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Odisha State Board BSE Odisha 7th Class English Solutions Follow-Up Lesson 7 Silver Textbook Exercise Questions and Answers.

BSE Odisha Class 7 English Solutions Follow-Up Lesson 7 Silver

BSE Odisha 7th Class English Follow-Up Lesson 7 Silver Text Book Questions and Answers

Session – 1

I. Pre-Reading

• Socialization:
• You have read the poem Cobweb’. You have seen how a poem is made of a simple sight – the sight of a cobweb on telephone wires. The poem ends with moon — how the moon turns the cobweb into magic white. Now we’ll read. enjoy and see how again a poem is made out of a ver common sight — the moonlit night. Let’s read the poem.

( ତମେ ‘ବଢିଆଣୀ ଜାଲ’ କବିତାଟି ପଢ଼ିଲ । ଜାଣିଲ କିଭଳି ସାଧାରଣ ଦୃଶ୍ୟକୁ କବିତାରେ ସଜାଇ ଦିଆଯାଏ – ଯେପରି ଟେଲିଫୋନ ତାରକୁ ଘେରି ରହିଥ‌ିବା ବୁଢ଼ିଆଣୀ ଜାଲ । କବିତାଟି ଚନ୍ଦ୍ର ଶବ୍ଦରେ ପରିସମାପ୍ତ ହୋଇଛି – ଜହ୍ନ ଆଲୁଅ କିଭଳି ବୁଢ଼ିଆଣୀ ଜାଲକୁ ଯାଦୁକରୀ ଶୁଭ୍ର ରଙ୍ଗରେ ରଞ୍ଜିତ କରିଦେଉଛି । ବର୍ତ୍ତମାନ ଚାଲ ପଢ଼ିବା, ଉପଭୋଗ କରିବା ଏବଂ ଦେଖବା କିଭଳି ଏକ ସାଧାରଣ ଦୃଶ୍ୟରୁ ଆଉ ଏକ କବିତା ରଚନା କରାଯାଇଛି – ଚନ୍ଦ୍ରାଲୋକିତ ରଜନୀ । ଚାଲ କବିତାଟିକୁ ପାଠ କରିବା ।)

II. While Reading

Text

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

Slowly, silently, now the moon
Walks the night in her silver shoon;

This way, and that, she peers, and sees
Silver fruit upon silver trees;

One by one the casements catch
Her beams beneath the silvery thatch;

Couched in his kennel, like a log.
With paws of silver sleeps the dog;

From their shadowy cote the white breasts peep
Of doves in a silver-feathered sleep;

A harvest mouse goes scampering by.
With silver claws, and silver eye;

And moveless fish in the water gleam,
By silver reeds in a silver stream.

Walter de la Mare

ଧୀରେ ଧୀରେ ନୀରବରେ ଚାଲିଥାଏ ଜହ୍ନ
ରାତିସାରା ଚାଲିଥାଏ ଶୁଭ୍ର ଜୋତା ପରିଧାନ କରି;

ରାତ୍ରିସାରା ଚାଲିଥାଏ ଶୁଭ୍ର ଜୋତା ପରିଧାନ କରି;
ଏବଂ ଦେଖେ ଶୁଭ୍ର ଫଳସବୁ ସଫେଦ ବୃକ୍ଷରେ;

ଏକ ପରେ ଏକ ବାତାୟନ | ଝରକା ପଥରେ
ପ୍ରବେଶ ତା’ର କିରଣ ଚାଳକୁ ତ ସଫେଦ ବନାଇ;

ପଡ଼ି ରହିଥାଏ ତ କୁକୁର କାଠଗଣ୍ଡି ତୁଲ୍ୟ ତା’ କୋଠିରେ
ଶୋଇରହେ ଊର୍ଦ୍ଧ୍ୱମୁଖୀ ପଞ୍ଝା ତା’ ସଫେଦ ସଜରେ ବନେଇ;

