ahead.
9. The ninth symbol indicates that there is a narrow road ahead.
lO. The tenth symbol indicates that there is a road wideness ahead.
Il. The eleventh symbol indicates that there is a narrow pass ahead.
12. The twelfth symbol indicates that the road is slippery ahead.
13. The thirteenth symbol indicates that there is a loose grovel ahead.
14. The fourteenth symbol indicates that there is a cycle crossing ahead.
15. The fifteenth symbol indicates that there is a ¿ebra crossing for pedestrians ahead.
16. The sixteenth symbol indicates stopping the running vehicle.
17. The seventeenth symbol indicates giving way to other vehicles.
18. The eighteenth symbol indicates that the vehicles can go in one way only.
19. The nineteenth symbol indicates no entry of vehicles there.
20. The twentieth symbol indicates that the vehicles can go in one way only.
21. The twenty-first symbol indicates that vehicles can go in both directions but one way only.
22. The twenty-second symbol indicates that a right turn is prohibited here.
23. The twenty-third symbol indicates that a left turn is prohibited here.
24. The twenty-fourth symbol indicates a one-way pass to vehicles.
25. The twenty-fifth symbol indicates that a U-turn is prohibited here.
26. The twenty-sixth symbol indicates that overtaking is prohibited here.
27. The twenty-seventh symbol indicates that blowing horns are prohibited here.
28. The twenty-eighth symbol indicates that the speed limit is 65 km./h here.
29. The twenty-ninth symbol indicates that the vehicles must turn to the left here.
30. The thirtieth symbol indicates that the vehicles must go ahead here.
31. The thirty-first symbol indicates that the vehicles must turn right ahead.
32. The thirty-second symbol indicates that the vehicles must go ahead or turn right here.
33. The thirty-third symbol indicates that the vehicles must go ahead or turn to the left here.
34. The thirty-fourth symbol indicates that the vehicles must keep to the left here.
35. The thirty-fifth symbol indicates that the vehicles must blow their horns here.

(c) Imagine that you are a reporter of ‘The Times of India’ In Bhubaneswar. Write a report based on the incident/contents of the poem From the Formalin Jar’ for the newspaper. Suggest some ways to check road accidents, and advise young children to be careful about the trame rules. (ମନେକର ତୁମେ ‘The Times of India’ ଭୁବନେଶ୍ଵରର ଜଣେ ସାମ୍ବାଦିକ । ‘From the Formalin Jar’ କବିତାର ବିଷୟବସ୍ତୁ| ଘଟଣାବଳୀକୁ ଆଧାର କରି ଏକ ବିବରଣୀ ଲେଖ । ସଡ଼କ ଦୁର୍ଘଟଣା ନିୟନ୍ତ୍ରଣ ପାଇଁ କେତେକ ପ୍ରସ୍ତାବ ଦିଅ ଏବଂ ଯୁବକ ଯୁବତୀମାନଙ୍କୁ ରାସ୍ତାର ନିୟମ ପ୍ରତି ଯତ୍ନଶୀଳ ହେବାକୁ ଉପଦେଶ ଦିଅ ।)
Answer:

Date:………………
Bhubaneswar

To
The Editor
The Times of India
Bhubaneswar

A TRAGIC ROAD ACCIDENT

Saheed Nagar. Bhubaneswar, 3rd June: It was a wet monsoon day yet there was heavy traffic on the roads. Vicky, a young smart boy, rode his bike to his friend’s house in the next lane. Vicky was very sincere, but that day he did not wear his helmet for the first time. To his bad luck, a truck coming from Acharya Vihar at a high speed hit his bike and he got a severe head injury and died on the spot. The police rushed to the spot and send the body to the hospital. This tragic road accident happened due to the high speed of the truck and the carelessness of the boy. If he wore a helmet, he would not have lost his life. So young children should be careful about the traffic rules while riding on the road. They should obey traffic rules like wearing helmets, and driving a considerable speed. avoid talking on mobiles ut listening to music etc.

Rajendra,
the reporter.

(d) Work in groups of four and prepare placards/posters on ROAD SAFETY. Display them near the school notice board. (Your teacher will guide you.)
(ଚାରିଜଣିଆ ଦଳ ହୋଇ ‘ସଡ଼କ ନିରାପତ୍ତା’ ବିଷୟରେ ପ୍ଲାକାର୍ଡ|ପୋଷ୍ଟର ତିଆରି କର । ବିଦ୍ୟାଳୟର ନୋଟିସ୍ ବୋର୍ଡ ପାଖରେ ସେଗୁଡ଼ିକୁ ପ୍ରଦର୍ଶନ କର । (ତୁମ ଶିକ୍ଷକ ତୁମକୁ ମାର୍ଗ ଦର୍ଶନ କରାଇବେ ।))

BSE Odisha 10th Class English From the Formalin Jar Important Questions and Answers

Very Short A Objective Questions With Answers
Answer The Following Questions In A Word Or A Phrase.

Question 1.
Whose brilliant brain is in the jar?
Answer:
Vicky’s

Question 2.
Who is Vicky?
Answer:
a strong and stout (ହୃଷ୍ଟପୁଷ୍ଟ ) youth

Question 3.
How is Vicky’s brain sitting?
Answer:
with disdain

Question 4.
Why is Vicky’s brain kept in the jar?
Answer:
to promote knowledge gain

Question 5.
What knowledge should pupils grow?
Answer:
about the functions of the brain

Question 6.
How do people treat the brain on the formalin jar?
Answer:
staring at it

Question 7.
What is Vicky’s brain like?
Answer:
brilliant

Question 8.
What is the brain kept for?
Answer:
as a specimen

Question 9.
What does the brain feel to be addressed (ସମ୍ବୋଲ୍ଡ କରାଯିବାରୁ) a specimen (ନମୁନା) ?
Answer:
insulting

Question 10.
How many years passed since Vicky died?
Answer:
two years

Question 11.
What sort of boy was Vicky?
Answer:
smart, vigorous, and healthy

Question 12.
What was Vicky’s domain?
Answer:
computers and maths

Question 13.
Whom did Vicky adore?
Answer:
his grandmother

Question 14.
“Only I could explain.’ Here for who does 1’ stand for?
Answer:
Vicky’s brain

Question 15.
What did Vicky refrain from one day?
Answer:
wearing his helmet

Fill In The Blanks With Right Words.

1. Vicky’s brain was in _____________.
Answer:
a formalin jar

2. _____________was insulting.
Answer:
Vicky’s brain being kept as a specimen

3. The expression ‘sun, wind or rain’ means _____________.
Answer:
all times

4. Vicky’s motto was _____________.
Answer:
‘No pain. No gain’

5. Vicky’s love for _____________ was true and great.
Answer:
his parents

6. In college Vicky glanced (looked) secretly at _____________.
Answer:
Lorraine

7. Vicky’s heart thudded at _____________.
Answer:
the sight of Lorraine

8. Vicky went to his friend’s house on _____________.
Answer:
a wet rainy day

9. The word ‘mundane’ means _____________.
Answer:
ordinary

10. Vicky went to his friend on _____________.
Answer:
a bike

11. Vicky’s accident was _____________.
Answer:
gory and inhumane

12. At Vicky’s death his family _____________ with pain.
Answer:
crushed

13. _____________ should be given importance to save life.
Answer:
Drive Safe

14. _____________ can be banc (cause tragedy).
Answer:
Risky driving

15. In the phrase ‘a stray moment’ the word ‘stray’ means _____________.
Answer:
painful or inattentive

Multiple Choice Questions (Mcqs) With Answers.
Pick out the correct alternative.

Question 1.
Vicky’s brilliant brain is sifting in a formalin jar with _____________.
(A) anguish
(B) astonishment
(C) disdain
(D) hatred
Answer:
(C) disdain

Question 2.
Vicky’s brilliant brain in the jar feels/felt _____________.
(A) disgusted
(B) anguished
(C) insulted
(D) pleasure
Answer:
(C) insulted

Question 3.
Vicky was full of life and vigor. This means Vicky was lively, strong, and _____________.
(A) stout
(B) healthy
(C) vigourous
(D) courigcous
Answer:
(A) stout

Question 4.
Vicky loved his _____________ and grandmother.
(A) grandfather
(B) parents
(C) mother
(D) sister
Answer:
(B) parents

Question 5.
Vicky’s grandmother has/had _____________ on her body.
(A) red spots
(B) blackspots
(C) wrinkles
(D) stripes
Answer:
(C) wrinkles

Question 6.
_____________ was Vicky’s girlfriend.
(A) Norraine
(B) Florraine
(C) Glorraine
(D) Lorraine
Answer:
(D) Lorraine

Question 7.
One day Vicky set out for his friend’s house without wearing his _____________.
(A) spectacles
(B) wristwatch
(C) helmet
(D) forest dress
Answer:
(C) helmet

Question 8.
Vicky met with an accident gory and _____________.
(A) bloody
(B) inhumane
(C) bane
(D) pathetic
Answer:
(B) inhumane

Question 9.
Vicky’s heart went _____________ when he stole a glance at Lorraine.
(A) thud-thud
(B) dhak-dhak
(C) hit-bit
(D) sim-sim
Answer:
(A) thud-thud

Question 10.
According to the poet, risky driving can be _____________.
(A) bane
(B) inane
(C) crushed
(D) inhumane
Answer:
(A) bane

Question 11.
Precious (Valuable) life can’t go wasted if we take up _____________.
(A) risky drive
(B) safe drive
(C) slow drive
(D) speed drive
Answer:
(B) safe drive

Question 12.
“Only I could explain.” Here ‘I’ stands for _____________.
(A) Vicky
(B) Vicky’s friend
(C) Vicky’s brain
(D) Vicky’s father
Answer:
(C) Vicky’s brain

From the Formalin Jar Summary in English

Lead-In:
The poem ‘From the Formalin Jar’ was written in September 2013 to create awareness (ସଚେତନତା ସୃଷ୍ଟି କରିବା ପାଇଁ) about ‘Safe Driving’ (ନିରାପଦ ଗାଡ଼ିଚାଳନା) and other road safety rules in the community and published (ପ୍ରକାଶିତ ହୋଇଥିଲା।)online at youthspring net, a forum for nurturing youth well-being ( ଯୁବକଗୋଷ୍ଠୀର ମଙ୍ଗଳ ନିମନ୍ତେ ପ୍ରତିପୋଷଣ ପାଇଁ). This heart-breaking (ହୃଦୟ ବିଦାରକ) poem conveys (ପ୍ରତିଫଳିତ କରୁଛି) how a moment of simple carelessness (ସାମାନ୍ୟ ଯତ୍ନହୀନତାର ଏକ କ୍ଷଣ) cost (ମୂଲ୍ୟ ଦେବାକୁ ହେଲା) a young, lively and caring boy his precious life. In this poem “Vicky” is a fictional (କାଳ୍ପନିକ) character (ଚରିତ୍ର) created by the poet to warn people against breaking rules of road.

Stanzawise Explanation:
Stanza 1 (Lines 1 to 4)
Hi! I am Vicky’s brilliant brain
Sitting in a formalin jar with disdain;
On display to promote knowledge gain
Watching people stare at me again and again.

Gist: Hi! I am Vicky’s brilliant or sharpest brain. I am sitting in this formalin jar with strong contempt (hatred) preserved. I have been kept here by the doctor for the growth or promotion of knowledge of students. I am feeling insulted to be stared at me again and again.
ଅନୁବାଦ : ହେ ! ମୁଁ ହେଉଛି ଭିକିର ଚମତ୍କାର ମସ୍ତିଷ୍କ । ତୀବ୍ର ଘୃଣାପୂର୍ବକ ମୁଁ ଏକ ଫର୍ମାଲିନ୍ ଜାର୍ ବା କାଚପାତ୍ରରେ ସୁରକ୍ଷିତ ଭାବରେ ବସିଛି । ଛାତ୍ରଛାତ୍ରୀମାନଙ୍କର ଜ୍ଞାନ ବୃଦ୍ଧି ନିମିତ୍ତ ମୋତେ ଏଠାରେ ଡାକ୍ତରଙ୍କଦ୍ୱାରା ରଖାଯାଇଛି । ବାରମ୍ବାର ଲୋକମାନେ ମୋତେ ଏକ ଲୟରେ ଚାହିଁ ରହିବାକୁ ମୁଁ ନିଜକୁ ଅପମାନିତ ମନେ କରୁଛି !

Stanza – 2 (Lines 5-8)
How insulting to be called a ‘specimen’
Two years ago I was part of a living human;
I got no stroke, tumor or bugs within
Yet ended up in this jar, to my chagrin.

Gist: Really I feel greatly insulted by being called a specimen (example or model). Two years ago I (the brain) was a part of a living human. Fortunately, I had no stroke, tumor or bug. Yet I had to remain in the jar thanks to anger.
ଅନୁବାଦ : ସତରେ ଏହି କାଚପାତ୍ରରେ ଲୋକମାନେ ମୋତେ ଏକ ନମୁନା ରୂପେ ବିବେଚିତ କରିବା ମୋତେ ବହୁତ ଅପମାନିତ ମନେହେଉଛି । ଦୁଇ ବର୍ଷ ପୂର୍ବେ ମୁଁ (ମସ୍ତିଷ୍କ) ଏକ ଜୀବନ୍ତ ମାନବର ଏକ ଅଂଶ ଥିଲା । ସୌଭାଗ୍ୟବଶତଃ ମୁଁ ମସ୍ତିଷ୍କାଘାତ ବା ଟ୍ୟୁମର ବା ସଂକ୍ରାମକ ରୋଗରେ ଆକ୍ରାନ୍ତ ନ ଥିଲି, ତଥାପି ମୋର ବିଷାଦ ବା କ୍ରୋଧର କାରଣ ହେଉଛି ଯେ ମୋତେ ଏହି କାଚ ଜାର୍‌ରେ ଶେଷରେ ରହିବାକୁ ପଡ଼ିଲା ।

Stanza – 3 (Lines 9 – 12)
Vicky was a young boy, smart and sane
Full of life and vigor, sun, wind or rain;
Computers and maths were his domain
His motto in life was ‘No pain, No gain ’.

Gist: Vicky was a smart and healthy youth. He always looked vigorous and lively, no matter what happened. He was strong and an expert in computers and mathematics. His motto or moral principle in life was one must work hard to get success in life.
ଅନୁବାଦ : ଭିକି ଥୁଲା ଜଣେ ବୁଦ୍ଧିମାନ ଓ ସୁସ୍ଥ ଯୁବକ । ଯାହା ଘଟୁ ପଛେ ସେ ସର୍ବଦା କର୍ମଠ ଓ ପ୍ରାଣବନ୍ତ ଦେଖା ଯାଉଥିଲା । କମ୍ପ୍ୟୁଟର ଓ ଗଣିତରେ ସେ କୁଶଳୀ ଓଁ ଦକ୍ଷ ଥିଲା । ତା’ ଜୀବନର ନୀତି ବା ଆଦର୍ଶ ଥିଲା ଯେ ଜୀବନରେ ସଫଳତା ଲାଭ କରିବାପାଇଁ ଜଣଙ୍କୁ ପରିଶ୍ରମ କରିବାକୁ ପଡ଼ିବ ।

Stanza – 4 (Lines 13 – 16)
Vicky’s love for his parents was truly insane
And he adored his gran ’ma, wrinkles, and cane;
In college, when he stole a glance at Lorraine
Why his heart went thud-thud, only I could explain!

Gist: Vicky loved his parents immensely (very much). Besides, he (Vicky) adored (loved strongly) his pouchy (having lines in the body due to ripe old age) grandmother with a walking stick. While in college he had a quick and quiet look at his girlfriend Lorraine and his heart in panic (fear) would thud.
ଅନୁବାଦ : ଭିକି ତା’ରି ପିତାମାତାଙ୍କୁ ପ୍ରକୃତରେ ଖୁବ୍ ଭଲ ପାଉଥିଲା । ଏହାଛଡ଼ା ଆଶାବାଡ଼ି ସାହାଯ୍ୟରେ ଚାଲୁଥିବା କୁଞ୍ଚୁ ଚର୍ମଧାରୀ ନିଜର ଜେଜେମା’ଙ୍କୁ ମଧ୍ୟ ଖୁବ୍ ଭଲ ପାଉଥିଲା । କଲେଜରେ ଭିକି ଯେତେବେଳେ ତାହାର ସହପାଠିନୀ (ଲୋରେନ୍) ଉପରେ କ୍ଷିପ୍ର ଓ ଚୋରା ଚାହାଣି ପକାଉଥିଲା, ଅଜଣା ଭୟରେ ତା’ର ହୃଦୟରେ କମ୍ପନ ଜାତ ହେଉଥିଲା ।

Stanza – 5 (Lines 17 – 20)
On a wet monsoon day, for a task mundane
Vicky rode his bike to his friend in the next lane;
For once, from wearing his helmet he did refrain
God! He met with an accident gory and inhumane.

Gist: On a rainy day Vicky rode his bike to his friend living in the next lane to attend to an ordinary task. This was the first time in his life Vicky went out without a helmet. O, God! On the way, he faced a bloody and barbaric accident.
ଅନୁବାଦ : ଏକ ବର୍ଷଣମୁଖର ଦିନରେ ପରବର୍ତ୍ତୀ ଗଳିରେ ବାସ କରୁଥିବା ନିଜର ଜଣେ ବନ୍ଧୁ ଘରକୁ ଭିକି ଏକ ସାଧାରଣ କାର୍ଯ୍ୟରେ ବାଇକ୍ ଚଳାଇ ଯାଇଥିଲା । ଜୀବନରେ ପ୍ରଥମ ଥର ପାଇଁ ସେ ବିନା ହେଲ୍‌ମେଟ୍‌ରେ ଘରୁ ବାହାରିଥିଲା । ହେ ଭଗବାନ ! ସେ ରାସ୍ତାରେ ଏକ ରକ୍ତାକ୍ତ ଓ ନିର୍ଭୟ ଦୁର୍ଘଟଣାର ସମ୍ମୁଖୀନ ହେଲା ।

Stanza – 6 (Lines 21 – 24)
All it took was a stray moment of insane
A young life lost, a family crushed with pain;
Dear friends, take care; risky driving can be a bane
DRIVE SAFE – let your precious life not be in vain!

Gist: It was just an act of stupidity on the part of Vicky for a moment. A young life was cut short. His family was shocked (very much in surprise and grief). Taking the example of Vicky, dear riders are appealed not to resort to (take to) terrible and fast driving which can bring in lots of misery. You are requested to drive safely not to allow your precious life to go uselessly.
ଅନୁବାଦ : ଭିକି ପାଇଁ ଏହା ଏକ ନିର୍ବୋଧ ନିଷ୍ପତ୍ତି ଥିଲା । ଅକାଳରେ ଏକ ଯୌବନ ଚାଲିଗଲା । ତାହାର ମୃତ୍ୟୁରେ ପୂରା ପରିବାର ଦୁଃଖ ବା ଶୋକରେ ମ୍ରିୟମାଣ ହୋଇଗଲା । ଭିକିର ଉଦାହରଣକୁ ଦୃଷ୍ଟିରେ ରଖ୍ ପ୍ରିୟ ଚାଳକମାନଙ୍କୁ ତୀବ୍ର ବେଗରେ ଗାଡ଼ି ନ ଚଳାଇବାପାଇଁ ନିବେଦନ କରାଯାଉଛି, ନଚେତ୍ ଏହା ବହୁତ ଦୁଃଖ ବା ଶୋକ ଆଣି ଦେଇପାରେ । ତୀବ୍ର ବେଗରେ ଗାଡ଼ି ଚଳାଇ ମୂଲ୍ୟବାନ୍ ଜୀବନକୁ ବ୍ୟର୍ଥରେ ନ ଯିବାକୁ ଦେବାପାଇଁ ଆପଣମାନଙ୍କୁ ଅନୁରୋଧ ।

About The Poet:
Dr. Reeta S. Mani is a doctor by profession (ବୃତ୍ତିରେ ) and a writer by passion (ପ୍ରବୃତ୍ତିରେ ). She is a Neurovirologist (ସ୍ନାୟୁଭୂତାଣୁବିଦ୍) at the National Institite of Mental Health and Neurosciences (ସ୍ନାୟୁ ବିଜ୍ଞାନ), Bengaluru. Dr. Reeta has published (online and in print) several (ଅନେକ) short stories, poems, essays, and (ଭ୍ରମଣ ବୃତ୍ତାନ୍ତ ), including (ସହିତ) several short stories for children. She loves to weave health information into her stories to educate and foster ( ବଢ଼ାଇବାପାଇଁ) inquisitiveness (ଜିଜ୍ଞାସା ବା ଜାଣିବାର ଇଚ୍ଛା) in children.

Notes And Glossary:
brilliant brain — very smart or intelligent brain (ଅତି ବୁଦ୍ଧିମାନ ମସ୍ତିଷ୍କ)
formalin — a chemical compound of hydrogen, oxygen and carbon (ଉଦ୍‌ଜାନ, ଅମ୍ଳଜାନ ଓ ଅଙ୍ଗାରକର ଏକ ରାସାୟନିକ ଯୌଗିକ)
on display — the act of showing (ପ୍ରଦର୍ଶନ ନିମିତ୍ତ)
with disdain — with intensc hatred (ତୀବ୍ର ଘୃଣାର ସହିତ)
stare — look at somebody or something for a long time (ବହୁ ସମୟ ଧରି ଚାହିଁ ରହିବା)
specimen — sample (ନମୁନା)
stroke — brain attack (ମସ୍ତିଷ୍କାଘାତ)
tumour — a mas of abnormal cells grown in the body (ଅସ୍ଵାଭାବିକ କୋଷ ବୃଦ୍ଧିଜନିତ ମାଂସପିଣ୍ଡୁଳା )
bug — an illness caused by small organism such as bacteria (ବ୍ୟାକ୍ ଟେରି ଆଦ୍ଵାରା ସୃଷ୍ଟ ରୋଗ ବା ବ୍ୟାଧ୍) or an infectious illness (ଏକ ପ୍ରକାର ସଂକ୍ରାମକ ରୋଗ)
chagrin — anger or annoyance (କ୍ରୋଧ ବା ବିରକ୍ତି)
smart — well – dressed and neat (ପରିଷ୍କାର ଓ ସୁନ୍ଦର ପୋଷାକ ପରିହିତ)
You look very smart today.
sane — normal or sound (ସୁସ୍ଥ ବା ସ୍ଵାଭାବିକ)
The old fellow (man) looks sane today.
vigor — strength or energy (ଶକ୍ତି)
The lion has plenty of vigor.
domain — area of knowledge(ଜ୍ଞାନର କ୍ଷେତ୍ର)
The answer to this question is not in my mental domain.
motto — belief or ethic ନୀତି ବା ବିଶ୍ୱାସ)
“No pain. No gain” (କଷ୍ଟ କଲେ କୃଷ୍ଣ ମିଳେ is my motto.
insane — extreme or immense (ଅତ୍ୟନ୍ତ । ଅପାର)
truly — really (ପ୍ରକୃତରେ )
wrinkles — lines of loose skin (କୁଞ୍ଚୁତ ଚର୍ମ ବା ଚର୍ମରେ ଗାର ପଡ଼ିଥ‌ିବା)
That old woman has wrinkles all over her body.
cane — walking stick (ଆଶାବାଡ଼ି)
stole a glance — looked secretely (ଲୁଚିକରି ଦୃଷ୍ଟି ଦେଉଥବା)
went thud-thud — became panic and dull in sound (ଭୟରେ ହୃଦୟ ଧକ୍‌ ଧକ୍ ହେଉଥିଲା)
on a wet monsoon day — on a rainy day (ଏକ ବର୍ଷଣମୁଖର ଦିନରେ)
task — work (କାମ)
mundane — dull and ordinary (ସାଧାରଣ)
We are living an mundane existence
did refrain — stopped taking with (ନେବାରୁ ନିବୃତ୍ତ ରହିଲା)
gory — bloody (ରକ୍ତାକ୍ତ)
inhumane — unkind or gruesome (ଶୋଚନୀୟ ବା ନିର୍ଭୟ )
inane — stupidity (ବୋକାମି ) carelessness (ଯତ୍ନହୀନତା)
crushed with pain — shocked or went mourning ( ଦୁଃଖରେ ମ୍ରିୟମାଣ ହୋଇ ପଡ଼ିଲେ)
bane — fatal, curse (ମାରାତ୍ମକ ବା ଅଭିଶାପ)
Sometimes science proves to be a bane.
stray — separated (ଅଲ ଗା, ବିଚ୍ଛିନ୍ନ)
A stray dog (ବୁ ଲା କୁକୁର) is walking along the village road.
in vain — useless (ଅଦରକାରୀ |)
All his attempts (ଚେଷ୍ଟା) went in vain.

