CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b)

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Exercise 9(b)

Question 1.
2nC0 + 2nC2 + …… + 2nC2n = 22n-1 and 2nC1 + 2nC3 + ….. + 2nC2n-1 = 22n-1
Solution:
We know that
(1 + x)2n = 2nC0 + 2nC1x + 2nC2x2 + …..+ 2nC2nxn …(1)
Putting x = 1
We get putting x = – 1  we get
∴ (2nC0 + 2nC2 + 2nC4 + ….. + 2nC2n) – (2nC1 + 2nC3 + ….. + 2nC2n-1) = 0
 2nC0 + 2nC2 + ….. + 2nC2n
= 2nC1 + 2nC3 + ….. + 2nC2n-1
= \(\frac{2^{2 n}}{2}\) = 22n-1

Question 2.
Find the sum of
(i) C1 + 2C2 + 3C3 + ….. + nCn
Solution:
C1 + 2C2 + 3C3 + ….. + nCn
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b)

(ii) C0 + 2C1 + 3C2 + ….. + (n+1)Cn Hint: write (C0 + C1 + ….. + Cn) + (C1 + 2C2 + ….. + nCn) use (5) and exercise 1.
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 1

Question 3.
Compute \(\frac{(1+k)\left(1+\frac{k}{2}\right) \ldots\left(1+\frac{k}{n}\right)}{(1+n)\left(1+\frac{n}{2}\right) \ldots\left(1+\frac{n}{k}\right)}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 2

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b)

Question 4.
Show that
(i) C0C1 + C1C2 + C2C3 + ….. + Cn-1Cn = \(\frac{(2 n) !}{(n-1) !(n+1) !}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 3
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 4

(ii) C0Cr + C1Cr+1 + C2Cr+2 + ….. + Cn-rCn = \(\frac{(2 n) !}{(n-r) !(n+r) !}\) Hint : Proceed as in Example 13. Compare the coefficient of an-1 to get (i) and the coefficient of an-r to get (ii)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 5
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 6
∴ C0Cr + C1Cr+1 + C2Cr+2 + ….. + Cn-rCn = \(\frac{(2 n) !}{(n-r) !(n+r) !}\)

(iii) 3C0 – 8C1 + 13C2 – 18C3 + ….. + (n+1)th term = 0
Solution:
3C0 – 8C1 + 13C2 – 18C3 + ….. + (n+1)th term
= 3C0 – (5 + 3)C1 + (10 + 3) C2 – (15 + 3) C3 + …
= 3(C0 – C1 + C2 …) + 5 (- C1 + 2C2 – 3C3 …..)
But (1 – x)n =C0 – C1x + C2x2 – C3x3 + … + (-1)n xnCn … (1)
Putting x = 1 we get
C0 – C1 + C2 ….. + (-1)nCn
and differentiating (1) we get n(1 – x)n-1 (- 1)
= – C1 + 2C2x – 3C3x2 +…..
Putting x = 1 we get
– C1 + 2C2 – 3C3 + ….. = 0
∴ 3C0 – 8C1 +13C2 …… (n+1) terms = 0

(iv) C0n2 + C1(2 – n)2 +C2(4 – n)2 + ….. + Cn(2n – n)2 = n.2n
Solution:
C0n2 + C1(2 – n)2 +C2(4 – n)2 + ….. + Cn(2n – n)2 = n.2n
= n2(C0 + C1 + + Cn) + (22C1 + 42C2 + ….. + (2n)2Cn) – 4n (C1 + 2C2 + 3C3 + ….+ nCn)
= n2 . 2n + 4n(n+l)2n-2 – 4 n . n . 2n-1
= 2n-1 (2n2 + 2n2 + 2n – 4n2)
= 2n . 2n-1 = n2n

(v) C0 – 2C1 +3C2 + ….. + (-1)n(n+1)Cn = 0
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 7

(vi) C0 – 3C1 +5C2 + ….. + (2n+1)Cn = (n+1)2n
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 8

Question 5.
Find the sum of the following :
(i) C1 – 2C2 + 3C3 + ….. + n(-1)n-1Cn 
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 9

(ii) 1.2C2 + 2.3C3 + ….. + (n-1)nCn
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 10
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 11

(iii) C1 + 22C2 + 32C3 + ….. + n2Cn
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 12

(vi) C0 – \(\frac{1}{2}\)C1 + \(\frac{1}{3}\)C2 + ….. + (-1)n \(\frac{1}{n+1}\)Cn
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 13

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b)

Question 6.
Show that
(i) C12 + 2C22 +3C32 + ….. + nCn2 = \(\frac{(2 n-1) !}{\{(n-1) !\}^2}\)
Solution:

(ii) C2 + 2C3 + 3C4 + ….. + (n – 1)Cn = 1 + (n – 2)2n-1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 14

Question 7.
C1 – \(\frac{1}{2}\)C2 + \(\frac{1}{3}\)C3 + ….. +(-1)n+1 \(\frac{1}{n}\)Cn = 1 + \(\frac{1}{2}\) + …. + \(\frac{1}{n}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 15

Question 8.
C0C1 + C1C2 + ….. + Cn-1Cn = \(\frac{2^n \cdot n \cdot 1 \cdot 3 \cdot 5 \ldots(2 n-1)}{(n+1)}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 16

Question 9.
The sum \(\frac{1}{1 ! 9 !}+\frac{1}{3 ! 7 !}+\ldots+\frac{1}{7 ! 3 !}+\frac{1}{9 ! 1 !}\) can be written in the form \(\frac{2^a}{b !}\) find a and b.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) 17

Question 10.
(a) Using binominal theorem show that 199 + 299 + 399 + 499 + 599 is divisible by 5 (Regional Mathematical Olympiad, Orissa – 1987)
Solution:
199 + 299 + 399 + 499 + 599
= 1 + (5 – 3)99 + 399 + (5 – 1)99 + 599
= 1 + (59999C1598.31 + 99C2597.32 – …399) + 399 – (1 – 99C151 + 99C252 – …  599) + 599
= (3 × 59999C1598.31 + 99C2597.32 – ….. + 99C9851.398) + (99C15199C252 + …. – 99C98598) ….(1) which is divisible by 5 as each term is a multiple of 5.

(b) Using the same procedure show that 199 + 299 + 399 + 499 + 599 is also divisible by 3 so that it is actually divisible by 15.
Solution:
From Eqn. (1) above, it is clear that each term within the 1st bracket is divisible by 3 and the terms in the 2nd bracket are divisible by 99 and hence divisible by 3.
Each term in Eqn. (1) is divisible by 3. As it is divisible by 3 and 5, it is divisible by 3 × 5 = 15

BSE Odisha 6th Class English Solutions Follow-Up Lesson 2 Bamboo Curry

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 2 Bamboo Curry Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 2 Bamboo Curry

BSE Odisha 6th Class English Follow-Up Lesson 2 Bamboo Curry Text Book Questions and Answers

Session – 1 (ସୋପାନ – ୧):
I. Pre-Reading (ପଢ଼ିବା ପୂର୍ବରୁ):

→ You have read the Odia folk-tale “The Foolish Son-in-Law”. In other languages, there are similar stories. Let us read a similar Santal folk tale “Bamboo Curry”.
(ତୁମେ ଓଡ଼ିଆ ଲୋକକଥା ‘ନିର୍ବୋଧ ଜ୍ଵାଇଁ’’ ପଢ଼ିସାରିଛ । ଅନ୍ୟ ଭାଷାମାନଙ୍କରେ, ସେହି ଏକାପରି ଗପସବୁ ଅଛି । ଆସ ଆମେ ଏକାପରି ଏକ ସାନ୍ତାଳ ଲୋକକଥା ‘ବାଉଁଶ ତରକାରି’’ ପଢ଼ିବା ।)

BSE Odisha 6th Class English Solutions Follow-Up Lesson 2 Bamboo Curry

II. While-Reading (ପଢ଼ିବା ସମୟରେ):
Text (ବିଷୟବସ୍ତୁ):

SGP-1:
Read paragraphs 1-2 silently and answer the questions that follow.
(ଅନୁଚ୍ଛେଦ ୧ – ୨ କୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)
SP-1

1. Once a foolish Santal son-in-law went to his in-law’s place. His mother-in-law cooked delicious dishes for her son-in-law. One of the dishes was a curry made out of the bamboo shoot. The son-in-law liked it very much and asked his mother-in-law, “Mother, the curry is extremely delicious. What is the curry made from ?” Instead of answering his question, she pointed at the bamboo door. The son-in-law asked, “Is it from bamboo ?” “Yes son, the curry is made from bamboo and is, therefore, called “Bamboo Curry”.

SP-22. Next day, the son-in-law was about to leave for his home. The bamboo curry came to his mind. He thought of cooking bamboo curry at home. But they did not have bamboo. So he carried home the bamboo door of his in-laws’ house.