ଛାୟାଯୁକ୍ତ ପର ଦେହ ଉଙ୍କିମାରେ ଧବଳ ହିଆରେ
କପୋତଟି ଶୋଇ ରହେ ଧବଳ ପର-ପରଦାର ତଳେ;

ଅମଳ ବେଳର ମୂଷା ଧାଇଁଯାଏ ଦ୍ରୁତ ପଦଚାଳି
ରୂପେଲି ପଞ୍ଝା ଏବଂ ରୂପେଲି ଚାହାଣି ସମ୍ଭାଳି;

ନିସ୍ତେଜ ମାଛମାନେ ଜଳ ତଳେ ଚକ୍ରଚକ୍ କାତି ଝଲସାଇ
ଥାଆନ୍ତି ତ ରମେଲି ତଣ ଏବଂ ରୁପେଲି ଝରଣା ହସାଇ ।

Notes And Glossary
moon (ମୁନ୍) – ଜହ୍ନ
shoon (ସୁନ୍) – ଯୋତା
peer – ଖୋଜିବା ପରି
silver (ସିଲଭର) – ରୂପେଲି
casement (କେସ୍‌ମେଣୁ) – ଝରକା
beams (ବିମ୍ସ) – ଆଲୋକ | କରଣ
beneath (ବି) – ତଳେ
thatch (ଥ୍ୟାଚ୍) – ମୋଟା କେଶ
couched (କାଉଚ) – ଶୋଇ ରହିଥିଲା
kennel (କେନେଲ୍) – କୁକୁର ରହିବା ଘର
log (ଲଗ) – କାଠଗଣ୍ଡି
paws (ପଢ) – ଖାବଲ୍
breasts (ବ୍ରେଷ୍ଟସ୍) – ଛାତିସବୁ
Peep (ପିପ୍) – ଦେଖୁଛି |
doves (ଡୋଭସ୍) – କପୋତସବୁ
feather (ଫ୍ଲିଦର) – ପର/ଡେଣା
harvest (ହାରଭେଷ୍ଟ) – ଅମଳ/ଉତ୍ପାଦନ
mouse (ମାଉସ୍) – ମୂଷା
scamper (ସ୍କାମ୍ପର) – ତରବର ହୋଇ ଚାଲିଯିବା
claws (କ୍ଲଜ୍) – ଅପେକ୍ଷା କର
moveless (ମୁଭେସ୍) – ସ୍ଥିର
gleam (ଗ୍ଲିମ୍ ) – ଚକ୍‌କ୍ କରୁଥୁବା
reeds (ରିଡ଼ସ) – ଜଳାଶୟ କୂଳରେ ଥିବା ଘାସ

Follow the steps of the main lesson. ( ମୂଳପାଠ୍ୟର ସୋପାନଗୁଡ଼ିକୁ ଅନୁସରଣ କର ।)

Comprehension Questions

Question 1.
What is the poem about?
( କବିତାଟି କେଉଁ ବିଷୟରେ ?)
Answer:
The poem is about the moon.

Question 2.
How does the moon walk?
(ଜହ୍ନ କିପରି ପଦଚାଳନା କରେ ?)
Answer:
The moon walks very slowly and silently the whole night. She wears silver shoes while walking.

Question 3.
“Shoon” is an old word for “shoe”. Why does the shoe look like silver?
(‘Shoon’ ‘Shoe’ ର ପ୍ରାଚୀନ ଶବ୍ଦ । ଜୋତା କାହିଁକି ରୂପେଲି | ସଫେଦ ପ୍ରତୀୟମାନ ହୁଏ ?)
Answer:
The beams of the moon is silvery. So the moon appears to wear shoes of silver.

Question 4.
How is ‘peer’ slightly different from ‘see’ ? See the dictionary?
(‘Peer’ (ଉଙ୍କି ମାରିବା) ‘see’ (ଦେଖ‌ିବା) ଠାରୁ ଅଳ୍ପ ଟିକିଏ ଅଲଗା କିପରି ? ଶବ୍ଦକୋଷ ଦେଖୁ ସ୍ଥିର କର ।)
Answer:
Peer means look searchingly while see means perceive with eyes.