BSE Odisha 10th Class English Detailed Text

• Chapter 1 All Things Bright and Beautiful
• Chapter 2 A letter to God
• Chapter 3 The Solitary Reaper
• Chapter 4 At the High School
• Chapter 5 Village Song
• Chapter 6 Festivals of North-East India
• Chapter 7 The Flower-School
• Chapter 8 Air Pollution: A Hidden Menace
• Chapter 9 From the Formalin Jar
• Chapter 10 School’s Goodbye

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 7 Linear Inequalities Ex 7(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Exercise 7(c)

Solve the following systems of linear inequalities graphically.
Question 1.
2x – y ≥ 0, x – 2y ≤ 0, x ≤ 2, y ≤ 2 [Hint: You may consider the point (2, 2) to determine the SR of the first two inequalities.]
Solution:
2x – y ≥ 0
x – 2y ≤ 0
x ≤ 2
y ≤ 2
Step – 1: Let us draw the lines.
2x – y = 0, x – 2y = 0, x = 2, y = 2
2x – y = 0

X	0	1
y	0	2

x – 2y = 0

X	0	2
y	0	1

Step – 2: Let us consider point (1, 0) which does not line on any of these lines.
Putting x = 1, y = 0 in the inequations we get
2 ≥ 0 (True)
1 ≤ 0 (False)
1 ≤ 2 (True)
0 ≤ 2 (True)
Point (1, 0) satisfies all inequality except x – 2y < 0.
∴ Thus the shaded region is the solution region.

Question 2.
x – y < 1, y – x < 1
Solution:
x – y < 1
y – x < 1
Step – 1: Let us draw the dotted lines.
x – y = 1 and y – x = 1
x – y = 1 ⇒ y = x – 1

X	1	0
y	0	-1

y – x = 1 ⇒ y = x + 1

X	0	-1
y	1	0

Step – 2: Let us consider the point (0, 0) which does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
0 < 1 (True)
0 < 1 (True)
∴ (0, 0) satisfies both the inequations.
∴ Thus the shaded region is the feasible region.

Question 3.
x – 2y + 2 < 0, x > 0
Solution:
x – 2y + 2 < 0, x > 0
Step – 1: Let us draw the dotted line x – 2y + 2 = 0
⇒ y = \(\frac{x+2}{2}\)

X	-2	0
y	0	1

Step – 2: Let us consider the point (1, 0) that does not lie on the lines  putting x = 0, y = 0 in the inequation, we get
2 < 0 (false)
1 > 0 (True)
⇒ (1, 0) satisfies x > 0 and does not satisfy x – 2y + 2 < 0.
∴ Thus the shaded region is the solution region.

Question 4.
x – y + 1 ≥ 0, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0
Solution:
x – y + 1 ≥ 0
3x + 4y ≤ 12
x ≥ 0, y ≥ 0
Step – 1: Let us draw the lines.
x – y + 1 = 0
3x + 4y = 12
Now, x – y + 1 = 0 ⇒ y = x + 1

X	0	-1
y	1	0

3x + 4y = 12

X	4	0
y	0	3

Step – 2: Let us consider the point (0, 0) which does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
1 ≥ 0 (True)
0 ≤ 12 (False)
∴ (0, 0) satisfies both the inequations and x > 0, y > 0 is the first quadrant.
∴ Thus the shaded region is the solution region.

Question 5.
x + y > 1, 3x – y < 3, x – 3y + 3 > 0
Solution:
x + y > 1
3x – y < 3
x – 3y + 3 > 0
Step – 1: Let us draw the lines.
x + y = 1
3x – y = 3
x – 3y + 3 = 0
Now x + y = 1
⇒ y = 1 –  x

X	1	0
y	0	1

3x – y = 3
⇒ y = 3x – 3

X	1	0
y	0	-3

x – 3y + 3 = 0
⇒ y = \(\frac{x+3}{3}\)

X	-3	0
y	0	1

Step – 2: Let us consider the point (0, 0) that does not lie on these lines. Putting x = 0, y = 0 in the inequations we get,
0 > 1 (False)
0 < 3 (True)
3 > 0 (True)
Thus (0, 0) satisfies 3x – y < 3 and x – 3y + 3 > 0 but does not satisfy x + y > 1
∴ The shaded region is the solution region.

Question 6.
x > y, x < 1, y > 0
Solution:
x > y, x < 1, y > 0
Step – 1: Let us draw the dotted lines.

Step – 2: Let us consider a point (2, 1) that does not lie on any of the lines.
Putting x = 2, y = 2 in the inequations
we get,
2 > 1 (True)
2 < 1 (False)
1 > 0 (True)
⇒ (2, 1) satisfies x > y and y > 0 but does not satisfy x < 1.
∴ Thus the shaded region is the solution region.

Question 7.
x < y, x > 0, y < 1
Solution:
x < y
x > 0
y < 1
Step – 1: Let us draw the dotted lines.
x = y
x = 0
and y = 1

Step – 2: Let us consider point (1, 0) that does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
1 < 0 (False)
1 > 0 (True)
0 < 1 (True)
Clearly (1, 0) satisfies x > 0, y < 1 but does not satisfy x < y.
∴ The shaded region is the solution region.

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Exercise 6(b)

Question 1.
If Z₁ and Z₂ are two complex numbers then show that
\(\begin{aligned}
& \left|1-z_1 \overline{z_2}\right|^2-\left|z_1-z_2\right|^2 \\
& =\left(1-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right)
\end{aligned}\)
Solution:
Let z₁ and z₂ be two complex numbers.
Let z₁ = a + ib, z₂ = c + id

Question 2.
If a, b, c are complex numbers satisfying a + b + c = 0 and a² + b² + c² = 0 then show that |a| = |b| = |c|
Solution:
Let a + b + c = 0 and a² + b² + c² = 0
Then (a + b + c)² = 0
⇒ a² + b² + c² + 2ab + 2bc + 2ca = 0
⇒ 2(ab + bc + ca) = 0
⇒ ab + bc = – ca
⇒ b(a + c) = – ca
⇒ b(- b) = – ca [a + b + c = 0]
⇒ b² = ca
⇒ b³ = abc
Similarly it can be shown that a³ = abc and c³ = abc
Thus a³ = b³ = c³
⇒ |a³|= |b³| = |c³|
⇒ |a|³ = |b|³ = |c|³
⇒ |a| = |b| = |c|

Question 3.
What do the following represent?
(i) { z : |z – a| + |z + a| = 2c } where |a| < c
Solution:
{ z : |z – a| + |z + a| = 2c } where |a| < c    …….(1)
Here z is a complex number.
Let z = x + iy.
∴ (x, y) is the point corresponding to the complex number z?
Let ‘a’ and ‘ – a’ be two fixed points.
∴ Eqn. (1) implies that the sum of the distances of the point (x, y) from two points la and a’ is constant i.e. 2c
∴ The locus is an ellipse.

(ii) {z : |z – a| – |z + a| = c }
Solution:
Here {z : |z – a| – |z + a| = c } implies that, the difference of the distances of the point (x, y) from two fixed points ‘ – a’ and ‘a’ is a constant i.e. c.
So the locus is a hyperbola.

(iii) What happens in (i) |a| > c?
Solution:
In(i), if |a| > c. then there is no locus. But if |a| = c, then the locus reduces to a straight line.

Question 4.
Given cos α + cos β + cos γ = sin α + sin β + sin γ = 0 Show that cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)
Solution:
Let a = cos α + i sin α,
b = cos β + i sin β
c = cos γ + i sin γ
∴ a + b + c = ( cos α + cos β + cos γ) + i ( sin α + sin β + sin γ)
= 0 + i0 = 0
∴ a³ + b³ + c³ – 3 abc
= (a + b + c )( a² + b² + c² – ab – bc – ca) = 0
or, a³ + b³ + c³ = 3 abc
or, ( cos α + i sin α)³ + (cos β + i sin β)³ + ( cos γ + i sin γ)³
= 3( cos α + i sin α) (cos β + i sin β) ( cos γ + i sin γ)
or, cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + i sin 3γ
= 3[cos (α + β + γ) + i sin (α + β + γ)]
or, (cos 3α + cos 3β + cos 3γ) + i (sin 3α + sin 3β + sin 3γ)
= 3 cos (α + β + γ) + i 3 sin (a + β + γ)
∴ cos 3α + cos 3β + cos 3γ
= 3 cos (α + β + γ) and sin 3α + sin 3β + sin 3γ
= 3 sin (α + β + γ)

Question 5.
Binomial theorem for complex numbers. Show that (a+b)ⁿ = a^{n n}C₁a^n-1b +  …..+ ⁿc_ra^n-rb^r + …..+ bⁿ where a,b ∈ C and n, rule of multiplication of complex numbers and the relation ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r)
Solution:
Let a and b be two complex numbers
Let a = α₁ + iβ₁, b = α₂ + iβ₂
(a + b)¹ = (α₁ + iβ₁ + α₂ + iβ₂)¹
= α₁ + iβ₁ + α₂ + iβ₂
= (α₁ + iβ₁)¹ + ¹C₁(α₁ + iβ₁)^1-1(α₁ + iβ₁)¹
= a¹ + ¹C₁ a^1-1 b¹
∴ P₁ is true
Let P_k be true
i.e., (a + b)^k = a^k + ^kC₁ a^k-1 b¹ + … + b^k
where a.b ∈ C
Now ( a + b)^k+1 = (a + b)^k (a + b)¹
= (a^k + ^kC₁ a^k-1b¹ +…+ b^k) (a + b)
= a^k-1 + ba^k+ ^kC₁ a^kb + ^kC₁a^k-1b² + … + b^k+1
= a^k+1 + a^kb(^kC₁+ 1)+ …+ b^k+1
= a^k+1 + ^k+1C₁a^kb¹ + ^k+1 C₂a^k-1b² +… + b^k+1
∴ P_k+1 is true
∴ P_n is true for all values of n ∈ N

Question 6.
Use the Binomial theorem and De Moiver’s theorem to show
cos 3θ = 4 cos ³ θ – 3 cos θ,
sin 3θ = 3 sin θ – 4 sin ³ θ
Express cos nθ as a sum of the product of powers of sin θ and cos θ. Do the same thing for sin nθ.
Solution:
We have (cos θ + i sin θ)³
= cos 3θ + i sin 3θ       …..(1)
But by applying the Binomial theorem, we have
(cos θ + i sin θ)³
= cos ³ θ + ³C₁ (cos θ)^3-1 (i sin θ)¹ + ³C₂ (cos θ)^3-2  (i sin θ)² + (i sin θ)³
= cos3θ + 3i cos2θ sin θ + 3i² cos θ sin² θ + i³ sin³ θ
= (cos³ θ – 3 cos θ sin² θ) + i(3 cos² θ sin θ – sin³ θ)
∴ cos ³ θ =cos³ θ – 3 cos θ(1 – cos² θ)
= cos³ θ – 3 cos θ + 3 cos ³ θ
= 4 cos³ θ – 3 cos θ and
sin 3θ = 3 cos² θ sin³ θ – sin³ θ
= 3 (1 – sin² θ) sin θ – sin³ θ
= 3 sin θ – 3 sin³ θ – sin³ θ
= 3 sin θ – 4 sin³ θ (Proved)
Again, (cos θ + i sin θ)ⁿ
= cos nθ + i sin nθ         ….(3)
Also, (cos θ + i sin θ)ⁿ
= cosⁿ θ + ⁿC₁ cos^n-1 θ ( i sin θ) + ⁿc₂ cos ^n-2 θ (i sin θ)² + …+ (i sin θ)
= cosⁿ θ – ⁿC₂ cos^n-2 θ sin² θ + ⁿC₄ cos^n-4 θ sin⁴ θ – …) + i (ⁿC₁ cos^n-1 θ sin θ – ⁿC₃ cos^n-3 θ sin ³ θ) + ⁿC₅ cos^n-5 θ sin⁵ θ – …)    …..(4)
Equating real part and imaginary parts in (1) and (3), we have
cos nθ = cosⁿ θ – ⁿC₂ cos^n-2 θ × sin² θ + ⁿC₄ cos^n-4 θ sin⁴ θ …
and sin nθ = ⁿC₁ cos^n-1 θ sin θ – ⁿC₃ cos^n-3 θ sin³ θ + ⁿC₅ cos^n-5 θ sin⁵ θ…

Question 7.
Find the square root of
(i) – 5 + 12 √-1
Solution:

(ii) – 11 – 60 √-1
Solution:
Let \(\sqrt{-11-60 \sqrt{-1}}\) = x + iy
Squaring both sides we get
– 11 – 60i = (x + iy)²

(iii) – 47 + 8 √-1
Solution:

As 2ab = 8 > 0, a and b have the same sign.
∴ \(\sqrt{-47+8 i}\)
\(=\pm\left(\sqrt{\frac{\sqrt{2273}-47}{2}}+i \frac{\sqrt{2273}+47}{2}\right)\)

(iv) – 8 + √-1
Solution:

(v) a² – 1 +2a √-1
Solution:
a² – 1 + 2a √-1
= a² + i² + 2ai = (a + i)²
∴ \(\sqrt{a^2-1+2 a \sqrt{-1}}\) = ±(a + i)

(vi) 4ab – 2 (a² – b²) √-1
Solution:
4ab – 2 (a² – b²) √-1
= (a + b)² – (a – b)² – 2(a² – b²)i
= (a + b)² + (a – b)²i² – 2(a + b)(a – b)i
= (a + b) – i(a – b)²
∴ \(\sqrt{4 a b-2\left(a^2-b^2\right) \sqrt{-1}}\)
= ±[(a + b) – i (a – b)]

Question 8.
Find the values of cos72° ….
Solution:
Let 18° = θ then 5θ = 90°
⇒ 3θ = 9θ – 2θ
⇒ cos 3θ = cos (9θ – 2θ) = sin 2θ
⇒ 4 cos³ θ – 3 cos θ = 2 sin θ cos θ
⇒ 4 cos² 0 – 3 = 2 sin 0 (∴ cos θ = cos 18° ≠ 0)
⇒ 4(1 – sin² θ) – 3 = 2 sin θ
⇒ 4 sin² θ + 2 sin θ – 1 = 0
⇒ sin θ = \(\frac{-2 \pm \sqrt{4+16}}{2 \times 4}=\frac{-1 \pm \sqrt{5}}{4}\)
⇒ sin 18° = \(\frac{-1+\sqrt{5}}{4}=\frac{\sqrt{5}-1}{4}\)
(∴ 18° is a cut)
Now cos 72° = cos (90° – 18°)
= sin 18° = \(\frac{\sqrt{5}-1}{4}\)
For other methods refer 148 pages of the text book.

Question 9.
Find the value of cos 36°.
Solution:
We have 36° = \(\frac{\pi^0}{5}\)
∴ cos 36° = cos \(\frac{\pi}{5}\)
Let α = cos \(\frac{\pi}{5}\) + i sin \(\frac{\pi}{5}\)
be the root of the equation x⁵ + 1 = 0
Again, if x⁵ + 1 = 0
or, x⁵ = – 1 = cosπ + i sinπ
or, x = (cos π + i sin π )^1/5
= [cos (π + 2kπ)+ i sin (π + 2kπ)]^1/5
or, x = cos \(\frac{(2 k+1) \pi}{5}+i \sin \frac{(2 k+1) \pi}{5}\)
where 2kπ is the period of sine and cosine and k = 0, 1, 2, 3, 4
∴ The eqn x⁵ + 1 = 0 has 5 roots out of which -1 is one root which corresponds to k = 2
Again,
x⁵ + 1 = (x + 1)(x⁴ – x³ + x² – x + 1)
So their 4 roots will be obtained on solving the eqn.
x⁴ – x³ + x² – x + 1 = 0
we have, x⁴ – x³ + x² – x + 1 = 0
or, x² – x + 1 – \(\frac{1}{x}+\frac{1}{x^2}\) = 0
(Dividing both sides by x²)

Re α i.e. cos \(\frac{\pi}{5}=\frac{1+\sqrt{5}}{5}=\frac{\sqrt{5}+1}{4}\)
and cos.108° = \(\frac{1-\sqrt{5}}{4}\)

Question 10.
Evaluate cos \(\frac{2 \pi}{17}\) using the equation x¹⁷ – 1 = 0
Solution:
x¹⁷ – 1 = 0
or, x¹⁷ = 1 = cos 0° + i sin 0°
= cos (0 + 2kπ) + i sin (0 + 2kπ)
= cos 2kπ + i sin 2kπ
or x = (cos 2kπ + i sin 2kπ)^1/17
= cos \(\frac{2k \pi}{17}\) + i sin \(\frac{2k \pi}{17}\)
If k = 1, x = cos \(\frac{2 \pi}{17}\) + i sin \(\frac{2 \pi}{17}\)
If k = 0 , x = 1
As x¹⁷ – 1 = (x – 1) (x¹⁶ + x¹⁵ + …. +1)
So one root of the eqn. x¹⁷ – 1 = 0 is 1 and all other roots are the roots of the eqn.
x¹⁶ + x¹⁵ + ….+ 1 = 0
∴ The value of cos \(\frac{2 \pi}{17}\) can be found from the roots of the eqn.  (1)

Question 11.
Solve the equations.
(i) z⁷ = 1
Solution:
z⁷ = 1 = cos 0 + i sin 0
= cos (0 + 2kπ) + i sin (0 + 2kπ)
= cos 2kπ + i sin 2kπ
∴ z = (cos 2kπ + i sin 2kπ)^1/7
= cos \(\frac{2k \pi}{7}\) + i sin \(\frac{2k \pi}{7}\)
where k = 0, 1, 2, 3, 4, 5, 6

(ii) z³ = i
Solution:

(iii) z⁶ = – i
Solution:

(iv) z³ = 1 + i
Solution:

Question 12.
If sin α + sin β + cos γ = 0
= cos α + cos β + cos γ = 0
Show that
(i) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)
Solution:
Refer to Q. No. 4

(ii) sin² α + sin² β + sin² γ = cos² α + cos² β + cos² γ =3/2
Solution:
Let x = cos α + i sin α,
y = cos β + i sin β
z = cos β + sin β
∴ x + y + z = (cos α + cos β + cos γ) + i ( sin α + sin β + sin γ) = 0 +i0= 0
∴ xy + yz + zx = xyz (\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)) = 0
Since \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) = cos α – i sin α + cos β – i sin β + cos γ – i sin γ
= (cos α + cos β + cos γ) – i (sin α + sin β + sin γ) = 0 – i0 =0
∴ (x + y + z)² = x ² + y² +z² + 2(xy + yz + zx)
= x² + y² + z² + 0 = x² + x² + z²
or 0 = x² + y² + z²
∴ x² +y² + z² =0
or, (cos α + i sin α)² + (cos β + i sin β)² + ( cos γ + i sin γ)² = 0
or, cos 2α + i sin 2α + cos 2β + i sin 2β + cos 2γ + i sin 2γ = 0
or, (cos 2α + cos 2β + cos 2γ) + i(sin 2α + sin 2β + sin 2γ) = 0
∴ cos 2α + cos 2β + cos 2γ = 0
or, cos² α – sin² α + cos² β – sin² β + cos² γ – sin² γ =0
or, (cos² α + cos² β + cos² γ) = (sin² α + sin² β + sin² γ)
But cos² α + sin² α + cos² β + sin² β + cos² γ + sin² γ = 1 + 1 + 1 = 3
∴ cos² α + cos² β + cos² γ = sin² α + sin² β + sin² γ = 3/2   (Proved)

Question 13.
If x + \(\frac{1}{x}\) = 2 cos θ
Show that \(x^n+\frac{1}{x^n}\) = 2 cos nθ

Question 14.
x_r = cos a^r + i sin a^r
r =1, 2, 3 and x₁ + x₂ + x₃ = 0 Show that \(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\) = 0
Solution:

Question 15.
Show that \(\left(\frac{1+\sin \theta+i \cos \theta}{1+\sin \theta-i \cos \theta}\right)^n\) = \(\cos \left(\frac{n \pi}{2}-n \theta\right)+i \sin \left(\frac{n \pi}{2}-n \theta\right)\)
Solution:

Question 16.
If α and β are roots x² – 2x + 4 = 0 then show that \(\alpha^n+\beta^n=2^{n+1} \cos \frac{n \pi}{3}\)
Solution:
we have x² – 2x + 4 = 0

= \(2^n \times 2 \cos \frac{n \pi}{3}=2^{n+1} \cos \frac{n \pi}{3}\)

Question 17.
For a positive integer n show that
(i) (1 + i)ⁿ + (1 – i)ⁿ = \(2^{\frac{n+2}{2}} \cos \frac{n \pi}{4}\)
Solution:

(ii) (1 + i√3)ⁿ + (1 – i√3)ⁿ = \(2^{n+1} \cos \frac{n \pi}{3}\)
Solution:

Question 18.
Let x + \(\frac{1}{x}\) = 2 cos α, y + \(\frac{1}{y}\) = 2 cos β, z + \(\frac{1}{z}\) = 2 cos γ. Show that
(i) 2 cos (α + β + γ) = xyz + \(\frac{1}{xyz}\)
Solution:
We can take x = cos α + i sin α
y = cos β + i sin β, z = cos γ + i sin γ
∴ xyz = (cos α + i sin α ) (cos β + i sin β ) (cos γ + i sin γ)
= cos (α + β + γ) – i sin (α + β + γ)
∴ \(\frac{1}{xyz}\) = cos (α + β + γ) – i sin(α + β + γ)
∴ xyz + \(\frac{1}{xyz}\) = 2 cos(α + β + γ)

(ii) 2 cos (pα + qβ + rγ) = \(x^p y^q z^r+\frac{1}{x^p y^q z^r}\)
Solution:

Question 19.
Solve x⁹ + x⁵ – x⁴ = 1
Solution:
x⁹ + x⁵ – x⁴ = 1
or, x⁵(x⁴ + 1) – (x⁴ + 1) = 0
or, (x⁵ – 1) (x⁴ + 1) = 0
x⁴ + 1 = 0 and x⁵ – 1 = 0
x⁴ = – 1 = cos π + i sin π
= cos (π + 2nπ) + i sin (π + 2nπ)
∴ x = [cos (2n + 1) π + 1 sin (2n + 1) π]^1/4
= \(\cos \frac{2 n+1 \pi}{4}+i \sin \frac{2 n+1 \pi}{4}\)
for n = 0, 1, 2, 3
Again, x⁵ – 1 = 0 or, x⁵ = 1
or, x⁵ = cos 0 + i sin 0
= cos 2nπ + i sin 2nπ
or, x = (cos 2nπ + i sin 2nπ)^1/5
= \(\cos \frac{2 n \pi}{5}+i \sin \frac{2 n \pi}{5}\)
Where n = 0, 1, 2, 3, 4.