ଓଡ଼ିଆ ଅନୁବାଦ :
(୧). ଏକଦା (ଥରେ) ଜଣେ ନିର୍ବୋଧ (ମୂର୍ଖ) ସାନ୍ତାଳ ଜ୍ଵାଇଁ ତା’ର ଶ୍ୱଶୁର ଘରକୁ ଗଲା । ତା’ର ଶାଶୁ ତା’ର ଜ୍ଵାଇଁ ପାଇଁ ସୁସ୍ୱାଦୁ ଖାଦ୍ୟ ରୋଷେଇ କଲା । ଖାଦ୍ୟଗୁଡ଼ିକ ମଧ୍ୟରୁ ଗୋଟିଏ ଥିଲା ବାଉଁଶ ଗଜାରେ ତିଆରି ହୋଇଥାଏ ତରକାରି । ଜ୍ଵାଇଁ ଜଣଙ୍କ ଏହାକୁ ବହୁତ ପସନ୍ଦ କଲା ଏବଂ ତା’ର ଶାଶୁକୁ ପଚାରିଲା, ‘‘ମାଆ, ତରକାରି ବହୁତ ସୁଆଦିଆ ହୋଇଛି । ତରକାରି କେଉଁଥିରେ ତିଆରି ହୋଇଛି ?’’ ତା’ର ପ୍ରଶ୍ନର ଉତ୍ତର ଦେବା ପରିବର୍ତ୍ତେ, ସେ ବାଉଁଶ ଦରଜା (ତାଟି) ଆଡ଼କୁ ଆଙ୍ଗୁଳି ନିର୍ଦ୍ଦେଶ କରି ଠାରିଲେ । ଜ୍ଵାଇଁ ପଚାରିଲା, ‘ଏହା ବାଉଁଶରେ ତିଆରି ହୋଇଛି ?’’ ‘ହଁ ପୁଅ, ତରକାରିଟି ବାଉଁଶରେ ତିଆରି ହୋଇଛି ଏବଂ ଏଥିପାଇଁ ‘ବାଉଁଶ ତରକାରି’’ କୁହାଯାଏ ।’’
(୨) ତା’ପରଦିନ, ଜ୍ଵାଇଁ ତାଙ୍କ ଘରକୁ ବାହାରୁଥିଲେ । ତାଙ୍କର ମନକୁ ବାଉଁଶ ତରକାରି କଥା ଆସିଲା । କଥା ଭାବିଲେ । କିନ୍ତୁ ସେମାନଙ୍କର ବାଉଁଶ ନଥିଲା । (ତାଟି)କୁ ତା’ ନିଜ ଘରକୁ ବୋହିନେଲା । ସେ ତାଙ୍କ ନିଜ ଘରେ ବାଉଁଶ ତରକାରି କରିବା ତେଣୁ ସେ ତାଙ୍କ ଶ୍ଵଶୁର ଘରର ବାଉଁଶ ଦରଜା

BSE Odisha 6th Class English Solutions Follow-Up Lesson 2 Bamboo Curry

Comprehension Questions : (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ)

Question 1.
Who went to his father-in-law’s house?
(କିଏ ତା’ର ଶ୍ୱଶୁର ଘରକୁ ଗଲା ?)
Answer:
A foolish Santal son-in-law went to his father-in-law’s house.

Question 2.
What curry did his mother-in-law cook?
(ତା’ର ଶାଶୁ କି ତରକାରି ତିଆରି କଲା ?)
Answer:
His mother-in-law cooked “Bamboo Curry”.

Question 3.
Did he like it?
(ଏହାକୁ ସେ ପସନ୍ଦ କଲା କି ?)
Answer:
Surely (ନିଶ୍ଚୟ), he liked it very much.

Question 4.
Why did he carry home the bamboo door of his in-law’s house?
(ସେ କାହିଁକି ଘରକୁ ଶ୍ୱଶୁର ଘରର ବାଉଁଶ ଦରଜା ବୋହିନେଲା ?)
Answer:
He carried home the bamboo door of his in-laws’ house because he thought of cooking bamboo curry at home, but they did not have any bamboo.

Session – 2 (ସୋପାନ – ୨):
SGP-2:

  • Read paragraphs 3-4 silently and answer the questions that follow.
    (ଅନୁଚ୍ଛେଦ ୩ – ୪ କୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

3. Reaching home, he asked his wife to prepare the bamboo curry. He helped his wife in chopping the dry bamboo sticks. But the dry bamboo pieces did not get boiled. The pieces remained as hard and stiff as before. He asked his wife to put more water and boil.
4. That evening his in-laws came to their son-in-law’s house. The son-in-law offered them the bamboo curry. The in-laws laughed at their foolish son-in-law. They told him, “The bamboo curry is made from soft bamboo shoots and not from dry bamboo pieces”.

BSE Odisha 6th Class English Solutions Follow-Up Lesson 2 Bamboo Curry

ଓଡ଼ିଆ ଅନୁବାଦ :
(୩) ଘରେ ପହଞ୍ଚିସାରି, ସେ ତା’ର ସ୍ତ୍ରୀକୁ ବାଉଁଶ ତରକାରି ପ୍ରସ୍ତୁତ କରିବାକୁ କହିଲା । ସେ ତା’ର ସ୍ତ୍ରୀକୁ ସେହି ଶୁଖୁଲା ବାଉଁଶ କାଠିଗୁଡ଼ିକୁ ଖଣ୍ଡ ଖଣ୍ଡ କରି କାଟିବାରେ ସାହାଯ୍ୟ କଲା । କିନ୍ତୁ ସେହି ଶୁଖୁ ବାଉଁଶ ଖଣ୍ଡଗୁଡ଼ିକ ସିଝିଲା ନାହିଁ । ସେହି ବାଉଁଶ କାଠି ଖଣ୍ଡଗୁଡ଼ିକ ପୂର୍ବପରି କଠିନ (ଟାଣ) ହୋଇ ରହିଲା । ସେ ତା’ର ସ୍ତ୍ରୀକୁ ଅଧିକ ପାଣି ଦେଇ ସିଝାଇବାକୁ କହିଲା ।
(୪) ସେହିଦିନ ସନ୍ଧ୍ୟାରେ ତା’ର ଶାଶୁ-ଶ୍ଵଶୁର ସେମାନଙ୍କର ଜ୍ୱାଇଁ ଘରକୁ ଆସିଲେ । ଜ୍ଵାଇଁ ଜଣଙ୍କ ତାଙ୍କୁ ବାଉଁଶ ତରକାରି ଖାଇବାକୁ ଦେଲେ । ଶାଶୁ-ଶ୍ଵଶୁର ତାଙ୍କର ନିର୍ବୋଧ (ବୋକା) ଜ୍ୱାଇଁ ଆଡ଼କୁ ଅନାଇ ହସିଲେ । ସେମାନେ ତାକୁ କହିଲେ, ‘ବାଉଁଶ ତରକାରି କୋମଳ (ନରମ) ବାଉଁଶ ଗଜାରେ ତିଆରି ହୁଏ କିନ୍ତୁ ଶୁଖିଲା ବାଉଁଶ କାଠି ଖଣ୍ଡଗୁଡ଼ିକରେ ନୁହେଁ ।

Comprehension Questions : (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ)

Question 1.
Who did he ask to prepare bamboo curry?
(ସେ କାହାକୁ ବାଉଁଶ ତରକାରି ପ୍ରସ୍ତୁତ କରିବାକୁ କହିଲା ?)
Answer:
He asked his wife to prepare bamboo curry.

Question 2.
How did he help his wife ?
(କିପରି ସେ ତା’ ସ୍ତ୍ରୀକୁ ସାହାଯ୍ୟ କଲା ?)
Answer:
He helped his wife in chopping the dry bamboo sticks.

Question 3.
When the bamboo did not boil what did he ask his wife to do?
(ଯେତେବେଳେ ବାଉଁଶ ଖଣ୍ଡଗୁଡ଼ିକ ସିଝିଲା ନାହିଁ, ସେ ତା’ସ୍ତ୍ରୀକୁ କ’ଣ କରିବାକୁ କହିଲା ?)
Answer:
When the bamboo pieces did not boil, he asked his wife to put more water and boil.

Question 4.
Who came to his house?
(କିଏ ଜ୍ଵାଇଁ ଘରକୁ ଆସିଲେ ?)
Answer:
His in-laws came to his house.

Question 5.
Where from is the bamboo curry made?
(କେଉଁଥୁରୁ ବାଉଁଶ ତରକାରି ପ୍ରସ୍ତୁତ ହୁଏ ?)
Answer:
The bamboo curry is made from soft bamboo shoots.

BSE Odisha 6th Class English Solutions Follow-Up Lesson 2 Bamboo Curry

Session – 3 (ସୋପାନ – ୩):
III. Post-Reading (ପଢ଼ିସାରିବା ପରେ):

1. Writing (ଲେଖ୍) :
(a) Answer the following questions.
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

(i) What is the story about ?
(ଗପଟି କାହା ଉପରେ ଆଧାରିତ ?)
Answer:
The story is about a foolish Santal son-in-law.

(ii) What curry did the mother-in-law prepare?
(କେଉଁ ତରକାରି ଶାଶୁ ତିଆରି କଲେ ?)
Answer:
The mother-in-law prepared “Bamboo Curry”.

(iii) Is the son-in-law foolish ? Why ?
(ଜ୍ଵାଇଁଟି ନିର୍ବୋଧ ଥିଲା କି ? କାହିଁକି ?)
Yes, the ____________ because he asked ____________ out of dry bamboo.
Answer:
Yes, the son-in-law was very foolish because he asked his wife to prepare bamboo curry out of dry bamboo.

BSE Odisha 6th Class English Solutions Follow-Up Lesson 2 Bamboo Curry

Word Note (ଶବ୍ଦାର୍ଥ):
(The words / phrases have been defined mostly on contextual meanings.) (ଶବ୍ଦ । ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକ ଅଧ୍ଵଂଶତଃ ପ୍ରସଙ୍ଗଗତ ଅର୍ଥ ଉପରେ ନିର୍ଭର କରି ବ୍ୟାଖ୍ୟା କରାଯାଇଛି ।)

ashamed – feeling shame, ଲଜ୍ଜା ଅନୁଭବ
bamboo curry – a kind of dish (food) made out of young bamboo plants ବାଉଁଶ ଗଜା ତରକାରି
bamboo shoot- new-young bamboo plants, ବାଉଁଶ ଗଜା
chopping- cutting into small pieces, ଖଣ୍ଡ ଖଣ୍ଡ କରି କାଟିବା ସ୍ବାଦିଷ୍ଟ,ସୁସ୍ବାଦୁ
delicious- tasty (food), ସ୍ଵାଦିଷ୍ଟ, ସୁସ୍ୱାଦୁ
dishes- food items, curry, ତରକାରି, ସ୍ଵାଦିଷ୍ଟ ବ୍ୟଞ୍ଜନ
folk-tale- popular story of a community, କଥୁତଳ୍ପ
gentlest- very kind (behaviour) ବହୁତ ଦୟାଳୁ (ଆଚରଣ)
heaved a great sign of relief- feel relieved,ଆରାମ ଅନୁଭବ କର |
high sounding words- difficult words, କଠିନ ବା ବଡ଼ ବଡ଼ ଶବ୍ଦ
impolite- not good behaviour, rude, ଭଲ ବ୍ୟବହାର ନୁହେଁ
lamb- young sheep, ଛୋଟ ମେଣ୍
offered- gave, served, ଦେଲେ, (ଖାଦ୍ୟ) ପରିବେଷଣ କଲେ
piled high – kept (things) in a heap, ଗଦା, ସ୍ତୂପ
plucking- collecting (from a tree) ସଂଗ୍ରହ (ଏକ ଗଛରୁ)
pointed – showed hand towards (bamboo door), ହାତ ଦେଖାଇ ନିର୍ଦ୍ଦେଶିତ କଲେ
preferred – chose, ପସନ୍ଦ କଲେ ବା ଆଗ୍ରହ ଦେଖାଇଲେ
quack – a self claimed ignorant practitioner, ଶଠ ବଇଦ | ଠକ ବଇଦ
smeared- spread something (substance) on body, ବୋଳି ହୋଇଗଲା, ଲାଗିଗଲା
stiff- hard, କଠିନ
thought of- got an idea, ଗୋଟିଏ ଉପାୟ ଚିନ୍ତା କଲେ
thrashed- beat, ମାଡ଼ିଦେଲେ, ପିଟିଲେ

BSE Odisha 6th Class English Solutions Follow-Up Lesson 5 The Crab and the Fox

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 5 The Crab and the Fox Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 5 The Crab and the Fox

BSE Odisha 6th Class English Follow-Up Lesson 5 The Crab and the Fox Text Book Questions and Answers

Session – 1 (ସୋପାନ – ୧):
I. Pre-Reading (ପଢ଼ିବା ପୂର୍ବରୁ):

→ The teacher finds an activity to introduce the topic. S/he may use the pictures in the text for the purpose.
(ବିଷୟକୁ ଉପସ୍ଥାପନ କରିବାପାଇଁ ଶିକ୍ଷକ ଏକ କାର୍ଯ୍ୟ ସ୍ଥିରକରିବେ । ସେ (ପୁ/ସ୍ତ୍ରୀ) ବିଷୟବସ୍ତୁରେ ଥିବା ଛବିଗୁଡ଼ିକୁ ଏହି ଉଦ୍ଦେଶ୍ୟରେ ବ୍ୟବହାର କରିପାରନ୍ତି ।)
Follow up

II. While-Reading (ପଢ଼ିବା ସମୟରେ):
→ Follow the three steps-teacher’s reading aloud two times followed by silent reading by the students.
(ତିନୋଟି ପର୍ଯ୍ୟାୟକୁ ଅନୁସରଣ କର – ଶିକ୍ଷକଙ୍କର ଦୁଇଥର ବଡ଼ପାଟିରେ ପଢ଼ିବା ଓ ଶିକ୍ଷାର୍ଥୀମାନେ ନୀରବରେ ତାଙ୍କ ପରେ ପଢ଼ିବା ।)

BSE Odisha 6th Class English Solutions Follow-Up Lesson 5 The Crab and the Fox

TEXT (ବିଷୟବସ୍ତୁ)

  • Read the poem silently and answer the questions that follow.
    (କବିତାଟିକୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

It was a very cool night
And there was no crab in sight.
The fox looked for one
Follow up 1
But there was none.
“Where did they go?”
Not even one in sight!
They must be in their holes
If I’m right.”
Follow up 2
Thinking so he got ready
(The hunger too made him greedy)
To go in search of a crab hole
And he straightened his tail like a pole.
Inside the hole his bushy tail he pushed
The crab, he thought, it slightly missed.
He waited long for a sweet pull
But for long there was none at all.
Finally, he pulled out his tail
But the crab was inside he could smell.
So he changed his plan and called “Brother Crab,
Let’s some song and dance have.
The weather calls for such merrymaking
What is life without dancing and singing ?”
The crab well understood
The fox’s real mood.
Thought he to himself ‘Am I a fool ?’
Follow up 3
And answered from his hole :
“Who is going to sing and dance
In such weather cool?
I’ll rather eat and sleep well
in my cozy little hole.”

BSE Odisha 6th Class English Solutions Follow-Up Lesson 5 The Crab and the Fox

କବିତାଟିର ଓଡ଼ିଆ ଉଚ୍ଚାରଣ :
ଇଟ୍ ୱାଜ୍ ଏ ଭେରି କୁଲ୍ ନାଇଟ୍
ଆଣ୍ଡ୍ ଦେୟାର୍ ୱାଜ୍ ନୋ କ୍ରାବ୍‌ ଇନ୍ ସାଇଟ୍ ।
ଦି ଫକ୍ସ ଲୁକ୍‌ ଫର୍ ୱାନ୍
ବଟ୍ ଦେୟାର ୱାଜ୍ ନନ୍ ।
‘‘ହୋୟାର୍ ଡିଡ଼୍ ଦେ ଗୋ ?’’
ନଟ୍ ଇଭିନ୍ ୱାନ୍ ଇନ୍ ସାଇଟ୍ !
ଦେ ମଷ୍ଟ ବି ଇନ୍ ଦେୟାର୍ ହୋଲ୍‌ସ୍
ଇଫ୍ ‘ଆଇ’ମ୍ ରାଇଟ୍ ।’’
ଥଙ୍କିଙ୍ଗ୍ ସୋ ହି ଗଟ୍ ରେଡ଼ି
(ଦି ହଙ୍ଗର୍ ଠୁ ମେଡ଼୍ ହିମ୍ ଗ୍ରିଡ଼ି)
ଟୁ ଗୋ ଇନ୍ ସର୍ଚ ଅଫ୍ ଏ କ୍ରାବ୍ ହୋଲ୍
ଆଣ୍ଡ ହି ଷ୍ଟେଟେଡ୍ ହିଜ୍ ଟେଲ୍ ଲାଇକ୍ ଏ ପୋଲ୍ ।
ଇସାଇଡ୍ ଦି’ ହୋଲ୍ ହିଜ୍ ବୁସି ଟେଲ୍ ହି ପୁସ୍‌
ଦି କ୍ରାବ୍, ହି ଥଟ୍, ଇଟ୍ ସ୍ଲାଇଟ୍‌ଲି ମିସିଡ୍ ।
ହି ୱେଟେଡ଼ ଲଙ୍ଗ୍ ଫର୍ ଏ ସୁଇଟ୍ ପୁଲ୍
ବଟ୍ ଫର୍ ଲଙ୍ଗ୍ ଦେୟାର୍ ୱାଜ୍ ନନ୍ ଆଟ୍ ଅଲ୍ ।
ଫାଇନାଲି ହି ପୁଲ୍‌ ଆଉଟ୍ ହିଜ୍ ଟେଲ୍
ବଟ୍ ଦି’ କ୍ରାବ୍ ୱାଜ୍ ଇନ୍‌ସାଇଡ୍ ହି କୁଡ଼ ସ୍କେଲ୍ ।
ସୋ ହି ଚେଞ୍ଜେଡ଼୍ ହିଜ୍ ପ୍ଲାନ୍ ଆଣ୍ଡ କଲ୍‌ ‘ବ୍ରଦର୍‌ କ୍ରାନ୍’’,
ଲେଟ୍ସ ସମ୍ ସଙ୍ଗ୍ ଆଣ୍ଡ୍ ଡ୍ୟାନ୍ସ ହାଭ୍ ।
ଦି ୱେଦର୍ କଲସ୍ ଫର୍ ସବ୍ ମେରିମେକିଙ୍ଗ୍
ଦ୍ଵାଟ୍ ଇଜ୍ ଲାଇଫ୍ ଉଇଦାଉଟ୍ ଡ୍ୟାନ୍‌ସିଙ୍ଗ୍ ଆଣ୍ଡ ସିଙ୍ଗିଙ୍ଗ୍ ?’’
ଦି କ୍ରାବ୍‌ ୱେଲ୍ ଅଣ୍ଡରଷ୍ଟୁଡ୍
ଦି ଫକ୍ସ’ସ୍ ରିଅଲ୍ ମୁଡ଼ ।
ଥଟ୍ ହି ଟୁ ହିମ୍‌ସେଲୁ ‘ଆମ୍ ଆଇ ଏ ଫୁଲ୍ ?’
ଆଣ୍ଡ ଆନ୍‌ସର୍‌ଡ୍‌ ଫ୍ରମ୍ ହିଜ୍ ହୋଲ୍ :
“ହୁ ଇଜ୍ ଗୋଇଙ୍ଗ୍ ଟୁ ସିଙ୍ଗ୍ ଆଣ୍ଡ୍ ଡ୍ୟାନ୍ସ,
ଇନ୍ ସଚ୍ ଏ ୱେଦର୍ କୁଲ୍ ?
ଆଇ’ଲ୍ ନ୍ୟାଦର୍ ଇଟ୍‌ ଆଣ୍ଡ ସ୍କ୍ରିପ୍ ୱେଲ୍
ଇନ୍ ମାଇଁ କୋଜି ଲିଟିଲ୍ ହୋଲ୍ ।’’