Question 20.
Find the general value of θ if (cos θ + i sin θ) (cos 2θ + i sin 2θ),…..(cos nθ + i sin nθ) =1
Solution:

Question 21.
If z = x + iy show that |x| + |y| ≤ √2 |z|
Solution:
z = x + iy
∴ |z| = \(\sqrt{x^2+y^2}\)
∴ |z| = x² + y²
We have (|x| – |y|)² ≥ 0
⇒ |x|² + |y|² -2|x||y|> 0
⇒ 2(|x|² + |y|²) – 2|x||y| ≥ |x|² + |y|²
⇒ 2(|x|² + |y|²) ≥ |x|² + |y|² + 2|x||y|
⇒ 2|z|² ≥ (|x| + |y|)²
⇒ √2|z| ≥ |x| + |y|
⇒ |x| + |y| ≤ √2|z|

Question 22.
Show that
Re (Z₁Z₂) = Re z₁, Re z₂ – Im z₁, Im z₂
Im (Z₁Z₂) = Re z₁, Im z₂ + Re z₂ Im z₁
Solution:
Let z₁ = a + ib, z₂ = c + id
∴ z₁, z₂ = (a + ib) (c + id)
= ac + iad + ibc + i²bc
= (ac – bd) + i (ad + be)
∴ Re (z₁, z₂) = ac – bd – Re z₁. Re z₂
– Im z₁,. Im z₂
Again, Im z₁, z₂ = ad + be
= Re z₁. Im z₂ + Im z₁,. Re z₂

Question 23.
What is the value of arg ω + arg ω²?
Solution:
arg ω = arg ω² = arg (ω • ω²)
= arg (ω³) = arg (1) = 2nπ
∴ The principal agrument = 0.

Question 24.
If |z₁| ≤ 1, |z₂| ≤ 1 show that \(\left|1-z_1 \overline{z_2}\right|^2-\left|z_1-z_2\right|^2=\left(1-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right)\) Hence and otherwise show that. \(\left|\frac{z_1-z_2}{1-z_1 z_2}\right|<1 \text { if }\left|z_1\right|<1,\left|z_2\right|<1\)
Solution:

Question 25.
If z₁² + z₂² + z₃² – z₁z₂ – z₂z₃ – z₃z₄ = 0 Show that |z₁ – z₂| = |z₂ – z₃| = |z₃ – z₁|
Solution:
Let z₁² + z₂² + z₃² – z₁z₂ – z₂z₃ – z₃z₄ = 0
⇒ 2z₁² + 2z₂² + 2z₃² – 2z₁z₂ – 2z₂z₃ – 2z₃z₄ = 0
⇒ (z₁ – z₂)² + (z₂ – z₃)² + (z₃ – z₁)² = 0
Put a = z₁ – z₂, b = z₂ – z₃, c = z₃ – z₁
Then a + b + c = 0 and a² + b² + c² = 0 As in Q2 we can show that |a| = |b| = |c|
⇒ |z₁ – z₂| = |z₂ – z₃| = |z₃ – z₁|

Question 26.
If |a| < |c| show that there are complex numbers z satisfying |z – a| = |z + a| = 2|c|
Solution:
Let z = x + iy
∴ |z – a| + |z + a| = 2c
or, |x + iy – a| + |x + iy + a| = 2c

Question 27.
Solve \(\frac{(1-i) x+3 i}{2+i}+\frac{(3+2 i) y+i}{2-i}=-i\) where x, y, ∈ R.
Solution:

Question 28.
If (1 + x + x²)ⁿ = p₀ + p₁x + p₂x² + …..+ p_2nx²ⁿ, then prove that p₀ + p₃ + p₆ + …..+ 3^n-1
Solution:
Given, (1 + x + x²)ⁿ = p₀ + p₁x + p₂x² + …..+ p_2nx²ⁿ
putting x = ω we get

Question 29.
Find the region on the Argand plane on which z satisfying
[Hint Arg (x + iy) =\(\frac{\pi}{2}\) = 0, y>0]
(i) 1 < |z – 2i| < 3
Solution:
Let z = x + iy
The given inequality is
1 < |x + i(y – 2)| < 3
⇒ \(1<\sqrt{x^2+(y-2)^2}<3\)

(ii) arg \(\left(\frac{z}{z+i}\right)=\frac{\pi}{2}\)
Solution:

As all are +ve we have
1 < x² + (y – 2)² < 9
x² + (y – 2)² < 9 is the region inside the circle with center (0, 2) and radius 1.
x² + (y – 2)² > 1 is the region outside the circle with center (0, 2) and radius.
∴ 1 < |z – 2i| < 3 is the region between two concentric circles with center (0, 2) and radius 1 and 3 which is shows below.

Odisha State Board BSE Odisha 9th Class English Solutions Chapter 7 Nine Gold Medals Textbook Exercise Questions and Answers.

BSE Odisha Class 9 English Solutions Chapter 7 Nine Gold Medals

BSE Odisha 9th Class English Nine Gold Medals Text Book Questions and Answers

E. Let’s Understand The Poem: (ଆସ କବିତାଟିକୁ ବୁଝିବା )
Read the poem silently and answer the following questions.

Question 1.
What is this poem about?
(ଏହି କବିତାଟି କେଉଁ ବିଷୟରେ ଆଧାରିତ ?)
Answer:
This poem is about an inspiring event of race in which nine differently-abled athletes took part in a “Special Olympic” event.

Question 2.
Who came from all over the country?
( ସମଗ୍ର ଦେଶରୁ କେଉଁମାନେ ଆସିଥିଲେ ? )
Answer:
Nine differently abled athletes came from all over the country.

Question 3.
Why did they come there?
(ସେମାନେ କାହିଁକି ଆସିଥିଲେ ? )
Answer:
The nine athletes came to take part in a Special Olympic event. They came to run for gold, silver, and bronze medals.

Question 4.
Who is usually awarded the three medals – gold, silver, and bronze?
( ସାଧାରଣତଃ କେଉଁମାନେ ସ୍ଵର୍ଣ୍ଣ, ରୌପ୍ୟ ଏବଂ କାଂସ୍ୟ ପଦକ ପ୍ରଦାନ କରାଯାଏ ? )
Answer:
Three athletes, who are placed in first, second, and third positions in an athletic event, are usually awarded the gold, silver, and bronze medals respectively.

Question 5.
What was the event for which they had come?
( ଯେଉଁଥିପାଇଁ ସେମାନେ ଆସିଥିଲେ ସେହି ପ୍ରତିଯୋଗିତାଟି କ’ଣ ଥିଲା ?)
Answer:
The athletes had come to take part in a hundred-yard race in a “Special Olympic” event.

Question 6.
What had they done before they came down to these games?
(ଏହି ଖେଳରେ ଯୋଗ ଦେବା ପୂର୍ବରୁ ସେମାନେ କ’ଣ କରିଥିଲେ ? )
Answer:
The athletes had undertaken many weeks and months of training before they came down to these games.

Question 7.
What did the spectators do around the old field ?
(ପୁରାତନ ଖେଳପଡ଼ିଆ ଚାରିପାଖରେ ଦର୍ଶକମାନେ କ’ଣ କରୁଥିଲେ ? )
Answer:
The spectators gathered around the old field and cheered the winners.

Question 8.
Why did they gather around the old field ?
(ସେମାନେ କାହିଁକି ପୁରାତନ ପଡ଼ିଆ ଚାରିପାଖରେ ଏକତ୍ର ହେଲେ ?)
Answer:
The spectators gathered around the old field to cheer on all the young athletes and enjoy the game.

Question 9.
Why did their excitement grow high?
(ସେମାନଙ୍କର ଉତ୍ସାହ ବଢ଼ିଗଲା କାହିଁକି ?)
Answer:
The excitement grew high because the final event of the day was approaching.

Question 10.
What were all lined up? For whom?
(କ’ଣସବୁ ପ୍ରସ୍ତୁତ ହୋଇଥିଲା ? କେଉଁମାନଙ୍କ ପାଇଁ ? )
Answer:
The blocks were all lined up for the athletes who would use them.

Question 11.
What was the event? (second line, third stanza)
[ପ୍ରତିଯୋଗିତାଟି କ’ଣ ଥୁଲା ? (ତୃତୀୟ ପଦର ଦ୍ବିତୀୟ ଧାଡ଼ି)]
Answer:
The event was that, the athletes were to run the hundred-yard dash.

Question 12.
How many athletes were there ?
( ସେଠାରେ କେତେ ଜଣ ଖେଳାଳି ଥିଲେ ?)
Answer:.
There were nine athletes in the competition.

Question 13.
Which word (in the third line, third stanza) says that the athletes had taken firm decision to win a medal?
(ତୃତୀୟ ପ୍ରଦର, ୩ୟ ଧାଡ଼ିରେ ଥ‌ିବା କେଉଁ ଶବ୍ଦଟି ଖେଳାଳିମାନେ ପଦକ ଜିତିବାର ଦୃଢ଼ସଂକଳ୍ପ ନେଇଛନ୍ତି ବୋଲି କହୁଛି ?)
Answer:
The word “resolved” in the third line, the third stanza says that the athletes had taken a firm decision to win a medal.

Question 14.
Where were those nine athletes?
( ସେହି ନଅ ଜଣ ଖେଳାଳି କେଉଁଠି ଥିଲେ ?)
Answer:
Those nine athletes were in back of the starting line.

Question 15.
What were they poised for ?
(କେଉଁଥିପାଇଁ ସେମାନେ ପ୍ରସ୍ତୁତ ଥିଲେ ?)
Answer:
The athletes were poised for the sound of the gun.

Question 16.
Why was the pistol exploded?
(ପିସ୍ତଲ କାହିଁକି ଗର୍ଜି ଉଠିଲା ?)
Answer:
The pistol exploded to give a signal to start running. ‘

Question 17.
What did the runners do?
(ଦୌଡ଼ାଳିମାନେ କ’ଣ କଲେ ?)
Answer:
When the pistol exploded, the runners charged ahead.

Question 18.
Who among the nine runners was unable to run?
(ଦୌଡ଼ାଳିମାନଙ୍କ ମଧ୍ୟରୁ କିଏ ଦୌଡ଼ିବାକୁ ଅସମର୍ଥ ଥିଲା ?)
Answer:
The youngest among the nine athletes was unable to run.

Question 19.
Why was he unable to run?
(ସେ କାହିଁକି ଦୌଡ଼ିବାକୁ ଅସମର୍ଥ ଥିଲା ?)
Answer:
The youngest among the nine athletes lost his balance and stumbled in the middle. He fell down and was unable to run.

Question 20.
Where did he fall?
(ସେ କେଉଁଠାରେ ପଡ଼ିଗଲା ?)
Answer:
The young athlete fell to the asphalt.

Question 21.
Why did he fall?
( ସେ କାହିଁକି ପଡ଼ିଗଲା ?)
Answer:
He fell down as he stumbled and lost his balance.

Question 22.
Who does ‘he’ in the fifth stanza stand for?
(ପଞ୍ଚମ ପଦରେ ‘he’ ଶବ୍ଦଟି କାହାପାଇଁ ବ୍ୟବହୃତ ହୋଇଛି ?)
Answer:
In the fifth stanza, he refers to the smallest athlete.

Question 23.
Why did the boy cry out in frustration and anguish?
(ପିଲାଟି କାହିକି ହତାଶା ଓ ଯନ୍ତ୍ରଣାରେ ଚିତ୍କାର କଲା ?)
Answer:
The boy cried out in frustration and anguish as he was unable to get up and run. All his dreams and efforts were spoiled in a moment.

Question 24.
What dashed in the dirt? What makes the poet say so?
(କ’ଣ ମାଟିରେ ମିଶିଗଲା ? କବି କାହିଁକି ଏପରି କହିଛନ୍ତି ?)
Answer:
All his dreams for the medals were dashed in the dirt. Because all his dreams and efforts came to an end. So the poet says all the hopes of the boy are dashed in the dirt.

Question 25.
What did the other runners do?
(ଅନ୍ୟ ଖେଳାଳିମାନେ କ’ଣ କଲେ ?)
Answer:
The other runners didn’t push ahead. They stopped running and turned back to help the young athlete.

Question 26.
How do you feel about their stopping here? Should athletes do like this?
(ଏଠାରେ ସେମାନେ ରହିଯିବାରୁ ତୁମେ କିପରି ଅନୁଭବ କରୁଛ ? ଖେଳାଳିମାନେ ଏଭଳି କରିବା ଉଚିତ କି ?)
Answer:
The athletes stopped running and turned back to help the young athlete stand on his feet. It conveys a message that humanity and fellow feeling is better than winning a race. It was a rare incident. Generally, athletes don’t do like this. But they should do this.

Question 27.
Who does ‘him’ stand for in the sixth stanza?
(ଷଷ୍ଠ ପଦରେ ‘him’ ଶବ୍ଦଟି କାହାପାଇଁ ଉଦ୍ଦିଷ୍ଟ ?)
Answer:
The word ‘him’ in the sixth stanza stands for the young athlete.

Question 28.
What had they done for the boy to complete the race?
(ବାଳକଟି ଦୌଡ଼ ପୂରା କରିବା ପାଇଁ ସେମାନେ ତା’ପାଇଁ କ’ଣ କଲେ ?)
Answer:
The other athletes stopped running and returned back to help the boy. They helped the boy stand on his feet and walked hand in hand to finish the race together.

Question 29.
Why did they go back to him?
(ସେମାନେ କାହିଁକି ତାହା ପାଖକୁ ଫେରିଗଲେ ?)
Answer:
All other athletes went back to help him stand on his feet.

Question 30.
What did they do first?
(ସେମାନେ ପ୍ରଥମେ କ’ଣ କଲେ ?)
Answer:
They returned to help their fellow athlete. They made him stand on his feet first.

Question 31.
What did they do next?
(ତା’ପରେ ସେମାନେ କ’ଣ କଲେ ?)
Answer:
Then all the athletes walked hand-in-hand to complete the race.

Question 32.
What happened to the hundred-yard dash?
(୧୦୦ ଗଜ ଦୌଡ଼ର କ’ଣ ହେଲା ?)
Answer:
The hundred-yard dash was reduced to a walk.

Question 33.
What did they carry with them?
(ସେମାନେ ସେମାନଙ୍କ ସହ କ’ଣ ବୋହି ନେଉଥିଲେ ? )
Answer:
They carried a banner with them.

Question 34.
What did the banner say?
(ବ୍ୟାନାର କ’ଣ କହୁଥିଲା ?)
Answer:
The banner said that it was “Special Olympics”.

Question 35.
What did the race end with?
(ଦୌଡ଼ର ସମାପ୍ତି କିପରି ଘଟିଲା ?)
Answer:
The race ended with a walk hand-in-hand and all nine athletes became the winner of the event.

Question 36.
How did they come to the finish line?
(ସେମାନେ finish lineକୁ କିପରି ଆସିଲେ ?)
Answer:
All the athletes walked hand-in-hand to come to the finish line.

Question 37.
What do you mean by “a standing ovation”?
(“a standing ovation’ର ଅର୍ଥ କ’ଣ ?)
Answer:
Here “a standing ovation” means the popular appreciation for the masterly performance the athletes had done.

Question 38.
How were their faces then? Were they all happy? Why?
(ସେତେବେଳେ ସେମାନଙ୍କର ମୁଖମଣ୍ଡଳ କିପରି ଥିଲା ? ସମସ୍ତେ ଖୁସି ଥିଲେ କି ? କାହିଁକି ?)
Answer:
Their faces were beaming then. They all won gold medals. They all were happy for the sympathy and love they showed to the young athlete.

Question 39.
What do you learn from such type of help of the athletes?
(ଖେଳାଳିମାନଙ୍କର ଏହି ପ୍ରକାର ସାହାଯ୍ୟରୁ ତୁମେ କ’ଣ ଶିକ୍ଷାଲାଭ କଲ ?)
Answer:
We learn that humanity, love, affection, fellow feeling, and cooperation are high above personal achievements.

F. Let’s Appreciate The Poem: (ଆସ କବିତାର ମୂଲ୍ୟାୟନ କରିବା)

Question 1.
What message does the poet convey in “Nine Gold Medals”?
(“Nine Gold Medals” କବିତାରେ କବି କି ବାର୍ତ୍ତା ଦେଇଛନ୍ତି ?)
Answer:
The poet conveys the message that humanity, love, affection, fellow feeling, and cooperation are better than any sort of competition or personal achievements.

Question 2.
What is important – winning a medal or helping a fellow brother in trouble?
(ପଦକ ଜିତିବା କିମ୍ବା ସାଙ୍ଗଭାଇକୁ ସାହାଯ୍ୟ କରିବା ଦୁଇଟିରୁ କେଉଁଟି ପ୍ରଧାନ ?)
Answer:
Helping a fellow brother in trouble is more important than winning a medal.

Question 3.
What is necessary for a person – to understand the emotions of others or to become self-centered?
(ଜଣେ ବ୍ୟକ୍ତି ପାଇଁ କେଉଁଟି ଆବଶ୍ୟକ – ଅନ୍ୟମାନଙ୍କ ମନୋଭାବ ବୁଝିବା କିମ୍ବା ସ୍ଵାର୍ଥକୈନ୍ଦ୍ରିକ ହେବା ?)
Answer:
Understanding the emotions of others is necessary for a person to become self-centered.

Question 4.
Explain – (ବ୍ୟାଖ୍ୟା କର) “He gave out a cry in frustration and anguish His dreams and efforts all dashed in the dirt.”
Answer:
The smallest athlete stumbled and lost his balance and fell to the ground. He cried in frustration and anguish as all his dreams and efforts were spoiled in a moment.

Question 5.
What do you mean by “nine beaming faces said more than these words ever will”?
(‘‘ନଅଟି ଉଜ୍ଜଳ ମୁଖମଣ୍ଡଳ ଶବ୍ଦରେ ଯାହା କହିହେବ ନାହିଁ ତା’ଠାରୁ ଅଧିକ କିଛି କହୁଥୁଲଶ’’ – ଏଥୁରୁ ତୁମେ କ’ଣ ବୁଝୁଛି ?)
Answer:
The nine athletes walked hand-in-hand to the finishing point and all were winners of the race. They were each awarded a gold medal for their empathy for each other. Their special deed justified the name ‘Special Olympic’ event. Their beaming faces told about the importance of humanity, brotherhood, and fellow feeling in this modem world.

G. Let’s Do The Activities: (ଆସ କାର୍ଯ୍ୟଗୁଡ଼ିକୁ କରିବା)

Question 1.
Let’s recite the poem :
(ଆସ, କବିତାଟିକୁ ଆବୃଦ୍ଧି କରିବା ।)
Answer:

• The teacher divides the class into five groups.
• S/he read aloud the poem two times with proper word stress and intonation.
• S/he read the poem aloud line by line and students repeat after him/her.
• Students read the. first line and the teacher reads aloud the second line and so on till the whole poem is complete.
• Each group reads a stanza and the rest of the groups repeat after them and so on.
• Finally, each group reads a stanza and other groups listen to them.
• The teacher invites some students to recite the poem individually.

Question 2.
Let’s match the stanza with their themes.
(ଆସ, ପଦଗୁଡ଼ିକୁ ସେଗୁଡ଼ିକର ବିଷୟବସ୍ତୁ ସହ ମେଳ କରିବା ।)
Match column ‘A’ with column ‘B’. ‘A’ contains the stanza numbers and ‘B’ contains the themes. Write the stanza numbers in the last column. One is done for you. (‘A’ ସ୍ତମ୍ଭ ସହିତ ‘B’ ସ୍ତମ୍ଭକୁ ମିଳାଅ ! ସ୍ତମ୍ଭ ‘A’ରେ ପଂକ୍ତି କ୍ରମ ସଂଖ୍ୟା ଏବଂ ସ୍ତମ୍ଭ ‘B’ରେ ବିଷୟବସ୍ତୁ ରହିଛି । ପଂକ୍ତିର କ୍ରମ ସଂ%: ଶେଷ ସ୍ତମ୍ଭରେ ଲେଖି । ଗୋଟିଏ ତୁମ ପାଇଁ କରିଦିଆଯାଇଛି ।)

Stanza Numbers	Themes of the stanzas
1	nine resolved athletes at the back of the starting line – ready for the event
2	athletes from all over the country – to run for the medals – attended long trainings
3	pistol exploded – signal given – running ahead – the smallest one lost control – fell down
4	spectators gathered – cheer – final event – highly excited	2
5	the other runners – came back – helped him stand
6	nine runners – joined hands – walked instead of running – banner – Special Olympic
7	end of the race – nine gold medals – faces – looked happy
8	cried in frustration and terrible pain – thought his dream shattered

Answer:

Stanza Numbers	Themes of the stanzas
1	nine resolved athletes at the back of the starting line – ready for the event	3
2	athletes from all over the country – to run for the medals – attended long trainings	1
3	pistol exploded – signal given – running ahead – the smallest one lost control – fell down	4
4	spectators gathered – cheer – final event – highly excited	2
5	the other runners – came back – helped him stand	6
6	nine runners – joined hands – walked instead of running – banner – Special Olympic	7
7	end of the race – nine gold medals – faces – looked happy	8
8	cried in frustration and terrible pain – thought his dream shattered	5

Question 3.
Let’s arrange the jumbled sentences to get the summary of the poem. The first and the last sentences are already in order. Write the number of the sentence in the box provided for each. (ଆସ କବିତାର ସାରାଂଶ ପାଇଁ ବାଦ୍ୟଗୁଡ଼ିକୁ ସଜାଇବା । ପ୍ରଥମ ଓ ଶେଷ ବାକ୍ୟ ପୂର୍ବରୁ ଠିକ୍ କ୍ରମରେ ରହିଛି । ବାକ୍ୟଗୁଡ଼ିକର କ୍ରମ ସଂଖ୍ୟା ପ୍ରଦତ୍ତ ବାକ୍ସରେ ଲେଖି ।)

• Nine athletes came from all over the country to take part in a running race. [ 1 ]
• The pistol exploded to signal a start.
• All of them were running ahead.
• They had already received training before coming there.
• But the youngest of them lost control and fell down.
• The blocks were all lined up for the running race and they were in the back of the starting line.
• Being frustrated, he cried in severe pain.
• They all joined hands and walked together to the end line holding a banner.
• The other runners, instead of running, came to him and helped him stand.
• They were all happy to help the youngest runner and all were awarded gold medals. [ 10 ]

Answer:

• Nine athletes came from all over the country to take part in a running race. [ 1 ]
• The pistol exploded to signal a start. [ 4 ]
• All of them were running ahead. [ 5 ]
• They had already received training before coming there. [ 2 ]
• But the youngest of them lost control and fell down. [ 6 ]
• The blocks were all lined up for the running race and they were in the back of the starting line. [ 3 ]
• Being frustrated, he cried in severe pain. [ 7 ]
• They all joined hands and walked together to the end line holding a banner. [ 9 ]
• The other runners, instead of running, came to him and helped him stand. [ 8 ]
• They were all happy to help the youngest runner and all were awarded gold medals. [ 10 ]

H. Let’s Write: (ଆସ ଲେଖୁବା )
Write a paragraph for each stanza of the poem. Go through the above matching activity before you begin to write. One paragraph is written to help you. (ପଦ୍ୟର ପ୍ରତ୍ୟେକ ପଂକ୍ତିପାଇଁ ଗୋଟିଏ ଅନୁଚ୍ଛେଦ ଲେଖ । ଲେଖିବା ପୂର୍ବରୁ ଉପରେ ପ୍ରଦତ୍ତ ମେଳକ କାର୍ଯ୍ୟକୁ ଅନୁଧ୍ୟାନ କର । ଗୋଟିଏ ଅନୁଚ୍ଛେଦ ତୁମକୁ ସାହାଯ୍ୟ କରିବା ପାଇଁ ଲେଖାଯାଇଛି ।)
Answer:

1. The poem “Nine Gold Medals’ has eight stanzas. The first stanza is about how athletes from all over the country came to take part in a running race. They had to run for medals. There were three kinds of medals – gold, silver, and bronze. They had attended long training before this event.
2. The second stanza is about how the spectators gathered around the old field to cheer the athletes. The final event of the day was approaching and excitement grew high.
3. The third stanza is about how nine resolved athletes stood at the back of the starting line. They were ready to run.
4. The fourth stanza is about how the pistol exploded giving the signal to run. All the athletes ran ahead, but the smallest one stumbled and lost balance, and fell down.
5. The fifth stanza is about how the smallest athlete cried in frustration and pain. His dream of winning a medal shattered in a moment.
6. The sixth stanza is about how all nine runners came back to help the smallest runner. They helped the boy to stand on his feet.
7. The seventh stanza is about how the nine athletes joined hands and walked instead of running to finish the race. They carry the banner of ‘Special Olympic7 with them.
8. The eighth stanza is about how the race ended. All were winners and awarded a gold medal each. Their faces were beaming and they were happy.