କବିତାର ଓଡ଼ିଆ ଅନୁବାଦ:
ଏହା ଏକ ବହୁତ ଶୀତଳ ରାତ୍ରି ଥିଲା
ଏବଂ କୌଣସି କଙ୍କଡ଼ା ଦୃଷ୍ଟିରେ ପଡୁନଥିଲା ।
କୋକିଶିଆଳ ଗୋଟାଏକୁ ଖୋଜୁଥିଲା
କିନ୍ତୁ ସେଠାରେ ଗୋଟିଏ ବି ନଥିଲା ।
‘ସେମାନେ କୁଆଡ଼େ ଗଲେ?’
ଏପରିକି ଗୋଟିଏବି ହେଲେ
ଦୃଷ୍ଟିରେ ପଡୁନାହାନ୍ତି !
ନିଶ୍ଚୟ ସେମାନେ ଥିବେ ସେମାନଙ୍କର ଗାତଗୁଡ଼ିକ ଭିତରେ
ଯଦି ମୁଁ ଠିକ୍ କହୁଥାଏ ।’
ଭାବି ସେ ପ୍ରସ୍ତୁତ ହୋଇଗଲେ |
(କ୍ଷୁଧା ମଧ୍ଯ ତାକୁ ଲୋଭୀ କରିଦେଲା)
ଅନ୍ଵେଷଣରେ ଯିବା ପାଇଁ ଗୋଟିଏ କଙ୍କଡ଼ା ଗାତ
ଏବଂ ସେ ସିଧା ବା ସଳଖ କରିଦେଲା ତା’ର ଲାଞ୍ଜକୁ ଗୋଟିଏ ଖୁଣ୍ଟ ପରି !
ଗାତ ଭିତରକୁ ତା’ର ବୁଦାଳିଆ ଲୋମଶ ଲାଞ୍ଚକୁ ସେ ଠେଲିଦେଲା
ସେ ଭାଙ୍ଗିଲା, କଙ୍କଡ଼ାଟିରେ ବାଜିବାରେ ଟିକେ ଭୁଲ୍ ହେଲା ।
ସେ ବହୁତ ସମୟ ଅପେକ୍ଷା କଲା ଗୋଟିଏ ମଧୁର ଟଣା ପାଇଁ
କିନ୍ତୁ ଦୀର୍ଘ ସମୟ ଧରି ସେପରି କିଛି ହେଲା ନାହିଁ ।
ଶେଷରେ ସେ ତା’ର ଲାଞ୍ଜକୁ ବାହାରକୁ ଟାଣି ଆଣିଲା
କିନ୍ତୁ କଙ୍କଡ଼ାଟି ଭିତରେ ଥ‌ିବାର ବାସନା ସେ ବାରିପାରିଲା ।
ତେଣୁ ସେ ତା’ର ଯୋଜନାକୁ ବଦଳାଇ ଦେଲା ଏବଂ ଡାକିଲା ‘କଙ୍କଡ଼ା ଭାଇ’’,
ଆସ ଆମେ କିଛି ଗୀତ ଏବଂ ନାଚ କରିବା ।
ଏପରି ମଉଜ କରିବାକୁ ପାଗ ଡାକୁଛି
ନାଚ ଓ ଗୀତ ବିନା ଜୀବନର ଅର୍ଥ କ’ଣ ଅଛି ?
କଙ୍କଡ଼ା ଭଲ ଭାବରେ ବୁଝିଗଲା
ସେ ନିଜକୁ ନିଜେ ଭାବିଲା, ‘ମୁଁ କ’ଣ ଏତେ ବୋକା ?”’
କୋକିଶିଆଳର ପ୍ରକୃତ ମନୋବୃତ୍ତି !
ସେ ନିଜକୁ ନିଜେ ଭାବିଲା, ‘ମୁଁ କ’ଣ ଏତେ ବୋକା ?”’
ଏବଂ ତା’ର ଗାତ ମଧ୍ୟରେ ଥାଇ ଉତ୍ତର ଦେଲା :
‘‘କିଏ ଯାଉଛି ଗୀତ ଗାଇବାକୁ ଓ ନାଚିବାକୁ
ଏପରି ଏକ ଶୀତଳ ପାଗରେ ?
ମୁଁ ବରଂ ଖାଇବି ଓ ଶୋଇବି ଭଲ ଭାବରେ
ମୋ’ର ଆରାମଦାୟକ ଛୋଟ ଗାତ ମଧ୍ଯରେ ।’’

BSE Odisha 6th Class English Solutions Follow-Up Lesson 5 The Crab and the Fox

Comprehension Questions (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ):

Question 1.
What is the story about?
(ଗପଟି କେଉଁ ବିଷୟରେ ଲେଖାଯାଇଛି ?)
Answer:
The story is about a crab and a fox.

Question 2.
What did the fox look for?
(କୋକିଶିଆଳ କ’ଣ ଖୋଜୁଥୁଲା ?)
Answer:
The fox looked for a crab.

Question 3.
Did he find one?
(ସେ ଗୋଟିଏ ହେଲେ କଙ୍କଡ଼ା ପାଇଲା କି ?)
Answer:
He did not find one.

Question 4.
Where did he push his tail? Why?
(ସେ ତା’ର ଲାଞ୍ଜକୁ କେଉଁଠାକୁ ଠେଲିଲା ? କାହିଁକି ?)
Answer:
He pushed his tail inside the hole. Because he wanted to pull the crab out of its hole.

Question 5.
How could he know that the crab was inside?
(କଙ୍କଡ଼ା ଭିତରେ ଅଛି ବୋଲି ସେ କିପରି ଜାଣିପାରିଲା ? )
Answer:
He could smell that the crab was inside the hole.

Question 6.
What was his new plan?
(ତା’ର ନୂଆ ଯୋଜନାଟି କ’ଣ ଥିଲା ?)
Answer:
His new plan was to make friends with the crab and to call him to come out of its hole for enjoying the fine weather.

Question 7.
Did the new plan work? Why?
(ନୂଆ ଯୋଜନାଟି କାମ କଲା କି ? କାହିଁକି ?)
Answer:
No, the new plan did not work well. Because the crab was cunning enough to the fox’s real mood.

BSE Odisha 6th Class English Solutions Follow-Up Lesson 5 The Crab and the Fox

Question 8.
Did the crab understand the intention of the fox?
(କୋକିଶିଆଳର ଉଦ୍ଦେଶ୍ୟକୁ କଙ୍କଡ଼ାଟି ବୁଝିପାରିଥିଲା କି ?)
Answer:
Yes, the crab understood the intention of the fox.

Question 9.
Did the crab come out of her hole?
(କଙ୍କଡ଼ାଟି ତା’ର ଗାତ ବାହାରକୁ ଆସିଥିଲା କି ?)
Answer:
No, the crab did not come out of her hole.

Question 10.
Who is clever?
(କିଏ ଚତୁର ଥିଲା ?)
Answer:
The crab was clever.

Session – 2 (ସୋପାନ – ୨):
III. Post-Reading (ପଢ଼ିସାରିବା ପରେ):

1. Writing (ଲେଖୁବା):
(a) Answer the following questions.
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

Question (i).
What is the story about?
(ଗପଟି କେଉଁ ବିଷୟରେ ଲେଖାଯାଇ ଅଛି ?)
Answer:
The story is about a crab and a fox.

Question (ii).
What did the fox look for?
(କୋକିଶିଆଳଟି କ’ଣ ଖୋଜୁଥୁଲା ?)
Answer:
The fox looked for a crab.

Question (iii).
Where did the fox push his tail?
(କୋକିଶିଆଳଟି ତା’ର ଲାଞ୍ଜକୁ କେଉଁଆଡ଼କୁ ଠେଲିଲା ?)
Answer:
The fox pushed his tail inside the crab hole.

BSE Odisha 6th Class English Solutions Follow-Up Lesson 5 The Crab and the Fox

Question (iv).
Who is clever?
(କିଏ ଚତୁର ଅଟେ ?)
Answer:
The crab is clever.

(b) Write the story by filling in the gaps:
(ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କରି ଗପଟିକୁ ଲେଖ ।)
(Question with Answer)
Once there lived a ____________ and a ____________. The fox looked for ____________. He____________ his tail inside the ____________. But the crab did not catch his ____________. The fox _____________his plan. He sang a song and called the crab to come out. But the ____________. She said, “Am I _________. I’ll _____________in my ____________.”
Answer:
Once there lived a crab and a fox. The fox looked for a crab. He pushed his tail inside the crab hole. But the crab did not catch his tail. The fox changed his plan. He sang a song and called the crab to come out. But the crab did not come out. She said, “Am I a fool ?” I’ll rather eat and sleep well in my cozy little hole.