BSE Odisha 9th Class English Nine Gold Medals Important Questions and Answers

Very Short A Objective Questions With Answers
Answer The Followings In A Sentence.

Question 1.
What is the poem ‘Nine Gold Medals’ about?
Answer:
The poem ‘Nine Gold Medals’ is about an inspiring running event that took place among nine disabled athletes.

Question 2.
Who came from all over the country?
Answer:
Disabled athletes came from all over the country.

Question 3.
Why did the disabled athletes come there?
Answer:
The disabled athletes came there to participate in a race competition.

Question 4.
Who has usually rewarded the three medals- gold, silver, and bronze?
Answer:
Usually, the first winner of the competition is rewarded with a gold medal and the second winner gets a silver medal and the winner who stands third gets a bronze medal.

Question 5.
What was the event for which they had come?
Answer:
They had come to participate in a running race competition in a special Olympic event.

Question 6.
What had they done before they came down to these games?
Answer:
They had trained themselves hard for many weeks and months before they came down to these games.

Question 7.
What did the spectators do around the Oldfield?
Answer:
The spectators gathered round the old field and cheered the athletes.

Question 8.
Why did they gather around the old field?
Answer:
They gathered round the old field to enjoy watching the games and cheered the athletes.

Question 9.
Why did their excitement grow up?
Answer:
Their excitement grew up because the final event of the day was approaching.

Question 10.
What were all lined up?
Answer:
The blocks were all lined up for the athletes who would use them.

Question 11.
What was the event?
Answer:
The event was a hundred-yard dash.

Question 12.
How many athletes were there?
Answer:
There were nine athletes in the competition.

Question 13.
Which word tells that the athletes had taken a firm decision to win a medal?
Answer:
The word ‘ resolved ’ in the third line of the third stanza tells that the athletes had taken a firm decision to win a medal.

Question 14.
Where were those nine athletes?
Answer:
Those nine athletes were at the back of the starting line.

Question 15.
What were they poised for?
Answer:
They were poised for the sound of the gun.

Answer The Following In A Word Or A Phrase.

Question 1.
Where did the athletes come from?
Answer:
all over the country

Question 2.
Why did they run for it?
Answer:
for gold, silver, and bronze medals

Question 3.
How many athletes were there in the race?
Answer:
nine athletes

Question 4.
Who gathered round the old field?
Answer:
the spectators

Question 5.
Where did the spectators assemble?
Answer:
around the old field

Question 6.
What were all lined up?
Answer:
the blocks

Question 7.
What was the event?
Answer:
a hundred-yard dash

Question 8.
What were the athletes poised for?
Answer:
the sound of the gun

Question 9.
What exploded to signal the beginning of the race?
Answer:
the gun

Question 10.
Who was unable to run?
Answer:
the youngest of the runners

Question 11.
Where did the young athlete fall?
Answer:
to the asphalt

Question 12.
Why did the young athlete cry out?
Answer:
in frustration and anguish

Question 13.
What did the athletes carry with them?
Answer:
a banner

Question 14.
What did the banner say?
Answer:
Special Olympics

Question 15.
How did the athletes go to the finishing point?
Answer:
holding their hands

Question 16.
What does the poem ‘ Nine Gold Medals ‘ say?
Answer:
humanity, not personal achievement

Question 17.
Who is the poem ‘ Nine Gold Medals ’ written by?
Answer:
David Roth

Question 18.
What does the expression ‘ anguish’ mean?
Answer:
mental pain

Question 19.
What do we call a person who takes part in sports?
Answer:
athlete

Question 20.
What does the expression ‘nine beaming faces’ refer to?
Answer:
nine happy disabled athletes

Fill In The Blanks.

1. The athletes came from ____________ the country.
Answer:
all over

2. The spectators ____________ round the old field.
Answer:
gathered

3. The hundred-yard ____________ and the race to be run.
Answer:
dash

4. They ____________ for the sound of the gun.
Answer:
poised

5. The ____________ were all lined up for the race.
Answer:
blocks

6. They turned back ____________ to help the young athlete.
Answer:
one by one

7. They were holding a ____________ that read “Special Olympics”.
Answer:
banner

8. Being frustrated. he cried is ____________.
Answer:
anguish

9. They all were awarded ____________.
Answer:
gold medal

10. His dreams and efforts all ____________ is the dirt.
Answer:
dashed

Multiple Choice Questions With Answers.

Question 1.
How many athletes were there in the race?
(A) two athletes
(B) five athletes
(C) nine athletes
(D) six athletes
Answer:
(C) nine athletes

Question 2.
Who gathered round the old field?
(A) the visitors
(B) the travelers
(C) the players
(D) the spectators
Answer:
(D) the spectators

Question 3.
What exploded to signal the beginning of the race?
(A) the gun
(B) the hell
(C) the trumpet
(D) the drum
Answer:
(A) the gun

Question 4.
Where did the young athlete fall?
(A) to the ground
(B) to the mud
(C) to the asphalt
(D) to the rocks
Answer:
(C) to the asphalt

Question 5.
What did the athletes carry with them?
(A) a leaflet
(B)a flag
(C) a banner
(D) none of these
Answer:
(C) a banner

Question 6.
What do we call a person who takes pan in sports?
(A) participant
(B) coach
(C) athlete
(D) winner
Answer:
(C) athlete

Question 7.
The spectators ____________ round the old field.
(A) assembled
(B) viewed
(C) gathered
(D) walked
Answer:
(C) gathered

Question 8.
The ____________dashandcheracetoberun.
(A) hundred-yard
(B) fifty-yard
(C) twenty-yard
(D) ten-yard
Answer:
(A) hundred-yard

Question 9.
They ____________ for the sound of the gun.
(A) poised
(B) waited
(C) expected
(D) awaited
Answer:
(A) poised

Question 10.
The ____________ were all lined up for the race.
(A) blocks
(B) players
(C) tracks
(D) athletes
Answer:
(A) blocks

Question 11.
They turned back ____________ to help the young athlete.
(A) simultaneously
(B) one by one
(C) in a row
(D) quickly
Answer:
(B) one by one

Question 12.
They were holding a ____________ that read “Special Olympics”.
(A) leaflet
(B) hag
(C) banner
(D) signboard
Answer:
(C) banner

Question 13.
Being frustrated, he cried ____________.
(A) pain
(B) despair
(C) anger
(D) anguish
Answer:
(D) anguish

Question 14.
They all were awarded ____________.
(A) silver medals
(B) bronze medals
(C) iron medals
(D) gold medals
Answer:
(D) gold medals

Question 15.
His dreams and efforts ____________ all in the dirt.
(A) vanished.
(B) crashed
(C) dashed
(D) washed
Answer:
(C) dashed

Nine Gold Medals Summary in English

Lead-In:
David Roth has sent a great message to humanity that the world now needs no more competition but more cooperation and collaboration. The authenticity of the incident offers a moral value. The present times need it badly when people forget to help one another in an unhealthy rat race in every aspect of life.
ଉପକ୍ରମ :
ଡେଭିଡ୍ ରୋଥ୍ ଆମକୁ କବିତା ମାଧ୍ୟମରେ ଗୋଟିଏ ସୁନ୍ଦର ନୈତିକ ଭାବ ପରିବେଷଣ କରିଛନ୍ତି । ବର୍ତ୍ତମାନର ପୃଥ‌ିବୀ ପ୍ରତିଯୋଗିତା ଅପେକ୍ଷା ପରସ୍ପର ସହଯୋଗୀ ମନୋଭାବ ଅଧିକ ଦରକାର କରୁଛି । ବର୍ତ୍ତମାନର ପୃଥ‌ିବୀବାସୀ ତାହା ସମ୍ପୂର୍ଣ୍ଣରୂପେ ଭୁଲି ଯାଇଛନ୍ତି । ସାହାଯ୍ୟର ହାତ ବଢ଼ାଇବା ପରିବର୍ତ୍ତେ ପ୍ରତିଯୋଗିତାରେ ଜୟଲାଭ କରିବାକୁ ଚାହୁଁଛନ୍ତି । କବିତା “Nine Gold Medals” ମାଧ୍ୟମରେ କବି ସୁନ୍ଦର ଭାବରେ ଏକଥା ବର୍ଣ୍ଣନା କରିଛନ୍ତି । କବିତାଟି ନୈତିକତାବୋଧକ ଏବଂ ସୃଜନଶୀଳ ।

Stanzawise Analysis:
Stanza – 1
The athletes had come from all over the country
To run for the gold, for the silver and bronze
Many weeks and months of training
All comings down to these games.
ଅନୁବାଦ :
ସମଗ୍ର ଦେଶରୁ ଖେଳାଳିମାନେ ଦୌଡ଼ ପ୍ରତିଯୋଗିତାରେ ଭାଗ ନେବାକୁ ଆସିଥିଲେ । କେବଳ ସ୍ଵର୍ଣ୍ଣ, ରୌପ୍ୟ ଏବଂ କାଂସ୍ୟ ପଦକ ଜିତିବାପାଇଁ ସେମାନେ ଆଗରୁ ସପ୍ତାହ ସପ୍ତାହ ଏବଂ ମାସ ମାସ ଧରି ପ୍ରଶିକ୍ଷଣ ନେଇଥିଲେ ।

Stanza – 2
The spectators gathered around the old field
To cheer on all the young women and men
The final event of the day was approaching
Excitement grew high to began.
ଅନୁବାଦ :
ଏକ ପୁରାତନ ଖେଳପଡ଼ିଆର ଚାରିପାଖରେ ଦର୍ଶକମାନେ ଏକତ୍ର ହେଲେ । ସମସ୍ତେ ଖେଳାଳିମାନଙ୍କୁ ଉତ୍ସାହିତ କରିବାକୁ ଆସିଥିଲେ । କାରଣ ଏହା ଥିଲା ଶେଷ ପର୍ଯ୍ୟାୟ ଖେଳ । ଦିବସର ଶେଷ ପ୍ରତିଯୋଗିତାର ସମୟ ନିକଟତର ହେଉଥିଲା । ଉତ୍ସାହ ଓ ଉଦ୍ଦୀପନା ବଢ଼ିବାରେ ଲାଗିଥିଲା ।

Stanza – 3
The blocks were all lined up for those who would use them
The hundred-yard dash and the race to be run
These were nine resolved athletes in back of the starting line
Poised for the sound of the gun.
ଅନୁବାଦ :
ଖେଳାଳିମାନଙ୍କ ବ୍ୟବହାର ପାଇଁ ସମସ୍ତ ଟ୍ରାକ୍ ପ୍ରସ୍ତୁତ ହୋଇସାରିଥିଲା । ଶହେ ଗଜ ଦୂରତା ଦୌଡ଼ିବାର ଥିଲା । ନଅ ଜଣ ସଂକଳ୍ପବଦ୍ଧ କ୍ରୀଡ଼ାବିତ୍ ଆରମ୍ଭରେଖା ପଶ୍ଚାତ୍‌ରେ ଦୌଡ଼ିବା ନିମନ୍ତେ ପ୍ରସ୍ତୁତ ହୋଇ ବନ୍ଧୁକ ଫୁଟିବା ଶବ୍ଦକୁ ଅପେକ୍ଷା କରିଥିଲେ ।

Stanza – 4
The signal was given, the pistol exploded
And so did the runners all charging ahead
But the smallest among them, he stumbled and staggered
And fell to the asphalt instead.
ଅନୁବାଦ :
ପିସ୍ତଲ ଗର୍ଜି ଉଠିଲା ଓ ଦୌଡ଼ିବାର ସଙ୍କେତ ଦିଆଗଲା । ଖେଳାଳିମାନେ ଆଗକୁ ଦୌଡ଼ିବାକୁ ଆରମ୍ଭ କଲେ । ମାତ୍ର ସମସ୍ତଙ୍କଠାରୁ ସାନ ପିଲାଟି ଝୁଣ୍ଟି ପଡ଼ିଲା ଓ ଭାରସାମ୍ୟ ହରାଇ ତଳେ ପଡ଼ିଗଲା ।

Stanza – 5
He gave out a cry in frustration and anguish
His dreams and his efforts all dashed in the dirt
But as sure as I’m standing here telling this story
The same goes for what next occurred.
ଅନୁବାଦ :
ପିଲାଟି ହତାଶାବୋଧ, ମାନସିକ ଓ ଶାରୀରିକ ଯନ୍ତ୍ରଣାରେ ଚିତ୍କାର କଲା । ତାହାର ସମସ୍ତ ସ୍ବପ୍ନ ଓ ଉଦ୍ୟମ ମୁହୂର୍ତ୍ତକରେ ମାଟିରେ ମିଶିଲା । ମୁଁ ସେଠାରେ ଛିଡ଼ା ହୋଇ ଦେଖୁଥୁଲି ଏବଂ ପରବର୍ତ୍ତୀ ଘଟଣା ଠିକ୍ ନିମ୍ନ ପ୍ରକାର ଥିଲା ।

Stanza – 6
The eight other runners pulled up on their heels
The ones who had trained for so long to compete
One by one they all turned around and went back to help him
And brought the young boy to his feet.
ଅନୁବାଦ :
ଅନ୍ୟ ଆଠଜଣ ଖେଳାଳି ଅଟକିଗଲେ ଯେଉଁମାନେ କି ପ୍ରତିଯୋଗିତା ନିମନ୍ତେ ଦୀର୍ଘଦିନର ପ୍ରଶିକ୍ଷଣ ନେଇଥିଲେ । ଆଉ ଆଗକୁ ନ ଯାଇ ଜଣ ଜଣ କରି ସମସ୍ତେ ସାହାଯ୍ୟ କରିବାକୁ ପଛକୁ ଫେରିଲେ ଏବଂ ପିଲାଟିକୁ ଠିଆ କରିବାରେ ସାହାଯ୍ୟ କଲେ ।

Stanza – 7
Then all the nine runners joined hands and continued
The hundred-yard dash now reduced to a walk
And a banner above that said (Special Olympics)
Could not have been more on the mark.
ଅନୁବାଦ :
ତା’ପରେ ସମସ୍ତ ନଅ ଜଣ ଦୌଡ଼ାଳି ହାତ ଧରାଧରି ହୋଇ ଆଗକୁ ବଢ଼ିଲେ । ଶହେ ଗଜର ଦୌଡ଼ ଚାଲିବାରେ ରୂପାନ୍ତରିତ ହେଲା । ଉପରେ ଲାଗିଥିବା bannerରେ ଲିଖ୍ ‘Special Olympics” ନାମକୁ ଏହା ଯଥାର୍ଥରେ ସାକାର କରୁଥିଲା ।

Stanza – 8
That’s how the race ended, with nine gold medals
They came to the finish line holding hands still
And a standing ovation and nine beaming faces
Said more than these words ever will
ଅନୁବାଦ :
ଏହିପରି ଭାବେ ଖେଳ ସମାପ୍ତ ହେଲା । ନଅଟି ସୁନା ପଦକ ସମସ୍ତ ଖେଳାଳିଙ୍କୁ ଦିଆଗଲା । ନଅ ଜଣ ଖେଳାଳିଙ୍କର ଉଜ୍ଜ୍ବଳ ମୁଖମଣ୍ଡଳ ଓ ଦର୍ଶକମାନଙ୍କର ଛିଡ଼ା ହୋଇ ସମ୍ଭାଷଣ ଏହାଠାରୁ ଆହୁରି ଅଧ୍ଵ କିଛି କହୁଥିଲା ।

About The Poet: (କବିଙ୍କ ସମ୍ପର୍କରେ )
David Roth was bom on 10 October 1954 in Bloomington, Indiana. He is an American rock vocalist, song composer, author, actor, and radio personality.  The present poem is based on an inspiring event of a race in which nine differently-abled athletes took part in a “Special Olympic” event. In the race when one runner fell to the ground, all others stopped and returned to help him stand on his feet, and they walked hand-in-hand to finish the race. All of them won and were awarded gold medals for their empathy for each other. The poem is highly educative in terms of value education in this competitive world. The poet highlights humanity which is high above personal achievements. It conveys a great message to humanity that the world now needs no more competition but more love, affection, fellow feeling, cooperation, and collaboration.

Notes And Glossary: (ଟିପ୍‌ପଣୀ ଓ କଠିନ ଶବ୍ଦାର୍ଥ)
athlete — a participant in a group of sports activities
training — practice on anything
spectators — persons watching an event
cheer — to encourage — ଉତ୍ସାହିତ କରିବା
approaching — drawing near — ପାଖେଇ ଆସୁଛି
excitement — a state of being excited — ଉତ୍ତେଜନା
blocks — two starting blocks on the ground where the manners push their feet against at the beginning of the race
resolved — determined — ସଂକଳ୍ପବଦ୍ଧ
poised — ready — ପ୍ରସ୍ତୁତ
signal — a sign to start — ସଙ୍କେତ
stumble — hit the feet against something — ଝୁଣ୍ଟି ପଡ଼ିବା
staggered — lost balance — ଭାରସାମ୍ୟ ହରାଇଲେ
instead — ଏହ ପରିବର୍ତ୍ତେ
medal — a gift for winning an event — ପଦକ
exploded — sounded — ଶବ୍ଦ ହେଲା
frustration — hopelessness — ନିରାଶାବୋଧ
occurred — happened — ଘଟିଲା |
reduce — cut down
ovation — an expression of popular
appreciation — ସ୍ବୀକୃତିସୂଚକ ସମ୍ମାନ
beaming — ଆଲୋକିତ
anguish — terrible mental pain — ଭୟଙ୍କର ମାନସିକ ଯନ୍ତ୍ରଣା |

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 7 Linear Inequalities Ex 7(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Exercise 7(b)

Solve graphically
Question 1.
x < y
Solution:
x < y
Step – 1: Let us draw the dotted line x = y

X	0	1
y	0	1

Step – 2: Let us take a point say (1, 0) which is not on the line. Putting x = 1, y = 0 in the equation we get 1 < 0 (false).
⇒ (1, 0) does not satisfy the inequality.
⇒ The solution is the half-plane that does not contain (1, 0)
Step – 3: The shaded region is the solution region.

Question 2.
3x + 4y ≥ 12
Solution:
3x + 4y ≥ 12
Step – 1: Let us draw the line 3x + 4y = 12

X	4	0
y	0	3

Step – 2: Let us consider the point (0, 0) which is not on the line. Putting x = 0, y = 0 in the inequality we have 0 ≥ 12 (false).
∴ (0, 0) does not satisfy the inequality.
⇒ The half-plane that does not contain (0, 0) is the solution region.
Step – 3: The shaded region is the solution region.

Question 3.
x – y > 0
Solution:
x – y > 0
Step – 1: Let us draw the dotted line x = y.

X	0	1
y	0	1

Step – 2: Let us consider (1, 0) which is not on the line.
Putting x = 1, y = 0 in the inequation we get 1 > 0 (True)
⇒ (1, 0) satisfies the inequation.
⇒ The half-plane containing (1, 0) is the solution region.

Question 4.
x + 2y – 5 ≤ 0
Solution:
x + 2y – 5 ≤ 0
Step – 1: Let us draw the line x + 2y – 5 = 0
⇒ y = \(\frac{5-x}{2}\)

X	5	1
y	0	2

Step – 2: Let us consider the point (0, 0) which does not lie on the line putting x = 0, y = 0 in the inequation we get – 5 < 0 (True).
⇒ The point satisfies the inequation.
⇒ The half-plane containing (0, 0) is the solution region.
Step – 3: The shaded region is the solution region.

Question 5.
7x – 4y < 14
Solution:
7x – 4y < 14
Step – 1: Let us draw the dotted line 7x – 4y = 14

X	2	6
y	0	7

Step – 2: Let us consider the point (0, 0) which does not lie on the line. Putting x = 0, y = 0 in the inequality we get 0 < 14 (True).
⇒ (0, 0) satisfies the inequation.
⇒ The half-plane including (0, 0) is the solution region.
Step – 3: The solution region is the shaded region.

Question 6.
x + 8y + 10 > 0
Solution:
x + 8y + 10 > 0
Step – 1: Let us draw the dotted line x + 8y + 10 = 0

X	-10	-2
y	0	-1

Step – 2: Let us consider the point (0, 0) which does not lie on the line. Putting x = 0, y = 0 in the inequation we get 10 > 0 (True)
⇒ (0, 0) satisfies the inequality.
⇒ The half-plane containing origin is the solution region.
Step – 3: The solution region’ is the shaded region.

Question 7.
5x + 6y < 12
Solution:
5x + 6y < 12
Step – 1: Let us draw the dotted line 5x + 6y = 12

x	-6	0	6
y	7	2	-3

Step – 2: Putting x = 0, y = 0 in the equation we get, 0 < 12 (True)
⇒ The (0, 0) satisfies the equation.
Step – 3: The shaded region is the required solution.

Question 8.
– 3x + y > 0
Solution:
Step – 1: Let us draw the dotted line – 3x + y = 0

x	0	1	-1
y	0	3	-3

Step – 2: Putting x – 1, y = 0 in the equation we get, – 3 > 0 (false)
∴ Point (1, 0) does not satisfy the in the equation.
Step – 3: The shaded half-plane is the solution.

Question 9.
3x + 8y > 24
Solution:
Step- 1: Let us draw the dotted graph of 3x + 8y = 24

x	8	0
y	0	3

Step- 2: Putting x = 0, y- 0 we get, 0 > 24 (false)
∴ 0, (0, 0) does not satisfy the in equality.

Question 10.
x + y > 1
Solution:
Step – 1: Let us draw the graph x + y = 1

x	1	0
y	0	1

Step – 2: Putting x = 0, y = 0 in the equation we get, 0 > 1 (false)
∴ 0 (0, 0) does not satisfy the in equation
Step – 3: The shaded region is the solution.

Question 11.
x ≤ 0
Solution:
Step – 1: Let us draw the graph x = 0

Step – 2: Putting x = – 1 we get, – 1 ≤ 0 (True)
Thus, the shaded region is the solution.

Question 12.
y > 5
Solution:
Step – 1: Let us draw the dotted graph of y = 5

x	0	1	-1
y	5	5	5

Step- 2: Putting x = 0, y = 0 we have 0 > 5 (false)
we 0(0, 0) does not satisfy. the inequality.
Step – 3: The shaded region is the solution.

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 7 Linear Inequalities Ex 7(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Exercise 7(a)

Question 1.
Determine whether the solution set is finite or infinite or empty:
(i) x < 1000, x ∈ N
Solution:
Finite

(ii) x < 1, x ∈ Z (set of integers)
Solution:
Infinite

(iii) x < 2, x is a positive integer.
Solution:
Finite

(iv) x < 1, x is a positive integer.
Solution:
Empty

Question 2.
Solve as directed:
(i) 5x ≤ 20 in positive integers, in integers.
Solution:
5x ≤ 20
⇒ \(\frac{5 x}{5} \leq \frac{20}{5}\)
⇒ x ≤ 4
If x is a positive integer, then the solution set is {1, 2, 3, 4}
If x is an integer, then the solution set is:
S = {x : x ∈ Z and x ≤ 4}
= { ….. -3, -2, -1, 0, 1, 2, 3, 4}

(ii) 2x + 3 > 15 in integers, in natural numbers.
Do you mark any difference in the solution sets?
Solution:
2x + 3 > 15
⇒ 2x + 3 – 3 > 15 – 3
⇒ 2x > 12
⇒ \(\frac{2 x}{2}>\frac{12}{2}\)
⇒ x > 6
If x ∈ Z, then the solution set is S = (x : x ∈ Z and x > 6}
= {7, 8, 9…… }
If x ∈ N. then the solution set is S = {x : x ∈ N and x > 6}
= {7, 8, 9…… }
Two solution sets are the same.