WORD NOTE (ଶବ୍ଦାର୍ଥ):
(The words/phrases have been defined mostly on contextual meanings.)
(ଶବ୍ଦ । ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକ ଅଧ୍ଯକାଂଶତଃ ପ୍ରସଙ୍ଗଗତ ଅର୍ଥ ଉପରେ ନିର୍ଭର କରି ବ୍ୟାଖ୍ୟା କରାଯାଇଛି ।)

dog — କୁକୁର
cat — ବିଲେଇ
thin — ପତଳା |
fat — ମେଦ
neither — ଏହା ନୁହେଁ କିମ୍ବା ତାହା ନୁହେଁ
pet — ଗୃହପାଳିତ ପଶୁ
mat — ଆସନ (ମସିଣା)
claimed — ଦାବି କଲେ
chased — ଗୋଡ଼ାଇଲା
retire — ବିଶ୍ରାମ ନେବା ବା ଶୋଇବା
hither — here, ଏଠାରେ
owner’s — ମାଲିକଙ୍କର,
cursed — ଅଭିଶାପିତ,
fate — ଭାଗ୍ୟ,
left — ବାମ,
As — ଯେପରି,
someone — କେହି ଜଣେ
pack — ପ୍ୟାକ୍ କରନ୍ତୁ |
sack — ଅଖା ବସ୍ତା
Hey – ହେ
gunny bag – ଛୋଟ ଅଖାଥଳି
grey — ଧୂସର ରଙ୍ଗ
wish — ଇଚ୍ଛା
religious — ଧାର୍ମିକ
carry — ବହନ କର
obey — ମାନ
nanny — ନାନୀ
funny — ମଜାଳିଆ
thought — ଭାବିଲା
rush — ଭିଡ଼, ଜନଗହଳି
Miss — ମିସ୍
hate — ଘୃଣା କରିବା
always — ସର୍ବଦା
late — ବିଳମ୍ବ, ଡେରି
cosy — ଉଷୁମ ଓ ଆରାମଦାୟକ
merry making — ହସଖୁସିରେ ତିଆରି
straightened — ସିଧା
fox — ଠେକୁଆ
crab — କଙ୍କଡ଼ା
greedy — ଲୋଭୀ
pole — ଖୁଣ୍ଟ

BSE Odisha 6th Class English Solutions Follow-Up Lesson 5 The Crab and the Fox

busy — ବୁଦାଳିଆ, ଲୋମଶ
pushed — ଠେଲି ହୋଇଗଲା
slightly — ଅଳ୍ପ ଟିକିଏ
missed — ମିସ୍
pull — ଟାଣନ୍ତୁ
smell — ଗନ୍ଧ ବା ବାସନା ଠଉରାଇବା
weather — ପାଣିପାଗ
understood — ବୋଧଗମ୍ୟ
real — ବାସ୍ତବ
mood — ମନ
himself – ତା’ ନିଜକୁ ନିଜେ
fool — ମୂର୍ଖ
hole — ଗାତ
cool — ଶୀତଳ
rather — ବରଂ
well — କୂପ

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a)

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Exercise 9(a)

Question 1.
The rows n = 6 and n = 7 in the pascal triangle have been kept vacant. Fill in the gaps
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a)

Question 2.
Write down the expansion of (a + b)8 using Pascal’s triangle.
Solution:
The row n = 8 in Pascal’s triangle is 1, 8, 28, 56, 70, 56, 28, 8, 1.
∴ (a + b)8 = a8 + 8a8-1b1 + 28a8-2b2 + 56a8-3b3 + 70a8-4b4 + 56a8-5b5 + 28a8-6b6 + 8a8-7b7 + b8
= a8 + 8a7b + 28a6b2 + 56a5b3 + 70a4b4 + 56a3b5 + 28a2b6 + 8ab7 + b8

Question 3.
Find the 3rd term in the expansion of \(\left(2 x^3-\frac{1}{x^6}\right)^4\) using rules of Pascal triangle.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 1
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 2

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a)

Question 4.
Expand the following :
(a) (7a + 3b)6
Solution:
(7a + 3b)6 = (7a)6 + 6C1(7a)6-1(3b)1 + 6C2(7a)6-2(3b)2 + ….. + (3b)676a6 + 6(7a)5(3b) + 15(7a4) × 9b2 + …. + 36b6
= 7a6 + 18 × 75a5b + 135 × 74a4b2 + ….. + 36b6

(b) \(\left(\frac{-9}{2} a+b\right)^7\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 3

(c) \(\left(a-\frac{7}{3} c\right)^4\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 4

Question 5.
Apply Binominal Theorem to find the value of (1.01)5.
Solution:
= 1 + 5C1(0.01)1 + 5C2(0.01)2 + 5C3(0.01)3 + 5C4(0.01)4 + (0.01)5
= 1 + 5 × 0.01 + 10(0.0001) + 10(0.000001) + 5(0.00000001) + 0.0000000001
= 1 + 0.05 + 0.001 + 0.00001 + 0.00000005 + 0.000000001
= 1.0510100501

Question 6.
State true or false.
(a) The number of terms in the expansion of \(\left(x^2-2+\frac{1}{x^2}\right)^6\) is equal to 7.
Solution:
False

(b) There is a term independent of both x and y in the expansion of \(\left(x^2+\frac{1}{y^2}\right)^9\)
Solution:
False

(c) The highest power in the expansion of \(x^{40}\left(x^2+\frac{1}{x^2}\right)^{20}\) is equal to 40.
Solution:
False

(d) The product of K consecutive natural numbers is divisible by K!
Solution:
True

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a)

Question 7.
Answer the following :
(a) If the 6th term in the expansion of (x + *)n is equal to nC5xn-10 find *
Solution:
Let the 6th term (x + y)n is nC5xn-10
nC5xn-5y5 = nC5xn-10 = nC5xn-5.x-5
⇒ y5 = x-5 = \(\frac{1}{x^5}\)
∴ y = \(\frac{1}{x}\) . Hence * = \(\frac{1}{x}\)

(b) Find the number of terms in the expansion of (1 + x)n (1 – x)n.
Solution:
(1 + x)n (1 – x)n = (1 – x2)n
∴ The number of terms in this expansion is (n + 1)

(c) Find the value of \(\frac{{ }^n \mathrm{C}_{r-1}}{{ }^n \mathrm{C}_r}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 5

(d) How many terms in the expansion of \(\left(\frac{3}{a}+\frac{a}{3}\right)^{10}\) have positive powers of a? How many have negative powers of a?
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 6

Question 8.
Find the middle term(s) in the expansion of the following.
(a) \(\left(\frac{a}{b}+\frac{b}{a}\right)^6\)
Solution:
Here there is only one middle term i.e. the 4th term.
∴ 4th term i.e. (3 + 1)th term in the expansion of
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 7

(b) \(\left(x+\frac{1}{x}\right)^9\)
Solution:
Here there are two middle terms i.e. 5th and 6th terms.
∴ 5th term in the above expansion is
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 8

(c) \(\left(x^{\frac{3}{2}}-y^{\frac{3}{2}}\right)^8\)
Solution:
Here there is only one middle term i.e. 5th term.
∴ 5th term i.e. (4 + 1)th term in the expansion of
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 9

Question 9.
Find the 6th term in the expansion of \(\left(x^2+\frac{a^4}{y^2}\right)^{10}\)
Solution:
6th term i.e. (5 + 1)th term in the expansion of
(x2 + \(\frac{a^4}{y^2}\))10 is 10C5(x2)10-5 (\(\frac{a^4}{y^2}\))5
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 10

Question 10.
(a) Find the fifth term in the expansion of (6x – \(\frac{a^3}{x}\))10
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 11

(b) Is there a term independent of x? If yes find it out.
Solution:
Let (r + 1)th term in the expansion of  (6x – \(\frac{a^3}{x}\))10 is independent of x.
∴ (r + 1)th term = 10Cr(6x)10-r(\(\frac{-a^3}{x}\))r
= 10Cr610-rx10-r(-1)ra3rx-r
= (-1)r 10Cr610-ra3rx10-2r
∴ x10-2r = 1 = x0
or, 10 – 2r = 0 or, r = 5
∴ 6th term is term independent of x in the above expansion and the term is (-1)5 10C5610-5a3.5
= – 10C565a15 = – 252 × 65a15

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a)

Question 11.
(a) Find the coefficient of \(\frac{1}{y^{10}}\) in the expansion of \(\left(y^3+\frac{a^7}{y^5}\right)^{10}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 12

(b) Does there exist a term independent of y in the above expansion?
Solution:
Let (r + 1)th term is independent of y.
∴ y30-8r = 1 = y0 or, 30 – 8r = 0
or, r = \(\frac{30}{8}\) = \(\frac{15}{4}\) which is not possible as r∈ N or zero.
∴ There is no term in the expansion of \(\left(y^3+\frac{a^7}{y^5}\right)^{10}\) which is independent of y.

Question 12.
(a) Find the coefficient of x4 in the expansion of (1 + 3x + 10x2)(x + \(\frac{1}{x}\))10
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 13

(b) Find the term independent of x in the above expansion.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 14

Question 13.
Show that the coefficient of am and an in expansion of (1 + a)m+n are equal.
Solution:
(m + 1)th and (n + 1)th terms in the expansion of (1 + a)m+n are m+nCmam and m+nCnan
∴ The coefficient of am and an are m+nCm and m+nCn which are equal.

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a)

Question 14.
An expression of the form (a + b + c + d + …. ) consisting of a sum of many distinct symbols called a multinomial. Show that (a + b + c)n is
the sum of all terms of the form \(\frac{\boldsymbol{n} !}{\boldsymbol{p} ! \boldsymbol{q} ! \boldsymbol{r} !}\) apbqcr where p, q and r range over all positive triples of non-negative integers such that p + q + r = n.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) 15

Question 15.
State and prove a multinominal Theorem.
Solution:
Multinominal Theorem:
(P1 + P2 + ……. + Pm)n
\(=\sum \frac{n !}{n_{1} ! n_{2} ! \ldots n_{m} !} p_1^{n_1} p_2^{n_2} \ldots p_m^{n_m}\)
where n1 + n2 + ……. + nm = n
The proof of this theorem is beyond the syllabus.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(b)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 8 Permutations and Combinations Ex 8(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Exercise 8(b)

Question 1.
Find the number of ways in which 5 different books can be arranged on a shelf.
Solution:
The number of ways in which 5 different books can be arranged on a shelf is 5! = 5. 4. 3. 2. 1 = 120

Question 2.
Compute nPr for
(i) n = 8, r = 4
Solution:
nPr = \(\frac{n !}{(n-r) !}=\frac{8 !}{(8-4) !}\)
\(=\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 . !}{4 !}\) = 8.7.6.5 = 1680

(ii) n = 10, r = 3
Solution:
n = 10, r = 3
nPr = \(\frac{n !}{(n-r) !}=\frac{10 !}{7 !}\)