(iii) 5x + 7 < 32 in integers, in non-negative integers.
Solution:
5x + 7 < 32
⇒ 5x + 7 – 7 < 32 – 7
⇒ 5x < 25
⇒ \(\frac{5 x}{5}<\frac{25}{5}\)
⇒ x < 5
If x ∈ Z, then the solution set is S = { x : x ∈ Z and x < 5 }
= {…..-3, -2, -1, 0, 1, 2, 3, 4}
If x is a non-negative solution then the solution set is S = {x : x is a non-negative integer < 5}
= (0, 1,2, 3,4}

(iv) -3x – 8 > 19, in integers, in real numbers.
Solution:
– 3x – 8 > 19
⇒ – 3 x – 8 + 8 > 19 + 8
⇒ – 3x > 27
⇒ \(\frac{-3 x}{-3}<\frac{27}{-3}\)
⇒ x < – 9
If x ∈ Z, then the solution set is S = (x : x ∈ Z and x < – 9}
= { ……..- 11, – 10}
If x ∈ R then the solution set is S = {x : x ∈ R and x < – 9}
= (∞, – 9)

(v) |x – 3| < 11, in N and in R.
Solution:
|x – 3| < 11
⇒ – 1 < x – 3 < 11
⇒ – 11 + 3 < x – 3 + 3 < 11+3
⇒ – 8 < x < 14
If x ∈ N the solution set is S = {1, 2, 3, 4, 5……..12, 13}
If x ∈ R then the solution set is: S = {x : x ∈ R and – 8 < x < 14}
= (- 8, 14)

Question 3.
Solve as directed:
(i) 2x + 3 > x – 7 in R
Solution:
2x + 3 > x – 7
⇒ 2x – x > – 7 – 3
⇒  x > – 10
x ∈ R, the solution set is S = (x : x ∈ R and x > – 10} = (-10, ∞)

(ii) \(\frac{x}{2}+\frac{7}{3}\) <  3x – 1 in R
Solution:
\(\frac{x}{2}+\frac{7}{3}\) <  3x – 1
\(\frac{3 x+14}{6}\) <  3x – 1
⇒ 3x + 14 < 18x – 6
⇒ 3x – 18x < – 6 – 14
⇒ – 15x < – 20
⇒ \(\frac{-15 x}{-15}>\frac{-20}{-15}\)
⇒ x > \(\frac{4}{3}\)
If x ∈ R, the solution set is S = \(\left(\frac{4}{3}, \infty\right)=\left\{x: x \in R \text { and } x>\frac{4}{3}\right\}\)

(iii) \(\frac{x}{2}-\frac{x}{3}+\frac{x}{5} \leq \frac{11}{3}\) for non-negative real numbers.
Solution:
\(\frac{x}{2}-\frac{x}{3}+\frac{x}{5} \leq \frac{11}{3}\)
⇒ \(\frac{15 x-10 x+6 x}{30}\) ≤ \(\frac{11}{3}\)
⇒ 11x ≤ \(\frac{11}{3}\) × 30
⇒ 11x ≤ 110
⇒ x ≤ 10
If x is a non-negative real number then the solution set is S = {x : x ∈ R and 0 ≤ x ≤ 10}
= {0, 10}

(iv) 2(3x – 1) < 7x + 1 < 3 (2x + 1) for real values.
Solution:
2(3x – 1) < 7x + 1 < 3(2x + 1)
⇒ 6x – 2 < 7x + 1< 6x + 3
⇒ – 2 < x + 1 < 3
⇒ – 3 < x < 2
If x ∈ R, the solution set is S = (x : x ∈ R and -3 < x < 2}
= {-3, 2}

(v) 7(x – 3) ≤ 4 (x + 6), for non-negative integral values.
Solution:
7(x – 3) ≤ 4(x + 6)
⇒ 7x – 21 ≤ 4x + 24
⇒ 7x – 4x ≤ 24 + 21
⇒ 3x ≤ 45
⇒ x ≤ 9
If x is a non-negative integer the solution set is S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(vi) Convert to linear inequality and solve for natural numbers: (x – 2) (x – 3) < (x + 3) (x – 1)
Solution:
(x – 2) (x – 3) < (x + 3) (x – 1)
⇒ x² – 5x + 6  <  x² + 2x – 3
⇒ – 5x + 6 < 2x – 3
⇒ – 5x – 2x < – 3 – 6
⇒ – 7x < – 9
⇒ x > \(\frac{9}{7}\)
If x ∈ N, the solution set is S = {2, 3, 4 }

(vii) Solve in R, \(\frac{x}{2}\) + 1 ≤ 2x – 5 < x. Also, find its solution in N.
Solution:
\(\frac{x}{2}\) + 1 ≤ 2x – 5 < x
⇒ \(\frac{x}{2}\) +1 ≤ 2x – 5 and 2x – 5 < x
⇒ \(\frac{x}{2}\) – 2x ≤ – 5 – 1 and x < 5
⇒ \(\frac{-3x}{2}\) ≤ – 6 and x < 5
⇒ – 3x ≤ – 12 and x < 5
⇒ x ≥ 4 and x < 5
⇒ 4 ≤ x < 5
If x ∈ R, the solution set is S = {x : x ∈ R and 4 < x < 5}
= {4, 5}
If x ∈ N, the solution set is S = { 4 }

(viii) Solve in R and also in Z: \(\frac{3 x+1}{5} \geq \frac{x+2}{3}-\frac{5-3 x}{5}\)
Solution:
\(\frac{3 x+1}{5} \geq \frac{x+2}{3}-\frac{5-3 x}{5}\)
⇒ \(\frac{3 x+1}{5} \geq \frac{5 x+10-15+9 x}{15}\)
⇒ 3x + 1 ≥ \(\frac{14 x-5}{3}\)
⇒ 9x + 3 ≥ 14x – 5
⇒ 9x – 14x ≥ – 5 – 3
⇒ – 5x ≥ – 8
⇒ x ≤ \(\frac{8}{5}\)
If x ∈ R, then the solution set is S = (x : x ∈ R and x ≤ \(\frac{8}{5}\)}
= (- ∞, \(\frac{8}{5}\))
If x ∈ Z, then the solution set is S = { x : x ∈ Z and x ≤ \(\frac{8}{5}\)}
= {……. -3, -2, -1, 0, 1}

Question 4.
Solve |x – 1| >1 and represent the solution on the number line.
[Exhaustive hints: By definition of modulus function
For x – 1 ≥ 0 or x ≥ 1, |x – 1| > 1
⇔ x – 1 > 1 ⇔ x > 2 ⇔ x ∈ (2, ∞)
For x- 1 < 0 or x < 1, |x – 1| > 1
⇔ – (x – 1) > 1
⇔ x – 1 < -1 (multiplication by -1 reverses the inequality)
⇔ x < 0 ⇔ x ∈ ( -∞, 0)
∴ The solution set is the Union,
(-∞, 0) ∪ (2, ∞) Show this as two disjoint open intervals on the number line, i.e., real line.]
Solution:
|x – 1| > 1
⇒ – 1 > x – 1 > 1
⇒ 0 > x > 2
⇒ x < 0 and x > 2
∴ The solution set is S = {x : x ∈ R, x < 0 and x > 2}
= (-∞, 0) ∪ (2, ∞)
We can show this solution in the number line as

Question 5.
Solve in R and represent the solution on the number line.
(i) |x – 5| < 1
Solution:
|x – 5| < 1
⇒ – 1< x – 5 < 1
⇒ 4 < x < 6
If x ∈ R, then the solution set is S = (4, 6)
We can represent the solution on the number line as

(ii) \(\frac{x}{5}<\frac{2 x+1}{3}+\frac{1-3 x}{6}\)
Solution:
\(\frac{x}{5}<\frac{2 x+1}{3}+\frac{1-3 x}{6}\)
⇒ \(\frac{x}{5}<\frac{4 x+2+1-3 x}{6}\)
⇒ \(\frac{x}{5}<\frac{x+3}{6}\)
⇒ 6x < 5x + 15
⇒ x < 15
If x ∈ R, the solution set is S = (-∞, 5)
We can represent the solution on the number line as

(iii) 2x + 1 ≥ 0
Solution:
2x + 1 ≥ 0
⇒ 2x ≥ -1
⇒ x ≥ -1/2
If x ∈ R, then the solution set is S = [\(-\frac{1}{2}\), ∞]
We can represent the solution on the number line as

(iv) \(\frac{x-1}{2} \leq \frac{x+1}{3}<\frac{3 x-1}{6}\)
Solution:
\(\frac{x-1}{2} \leq \frac{x+1}{3}<\frac{3 x-1}{6}\)
⇒ 3x – 3 ≤ 2x + 2 < 3x – 1
⇒ 3x – 3 ≤ 2x + 2 and 2x + 2 < 3x – 1
⇒ x ≤ 5 and – x < – 3
⇒ x ≤ 5 and x > 3
⇒ 3 < x ≤ 5
If x ∈ R, the solution set is S = {3, 5}
We can represent the solution on the number line as

Question 6.
In a triangle, ABC; AB, BC, and CA are x, 3x + 2, and x + 4 units respectively where x ∈ N. Find the length of its sides. (Hint: Apply triangle-inequality).
Solution:
Given AB = x
BC = 3x + 2
and CA = x + 4
Now AB + AC > BC (Triangle inequality)
⇒ x + x + 4 > 3x + 2
⇒ 2x + 4 > 3x + 2
⇒ – x > – 2
⇒ x < 2
As x ∈ N we have x = 1
The sides of triangle ABC are
AB = 1 unit
BC = 5 units
and CA = 5 units

Question 7.
The length of one side of a parallelogram is 1 cm. shorter than that of its adjacent side. If its perimeter is at least 26 c.m., find the minimum possible lengths of its sides.
Solution:
Let the longer side = x cm
∴ The smaller side = (x – 1) cm
Perimeter = 2(x + (x – 1)) = 4x – 2 cm
According to the question
4x – 2 ≥ 26
⇒ 4x ≥ 28
⇒ x > 7
The minimum value of x = 7.
∴ The minimum length of the sides is 7cm and 6 cm.

Question 8.
The length of the largest side of a quadrilateral is three times that of its smallest side. Out of the other two sides, the length of one is twice that of the smallest and the other is 1 cm. longer than the smallest. If the perimeter of the quadrilateral is at most 36 c.m., then find the maximum possible lengths of its sides.
Solution:
Let the smallest side = x cm.
Largest side = 3 times x = 3x cm.
The other two sides are 2x cm and x + 1 cm.
⇒ The perimeter = x + 3x + 2x + x + 1
= 7x + 1 cm
According to the question:
7x + 1 ≤ 36
⇒ 7x ≤ 35
⇒ x ≤ 5
Maximum value of x = 5
∴ The maximum possible length of sides are x = 5 cm, 3x = 15 cm, 2x = 10 cm, and x + 1 = 6 cm.

Question 9.
Find all pairs of consecutive odd numbers each greater than 20, such that their sum is less than 60.
Solution:
Let two consecutive odd numbers are
2n – 1 and 2n + 1
Now 2n – 1 > 20 and 2n + 1 > 20
But their sum = 2n – 1 + 2n + 1
= 4n < 60
⇒ n < 15
for n = 14 two numbers are 27, 29
for n = 13 two numbers are 25, 27
for n = 12 two numbers are 23, 25
for n = 11 two numbers are 21, 23
∴ All pairs are 21, 23; 23, 25; 25, 27 and 27, 29

Question 10.
Find all pairs of even numbers each less than 35, such that their sum is at least 50.
Solution:
Let two even numbers be x and y.
According to the question
x < 35, y < 35 and x + y ≥ 50
⇒ x ≤ 34, y ≤ 34 and x + y ≥ 50
⇒ x + y ≤ 70, x + y ≥ 50
⇒ 50 ≤ x + y ≤ 70
If x + y = 50 the numbers are {34, 16}, {32, 18}, {30, 20}, {28, 22}, {26, 24}
If x + y = 52 the numbers are {34, 18}, {32, 20}, {30, 22}, {28, 24}, {26, 26}
If x + y = 34 the numbers are {34, 20}, {32, 22}, {30, 24}
If x + y = 56 the numbers are {34, 22}, {32, 24}, {30, 26}, {28, 28}
If x + y = 58 the numbers are {34, 24}, {32, 26}, {30, 28}
If x + y = 60 the numbers are {34, 26}, {32, 28}, {30, 30}
If x + y = 62 the numbers are {34, 28}, {32, 30}
If x + y = 64 the numbers are {34, 30}, {32, 32}
If x + y = 68 the numbers are {34, 34}

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Ex 3(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Exercise 3(a)

Question 1.
Compute the product A × B when
(i) A = {0} = B
(ii) A = {a, b}, B = {a, b, c}
(iii) A = Z, B = Φ
Solution:
(i) A = {0} = B
∴ A × B = {(0, 0)}

(ii) A = {a, b}, B = {a, b, c}
∴ A × B = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c)}

(iii) A = Z, B = Φ ∴ AxB = Φ

Question 2.
If |A| = m, |B| = n, what can you say about
(i) |A × B| (ii) |P(A) × P(B)|
Solution:
If |A| = m. |B| = n then

(i) lA × B| = mn.

(ii) |P(A)| = 2^m . |P(B)| = 2ⁿ
∴  |P(A) × P(B)| =2^m × 2ⁿ = 2^m+n

Question 3.
Find x, y if
(i) (x, y) = (-3, 2)
(ii) {x + y, 1) = (1, x – y)
(iii) (2x + y, 1) = (x, 2x + 3y)
Solution:

(i) ∴ x = – 3, y = 2

(ii) ∴ x + y = 1, x – y = 1
∴ 2x = 2 or, x = 1
∴ y=0

(iii) ∴ 2x + y = x, 1 = 2x + 3y
∴ {x + y = 0} × 2
2x + 3y = 1
–      –      –
∴ – y = – 1 or, y = 1
∴ x = – 1

Question 4.
If, A × B = B × A then what can you
Solution:
If A × B = B × A then A = B

Question 5.
|A × B| = 6. If ( -1, y ), (1, x), (0, y) are in A × B. Write other elements in A × B, where x ≠ y.
Solution:
Let |A × B| = 6 and (-1, y) (1, .x) (0, y) ∈ A × B
⇒ -1, 1, 0 ∈ A and x, y ∈ B
As |A × B| = 6 and 3 × 2 = 6
We have A = {-1, 1, 0} and B = {x, y}
Thus other elements of A × B is (-1, x) , (1, y), (0, x)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 4 Trigonometric Functions Ex 4(d) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Exercise 4(d)

Question 1.
Fill in the blanks choosing the correct answer from the brackets.

(i) In Δ ABC, b =____________. (b cos B + c cos C, a cos A + c cos C, c cos A + a cos C)
Solution:
c cos A + a cos C

(ii) If a cot A = b cot B then Δ ABC is__________. (isosceles, right-angled, equilateral)
Solution:
isosceles

(iii) In Δ ABC if b sin C = c sin B = 2 then b sin C = ___________. (0, 1, 2)
Solution:
1

(iv) In Δ ABC if \(\frac{\cos \mathrm{A}}{a}=\frac{\cos \mathrm{B}}{b}=\frac{\cos \mathrm{C}}{c}\) then the tringle is_________ (equilateral, isosceles, scalene)
Solution:
equilateral

(v) If sin A = sin B and b = 1/2, then a = _______________. (2, 1/2, 1)
Solution:
a = 1/2

(vi) In Δ ABC if A = 60°, B = 45° a : b = __________. ( √2 : √3, √6 : 2, √3 : 2)
Solution:
√6 : 2

(vii) In Δ ABC if b² + c² < a² , then _________ angle is obtuse. (A, B, C)
Solution:
A

(viii) If a cos B = b cos A. then cos B = _____________. \(\left(\frac{c}{a}, \frac{a}{2 c}, \frac{c}{2 a}\right)\)
Solution:
cos B = \(\frac{c}{2 a}\)

(ix) If a – b cos C, then __________ angle is a right angle. (A, B, C)
Solution:
∠B is a right angle

(x) If a = 12, b = 7, C = 30°, then Δ = ______________. (42, 84, 21)
Solution:
Δ = 21

Question 2.
Prove that
(i) a sin A – b sin B = c sin (A – B)
Solution:
R.H.S. = c sin (A – B)
= 2R sin C sin (A – B)
= 2R sin (A + B) sin (A – B)
[∴ A + B + C = π or, A + B = π – C
or sin (A + B) = sin (π – C) sin C]
= 2R (sin² A – sin²  B)
= 2R sin A sin A – 2R sin B sin B
= a sin A – b sin B = L.H.S.

(ii) b cos B + c cos C = a cos (B – C)
Solution:
R.H.S. = a cos (B – C)
= 2R sin A cos (B – C)
= 2R sin (B + C) cos (B – C)
= R sin (B + C + B – C) + sin (B + C – B + C)}
= R [(sin 2B + sin 2C)
= R(2sin B cos B + 2 sin C cos C)
= 2R sin B cos B + 2R sin C cos C
= b cos B + c cos C = L.H.S.

(iii) If (a + b + c)(b + c – a) = 3bc, then A = 60°
Solution:

(iv) If \(\frac{b+c}{5}=\frac{c+a}{6}=\frac{a+b}{7}\) then sin A : sin B : sin C = 4 : 3 : 2
Solution:

(v) If A: B: C = 1 : 2 : 3 then sin A: sin B: sin C = 1 : 2
Solution:

(vi) If b² + c² – a² = bc, then A = 60°
Solution:
If b² + c² – a² = bc, then A = 60°
or, \(\frac{b^2+c^2-a^2}{2 b c}\) = 1/2 or, cos A = 1/2
or, A = 60°

(vii) If A : B: C = 1 : 2 : 7, then c: a = (√5 + 1) : (√5 – 1)
Solution:

But we know that \(\frac{\sin C}{\sin A}=\frac{c}{a}\)
∴ \(\frac{c}{a}=\frac{\sqrt{5}+1}{\sqrt{5}-1}\)

Question 3.
(i) If cos A = \(\frac{12}{13}\), cos B = \(\frac{5}{13}\) then find a : b.
Solution:

(ii) If a = 7, b = 3, c = 5 then find A.
Solution:

(iii) If a = 8, b = 6, c = 4 find tan \(\frac{B}{2}\)
Solution:

(iv) If \(\frac{a}{\sec A}=\frac{b}{\sec B}\) and a ≠ b then find C.
Solution:

(v) If a = 48, b = 35, ∠C = 60° then find c.
Solution:

In Δ ABC prove that (Q. 4 to Q. 26)

Question 4.
a sin (B – C) + b sin (C – A) + c sin (A – B) = 0
Solution:
a sin (B – C) + b sin (C – A) + c sin (A – B)
= 2R sin A sin (B – C) + 2R sin B sin (C – A) + 2R sin C sin (A – B)
= 2R [sin (B + C) sin (B – C) + sin (C + A) sin (C – A) + sin (A + B) sin (A – B)]
= 2R[sin² B – sin² C + sin² C – sin² A + sin² A – sin² A]
= 2R x 0 = 0

Question 5.
\(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b \cos C-c \cos B}{b \cos C+c \cos B}\)
Solution:

Question 6.
\(\sum \frac{a^2 \sin (B-C)}{\sin (B+C)}=0\)
Solution:

Question 7.
a²(cos² B – cos² C) + b²(cos² C – cos² A) + c²(cos² A – cos² B) = 0
Solution:

Question 8.
\(\frac{b^2-c^2}{a^2} \sin 2 A+\frac{c^2-a^2}{b^2} \sin 2 B\) \(+\frac{a^2-b^2}{c^2} \sin 2 \mathrm{C}=0\)
Solution:

Question 9.
\(\frac{a^2\left(b^2+c^2-a^2\right)}{\sin 2 \mathrm{~A}}=\frac{b^2\left(c^2+a^2-b^2\right)}{\sin 2 \mathrm{~B}}\) \(=\frac{c^2\left(a^2+b^2-c^2\right)}{\sin 2 C}\)
Solution:

Question 10.
\(\Sigma \frac{\cos A}{\sin B \cdot \sin C}=2\)
Solution:

Question 11.
(a² – b² + c²) tan B = (a² + b² – c²) tan C
Solution:

Question 12.
(b² – c²) cot A + (c² – a²) cot B + (a² – b²) cot C = 0
Solution:

Question 13.
\(\frac{b+c}{a}=\frac{\cos \mathbf{B}+\cos C}{1-\cos A}\)
Solution:

Question 14.
\(\sum a^3 \sin (B-C)=0\)
Solution:

Question 15.
(b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c
Solution:
L.H.S. = (b + c) cos A + (c + a) cos B + (a + b) cos C
= b cos A + c cos A + a cos B + c cos B + a cos C + b cos C
= (b cos C+ c cos B) +(c cos A + a cos C) + (a cos B + b cos A)
= a + b + c = R.H.S.

Question 16.
2 (bc cos A + ca cos B + ab cos C) = a² + b² + c²
Solution:
2 (bc cos A + ca cos B + ab cos C) = a² + b² + c²
\(=2\left(b c \times \frac{\left(b^2+c^2-a^2\right)}{2 b c}+c a \times \frac{c^2+a^2-b^2}{2 c a}\right.\) \(\left.+a b \times \frac{a^2+b^2-c^2}{2 a b}\right)\)
= b² + c² – a² + c² + a² – b² + a² + b²– c²
= a² + b² + c²

Question 17.
a (b² + c²) cos A + b (c² + a²) cos B + c(a² + b²) cos C = 3 abc.
Solution:
a (b² + c²) cos A + b (c² + a²) cos B + c(a² + b²) cos C
= ab² cos A + ac² cos A + bc² cos B + ba² cos B + ca² cos C + cb² cos C
= ab² cos A + ba² cos B + ac² cos A + ca² cos C + bc² cos B + cb² cos C
= ab (b cos A + a cos B) + ac (c cos A + a cos C) bc (c cos B + b cos C)
= abc = abc + abc = 3abc

Question 18.
a³ cos (B – C) + b³ cos (C – A) + c³ cos (A – B) = 3 abc
Solution:
1st term of L.H.S. = a³ cos (B – C)
= a² a cos (B – C)
= a² . 2R sin A cos (B – C)
= 2a²R sin (B + C) cos (B- C)
= a²R [sin (B + C + B – C) + sin (B + C – B + C)]
= a² R (sin 2B + sin 2C)
= a²R [2 sin B cos B + 2 sin C cos C]
= a² [2R sin B cos B + 2R sin C cos C]
= a² (b cos B + c cos C)
Similarly, 2nd term
= b² (c cos C + a cos A) and
3rd term = c² (a cos A + b cos B)
∴ L.H.S.= a²b cos B+a²c cos C+b²c cos C + b²a cos A + c²a cos A + c²b cos B
= ab (a cos B + b cos A) + bc (b cos C + c cos B) + ca (c cos A + a cos C)
= abc + bca + cab = 3abc = R.H.S.

Question 19.
a (cos B + cos C) = 2(b + c) sin² \(\frac{A}{2}\)
Solution:
Refer to Q. N. 13.