(iii) n = 11, r = 0
Solution:
n = 11, r = 0
nPr = 11P0 = 1

Question 3.
Compute the following :
(i) \(\frac{10 !}{5 !}\)
Solution:
\(\frac{10 !}{5 !}\) = 10. 9. 8. 7. 6 = 30240

(ii) 5! + 6!
Solution:
5 ! + 6! = 5 ! + 6.5 !
= 5 ! (1 + 6) = 120. 7 = 840

(iii) 3! × 4!
Solution:
3 ! × 4 ! = 6 × 24 = 144

(iv) \(\frac{1}{8 !}+\frac{1}{9 !}+\frac{1}{10 !}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(b)

(v) 2!3! = 26 = 64
(vi) 23! = 8! =40320

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(b)

Question 4.
Show that 2.6.10 ……. to n factors = \(\frac{(2 n) !}{n !}\)
Solution:
2.6.10 ……. to n factors = \(\frac{(2 n) !}{n !}\)
CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(b) 1

Question 5.
Find r if P(20, r) = 13. P (20, r – 1).
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(b) 2

Question 6.
Find n if P(n, 4) = 12. P(n, 2).
Solution:
nP4 = 12 × nP2
or, \(\frac{n !}{(n-4) !}=12 \times \frac{n !}{(n-2) !}\)
or, (n – 2)! = 12 (n – 4)!
or, (n – 2) (n – 3) (n – 4)! = 12 (n – 4)!
or, (n – 2) (n – 3) = 12
or, n2 – 5n – 6 = 60
or, (n – 6) (n + 1) = 0
or, n = 6 – 1
Hence n = 6 as n is a natural number.

Question 7.
If P (n – 1, 3) : P (n + 1, 3) = 5: 12, Find n.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(b) 3

Question 8.
Find m and n  if P(m + n, 2) = 56, P(m – n, 2) = 12
Solution:
m+nP2 = 56, m-nP2 = 12
or, \(\frac{(m+n) !}{(m+n-2) !}=56, \frac{(m-n) !}{(m-n-2) !}=12\)
or, (m + n) (m + n – 1) = 8 × 7
(m – n) (m – n – 1) = 4 × 3
∴ (m + n) = 8, m – n = 4
∴ m = 6, n = 2

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(b)

Question 9.
Show that
(i) P(n, n) = P(n, n – 1) for all positive integers.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(b) 4

(ii) P(m, 1) + P(n, 1) = P(m + n, 1) for all positive integers m, n.
Solution:
mp1 + np1 = m+np1 ∀ m, n ∈ Z
∴ L.H.S.
= mp1 + np1 = m + n = m+np1 = R.H.S.
(∴ np1 = n)

Question 10.
How many two-digit even numbers of distinct digits can be formed with the digits 1, 2, 3, 4, and 5?
Solution:
Two-digit even numbers of distinct digits are to be formed with the digits 1, 2, 3, 4, and 5. Here the even numbers must end with 2 or 4. When 2 is placed in the unit place, the tenth place can be filled up by the other 4 digits in 4 different ways. Similarly, when 4 is placed in the unit place, the tenth place can be filled up in 4 different ways.
∴ The total number of two-digit even numbers = 4 + 4 = 8.

Question 11.
How many 5-digit odd numbers with distinct digits can be formed with the digits 0, 1, 2, 3, and 4?
Solution:
5-digit odd numbers are to be formed with distinct digits from the digits 0, 1, 2, 3, and 4. The numbers are to end with 1 or 3 and must not begin with 0.
The 5th place can be filled up by any one of 1 or 3 by 2 ways.

1st 2nd 3rd 4th 5th

Places
The 1st place can be filled by the rest 3 digits except 0 and the digit in 5th place.
The 2nd, 3rd, and 4th places can be filled up by the rest 3 digits in 3! ways.
So total no. of ways = 2 × 3 × 3 ! = 2 × 3 × 2 = 36 ways.

Question 12.
How many numbers, each less than 400 can be formed with the digits 1, 2, 3, 4, 5, and 6 if repetition of digits is allowed?
Solution:
Different numbers each less than 400 are to be formed with the digits 1, 2, 3, 4, 5, and 6 with repetition. Here the numbers are 1-digit, 2-digit, and 3-digit.
∴ The number of 1-digit numbers = 6.
The number of 2-digit numbers = 62 = 36.
The 3-digit number each less than 400 must begin with 1, 2, or 3. So the hundred’s place can be filled by 3 digits and ten’s and unit place can be filled by 6 digits each. So the number of 3 digit numbers = 3 × 6 × 6 = 108.
∴ The total number of numbers = 6 + 36 + 108 = 150.

Question 13.
How many four-digit even numbers with distinct digits can be formed out of digits 0, 1, 2, 3, 4, and 5, 6?
Solution:
Four-digit even numbers mean, they must end with 0, 4, 2, 6. When 0 is placed in last, the 1st place is filled by 1, 2, 3, 4, 5, 6, and the remaining 2 places can be filled by 5P2 ways.
The number of numbers ending with 0 = 5P2 × 6 = 120
Similarly, the number of numbers ending. with 2, 4 and 6 = 5P2  × 5 × 3 = 300
∴ The total number of numbers = 420.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(b)

Question 14.
How many integers between 100 and 1000 (both inclusive) consist of distinct odd digits?
Solution:
Integers are to be formed with distinct odd digits between 100 and 1000.
The numbers between 100 and 1000 are 3-digited.
The odd digits are 1, 3, 5, 1, and 9.
The number of distinct 3-digit odd numbers = 5P3 = 5.4.3 = 60.

Question 15.
An unbiased die of six faces, marked with the integers 1, 2, 3, 4, 5, 6, one on each face, thrown thrice in succession. What is the total number of outcomes?
Solution:
An unbiased die of six faces, marked with the integers 1, 2, 3, 4, 5, 6, one on each face is thrown thrice in succession.
∴ The total number of outcomes = 63 = 216.

Question 16.
What is the total number of integers with distinct digits that exceed 5500 and do not contain 0, 7, and 9?
Solution:
The integer exceeding 55000 must be 4-digited, 5-digited, 6-digited, and 7-digited, as there are seven digits i.e. 1, 2, 3, 4, 5, 6, and 8 to be used for the purpose.
In the 4-digit integers, when 1st place is filled by 5 and 2nd place by .6, the rest two places can be filled by the remaining 5 digits in 5P2 ways. Similarly, when 5 is in 1st place and 8 in 2nd place, the remaining 5 digits be used in 5P2 ways. So the number of 4-digit integers beginning with 5 is 2 × 5P = 40.
When 6 is placed in 1st place, the remaining 3 placed be filled by the remaining 6 digits in 6P3 ways. Similarly, when 8 is placed in 1 st place, the remaining 3 places be filled by the remaining 6 digits in 6P3 ways.
So the number of 4-digit integers starting with 6 and 8 is 2 × 6P3  = 240.
∴ The total number of 4-digit numbers is 40 + 240 = 280.
The number of 5-digit integers is 7P5 = 2520.
The number of 6-digit numbers is 7P6 = 5040 and the number of 7 -digit numbers is 7P6  = 5040.
∴ The total number of integers exceeding 5500 and not containing 0, 7, and 9 is 280 + 2520 + 5040 + 5040 = 12880.

Question 17.
Find the total number of ways in which the letters of the word PRESENTATION can be arranged.
Solution:
The number of letters in the word “PRESENTATION” is 12, out of which there are 2N’s, 2E’s, and 2T’s. So the total number of arrangements.
\(=\frac{12 !}{2 ! 2 ! 2 !}=\frac{1}{8}(12) !\)

Question 18.
Find the numbers of all 4-lettered words (not necessarily having meaning) that can be formed using the letters of the word BOOKLET.
Solution:
We have to form 4-lettered words using the letters of the word BOOKLET. The word contains 7 letters out of which there are 20’s. So there are 6 letters.
∴ The number of 4-lettered words = 7P46P4 = 7.6.5.4 – 6.5.4.3 = 480

Question 19.
In how many ways can 2 boys and3 girls sit in a row so that no two girls sit side by side?
Solution:
Two boys and 3 girls sit in a row so that no two girls sit side by side.
So the only possibility is boys should be situated in between the two girls. In between 3 girls there are 2 gaps where 2 boys will be site.
The girls will be arranged in 3! and boys in 2! ways.
∴ The total number of ways = 2! × 3!=2 × 6= 12

Question 20.
Five red marbles, four white marbles, and three blue marbles the same shape and size are placed in a row. Find the total number of possible arrangements.
Solution:
5 red, 4 white, and 3 blue marbles of the same size and shape are placed in a row.
∴ The total number of marbles is 12 out of which 5 are of one kind, 4 are of 2nd kind and 3 are of 3rd kind
∴ The total number of possible arrangements
\(=\frac{12 !}{5 ! 4 ! 3 !}=\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 3 \cdot 2}\)
= 12.11.10.3.7 = 27720.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(b)

Question 21.
Solve example 2.
Solution:
We have |A| = n, |B| = m
∴ The number of one-one functions from A to B is mPn = \(\frac{m !}{(m-n) !}\) when m> n.
If m = n, the number of one-one functions is \(\frac{m !}{(m-m) !}=\frac{m !}{0 !}\) = m! = n!
If m < n, then there is no possibility of one-one functions.

Question 22.
In how many ways can three men and three women sit at a round table so that no two men can occupy adjacent positions?
Solution:
Since no 2 men are to sit together, there are 4 places available for them corresponding to any one way of sitting of 2 men i.e., two places between the women and 2 places at two ends.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(b) 5

Let m1 be fixed m2 can sit in 2 places, w1 can sit in 3 places, m3 can sit in 1 place, w2 can sit in 2 places w3 can sit in 1 place.
∴ The total number of ways = 2 × 3 × 1 × 2 × 1 = 12

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

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

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

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

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

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c) 1
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

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c) 2
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

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

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

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

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c) 4
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.
CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c) 5
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
CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c) 6
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.