Question 20.
(b + c – a) tan \(\frac{A}{2}\) = (c + a – b) tan \(\frac{B}{2}\) = (a + b – c) tan \(\frac{C}{2}\)
Solution:

Question 21.
\((b+c-a)\left(\cot \frac{B}{2}+\cot \frac{C}{2}\right)\) \(=2 a \cot \frac{A}{2}\)
Solution:

Question 22.
(a – b)² cos² \(\frac{C}{2}\) + (a + b)² sin² \(\frac{C}{2}\) = c²
Solution:
L.H.S = (a – b)² cos² \(\frac{C}{2}\) + (a + b)² sin² \(\frac{C}{2}\)

Question 23.
1 – tan \(\frac{A}{2}\) tan \(\frac{B}{2}\) = \(\frac{c}{2}\)
Solution:

Question 24.
(b – c) cot \(\frac{A}{2}\) + (c – a) cot \(\frac{B}{2}\) + (a – b) cot \(\frac{C}{2}\) = 0
Solution:

Question 25.
cot A + cot B + cot C = \(\frac{a^2+b^2+c^2}{4 \Delta}\)
Solution:

Question 26.
a² cot A + b² cot B + c² cot C = 4Δ
Solution:

Question 27.
If \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\) then prove C = 60°.
Solution:

or, cos C = 1/2 or, ∠C = 60°

Question 28.
If a = 2b and A = 3B, find the measures of the angles of the triangle.
Solution:
If a = 2b and A = 3B, we have \(\frac{a}{b}\) = 2
or, \(\frac{\sin A}{\sin B}\) = 2
or, sin A = 2 sin B         …..(1)
Also  sin A = sin 3B (as a = 3B)    …..(2)
∴ From (1) and (2), we have
2 sin B = sin 3B = 3 sin B – 4 sin³ B
or, 4 sin³ B – sin B = 0
or, sin B(4 sin² B – 1) = 0
or, sin B = 0, 4 sin² B = 1
Now sin B = 0 ⇒ B = 0 (Impossible)
∴ sin² B = \(\frac{1}{2}\) or, sin B = ± \(\frac{1}{2}\)
If sin B = \(\frac{1}{2}\) then ∠B = 30°
∴ A = 3B = 3 x 30° = 90°
∠C = 60°

Question 29.
If a⁴ + b⁴ + c⁴ = 2c² (A² + b²), prove that m∠ACB = 45° or 135°.
Solution:
a⁴ + b⁴ + c⁴ = 2c² (a² + b²)
or, a⁴ + b⁴ + c⁴ + 2a²b² – 2b²c² – 2c²a² = 2a²b²
or, (a² + b² – c²)² = 2a²b²
or, a² + b² – c² = ± √2 ab
or, \(\frac{a^2+b^2-c^2}{2 a b}=\pm \frac{1}{\sqrt{2}}\)
or, cos C = ± \(\frac{1}{\sqrt{2}}\)
∴ ∠C = 45° or 135°.

Question 30.
If x² + x + 1, 2x + 1, and x² – 1 are lengths of sides of a triangle, then prove that the measure of the greatest angle is 120°.
Solution:

Question 31.
if cos B = \(\frac{\sin A}{2 \sin C}\), prove that the triangle is isosceles.
Solution:
cos B = \(\frac{\sin A}{2 \sin C}\)
⇒ \(\frac{c^2+a^2-b^2}{2 c a}=\frac{a}{2 c}\) ⇒ c² + a² – b² = a²
or, c² = b² or, c = b
∴ The triangle is isosceles.

Question 32.
If a tan A + b tan B = (a + b)tan \(\frac{1}{2}\) (A + B) prove that the triangle is isosceles.
Solution:

Question 33.
If (cos A + 2 cos C) : (cos A + 2 cos B) = sin B : sin C prove that the triangles are either isosceles or right-angled.
Solution:
\(\frac{\cos A+2 \cos C}{\cos A+2 \cos B}=\frac{\sin B}{\sin C}\)
⇒ cos A sin C = cos A sin B + 2 cos B sin B
⇒ cos A (sin B – sin C) + (sin 2B – sin 2c) = 0
⇒ cos A (sin B – sin C) + 2 cos (B + C) sin (B – C) = 0
⇒ cos A (sin B – sin C) – 2 cos A sin (B – C) = 0
(∴ cos (B + C) = cos (π – A) = – cos A)
⇒ cos A = 0 or sin B – sin C – 2 sin (B – C) = 0
cos A = 0 ⇒ A = 90°
i.e. the triangle is right-angled. sin B sin C – 2 sin (B – C) = 0

Question 34.
If cos A = sin B – cos C, prove that the triangle is right-angled.
Solution:
cos A = sin B – cos C
or, cos C + cos A = sin B

or, 2C = π
or, C = \(\frac{\pi}{2}\)

Question 35.
If a², b², c²  be in A.P., prove that cot A, cot B, cot C are also in A.P.
Solution:
If a², b², c²  be in A.P.
then b² – a² = c² – b²
or, 2b² = c² + a²
or, \(b^2=\frac{c^2+a^2}{2}\)
or, 2b² = c² + a² …..(1)
We have to prove that cot A, cot B, cot C are in A.P.
i.e. to prove cot B – cot A = cot C – cot B
i.e. 2 Cot B = cot C + cot A
∴ R.H.S. = cot C + cot A

Question 36.
If sin A: sin C = sin (A – B) : sin (B – C) prove that a², b², c² are in A.P.
Solution:
\(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\)
or, sin A sin (B – C) = sin C sin (A – B)
or, sin (B + C) sin (B – C) = sin (A + B) sin (A – B)
or, sin² B – sin² C = sin² A – sin² B
or, 2 sin² B = sin² C + sin² A
or, \(2 \frac{b^2}{4 \mathrm{R}^2}=\frac{c^2}{4 \mathrm{R}^2}+\frac{a^2}{4 \mathrm{R}^2}\)
or, 2b² = c² + a²
or, b² – a² = c² – b²
∴ a², b², c² are in A.P

Question 37.
If the side lengths a, b, and c are in A.P., then prove that cos \(\frac{1}{2}\) (A – C) = 2 sin \(\frac{1}{2}\) B.
Solution:
If a,b, and c are in A.P. then b – a – c – b or, 2b = c + a
We have to prove that

Question 38.
If the side lengths a, b, and c are in A.P., prove that cot \(\frac{1}{2}\) A, cot \(\frac{1}{2}\) B, cot \(\frac{1}{2}\) C are in A.P.
Solution:

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 2 Sets Ex 2(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 2 Sets Exercise 2(b)

Question 1.
An examination was conducted in physics, chemistry, and mathematics. If P.C.M. denotes respectively the sets of students who passed in Physics, Chemistry, and Mathematics, express the following sets using union, intersection, and different symbols.
(a) Set of candidates who passed in Mathematics and Chemistry, but not in Physics.
(b) Set of candidates who passed in all three subjects.
(c) Set of candidates who passed in Mathematics only.
(d) Set of candidates who failed in Mathematics, but passed in at least one subject.
(e) Set of candidates who passed in at least two subjects.
(f) Set of candidates who failed in one subject only.
Solution:
An examination was conducted in Physics, Chemistry, and Mathematics. P, C, and M denoted the set of students who passed Physics, Chemistry, and Mathematics, respectively. Then.
(a) Set of candidates who passed in Mathematics and Chemistry, but not in Physics (M ∩ C) – P.
(b) Set of candidates who passed in all three subjects M ∩ C ∩ P.
(c) Set of candidates who passed in Mathematics only M – C – P.
(d) Set of candidates who failed in Mathematics, but passed in at least one subject (P ∪ C) – M.
(e) Set of candidates who passed in at least two subjects.
(f) Set of candidates who failed in one subject only.
(P ∩ C – M) ∪ (P ∩ M – C) ∪ (M ∩ C – P)

Question 2.
What can you say about sets A and B if
(i) A ∪ B= Φ
(ii) A Δ B = Φ
(iii) A \ B = Φ
(iv) A \ B = A
(v) A ∩ B= U, Where U is the Universal set, A \ B = U?
Solution:
(i) if A ∪ B = Φ then A = Φ =B
(ii) A Δ B = Φ ⇒ A = B
(iii) A – B = Φ ⇒ A ⊆ B
(iv) A – B = A ⇒ B = Φ
(v) A ∩ B = U ⇒ A = B = U
(vi) A – B = U ⇒ A = U and B = Φ

Question 3.
Are differences and symmetric commutative? Give reason.
Solution:
The difference of the two sets is not commutative but the symmetric of the two sets is commutative.
Reason:
Let x ∈ A – B ⇔ x ∈ A ∧ x ∉ B
≠ x ∈ B ∧ x ∉ A ⇔ x ∈ b – A
A- B ≠ B – A
But if y ∈ A Δ ⇔ y ∈ (A-B) ∪ (B – A)
⇒ y ∈ (B – A) ∪ (A – B) ⇔ y ∈ B Δ A
∴ A Δ B = B Δ A.

Question 4.
If B ⊂ C, prove that A/B = A/C. Is this result true when a difference is replaced by a symmetric difference? Give reason.
Solution:
If B ⊂ C, then x ∈ A ⇒ x ∈ C
Now x ∈ A – C ⇔ x: x ∈ A ∧ x ∉ C
⇔ {x: x ∈ A ∧ x ∉ B}
⇔ {x: x ∈ A – C}
∴ A – C = A – B
but, A Δ B ≠ A Δ C.

Question 5.
Prove the following :
(i) (A\B)\C = (A\C)\B = A\(B ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∪ B)
(iii) A Δ (B Δ C) = (A Δ B) Δ C
(iv) A ⊂ B ⇔ B’ ⊂ A’ A ⇔ A’ ∪ B = U
⇔ B’ ∩ A = Φ, where U is the universal set.
(v) A ∪ B = U and A ∩ B = Φ
⇒ B = A’
(iv) A ∪ B = A for all A ⇒ B = Φ
Solution:

(i) Let x ∈ (A – B) – C    ……(1)
⇔ x ∈ A- B ∧ x ∉ C
⇔ (x ∈ A ∧ x ∉ B) ∧ x ∉ C   ……(2)
⇔ (x ∈ A ∧ x ∉ C) ∧ x ∉ B
⇔ x ∈ A – C ∧ x ∉ B
⇔ x ∈ (A – C) – B   ……(3)
∴ from (2), we have
x ∈ A ∧ ∉ B ∧ x ∉ C
⇔ x ∈ A ∧ (x ∉ B ∧ x ∉ C)
⇔ x ∈ A ∧ x ∉ B ∪ C
[∴ ~ (p ∨ q) = ~ p ∧ ~ q]
⇔ x ∈ A – (B ∪ C)   …….(4)
∴ From (1), (3), and (4), we have
(A – B) – C = (A – C)-B = A – (B ∪ C)

(ii) Let x ∈ A ∩ (B ∪ C)
⇔ x ∈ A ∧ x ∈ B Δ C
⇔ x ∈ A ∧ (x ∈ B – C ∨ x ∉ C – B)
⇔ x ∈ A ∧ (x ∈ B ∧ x ∉ C ∨ x ∈ C ∧ x ∉ B)
⇔ x ∈ A ∧ (x ∈ B ∧ x ∈ A ∧ C) ∨ x ∈ A ∧ (x ∈ B ∧ x ∈ B ∧ x ∈ A ∧ x ∉ B)
⇔ (x ∈ A ∩ B ∧ x ∉ A ∩ C) ∨ (x ∈ A ∩ C  ∨ x ∉ A ∩ B)
⇔ x ∈ (A ∩ B) – (A ∩ C) ∨ x ∈ (A ∩ C) – (A ∩ B)
⇔ x ∈ (A ∩ B) Δ (A ∩ C)   ……(2)
∴ From (1) and (2), we have
A ∩ (B Δ C) = (A ∩ B) Δ (A ∩ C)

(iii) A Δ (B Δ C)
= A ∪ (B Δ C)- A ∩ (B Δ C)
= A Δ (B ∪ C)- A Δ (B ∩ C)
[ ∴ A ∪ (B Δ C) = A Δ (B ∪ C)]
and A n (B Δ C) = A Δ (B ∩ C)
= (A Δ B) ∪ C- (A Δ B) ∩ C
[∴ A Δ (B ∪ C) = (A Δ B) ∪ C and A Δ (B ∩ C) = (A Δ B) ∩ C
= (A Δ B) Δ C
∴ A Δ (B Δ C)= (A Δ B) Δ C
(Proved)

(iv) If A ⊂ B then x ∈ B’ or ⇒ x ∉ B
⇒ x ∉ B ⇒ x ∈ A’ (∴ A ⊂ B)
∴ B’ ⊂ A’
Again, let y ∈ A ⇒ y ∉ A’ ⇒ y ∈ B’
(∵ B’ ⊂ A’)
⇒ y ∈ B ∴ A ⊂ B
∴ A ⊂ B ⇔ B’ ⊂ A’
∴ Again as A ⊂ B, we have
U = A ∪ B = B = U, where U is the universal set of A and B.
∴ A’= B – A ⇒ A’ ∪ B
= (B – A) ∪ B = B = U
∴ A ⊂ B ⇒ A’ ∪ B = U
Again A’ ∪ B = U
⇒ A ∩ (A’ ∪ B) = A ∩ U = A
⇒ (A ∩ A’) ∪ (A ∩ B) = A
⇒ Φ ∪ (A ∩ B) = A
⇒ A ∩ B = A ⇒ A ⊂ B
Lastly, B’ = U’ = Φ
∴ B’ ∩ A = Φ

(v) Let A ∪ B = U and A ∩ B = Φ
∴ Let x ∈ B ⇔ x ∉ B’ ⇔ x ∉ U – B
⇔ x ∉ A ⇔ x ∉ A’

(vi) As A ∪ B = A for all A
we have B ⊂ A for all A
∴ B ⊂ A even for A = Φ Thus B = Φ

Question 6.
Prove all the results of sections 1.13 and 1.14 that are started without proof.
Solution:
(i) A ∪ B = B ∪ A
Let x ∈ A ∪ B ⇔ x ∈ A ∨ x ∈ B
⇔ x ∈ B ∨ x ∈ B ⇔ x ∈ B ∪ A

(ii) A ∩ B = B ∩ A
Let x ∈ A ∩ B ⇔ x ∈ A ∧ x ∈ B
∴ A ∩ B = B ∩ A

(iii) A ∩ (B ∪ C)
= (A ∩ B) ∪ (A ∩ C)
Let x ∈ A ∩ (B ∪ C)
⇔ x ∈ A ∧ x ∈ A ∪ C
⇔ x ∈ A ∧ (x ∈ A ∨ x ∈ C)
⇔ (x ∈ A ∩ B ∨ x ∈ A ∩ C)
⇔ (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈C)
⇔ x ∈ (A ∩ B) ∪ (A ∩ C)
∴ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Question 7.
Prove that
(i) \(\mathbf{A}-\bigcup_{i=1}^n \mathbf{B}_i=\bigcap_{i=1}^n\left(\mathbf{A}-\mathbf{B}_i\right)\)
Solution:
Let x ∈ \(A-\bigcup_{i=1}^n B_i \Rightarrow x \in A \wedge x \notin \bigcup_{i=1}^n B_i\)
⇔ x ∈ A  ∧ x ∉(B₁ ∪ B₂ ∪….∪ B_n )
⇔ x ∈ A  ∧ (x ∉ B₁ ∧ x ∉ B₂ ∧…..∧ x ∉ B_n )
⇔ (x ∈ A  ∧ x ∉ B₁ ) ∧ (x ∈ A ∧ x ∉ B₂ ) ∧….∧ (x ∈ A  ∧ x ∉ B_n )
⇔ x ∈ A – B₁ ∧ x ∈ A – B₂ ∧……..∧ x ∈ A – B_n
⇔ x ∈ (A – B₁ ) ∩ (A – B_i ) ∩…..∩ (A – B_n )
⇔ \(x \in \bigcap_{i=1}^n\left(\mathrm{~A}-\mathrm{B}_{\mathrm{i}}\right)\)
∴ \(\mathrm{A}-\cup_{i=1}^n \mathrm{~B}_{\mathrm{i}}=\bigcap_{i=1}^n\left(\mathrm{~A}-\mathrm{B}_{\mathrm{i}}\right)\)

(ii) ∴ \(\mathbf{A}-\bigcap_{i=1}^n \mathbf{B}_i=\bigcup_{i=1}^n\left(\mathbf{A}-\mathbf{B}_i\right)\)
Solution:
Let x ∈ \(A-\bigcap_{i=1}^n B_i\)
⇔ x ∈ A ∧ x ∈ \(\bigcap_{i=1}^n \mathrm{~B}_{\mathrm{i}}\)
⇔ x ∈ A ∧ x ∉ (B₁ ∩ B₂ ∩….∩ B_n )
⇔ x ∈ A ∧ (x ∉ B₁ ∨ x ∉ B₂ ∨….∨ x ∉ B_n )
⇔ (x ∈ A ∧ x ∉ B₁ ) ∨ (x ∈ A ∧ x ∉ B₂ ) ∨….∨ (x ∈ A ∧ x ∉ B_n )
⇔ x ∈ A – B₁ ∨ x ∈ A – B₂ ∨……..∨ x ∈ A – B_n
⇔ x ∈ (A – B₁ ) ∪ (A – B₂ )……(A – B_n )
⇔ x ∈ \(\cup_{i=1}^n\left(\mathrm{~A}-\mathrm{B}_{\mathrm{i}}\right)\)
∴ \(\mathrm{A} \bigcap_{i=1}^n \mathrm{~B}_1=\bigcup_{u=1}^n\left(\mathrm{~A}-\mathrm{B}_i\right)\)

Question 8.
Prove that |A ∪ B ∪ C|
Solution:
= |A| + |B| + |C| + |A ∩ B ∩ C| – |A ∩ B| – |B ∩ C| – |C ∩ A|
L.H.S. =|A ∪ B ∪ C| = |A ∪ D|
where D = B ∪ C
= |A| + |D| – |A ∩ D|
( ∵ |A ∪ B| =|A| + |B| – |A ∩ B|)
= |A| + |B ∪ C| – | A ∩ (B ∪ C)|
= |A| + |B| + |C| – |B ∩ C| – |(A ∩ B) ∪ (A ∩ C)|
= |A| + |B| + | C |- |B ∩ C| – [|A ∩ B| + |A ∩ C| – |(A ∩ B) ∩ (A ∩ C)|]
= |A| + |B| + |C| – |B ∩ C| – |A∩ B| – |A ∩ C| + |A ∩ B ∩ Cl
= |A| + |B| + |C| – | A ∩ Bl – |B ∩ C| – |C ∩ A| + |A ∩ B  ∩ Cl = R.H.S.

Question 9.
If X and Y are two sets such that X ∪ Y has 20 objects, X has 10 objects and Y has 15 objects; how many objects does X ∩ Y have?
Solution:
Given |X ∪ Y| = 20
|X| = 10
|Y| = 15
We know that |X ∪ Y|
= |X| + |Y| – |X ∩ Y|
⇒ 20 = 10 + 15 – |X ∩ Y|
⇒ |X ∩ Y| = 25 -20 = 5
∴ X ∩ Y has 5 elements.

Question 10.
In a group of 450 people, 300 can speak Hindi and 250 can speak English. How many people can speak both Hindi and English?
Solution:
Let H = The set of people who can speak Hindi
E = The set of people who can speak English.
According to the question we have
|H ∪ E| = 450, |H| = 300,
|E| = 250
We want to find 1 H ∩ E
|H ∩ E| = |H| + |E|-  |H ∪ E|
= 300 + 250 – 450 = 100
∴ 100 people can speak both Hindi and English.

Question 11.
In a group of people,37 like coffee, 52 like tea and each person in the group likes at least one of the two drinks. 19 people like both tea and
coffee, how many people are in the group?
Solution:
Let T = The set of persons who like Tea. ,
C = The set of persons who like coffee According to the question
|C| = 37, |T| = 52 and |T ∩ C| = 19
Total number of persons in the group
= |T ∩ C| = |T| + |C| – |T ∩ C|
= 37 + 52 – 19 = 70

Question 12.
In a class of 35 students, each student likes to play either cricket or hockey. 24 students like to play cricket and 5 students like to play both games; how many students play hockey?
Solution:
Let C = Set of students like to play cricket
H = The set of students like to play Hockey.
According to the question
|C ∪ H| = 35
|C| = 24, |C ∩ H| = 5
Now |C ∪ H| = |C| + |H| – |C ∩ H|
⇒ 35 = 24 + |H| – 5
⇒ |H| = 16
16 students like to play Hockey.

Question 13.
In a class of 400 students, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice or orange juice.
Solution:
Let A = The set of students take apple juice
O = The set of students take orange juice
According to the question
|A| = 100, |O| = 150 and
|A ∩ O| = 75
∴ Number of students take at least one of the juice = |A ∪ O|
= |A| + |O| – |A ∩ O|
= 100 + 150 – 75 = 175
Total number of students
= |U| = 400
Number of students taking neither of these juice
= |U| – |A ∪ O|
= 400 – 175 = 225

Question 14.
In a group of 65 people, 40 like cricket, and 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Solution:
Let C = The set of persons like cricket
T = The set of people who like tennis.
According to the question

|C| = 40 , |C ∩ T| = 10 and
|C ∪ T| = 65
A number of people like tennis only but not cricket = |C ∪ T| – |C|
= 65 – 40 = 15
Number of persons like tennis
= |C ∩ T| – |C| + |C ∩ T|
= 65 – 40 + 10 = 25

Question 15.
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12
people liked products C and A, 14 people liked products B and C and 8 liked all the three products, find how many liked products C only.
Solution:
Let E = Set of persons like product A
F = Set of persons like product B
G = Set of persons like product C

According to the question
a + b + d + e = 21
c + b + f + e = 26
g + f + d + e = 29
b + e = 14
f + e = 14
d + e = 12
e = 8
⇒ e = 8    g = 11
d = 4        c = 6
f = 6          a = 3
b = 6
Number of persons like product C only
O = g = 11

Odisha State Board CHSE Odisha Class 12 Psychology Solutions Unit 4 Long Answer Questions Part-3.

CHSE Odisha 12th Class Psychology Unit 4 Long Answer Questions Part-3

Long Questions With Answers

Question 1.
What are the different types of psychotherapy? On what basis are they classified?
Answer:
Different types of psychotherapy are:

• Psychodynamic therapy
• Behaviour therapy
• Humanistic-existential therapy
• Biomedical therapy

Also, there are many alternative therapies such as yoga, meditation, acupuncture, herbal remedies etc.
Basis of classification of psychotherapy:

On the cause which has led to the problem:
Psychodynamic therapy is of the view that intrapsychic conflicts, i.e. the conflicts that are present within the psyche of the person, are the source of psychological problems.

On how did the cause come into existence:
The psychodynamic therapy, unfulfilled desires of childhood and unresolved childhood fears lead to intrapsychic conflicts.

What is the chief method of treatment?
Psychodynamic therapy uses the methods of free association and reporting of dreams to elicit the thoughts and feelings of the client.

What is the nature of the therapeutic relationship between the client and the therapist?
Psychodynamic therapy assumes that the therapist understands the client’s intrapsychic conflicts better than the client and hence it is the therapist who interprets the thoughts and feelings of the client to her/him so that s/he gains an understanding of the same.

What is the chief benefit to the client?
Psychodynamic therapy values emotional insight as the important benefit that the client derives from the treatment. Emotional insight is present when the client understands her/his conflicts intellectually; is able to accept the same emotionally, and is able to change her/his emotions towards the conflicts.

On the duration of treatment:
Hie duration of classical psychoanalysis may continue for several years. However, several recent versions of psychodynamic therapies are completed in 10—15 sessions.

Question 2.
A therapist asks the client to reveal all her/his thoughts including early childhood experiences. Describe the technique and type of therapy being used.
Answer:
In this case psychodynamic, therapy is used in the treatment of the client. Since the psychoanalytic approach views intrapsychic conflicts to be the cause of the psychological disorder. The first step in the treatment is to elicit this intrapsychic conflict. Psychoanalysis has invented free association and dream interpretation as two important methods for eliciting intrapsychic conflicts.

The free association method is the main method for understanding the client’s problems. Once a therapeutic relationship is established, and the client feels comfortable, the therapist makes her/him lie down on the couch, close her/his eyes and asks her/him to speak whatever comes to mind without censoring it in any way. The client is encouraged to freely associate one thought with another, and this method is called the method of free association.

The censoring superego and the watchful ego are kept in abeyance as the client speaks whatever comes to mind in an atmosphere that is relaxed and trusting. As the therapist does not interrupt, the free flow of ideas, desires and conflicts of the unconscious, which had been suppressed by the ego, emerges into the conscious mind. This free uncensored verbal narrative of the client is a window into the client’s unconscious to which the therapist gains access.