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

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

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|

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

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 = an nC1an-1b +  …..+ ncran-rbr + …..+ bn where a,b ∈ C and n, rule of multiplication of complex numbers and the relation nCr + nCr-1 = n+1Cr)
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 + 1C11 + iβ1)1-11 + iβ1)1
= a1 + 1C1 a1-1 b1
∴ P1 is true
Let Pk be true
i.e., (a + b)k = ak + kC1 ak-1 b1 + … + bk
where a.b ∈ C
Now ( a + b)k+1 = (a + b)k (a + b)1
= (ak + kC1 ak-1b1 +…+ bk) (a + b)
= ak-1 + bak+ kC1 akb + kC1ak-1b2 + … + bk+1
= ak+1 + akb(kC1+ 1)+ …+ bk+1
= ak+1 + k+1C1akb1 + k+1 C2ak-1b2 +… + bk+1
∴ Pk+1 is true
∴ Pn 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 cosn-1 θ ( i sin θ) + nc2 cos n-2 θ (i sin θ)2 + …+ (i sin θ)
= cosn θ – nC2 cosn-2 θ sin2 θ + nC4 cosn-4 θ sin4 θ – …) + i (nC1 cosn-1 θ sin θ – nC3 cosn-3 θ sin 3 θ) + nC5 cosn-5 θ sin5 θ – …)    …..(4)
Equating real part and imaginary parts in (1) and (3), we have
cos nθ = cosn θ – nC2 cosn-2 θ × sin2 θ + nC4 cosn-4 θ sin4 θ …
and sin nθ = nC1 cosn-1 θ sin θ – nC3 cosn-3 θ sin3 θ + nC5 cosn-5 θ sin5 θ…

Question 7.
Find the square root of
(i) – 5 + 12 √-1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 1

(ii) – 11 – 60 √-1
Solution:
Let \(\sqrt{-11-60 \sqrt{-1}}\) = x + iy
Squaring both sides we get
– 11 – 60i = (x + iy)2
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 2

(iii) – 47 + 8 √-1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 3
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:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 4

(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)]

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(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)
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 5
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:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 6
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 7

(iii) z6 = – i
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 8

(iv) z3 = 1 + i
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 9

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)

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

Question 13.
If x + \(\frac{1}{x}\) = 2 cos θ
Show that \(x^n+\frac{1}{x^n}\) = 2 cos nθ
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 10

Question 14.
xr = cos ar + i sin ar
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:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 11
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 12

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:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 13
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 14

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
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 15
= \(2^n \times 2 \cos \frac{n \pi}{3}=2^{n+1} \cos \frac{n \pi}{3}\)

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:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 16

(ii) (1 + i√3)n + (1 – i√3)n = \(2^{n+1} \cos \frac{n \pi}{3}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 17

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:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 18

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.

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

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:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 19

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:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 20
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 21

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
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 22
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 23

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:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 24

Question 28.
If (1 + x + x2)n = p0 + p1x + p2x2 + …..+ p2nx2n, then prove that p0 + p3 + p6 + …..+ 3n-1
Solution:
Given, (1 + x + x2)n = p0 + p1x + p2x2 + …..+ p2nx2n
putting x = ω we get
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 25

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

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:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 26
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.
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 27

BSE Odisha 7th Class English Solutions Test-2

Odisha State Board BSE Odisha 7th Class English Solutions Test-2 Textbook Exercise Questions and Answers.

BSE Odisha Class 7 English Solutions Test-2

BSE Odisha 7th Class English Test-2 Text Book Questions and Answers

The figures in the right-hand margin indicate the marks for each question.
1. Your teacher will dictate ten words. Listen to him/her and write. [10]

text 2
Answer:
drowsy lullaby leap
mob stones heavier
weaver color
already float

BSE Odisha 7th Class English Solutions Test-2

2. Given below are some words. Your teacher will read aloud ten of them. Tick those s/he reads aloud. [10]
the study, family, paddy, wood, quack, worship, bow, arrow, collector, respect, grow, float, change, ride, spring.
Answer:
[Students tick the words that their teacher read aloud]

3. Your teacher will read aloud a paragraph. Listen to him/her and fill in the gaps. [10]
One day the ___________ man took ___________to the poor people living in ___________down the hills. He had in his mind to show his son how ___________he was in contrast to the ___________. He thought this would work like ______________.
Answer:
One day the rich man took his, son to the poor people living in small huts down the hills. He had in his mind to show his son how rich he was in contrast to the poor farmers. He thought this would work like medicine for his son’s sadness.

4. Your teacher will dictate ten names of persons. Listen to him/her and write in the space provided. [10]
text 2 Q4
Answer:
ବିରାକିଶୋର ଦାସ |  – Birakishore Das
ନିଲମାନି ମାର୍ଗ        – Nilamani Routray
ମଧୁସୂଦନ ଦାସ       – Madhusudan Das
ଗୋପାଳକୃଷ୍ଣ ଗୋଖଲେ | – Gopalkrushna Gokhle
ତିଲ୍କା ମାଜି             – Tilka Majhi
ଲକ୍ଷ୍ମଣ ନାୟକ |       – Laxman Nayak
ସୁରେନ୍ଦ୍ର ସାଇ          – Surendra Sai
ପଦ୍ମଲୋଚନ ବେହେରା | – Padmalochan Behera
ଲିଙ୍ଗରାଜ ନନ୍ଦା |      – Lingaraj Nanda
ରାଧନାଥ ରୟ        – Radhanath Roy

5. Your teacher will dictate ten names of persons. Listen to him/her and write in the space provided. [10]
text 2 Q5
Answer:
ବରିପଡା |  – Baripada
କେନ୍ଦୁଝର   – Keonjhar
କାକଟପୁର  – Kakatpur
ମୁମ୍ବାଇ |     – Mumbai
ରାୟାଗଡା | – Raygada
ବୋରିଗୁମା  – Boriguma
ସମ୍ବଲପୁର   – Sambalpur
କୋଲକାତା – Kolkata
ବେଙ୍ଗାଲୁରୁ  – Bengaluru
ଆହ୍ଲାବାଦ  – Allahabad

BSE Odisha 7th Class English Solutions Test-2

6. Match the pair of words that sound alike at the end.
Question 6
Answer:
Question 6.1

7. Order the letters to make meaningful words.
melca, mio, tresof, bitrba, cajkla, reed, veirr
Answer:
melca — camel
Inio — lion
tresof — forest
bitrba — rabbit
cajkla — jackal
reed — deer
veirr — river

BSE Odisha 7th Class English Solutions Test-2

8. Read the paragraph and answer the questions in complete sentences.[20]
There was a deep forest. In that deep forest lived a rabbit. One moonlit night the rabbit was walking happily near that forest. On his way, he came across a well. He looked into the well, and to his surprise, saw a big white ball. The white ball was floating on the water. The ball was nothing, but the reflection of the moon. But he thought it was a big cake.

(i) Where did the rabbit live?
Answer:
The rabbit lived in a deep forest.

(ii) Where was the rabbit walking?
Answer:
The rabbit was walking near the forest.

(iii) When was he walking near the forest?
Answer:
He was walking near the forest on a moonlit night.

(iv) What did he come across on his way?
Answer:
He came across a well on his way.

(v) What did he see in the well?
Answer:
He saw a big white ball floating on the water of the well.

(vi) What did he think?
Answer:
He thought that the big white ball was a big cake.

(vii) What was it?
Answer:
It was the reflection of the moon.

(viii)Was the rabbit clever? How do you know?
Answer:
No, the rabbit was not clever. Because he thought the reflection of the moon on the water was a big cake.

(ix) In the last line ‘it’ is used for
Answer:
In the last line ‘it’ is used for the reflection of the moon.

(x) What looked like a cake?
Answer:
The reflection of the moon on the water of the well looked like a cake.

BSE Odisha 7th Class English Solutions Test-2

9. Read the following poem and answer the questions in complete sentences. [14]

White sheep, white sheep
On a blue hill,
When the winds stop
You all stand still.

You all run away
When the winds blow.
White sheep, white sheep,
Where do you go?

Question 1.
What is the poem about?
Answer:
The poem is about clouds floating in the sky.

Question 2.
How many stanzas are there in this poem?
Answer:
There are two stanzas in this poem.

Question 3.
Where are the white sheep?
Answer:
The white sheep are on the blue hill.

Question 4.
When do they stand still?
Answer:
When the winds stop, they stand still.

Question 5.
When do they run away?
Answer:
When the winds blow, they run away.

Question 6.
Who is asking “Where do you go ?”
Answer:
The poet is asking “Where do you go ?”

Question 7.
Who is compared to the white sheep?
Answer:
The white clouds floating in the blue sky are compared to white sheep.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c)

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

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

Question 1.

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

(i) The number of solutions of  2 sin θ – 1 = 0 is__________. (one, two, infinite)
Solution:
Infinite

(ii) If cos α = cos β, then α + β = ____________. (0, π, 2π)
Solution:

(iii) The number of solution(s) of 2 sin θ + 1 = 0 is__________. (zero, two, infinite)
Solution:
Zero

(iv) If tan θ = tan α and 90° < α < 180°, then θ can be in ____________quadrant. (1st, 3rd, 4th)
Solution:
4th

(v) If tan x. tan 2x. tan 7x = tan x + tan 2x + tan 7x, then x = _____________. (\(\frac{\pi}{4}, \frac{\pi}{5}, \frac{\pi}{10}\))
Solution:
\(\frac{\pi}{10}\)

(vi) For_____________value of θ, sin θ + cos θ = √2. (\(\frac{\pi}{4}, \frac{\pi}{2}, \frac{\pi}{3}\))
Solution:
\(\frac{\pi}{4}\)

(vii) The number of values of x for which cos2 x = 1 and x2 ≤ 4 is______________. (1, 2, 3)
Solution:
1

(viii) In the 1st quadrant the solution of tan2 θ = 3 is_____________. (\(\frac{\pi}{2}, \frac{\pi}{3}, \frac{\pi}{4}\))
Solution:
\(\frac{\pi}{3}\)

(ix) The least positive value of θ for which 1 + tan θ = 0 and √2 cos θ + 1 = 0 is___________. (\(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}\))
Solution:
\(\frac{3 \pi}{4}\)

(x) the least positive value of x for which tan 3x = tan x is______________. (\(\frac{\pi}{2}, \frac{\pi}{3}, \pi\))
Solution:
\(\frac{\pi}{2}\)

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c)

Question 2.
Find the principal solution of the following equations:
(i) sin θ = sin 2θ
Solution:
sin θ = sin 2θ
or, 2θ = nπ + (-1)n θ
or, 2π – (-1)n θ = nπ
or, θ = \(\frac{n \pi}{2-(-1)^n}\)
when n = 0, θ = 0
when n = 1, θ = \(\frac{\pi}{3}\)
when n = 2, θ = 2π
when n = 3, θ = π
when n = 4, θ = 4π
when n = 5, θ = \(\frac{5 \pi}{3}\)
∴ The principal solution are 0, \(\frac{\pi}{3}\), π, \(\frac{5 \pi}{3}\)

(ii) √3 sin θ – cos θ = 2
Solution:
√3 sin θ – cos θ = 2
or, \(\frac{\sqrt{3}}{2}\) sin θ – 1/2 cos θ = 1
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c)
which is the only principal solution.

(iii) cos2 θ + sin θ + 1 = 0
Solution:
cos2 θ + sin θ + 1 = 0
or, 1 – sin2 θ + sin θ + 1 = 0
or,  sin2 θ – sin θ + 2 = 0
or, sin2 θ – 2 sin θ + sin θ – 2 = 0
or, sinθ(sinθ – 2) + (sinθ – 2) = 0
or, (sinθ – 2) (sinθ + 1) = 0
∴ sinθ = 2, sinθ = – 1
= sin \(\left(-\frac{3 \pi}{2}\right)\) or, θ = – \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)
∴ The principal solution is \(\frac{3 \pi}{2}\).

(iv) sin 4x + sin 2x = 0
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 1
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 2

(v) sin x + cos x = \(\frac{1}{\sqrt{2}}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 3

Question 3.
Find the general solution of the following equations:
(i) cos 2x = θ
Solution:
cos 2x = θ
or, 2x = (2n + 1)\(\frac{\pi}{2}\)
or, x = (2n + 1)\(\frac{\pi}{4}\), n∈Z

(ii) sin(x° + 40°) = \(\frac{1}{\sqrt{2}}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 4

(iii) sin 5θ = sin 3θ
Solution:
sin 5θ = sin 3θ
or, 5θ = nπ + (-1)n
or, 5θ – (-1)n 3θ = nπ
or, θ[5 – (-1)n3] = nπ
or, θ = \(\frac{n \pi}{5-(-1)^n 3}\)

(iv) tan ax = cot bx
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 5

(v) tan2 3θ = 3
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 6

Question 4.
Solve the following:
(Hints : cos x ≠ 0 and sin2 x- sin x + 1/2 = 0)
(i) tan2 x + sec2 x = 3
Solution:
tan2 x + sec2 x = 3
or, tan2 x + 1 + tan2 x = 3
or, 2tan2 x = 2
or, tan2 x = 1
or, tan x = ± 1 = tan \(\left(\pm \frac{\pi}{4}\right)\)
∴ x = nπ ± \(\frac{\pi}{4}\)

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c)

(ii) 4 sin2 x + 6 cos2 x = 5
Solution:
4 sin2 x + 6 cos2 x = 5
or, 4 sin2 x + 6(1 – sin2 x) = 5
or, 4 sin2 x + 6 – 6 sin2 x = 5
or, 6- 2 sin2 x = 5
or, 2 sin2 x = 1
or, sin2 x = 1/2
or, sin x = ± \(\frac{1}{\sqrt{2}}\) = sin \(\left(\pm \frac{\pi}{4}\right)\)
or, x = nπ + (-1)n \(\left(\pm \frac{\pi}{4}\right)\)

(iii) 3 sin x + 4 cos x = 5
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 7

(iv) 3 tan x + cot x = 5 cosec x
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 8
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 9

(v) cos x + √3 sin x = √2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 10

(vi) sin 3x – 2 cos2 x = 0
Solution :
sin 2x – 2 cos2 x = 0
or, 2 sin x cos x – 2 cos2 x = 0
or, 2 cos x(sin x – cos x) = 0
∴ cos x = 0, sin x = cos x
∴ x = (2n + 1)\(\frac{\pi}{2}\), tan x = 1 = tan \(\frac{\pi}{4}\)
or, x = nπ + \(\frac{\pi}{4}\)

(vii) sec θ + tan θ = √3
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 11
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 12
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 13

(viii) cos 2θ – cos θ = sin θ – sin 20
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 14
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 15

(ix) sin θ + sin 2θ + sin 3θ + sin 4θ = 0
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 16
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 17

(x) cos 2x° + cos x° – 2 = 0
Solution:
cos 2x° + cos x° – 2 = 0
or, 2 cos2 x° – 1 + cos x° – 2 = 0
or, 2 cos2 x° + cos x° – 3 = 0
or 2 cos2 + 3cos x° – 2cos x°- 3 = 0
or, cos x°(2 cos x° + 3) – 1(2 cos x° + 3) = 0
or, (2 cos x° + 3)(cos x° – 1) = 0
∴ cos x° = 1 = cos 0°
∴ x° = 2nπ ± 0 = 2nπ
or, \(\frac{\pi x}{180}\) = 2nπ
or, x = 360 n
Again 2 cos x° + 3 = 0
⇒ cos x° = – 3/2 which has no solution.
Hence x = 360 n.

(xi) tan θ + tan 2θ = tan 3θ
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 18

(xii) tan θ + tan (\(\theta+\frac{\pi}{3}\)) + tan (\(\theta+\frac{2\pi}{3}\)) = 3
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 19

(xiii) cot2 θ – tan2 θ = 4 cot 2θ
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 20

(xiv) cos 2θ = \((\sqrt{2}+1)\left(\cos \theta-\frac{1}{\sqrt{2}}\right)\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 21
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 22

(xv) sec θ – 1 = \((\sqrt{2}-1)\) tan θ
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 23
⇒ θ = 2nπ + \(\frac{\pi}{4}\)

(xvi) 3cot2 θ – 2 sin θ = 0
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 24

(xvii) 4 cos x. cos 2x . cos 3x = 1
Solution:
4 cos x cos 2x cos 3x = 1
⇒ 2 cos x cos 2x. 2 cos 3x = 1
⇒ (cos 3x + cos x) 2 cos 3x = 1
⇒ 2 cos2 3x + 2 cos 3x cos x = 1
⇒ 2 cos2 3x – 1 + cos 4x + cos 2x = 0
⇒ cos 6x + cos 4x + cos 2x = 0
⇒ cos 6x + cos 2x + cos 4x = 0
⇒ 2 cos 4x cos 2x + cos 4x = 0
⇒ cos 4x (2 cos 2x + 1) = 0
⇒ cos 4x = 0, cos 2x = – 1/2
cos 4x = 0 ⇒ 4x = (2n + 1) \(\frac{\pi}{2}\)
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 25

(xviii) cos 3x – cos 2x = sin 3x
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 26
⇒ 1 – 2 sin x cos x = y2
∴ Equation (1) reduces to
1 – 2(1 – y2) + y = 0
⇒ 2y2 + y – 1 = 0
⇒ (2y- 1) (y + 1) = 0
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 27
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 28

(xix) cos x + sin x = cos 2x + sin 2x
Solution:
cos x + sin x – cos 2x + sin 2x
[Refer (viii)]

(xx) tan x + tan 4x + tan 7x = tan x. tan 4x. tan 7x
Solution:
tan x + tan 4x + tan 7x = tan x tan 4x tan 7x
or, tan x + tan 4x
= – tan 7x + tan x tan 4x tan 7x
= – tan 7x (1 – tan x tan 4x)
or, \(\frac{\tan x+\tan 4 x}{1-\tan x \tan 4 x}\) = – tan 7x
or, tan (x + 4x) = tan (π – 7x)
or, tan 5x = tan (π – 7x)
or, 5x = nπ + π – 7x
or, 12x = π(n + 1)
or, x = \(\frac{\pi(n+1)}{12}\), n∈Z

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c)

(xxi) 2(sec2 θ + sin2 θ) = 5
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c) 29

(xxii) \((\cos x)^{\sin ^2 x-\frac{3}{2} \sin x+\frac{1}{2}}=1\)
Solution:
\((\cos x)^{\sin ^2 x-\frac{3}{2} \sin x+\frac{1}{2}}=0\)
As cos x ≠ 0.
we have sin2 – \(\frac{3}{2}\) sin x + \(\frac{1}{2}\) = 0
∴ 2 sin2 x – 3 sin x + 1 = 0
or, 2 sin2 x – 2 sin x – sin x + 1 = 0
or, (2sin x – 1)(sin x – 1) = 0
∴ sin x = \(\frac{1}{2}\) or, sin x = 1
But as cos x ≠ 0, we have sin x ≠ 1
∴ sin x = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
∴ x = nπ + (-1)n \(\frac{\pi}{6}\), n∈Z

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.