Along with this technique, the client is asked to write down her/his dreams upon waking up. Psychoanalysts look upon dreams as symbols of the unfulfilled desires present in the unconscious. The images of dead dreams are symbols which signify intrapsychic forces. Dreams use symbols because they are indirect expressions and hence would not alert the ego.

If the unfulfilled desires are expressed directly, the ever-vigilant ego would suppress them and that would leads to anxiety. These symbols are interpreted according to an accepted convention of translation as indicators of unfulfilled desires and conflicts.

Question 3.
Discuss the various techniques used in behaviour therapy.
Answer:
Various techniques used in behaviour therapy:
A range of techniques is available for changing behaviour. The principles of these techniques are to reduce the arousal level of the client, alter behaviour through classical conditioning or operant conditioning with different contingencies of reinforcements, as well as to use vicarious learning procedures, if necessary. Negative reinforcement and aversive conditioning are the two major techniques of behaviour modification.

Negative reinforcement refers to following an undesired response with an outcome that is gainful or not liked. For example, one learns to put on woollen clothes, bum firewood or use electric heaters to avoid the unpleasant cold weather. One learns to move away from dangerous stimuli because they provide negative reinforcement.

Aversive conditioning refers to the repeated association of an undesired response with an aversive consequence. For example, an alcoholic is given a mild electric shock and asked to smell the alcohol. With repeated pairings, the smell of alcohol is aversive as the pain of the shock is associated with it and the person will give up alcohol.

Positive reinforcement is given to increase the deficit if adaptive behaviour occurs rarely. For example, if a child does not do homework regularly, positive reinforcement may be used by the child’s mother by preparing the child’s favourite dish whenever s/he does homework at the appointed time. The positive reinforcement of food will increase the behaviour of doing homework at the appointed time.

The token economy in which persons with behavioural problems can be given a token as a reward every time a wanted behaviour occurs. The tokens are collected and exchanged for a reward such as an outing for the patient or a treat for the child. Unwanted behaviour can be reduced and waited behaviour can be increased simultaneously through differential reinforcement.

Positive reinforcement for the wanted behaviour and negative reinforcement for the unwanted behaviour attempted together may be one such method. The other method is to positively reinforce the wanted behaviour and ignore the unwanted behaviour. The latter method is less painful and equally effective. For example, let us consider the case of a girl who sulks and cries when she is not taken to the cinema when she asks.

The parent is instructed to take her to the cinema if she does not cry and sulk but not to take her if she does. Further, the parent is instructed to ignore the girl when she cries and sulks. The wanted behaviour of politely asking to be taken to the cinema increases and the unwanted behaviour of crying and sulking decreases.

Question 4.
Explain with the help of an example how cognitive distortions take place.
Answer:
Cognitive distortions are ways of thinking which are general in nature but which distort reality in a negative manner. These patterns of thought are called dysfunctional cognitive structures. They lead to errors of cognition about social reality. Aaron Beck’s theory of psychological distress states that childhood experiences provided by the family and society develop core, schemas or systems, which include beliefs and action patterns in the individual.

Thus, a client, who was neglected by the parents as a child, develops the core schema of “I am not wanted”. During the course of their life, a critical incident occurs in her/his life. S/he is publicly ridiculed by the teacher in school. This critical incident triggers the core schema of “I am not wanted” leading to the development of negative automatic thoughts. Negative thoughts are persistent irrational thoughts such as “nobody loves me”, “I am ugly”, “l am stupid”, “I will not succeed”, etc.

Such negative automatic thoughts are characterised by cognitive distortions. Repeated occurrence of these thoughts leads to the development of feelings of anxiety and depression. The therapist uses questioning, which is a gentle, non-threatening disputation of the client’s beliefs and thoughts. Examples of such questions would be, “Why should everyone love you?”, “What does it mean to you to succeed?” etc.

Question 5.
Which therapy encourages the client to seek personal growth and actualise their potential? Write about the therapies which are based on this principle.
Answer:
Humanistic-existential therapy encourages the client to seek personal growth and actualise their potential. It states that psychological distress arises from feelings of loneliness, alienation, and an inability to find meaning and genuine fulfilment in life.
The therapies which are based on this principle are:

Existential therapy:
There is a spiritual unconscious, which is the storehouse of love, aesthetic awareness, and values of life. Neurotic anxieties arise when the problems of life are attached t6 the physical, psychological or spiritual aspects of one’s existence. Frankl emphasised the role of spiritual anxieties in leading to meaninglessness and hence it may be called existential anxiety, i.e. neurotic anxiety of spiritual origin.

Client-centred therapy:
Client-centred therapy was given by Carl Rogers. He combined scientific rigour with the individualised practice of client-centred psychotherapy. Rogers brought into psychotherapy the concept of self, with freedom and choice as the core of one’s being. The therapy provides a warm relationship in which the client can reconnect with her/his disintegrated feelings. The therapist shows empathy, i.e. understanding the client’s experience as if it were her/his own, is warm and has unconditional positive regard, i.e. total acceptance of the client as s/he is. Empathy sets up an emotional resonance between the therapist and the client.

Gestalt therapy:
The German word gestalt means ‘whole’. This therapy was given by Frederick (Fritz) Peris together with his wife Laura Peris. The goal of gestalt therapy is to increase an individual’s self-awareness and self-acceptance. The client is taught to recognise the bodily processes and die emotions that are being blocked out from awareness. The therapist does this by encouraging the client to act out fantasies about feelings and conflicts. This therapy can also be used in group settings.

Question 6.
What are the factors that contribute to healing in psychotherapy? Enumerate some of the alternative therapies.
Answer:
Factors Contributing to Healing in Psychotherapy are:

A major factor in healing is the techniques adopted by the therapist and the implementation of the same with the patient/client. If the behavioural system and the CBT school are adopted to heal an anxious client, the relaxation procedures and the cognitive restructuring largely contribute to the healing.

The therapeutic alliance, which is formed between the therapist and the patient/ client, has healing properties, because of the regular availability of the therapist and the warmth and empathy provided by the therapist.

At the outset of therapy, while the patient/client is being interviewed in the initial sessions to understand the nature of the problem, s/he unburdens the emotional problems being faced. This process of emotional unburdening is known as catharsis and it has healing properties.

There are several non-specific factors associated with psychotherapy. Some of these factors are attributed to the patient/client and some to the therapist. These factors are called non-specific because they occur across different systems of psychotherapy and across .different clients/patients and different therapists. Non-specific factors attributable to the client/patient are the motivation for change, the expectation of improvement due to the treatment, etc.

These are called patient variables. Non-specific factors attributable to the therapist are positive nature, absence of unresolved emotional conflicts, presence of good mental health, etc. These are called therapist variables. Some of the alternative therapies are Yoga, meditation, acupuncture, herbal remedies etc.

Question 7.
What are the techniques used in the rehabilitation of the mentally ill?
Answer:
The treatment of psychological disorders has two components, i.e. reduction of symptoms, and improving the level of functioning or quality of life. In the case of milder disorders such as generalised anxiety, reactive depression or phobia, reduction of symptoms is associated with an improvement in the quality of life. However, in the case of severe mental disorders such as schizophrenia, reduction of symptoms may not be associated with an improvement in the quality of life.

Many patients suffer from negative symptoms such as disinterest and lack of motivation to do work or interact with people. The aim of rehabilitation is to empower the patient to become a productive member of society to the extent possible. In rehabilitation, the patients are given occupational therapy, social skills training, and vocational therapy. In occupational therapy, the patients are taught skills such as candle making, paper bag making and weaving to help them to form a work discipline.

Social skills. training helps the patients to develop interpersonal skills through role play, imitation and instruction. The objective is to teach the patient to function in a Social group. Cognitive retraining is given to improve the basic cognitive functions of attention, memory and executive functions. After the patient improves sufficiently, vocational training is given wherein the patient is helped to gain the skills necessary to undertake productive employment.

Question 8.
How would a social learning theorist account for a phobic fear of lizards/ cockroaches? How would a psychoanalyst account for the same phobia?
Answer:
Systematic desensitisation is a technique introduced by Wolpe for treating phobias or irrational fears. The client is interviewed to elicit fear-provoking situations and together with the client, the therapist prepares a hierarchy of anxiety-provoking stimuli with the least anxiety-provoking stimuli at the bottom of the hierarchy. The therapist relaxes the client and asks the client to think about the least anxiety-provoking situation.

The client is asked to stop thinking of the fearful situation if the slightest tension is felt. Over sessions, the client is able to imagine more severe fear-provoking situations while maintaining relaxation. The client gets systematically desensitised to the fear.

Question 9.
Should Electroconvulsive Therapy (ECT) be used in the treatment of mental disorders?
Answer:
Yes, Electro-convulsive Therapy (ECT) can be used in the treatment of mental disorders. Electroconvulsive Therapy (ECT) is another form of biomedical therapy. Mild electric shock is given via electrodes to the brain of the patient to induce convulsions. The shock is given by the psychiatrist only when it is necessary for the improvement of the patient. ECT is not a routine treatment and is given only when drugs are not effective in controlling, the symptoms of the patient.

Question 10.
What kind of problems is cognitive behaviour therapy best suited for?
Answer:
Cognitive behaviour treatment best suited for a wide range of psychological disorders such as anxiety, depression, panic attacks, borderline personality, etc. adopts a bio-CBT psychosocial approach to the delineation of psychopathology. It combines cognitive therapy with behavioural techniques.

Question 11.
What is the nature and process of therapeutic approaches?
Answer:
Psychotherapy is a voluntary relationship between the one seeking treatment or the client and the one who treats the therapist. The purpose of the relationship is to help the client to solve the psychological problems faces by her or him. The relationship is conducive to building the trust of the client so that problems may be freely discussed.

Psychotherapies aim at changing maladaptive behaviours, decreasing the sense of personal distress and helping the client to adapt better to her/his environment. The inadequate marital, occupational and social adjustment also requires that major changes be made in an individual’s personal environment. All psychotherapeutic approaches have the following characteristics:

• there is the systematic application of principles underlying the different theories of therapy.
• persons who have received practical training under expert supervision can practice psychotherapy and not everybody. An untrained person may unintentionally cause more harm than good.
• the therapeutic situation involves a therapist and a client who seeks and receives help for her/his emotional problems (this person is the focus of attention in the therapeutic process).
• the interaction of these two persons — the therapist and the client— results in the consolidation/formation of the therapeutic relationship. This is a confidential, interpersonal and dynamic relationship.

This human relationship is central to any sort of psychological therapy and is the vehicle for change. All psychotherapies aim at a few or all of the following goals :

• Reinforcing the client’s resolve for betterment.
• Lessening emotional pressure.
• Unfolding the potential for positive growth.
• Modifying habits.
• Changing thinking patterns
• Increasing self-awareness.
• Improving interpersonal relations and communication.
• Facilitating decision-making.
• Becoming aware of one’s choices in life.
• Relating to one’s social environment in a more creative and self-aware manner.

Question 12.
What is the relationship between the client and therapist?
Answer:
Therapeutic Relationship :
The special relationship between the client and the therapist is known as the therapeutic relationship or alliance. It is neither a passing acquaintance nor a permanent and lasting relationship. There are two major components of a therapeutic alliance. The first component is the contractual nature Of the relationship in which two willing individuals, the client and the therapist, enter into a partnership which aims at helping the client overcome her/his problems.

The second component of the therapeutic alliance is the limited duration of the therapy. This alliance lasts until the client becomes able to deal with her/his problems and take control of her/ his life. This relationship has several unique properties. It is a trusting and confiding relationship. The high level of trust enables the client to unburden herself/himself to the therapist and confide her/his psychological and personal problems to the latter.

The therapist encourages this by being accepting, empathic, genuine and warm to the client. The therapist conveys by her/his words and behaviours that s/he is not judging the client and will continue to show the same positive feelings towards the client even if the client is rude or confides in all the ‘wrong’ things that s/he may have done or thought about. This is the unconditional positive regard that the therapist has for the client. The therapist has empathy for the client.

Empathy:
Empathy is different from sympathy and intellectual understanding of another person’s situation. Iii sympathy, one has compassion and pity towards, the .suffering of another but is not able to feel like the other person. Intellectual understanding is cold in the sense that the person is unable to feel like the other person and does not feel sympathy either. On the other hand, empathy is present when one is able to understand the plight of another person and feel like the other person.

It means understanding things from the other person’s perspective, i.e. putting oneself in the other person’s shoes. Empathy enriches the therapeutic relationship and transforms it into a healing relationship. The therapeutic alliance also requires that the therapist must keep strict confidentiality of the experiences, events, feelings or thoughts disclosed by the client. The therapist must not exploit the trust and confidence of the client in any way. Finally, it is a professional relationship and must remain so.

Question 13.
Write the types of therapies.
Answer:
Though all psychotherapies aim at removing human distress and fostering effective behaviour, they differ greatly in concepts, methods, and techniques. Psychotherapies may be classified into three broad groups, viz. the psychodynamic, behaviour sad existential psychotherapies. In terms of chronological order, psychodynamic therapy emerged first followed by behaviour therapy while existential therapies which are also called the third force, emerged last. The classification of psychotherapies is based on the following parameters:

What is the cause, which has led to the problem?
Psychodynamic therapy is of the view that intrapsychic conflicts, i.e. the conflicts that are present within the psyche of the person, are the source of psychological problems. According to behaviour therapies, psychological problems arise due to faulty learning of behaviours and cognitions. Existential therapies postulate that questions about the meaning of one’s life and existence are the cause of psychological problems.

How did the cause come into existence?
In psychodynamic therapy, unfulfilled desires of childhood and unresolved childhood fears lead to intrapsychic conflicts. Behaviour therapy postulates that faulty conditioning patterns, faulty learning, and faulty thinking and beliefs lead to maladaptive behaviours that, in turn, lead to psychological problems. Existential therapy places importance on the present. It is the current feelings of loneliness, alienation, a sense of the futility of one’s existence, etc., which cause psychological problems.

What is the chief method of treatment?
Psychodynamic therapy uses the methods of free association and reporting of dreams to elicit the thoughts and feelings of the client. This material is interpreted to the client to help her/him to confront and resolve the conflicts and thus overcome problems. Behaviour therapy identifies faulty conditioning patterns and sets up alternate behavioural contingencies to improve behaviour.

The cognitive methods employed in this type of therapy challenge the faulty thinking patterns of the client to help her/him overcome psychological distress. Existential therapy provides a therapeutic environment which is positive, accepting and non-judgmental. The client is able to talk about the problems and the therapist acts as a facilitator. The client arrives at the solutions through a process of personal growth.

What is the nature of the therapeutic relationship between the client and the therapist?
Psychodynamic therapy assumes that the therapist understands the client’s intrapsychic conflicts better than the client and hence it is the therapist who interprets the. thoughts and feelings of the client to her/him so that s/he gains an understanding of the same. Behaviour therapy assumes that the therapist is able to discern the faulty behaviour and thought patterns of the client.

It further assumes that the therapist is capable of finding out the correct behaviour and thought patterns, which would be adaptive for the client. Both psychodynamic and behaviour therapies assume that the therapist is capable of arriving at solutions to the client’s problems. In contrast to these therapies, existential therapies emphasise that the therapist merely provides a warm, empathic relationship in . which the client feels secure to explore the nature and causes of her/his problems by herself/ himself.

What is the chief benefit to the client?
Psychodynamic therapy values emotional insight as the important benefit that the client derives from the treatment. Emotional insight is present when the client understands her/his conflicts intellectually; is able to accept the same emotionally and is able to change her/his emotions towards the conflicts. The client’s symptoms and distresses reduce as a consequence of this emotional insight.

Behaviour therapy considers changing faulty behaviour and thought patterns to adaptive ones as the chief benefit of the treatment. Instituting adaptive or healthy behaviour and thought patterns ensures the reduction of distress and the removal of symptoms. Humanistic therapy values personal growth as the chief benefit. Personal growth is the process of gaining an increasing understanding of oneself and one’s aspirations, emotions and motives.

What is the duration of treatment?
The duration of classical psychoanalysis may continue for several years. However, several recent versions of psychodynamic therapies are completed in 10-15 sessions. Behaviour and cognitive behaviour therapies as well as existential therapies are shorter and are completed in a few months. Thus, different types of psychotherapies differ on multiple parameters.

However, they all share the common method of providing treatment for psychological distress’ through psychological means. The therapist, the therapeutic relationship, and the process of therapy become the agents of change in the client leading to the alleviation of psychological distress. The process of psychotherapy begins by formulating the client’s problem.

Odisha State Board CHSE Odisha Class 12 Psychology Solutions Unit 4 Long Answer Questions Part-2.

CHSE Odisha 12th Class Psychology Unit 4 Long Answer Questions Part-2

Long Questions With Answers

Question 1.
Write the classification of biological disorders.
Answer:
In order to understand psychological disorders, we need to begin by classifying them. A classification of such disorders consists of a list of categories of specific psychological disorders grouped into various classes on the basis of some shared characteristics. Classifications are useful because they enable users like psychologists, psychiatrists and social workers to communicate with each other about the disorder and help in understanding the causes of psychological disorders and the processes involved in their development and maintenance.

The American Psychiatric Association (APA) has published an official manual describing and classifying various kinds of psychological disorders. The current version of it, the Diagnostic and Statistical Manual of Mental Disorders, IV Edition (DSM-IV), evaluates the patient on five axes or dimensions rather than just one broad aspect of ‘mental disorder’. These dimensions relate to biological, psychological, social and other aspects.

The classification scheme officially used in India and elsewhere is the tenth revision of the International Classification of Diseases (ICD-10), which is known as the ICD-10 Classification of Behavioural and Mental Disorders. It was prepared by the World Health Organisation (WHO). For each disorder, a description of the main clinical features or symptoms and of other associated features including diagnostic guidelines is provided in this scheme.

Question 2.
What are the approaches to understanding abnormal behaviour?
Answer:
In order to understand something as complex as abnormal behaviour, psychologists use different approaches. Each approach in use today emphasises a different aspect of human behaviour and explains and treats abnormality in line with that aspect. These approaches also emphasise the role of different factors such as biological, psychological and interpersonal and socio-cultUral factors.

We will examine some of the approaches which are currently being used to explain abnormal behaviour. Biological factors influence all aspects of our behaviour. A wide range of biological factors such as faulty genes, endocrine imbalances, malnutrition, injuries and other conditions may interfere with the normal development and functioning of the human body. These factors may be potential causes of abnormal behaviour. We have already come across the biological model.

According to this model, abnormal behaviour has a biochemical or physiological basis. Biological researchers have found that psychological disorders are often related to problems in the transmission of messages from one neuron to another. You have studied in Class XI, that a tiny space called a synapse separates one neuron from the next and the message must move across that space.

When an electrical impulse reaches a neuron’s ending, the nerve ending is stimulated to release a chemical, called a neurotransmitter. Studies indicate that abnormal activity by certain neurotransmitters can lead to specific psychological disorders. Anxiety disorders have been linked to low activity of the neurotransmitter gamma-aminobutyric acid (GABA) schizophrenia to the excess activity of dopamine, and depression to low activity of serotonin.

Genetic factors have been linked to mood disorders, schizophrenia, mental retardation and other psychological disorders. Researchers have not, however, been able to identify the specific genes that are the culprits. It appears that in most cases, no single gene is responsible for a particular behaviour or a psychological disorder. In fact, many genes combine to help bring about our various behaviours and emotional reactions, both functional and dysfunctional.

Although there is sound evidence to believe that genetic/ biochemical factors are involved in mental disorders as diverse as schizophrenia, depression, anxiety, etc. and biology alone cannot account for most mental disorders. There are several psychological models which provide a psychological explanation of mental disorders. These models maintain that psychological and interpersonal factors have a significant role to play in abnormal behaviour.

These factors include maternal deprivation (separation from the mother, or lack of warmth and stimulation during early years of life), faulty parent-child relationships (rejection, overprotection, over permissiveness, faulty discipline, etc.), maladaptive family structures (inadequate or disturbed family) and severe stress. The psychological models include the psychodynamic, behavioural, cognitive and humanistic-existential models.

The psychodynamic model is the oldest and most famous of the modern psychological models. You have already read about this model in Chapter 2 on Self and Personality. Psychodynamic theorists believe that behaviour, whether normal or abnormal, is determined by psychological forces within the person of which s/he is not consciously aware. These internal forces are considered dynamic, i.e. they interact with one another and their interaction gives shape to behaviour, thoughts and emotions.

Abnormal symptoms are viewed as the result of conflicts between these forces. This model was first formulated by Freud who believed that three central forces shape personality — instinctual needs, drives and impulses (id), rational thinking (ego), and moral standards (superego). Freud stated that abnormal behaviour is a symbolic expression of unconscious mental conflicts that can be generally traced to early childhood or infancy.

Another model that emphasises the role of psychological factors is the behavioural model. This model states that both normal and abnormal behaviours are learned and psychological disorders are the result of learning maladaptive ways of behaving. The model concentrates on behaviours that are learned through conditioning and propose that what has been learned can be unlearned.

Learning can take place by classical conditioning (temporal association in which two events repeatedly occur close together in time), operant conditioning (behaviour is followed by a reward), and social learning (learning by imitating others’ behaviour). These three types of conditioning account for behaviour, whether adaptive or maladaptive. Psychological factors are also emphasised by the cognitive model. This model states that abnormal functioning can result from cognitive problems.

People may hold assumptions and attitudes about themselves that are irrational and inaccurate. People may also repeatedly think in illogical ways and makeover generalisations, that is, – they may draw broad, negative conclusions on the basis of a single insignificant event. Another psychological model is the humanistic-existential model which focuses on broader aspects of human existence.

Humanists believe that human beings are born with a natural tendency to be friendly, cooperative and constructive, and are driven to self-actualise, i.e. to fulfil this potential for goodness and growth. Existentialists believe that from birth we have total freedom to give meaning to our existence or to avoid that responsibility. Those who shirk from this responsibility would live empty, inauthentic and dysfunctional lives.

In addition to the biological and psychosocial factors, socio-cultural factors such as war and violence, group prejudice and discrimination, economic and employment problems and rapid social change, put stress on most of us and cafes also lead to psychological problems in some individuals. According to the sociocultural model, abnormal behaviour is best understood in light of the social and cultural forces that influence an individual.

As behaviour is shaped by societal forces, factors such as family structure and communication, social networks, societal conditions and societal labels and roles become more important. It has been found that certain family systems are likely to produce abnormal functioning in individual members. Some families have an enmeshed structure in which the members are over involved in each other’s activities, thoughts and feelings.

Children from this kind of family may have difficulty in becoming independent in life. The broader social networks in which people operate include their social and professional relationships. Studies have shown that people who are isolated and lack social support, i.e. strong and fulfilling interpersonal relationships in their lives are likely to become more depressed and remain depressed longer than those who have good friendships.

Socio-cultural theorists also believe that abnormal functioning is influenced by the societal labels and roles assigned to troubled people. When people break the norms of their society, they are called deviant and ‘mentally ill’. Such labels tend to stick so that the person may be viewed as ‘crazy’ and encouraged to act sick. The person gradually learns to accept and play the sick role, and functions in a disturbed manner.

In addition to these models, one of the most widely accepted explanations of abnormal behaviour has been provided by the diathesis-stress model. This model states that psychological disorders develop when a diathesis (biological predisposition to the disorder) is set off a stressful situation. This model has three components. The first is the diathesis or the presence of some biological aberration which may be inherited.

The second component is that the diathesis may carry a vulnerability to developing a psychological disorder. This means that the person is ‘at risk’ or ‘predisposed’ to develop the disorder. The third component is the presence of pathogenic stressors, i.e. factors/stressors that may lead to psychopathology. If such “at risk” persons are exposed to these stressors, their predisposition may actually evolve into a disorder. This model has been applied to several disorders including anxiety, depression, and schizophrenia.

Question 3.
What are the major psychological disorders?
Answer:
Anxiety Disorders:
One day while driving home, Deb felt his heart beating rapidly, he started sweating profusely and even felt short of breath. He was so scared that he stopped the car and stepped out. In the next few months, these attacks increased and now he was hesitant to drive for fear of being caught in traffic during an attack. Deb started feeling that he had gone crazy and would die. Soon he remained indoors and refused to move out of the house.

We experience anxiety when we are waiting to take an examination or visit a dentist, or even give a solo performance. This is normal and expected and even motivates us to do our tasks well. On the other hand, high levels of anxiety that are distressing and interfere with effective functioning indicate the presence of an anxiety disorder— the most common category of psychological disorders. Everyone has worries and fears.

The term anxiety is usually defined as a diffuse, vague, very unpleasant feeling of fear and apprehension. The anxious individual also shows combinations of the following symptoms: rapid heart rate, shortness of breath, diarrhoea, loss of appetite, fainting, dizziness, sweating, sleeplessness, frequent urination and tremors. There are many types of anxiety disorders (see Table 4.2).

They include generalised anxiety disorder, which consists of prolonged, vague, unexplained and intense fears that are not attached to any particular object. The symptoms include worry and apprehensive feelings about the future; hypervigilance, which involves constantly scanning the environment for dangers. It is marked by motor tension, as a result of which the person is unable to relax, is restless and visibly shaky and tense.

Another type of anxiety disorder is panic disorder, which consists of recurrent anxiety attacks in which the person experiences intense terror. A panic attack denotes an abrupt surge of intense anxiety rising to a peak when thoughts of particular stimuli are present. Such thoughts occur in an unpredictable manner. The clinical features include shortness of breath, dizziness, trembling, palpitations, choking, nausea, chest pain or discomfort, fear of going crazy, losing control or dying.

You might have met of heard of someone who was afraid to travel in a lift or climb to the tenth floor of a building or refused to enter a room if s/he saw a lizard. You may have also felt it yourself or seen a friend unable to speak a word of a well-memorised and rehearsed speech before an audience. These kinds of fears are termed as phobias. People who have phobias have irrational fears related to specific objects, people, or situations. Phobias often develop gradually or begin with a generalised anxiety disorder. Phobias can be grouped into three main types, i.e. specific phobias, social phobias and agoraphobia.

Specific phobias:
Specific phobias are the most commonly occurring type of phobia. This group includes irrational fears such as intense fear of a certain type of animal, or of being in an enclosed space. Intense and incapacitating fear and embarrassment when dealing with others characterises social phobias.

Agoraphobia:
Agoraphobia is the term used when people develop a fear of entering unfamiliar situations. Many agoraphobics are afraid of leaving their homes. So their ability to carry out normal life activities is severely limited. Have you ever noticed someone washing their hands every time they touch something, or washing even things like coins, or stepping only within the patterns on the floor or road while walking.

People affected by the obsessive-compulsive disorder are unable to control their preoccupation with specific ideas or are unable to prevent themselves from repeatedly carrying out a particular act or series of acts that affect their ability to carry out normal activities.

Obsessive behaviour:
Obsessive behaviour is the inability to stop thinking about a particular idea or topic. The person involved/often finds these thoughts to be unpleasant and shameful.

Compulsive behaviour:
Compulsive behaviour is the need to perform certain behaviours over and over again. Many compulsions deal with counting, ordering, checking, touching and washing. Very often people who have been caught in a natural disaster (such as a tsunami) or have been victims of bomb blasts by terrorists, or been in a serious accident or in a war-related situation, experience posttraumatic stress disorder (PTSD). PTSD symptoms vary widely but may include recurrent dreams, flashbacks, impaired concentration and emotional numbing.

Somatoform Disorders:
These are conditions in which there are physical symptoms in the absence of physical disease. In somatoform disorders, the individual has psychological difficulties and complains of physical symptoms, for which there is no biological cause. Somatoform disorders include pain disorders, somatisation disorders, conversion disorders, and hypochondriasis.

Pain disorders:
Pain disorders involve reports of extreme and incapacitating pain, either without any identifiable biological symptoms or greatly in excess of what might be expected to accompany biological symptoms. How people interpret pain influences their overall adjustment. Some pain sufferers can learn to use active coping, i.e. remaining active and ignoring the pain. Others engage in passive coping, which leads to reduced activity and social withdrawal.

Patients with somatisation disorders have multiple recurrent or chronic bodily complaints. These complaints are likely to be presented in a dramatic and exaggerated way. Common complaints are headaches, fatigue, heart palpitations, fainting spells, vomiting, and allergies. Patients with this disorder believe that they are sick, provide long and detailed histories of their illness and take large quantities of medicine.

The symptoms of conversion disorders are the reported loss of part or all of some basic body functions. Paralysis, blindness, deafness and difficulty in walking are generally among the symptoms reported. These symptoms often occur after a stressful experience and may be quite sudden.

Hypochondriasis:
Hypochondriasis is diagnosed if a person has a persistent belief that s/he has a serious illness, despite medical reassurance, lack of physical findings, and failure to develop the disease. Hypochondriacs have an obsessive preoccupation and concern with the condition of their bodily organs, and they continually worry about, their health.

Question 4.
Write the major anxiety disorders.
Answer:
Generalised Anxiety Disorder:
prolonged, vague, unexplained and intense fears that have no object, accompanied by hypervigilance and motor tension.

Panic Disorder:
frequent anxiety attacks characterised by feelings of intense terror arid dread; unpredictable ‘panic attacks’ along with physiological symptoms like breathlessness, palpitations, trembling, dizziness, and a sense of losing control or even dying.

Phobias :
irrational fears related to specific objects, interactions with others, and unfamiliar situations.

Obsessive-compulsive Disorder :
being preoccupied with certain thoughts that are viewed by the person to be embarrassing or shameful, and being unable to check the impulse to repeatedly carry out certain acts like checking, washing, counting, etc.

Post-traumatic Stress Disorder (PTSD) :
recurrent dreams, flashbacks, impaired concentration and emotional numbing followed by a traumatic or stressful event like a natural disaster, serious accident, etc.

Question 5.
What is dissociative disorders?
Answer:
Dissociative Disorders: Dissociation can be viewed as a severance of the connections between ideas and emotions. Dissociation involves feelings of unreality, estrangement, depersonalisation, and sometimes a loss or shift of identity. Sudden temporary alterations of consciousness that blot out painful experiences are a defining characteristic of dissociative disorders.

Four conditions are included in this group: dissociative amnesia, dissociative fugue, dissociative identity disorder, and depersonalisation. Salient features of somatoform and dissociative disorders are given.

Salient Features of Somatoform and Dissociative Disorders
Dissociative Disorders

Dissociative amnesia:
The person is unable to recall important, personal information often related to a stressful and traumatic report. The extent of forgetting is beyond normal.

Dissociative fugue:
The person suffers from a rare disorder that combines amnesia with travelling away from a stressful environment.

Dissociative identity (multiple personalities) :
The person exhibits two or more separate and contrasting personalities associated with a history of physical abuse.

Somatoform Disorders
Hypochondriasis:
A person interprets insignificant symptoms as signs of a serious illness despite repeated medical evaluations that point to no pathology disease.

Somatisation :
A person exhibits vague and recurring physical/bodily symptoms such as pain, acidity, etc., without any organic cause.

Conversion :
The person suffers from a loss or impairment of motor or sensory function (e.g., paralysis, blindness, etc.) that has no physical cause but may be a response to stress and psychological problems.

Dissociative amnesia:
Dissociative amnesia is characterised by extensive but selective memory loss that has no known organic cause (e.g., head injury). Some people cannot remember anything about their past. Others can no longer recall specific events, people, places, Or objects, while their memory for other events remains intact. This disorder is often associated with overwhelming stress.

Dissociative fugue:
Dissociative fugue has, as its essential feature, an unexpected travel away from home and the workplace, the assumption of a new identity, and the inability to recall the previous identity. The fugue usually ends when the person suddenly ‘wakes up’ with no memory of the events that occurred during the fugue.

Dissociative identity disorder:
Dissociative identity disorder often referred to as multiple personalities, is the most dramatic of the dissociative disorders. It is often associated with traumatic experiences in childhood. In this disorder, the person assumes alternate personalities that may or may not be aware of each other.

Depersonalisation:
Depersonalisation involves a dreamlike state in which the person has a sense of being separated both from self and from reality. In depersonalisation, there is a change of self-perception, and the person’s sense of reality is temporarily lost or changed.

Question 6.
What is mood disorders?
Answer:
Mood disorders are characterised by disturbances in mood or prolonged emotional state. The most common mood disorder is depression, which covers a variety of negative moods and behavioural changes. Depression can refer to a symptom Oi a disorder. In day-to-day life, we often use the term depression to refer to normal feelings after a significant loss, such as the break-up of a relationship, or the failure to attain a significant goal. The main types of mood disorders include depressive, manic dead bipolar disorders.

Major depressive disorder:
Major depressive disorder is defined as a period of depressed mood and/or loss of interest or pleasure in most activities, together with other symptoms which may include a change in body weight, constant sleep problems, tiredness, inability to think clearly, agitation, greatly slowed behaviour and thoughts of death and suicide. Other symptoms include excessive guilt or feelings of worthlessness.

Factors Predisposing towards Depression :
Genetic makeup or heredity is an important risk factor for major depression and bipolar disorders. Age is also a risk factor. For instance, women are particularly at risk during young adulthood, while for men the risk is highest in early middle age. Similarly, gender also plays a great role in this differential risk addition. For example, women in comparison to men are more likely to report a depressive disorder.

Other risk factors are experiencing negative life events and a lack of social support. Another less common mood disorder is mania. People suffering from mania become euphoric (‘high’), extremely active, excessively talkative, and easily distractible. Manic episodes rarely appear by themselves; they usually alternate with depression. Such a mood disorder, in which both mania and depression are alternately present, is sometimes interrupted by periods of normal mood.

This is known as a bipolar mood disorder. Bipolar mood disorders were earlier referred to as manic-depressive disorders. Among the mood disorders, the lifetime risk of a suicide attempt is highest in case of bipolar mood disorders. Several risk factors in addition to the mental health status of a person predict the likelihood of suicide. These include age, gender, ethnicity, or race and recent occurrence of serious life events. Teenagers and young adults are as much at high risk for suicide, as those who are over 70 years.

Gender is also an influencing factor, i.e. men have a higher rate of contemplated suicide than women. Other factors that affect suicide rates are cultural attitudes toward suicide. In Japan, for instance, suicide is the culturally appropriate way to deal with feelings of shame and disgrace. Negative expectations, hopelessness, setting unrealistically high standards and being over-critical in self-evaluation are important themes for those who have suicidal, preoccupations.

Suicide can be prevented by being alert to some of the symptoms which include :

• changes in eating and sleeping habits
• withdrawal from friends, family and regular activities
• violent actions, rebellious behaviour, running away
• drug and alcohol abuse.
• marked personality change
• persistent boredom
• difficulty in concentration.
• complaints about physical symptoms, and
• loss of interest in pleasurable activities.
  However, seeking timely help from a professional counsellor/psychologist can help to prevent the likelihood of suicide.

Question 7.
What is Schizophrenic Disorders and state its symptoms?
Answer:
Schizophrenia is the descriptive term for a group of psychotic disorders in which personal, social and occupational functioning deteriorate as a result of disturbed thought processes, strange perceptions, unusual emotional states, and motor abnormalities. It is a debilitating disorder. The social and psychological costs of schizophrenia are tremendous, both to patients as well as to their families and society.

Symptoms of Schizophrenia:
The symptoms of schizophrenia can be grouped into three categories, viz. positive symptoms (i.e. excesses of thought, emotion and behaviour), negative symptoms (i.e. deficits of thought, emotion, and behaviour) and psychomotor symptoms.

Positive symptoms:
Positive symptoms are ‘pathological excesses’ or ‘bizarre addition?’ to a person’s behaviour. Delusions, disorganised thinking and speech, heightened perception and hallucinations, and inappropriate effects are the ones most often found in schizophrenia. Many people with schizophrenia develop delusions. A delusion is a false belief that is firmly held on inadequate grounds. It is not affected by rational argument and has no basis in reality.

Delusions of persecution:
Delusions of persecution are the most common in schizophrenia. People with this delusion believe that they are being plotted against, spied on, slandered, threatened, attacked Or deliberately victimised. People with schizophrenia may also experience delusions of reference in which they attach special and personal meaning to the actions of others or to objects and events.

Delusions of grandeur:
In delusions of grandeur, people believe themselves to be specially empowered persons and in delusions of control, they believe that their feelings, thoughts and actions are controlled by others. People with schizophrenia may not be able to think logically and may speak in peculiar ways. These formal thought disorders can make communication extremely difficult.

These include rapidly shifting from one topic to another so that the normal structure of thinking is muddled and becomes illogical (loosening of associations, derailment), inventing new words or phrases (neologisms), and persistent aid inappropriate repetition of the same thoughts (perseveration). Schizophrenics may have hallucinations, i. e. perceptions that occur in the absence of external stimuli.

Auditory hallucinations:
Auditory hallucinations are most common in schizophrenia. Patients hear sounds or voices that speak words, phrases and sentences directly to the patient (second-person hallucination) or talk to one another referring/to the patient as s/he (third-person hallucination). Hallucinations can also involve the other senses.

These include tactile hallucinations (i.e. forms of tingling, burning), somatic hallucinations (i.e. something happening inside the body such as a snake crawling inside one’s stomach), visual hallucinations (i.e. vague perceptions of colour or distinct visions of people or objects), gustatory hallucinations (i.e. food or drink taste strange), and olfactory hallucinations (i.e. smell of poison or smoke). People with schizophrenia also show inappropriate effects, i.e’. emotions that are unsuited to the situation.

Negative symptoms:
Negative symptoms are ‘pathological deficits’ and include poverty of speech, blunted and flat affect, loss of volition, and social withdrawal. People with schizophrenia show alogia or poverty of speech, i.e. a reduction in speech and speech content, felony people with schizophrenia show less anger, sadness, joy, and other feelings than most people do. Thus they have blunted effect Some show no emotions at all, a condition is known as flat affect. Also, patients with schizophrenia experience avolition or apathy and an inability to start or complete a course of action.

People with this disorder may withdraw socially and become totally focused on their own ideas and fantasies. People with schizophrenia also show psychomotor Symptoms. They move less spontaneously or make odd grimaces and gestures. These symptoms may take extreme forms known as catatonia. People in a catatonic stupor remain motionless and silent for long stretches of time. Some show catatonic rigidity, i.e. maintaining a rigid, upright posture for hours. Others exhibit catatonic posturing, i.e. assuming awkward, bizarre positions for long periods.

Question 8.
Write the: Sub-types of Schizophrenia.
Answer:
According to DSM-IV-TR, the sub-types of schizophrenia and their characteristics are:

• Paranoid type :
  Preoccupation with delusions or auditory hallucinations; no disorganised speech or behaviour or inappropriate affect.
• Disorganised type:
  Disorganised speech and behaviour; inappropriate or flat affect; no catatonic symptoms.
• Catatonic type :
  Extreme motor immobility; excessive motor inactivity; extreme negativism (i.e. resistance to instructions) or mutism (i.e. refusing to speak).
• Undifferentiated type :
  Does not fit any of the sub-types but meets symptom criteria.
• Residua] type:
  Has experienced at least one episode of schizophrenia; no positive symptoms but shows negative symptoms.

Question 9.
What is Behavioural and Developmental Disorders?
Answer:
There are certain disorders that are specific to children and if neglected can lead to serious consequences later in life. Children have less self-understanding and they have not yet developed a stable sense of identity nor do they have an adequate frame of reference regarding reality, possibility and value. As a result, they are unable to cope with stressful events which might be reflected in behavioural and emotional problems.

On the other hand, although their inexperience and lack of self-sufficiency make them easily upset by problems that seem minor to an adult, children typically bounce back more quickly. We will now discuss several disorders of childhood like Attention-deficit Hyperactivity Disorder (ADHD), Conduct Disorder, and Separation Anxiety Disorder. These disorders, if not attended to, can lead to more serious and chronic disorders as the child moves into adulthood.

Classification of children’s disorders has followed a different path than that of adult disorders. Achenbach has identified two factors, i.e. extermination and internalisation, which include the majority of childhood behaviour problems. The externalising disorders, or under-controlled problems, include behaviours that are disruptive and often aggressive and aversive to others in the child’s environment.

Internalising disorders, or overcontrolled problems, are those conditions where the child experiences depression, anxiety, and discomfort that may not be evident to others. There are several disorders in which children display disruptive or externalising behaviours. We will now focus on three prominent disorders, viz. Attention-deficit Hyperactivity Disorder (ADHD), Oppositional Defiant Disorder (ODD), and Conduct Disorder.

The two main features of (ADHD) are inattention and hyperactivity-impulsivity. Children who are inattentive find it difficult to sustain mental effort during work or play. They have a hard time keeping their minds on any one thing or in following instructions. Common complaints are that the child does not listen, cannot concentrate, does not follow instructions, is disorganised, easily distracted, forgetful, does not finish assignments and is quick, to lose interest in boring activities.

Children who are impulsive seem unable to control their immediate reactions or to think before they act. They find it difficult to wait or take turns and have difficulty resisting immediate temptations or delaying gratification. Minor mishaps such as knocking things over are common whereas more serious accidents and injuries can also occur. Hyperactivity also takes many forms. Children with (ADHD) are in constant motion. Sitting still through a lesson is impossible for them.

The child may fidget, squirm, climb and run around the room aimlessly. Parents and teachers describe them as ‘driven by a motor’, always on the go, and talking incessantly. Boys are four times more likely to be given this diagnosis than girls. Children with Oppositional Defiant Disorder (ODD) display age-inappropriate amounts of stubbornness, are irritable, defiant, disobedient, and behave in a hostile manner. Unlike ADHD, the rates of ODD in boys and girls are not very different.

The terms Conduct Disorder and Antisocial Behaviour refer to age-inappropriate actions and attitudes that violate family expectations, societal norms, and the personal or property rights of others. The behaviours typical of conduct disorder include aggressive actions that cause or threaten harm to people or animals, non-aggressive conduct that causes property damage, major deceitfulness or theft, and serious rule violations.

Children show many different types of aggressive behaviour, such as verbal aggression (i.e. name-calling, swearing), physical aggression (i.e. hitting, fighting), hostile aggression (i.e. directed at inflicting injury to others) and proactive aggression (i.e. dominating and bullying others without provocation). Internalising disorders include Separation Anxiety Disorder (SAD) and Depression. Separation anxiety disorder is an internalising disorder unique to children.

Its most prominent symptom is excessive anxiety or even panic experienced by children at being separated from their parents. Children with SAD may have difficulty being in a room by themselves, going to school alone, are fearful of entering hew situations, and cling to and shadow their parents’ every move. To avoid separation, children with SAD may fuss, scream, throw severe tantrums, or make suicidal gestures.

The ways in which children express and experience depression are related to their level of physical, emotional, and cognitive development. An infant may show sadness by being passive and unresponsive; a pre¬schooler may appear withdrawn and inhibited; a school-age child may be argumentative and combative, and a teenager may express feelings of guilt and hopelessness. Children may also have more serious disorders called Pervasive Developmental Disorders.

These disorders are characterised by severe and widespread impairments in social interaction and communication skills, and stereotyped patterns of behaviours, interests and activities. Autistic disorder or autism is one of the most common of these disorders. Children with autistic disorder have marked difficulties in social interaction and communication a restricted range of interests, and a strong desire for routine.

About 70 per cent of children with autism are also mentally retarded. Children with autism experience profound difficulties in relating to other people. They are unable to initiate social behaviour and seem unresponsive to other people’s feelings. They are unable to share experiences or emotions with others. They also show serious abnormalities in communication and language that persist over time.

Many autistic children never develop speech and those who do, have repetitive and deviant speech patterns. Children with autism often show narrow patterns of interest and repetitive behaviours such as lining up objects or stereotyped body movements such as rocking. These motor movements may be self-stimulatory such as hand flapping or self-injurious such as banging their head against the wall.

Question 10.
What is Substance-use Disorders?
Answer:
Addictive behaviour, whether it involves excessive intake of high-calorie food resulting in extreme obesity or involving the abuse of substances such as alcohol or cocaine, is one of the most severe problems being faced by society today. Disorders relating to maladaptive behaviours resulting from regular and consistent use of the substance involved are called substance abuse disorders.

These disorders include problems associated with using and abusing Such drugs as alcohol, cocaine and heroin, which alter the way people think, feel and behave. There are two sub-groups of substance-use disorders, i.e. those related to substance dependence and those related to substance abuse.

Insubstance dependence:
In substance dependence, there is an intense craving for the substance to which the person is addicted, and the person shows tolerance, withdrawal symptoms and compulsive drug-taking. Tolerance means that the person has to use more and more of a substance to get the same effect. Withdrawal refers to physical symptoms that occur when a person stops or cuts down on the use of a psychoactive substance, i.e. a substance that has the ability to change an individual’s consciousness, mood and thinking processes.

Insubstance abuse:
In substance abuse, there are recurrent and significant adverse consequences related to the use of substances. People who regularly ingest drugs damage their family and social relationships, perform poorly at work and create physical hazards. We will now focus on the three most common forms of substance abuse, viz. alcohol abuse and dependence, heroin abuse and dependence and cocaine abuse and dependence.

Alcohol Abuse and Dependence People who abuse alcohol drink large amounts regularly and rely on it to help Heroin Abuse and Dependence Heroin intake significantly interferes with social and occupational functioning. Most abusers further develop a dependence on heroin, revolving their lives around the substance, building up a tolerance for it and experiencing a withdrawal reaction when they stop taking it.

The most direct and stopping it results in feelings of depression, fatigue, sleep problems, irritability and anxiety. Cocaine poses serious dangers. It has dangerous effects on psychological functioning and physical well-being.

Question 11.
Describe the nature and scope of psychotherapy. Highlight the importance of therapeutic relationships in psychotherapy.
Answer:
Nature and scope of psychotherapy: Psychotherapy is a voluntary relationship between the one seeking treatment or the client and the one who treats or the therapist. The purpose of the relationship is to help the client to solve the psychological problems faces by her or him. The relationship is conducive for building the trust of the client so that problems may be freely discussed.

Psychotherapies aim at changing maladaptive behaviours, decreasing the sense of personal distress and helping the client to adapt better to her/his environment. The inadequate marital, occupational and social adjustment also requires that major changes be made in an individual’s personal environment.

AH, psychotherapies aim at a few or all of the following goals :

• Reinforcing the client’s resolve for betterment.
• Lessening emotional pressure.
• Unfolding the potential for positive growth.
• Modifying habits,
• Changing thinking patterns.
• Increasing self-awareness.
• Improving interpersonal relations and communication.
• Facilitating decision-making.
• Becoming aware of one’s choices in life.

Relating to one’s social environment in a more creative and self-aware manner. The special relationship between the client and the therapist is known as the
therapeutic relationship or alliance.

There are two major components of a therapeutic alliance:

• The first component is the contractual nature of the relationship in which two willing individuals, the client and the therapist, enter into a partnership that aims at helping the client overcome her/his problems.
• The second component of the therapeutic alliance is the limited duration of the therapy. This alliance lasts until the client becomes able to deal with her/his problems and take control of her/his life.

This relationship has several unique properties. It is a trusting and confiding relationship. The high level of trust enables the client to unburden herself/himself to the therapist and confide her/his psychological and personal problems to the latter. The therapist encourages this by being accepting, empathic, genuine, and warm to the client.

The therapist conveys by her/his words and behaviours that she is not judging the client and will continue to show the same positive feelings towards the client even if the client is rude or confides in all the wrong things that she may have done or thought about. The therapeutic alliance also requires that the therapist must keep strict confidentiality of the experiences, events, feelings, or thoughts disclosed by the client. The therapist must not exploit the trust and confidence of the Client in any way.