Question 5.
What does the moon see first?
(ଜହ୍ନ ପ୍ରଥମେ କ’ଣ ଦେଖେ ?)
Answer:
First, the moon sees the silver fruit upon silver trees.

Question 6.
Why do the fruit and tree look silvery?
(କାହିଁକି ଗଛର ଫଳ ଏବଂ ନିଜେ ଗଛ ରୂପେଲି ଦେଖାଯାଏ ?)
Answer:
The moon beam falls on the trees and fruit and turns them into silvery.

Question 7.
Casement’ is an old word for ’window’ and ’ beneath’ is an old word for ’below’. Where are the windows?
(‘Casement’ ‘window’ ର ଏବଂ ‘beneath’ ‘below’ ର ପ୍ରାଚୀନ ଶବ୍ଦ । ଝରକାସବୁ କେଉଁଠି ଥାଏ ? )
Answer:
The windows are below the silvery thatch.

Question 8.
Where does the dog sleep?
(କୁକୁରଟି କେଉଁଠି ଶୁଏ ? )
Answer:
The dog sleeps in its kennel.

Question 9.
The dog sleeps like a log- a piece of wood. What does this tell about the dog’s sleep?
(କୁକୁରଟି ଖଣ୍ଡେ କାଠଗଣ୍ଡି ଭଳି ଶୋଇ ରହିଥାଏ । ଏହା କୁକୁରର ସୁପ୍ତି (ଶୋଇବା) ବାବଦରେ କ’ଣ କହେ ?)
Answer:
“The dog sleeps like a log”. This expression shows that the dog is in deep sleep. It is seen like a log is lying.

Question 10.
Why do its paws look silvery?
(ତାହାର (କୁକୁରର) ପଞ୍ଝା କାହିଁକି ରୂପେଲି ଦିଶେ ?)
Answer:
When the white moonbeam falls on the paws of the dog, they look like silvery.

Question 11.
Can you guess why the dog does not look silvery?
(ଅନୁମାନ କରିପାରିବ କି କାହିଁକି କୁକୁର ରୂପେଲି ଦେଖାଯାଏ ନାହିଁ ?)
Answer:
The dog was inside the kennel but his paws were outside the kennel. So the moonbeams reflect on his paws not on its body. So the dog does not look silvery.

Question 12.
Where does the dove sleep?
(କପୋତ କେଉଁଠି ଶୁଏ ?)
Answer:
The dove sleeps in its shadowy cote.

Question 13.
The breast of the dove peeps out. Silver-feather sleep – the feather looks silvery. What about the head of the dove? Can you think how the dove is sleeping?
(କପୋତର ଛାତି ବାହାରକୁ ଉଙ୍କିମାରେ । ରୂପେଲି ପର ତଳେ ନିଦ୍ରା- ପରସବୁ ରୂପେଲି ଦେଖାଯାଏ । କମୋଦର ମଣ୍ଡ କିଭଳି ଦେଖାଯାଏ ? ଚିନ୍ତା କରିପାରଛ ତ କପୋତ କିଭଳି ଶୋଇଛି ? )
Answer:
The dove in its shadowy cote sleeps keeping its breast out. Its head is hidden inside its feather.

Question 14.
Who goes scampering by (running very fast)?
(କିଏ ତରବର ହୋଇ ଦୃତଗତିରେ ଚାଲିଯାଏ ? )
Answer:
A harvest mouse goes scampering by.

Question 15.
Everyone is asleep. Why is the mouse awake?
(ପ୍ରତ୍ୟେକ ଶୟନରତ । ମୂଷା କାହିଁକି ଜାଗ୍ରତ ଥାଏ ?)
Answer:
Everyone is asleep in the cool moon light, but the mouse is awake. Because it collects its food in silent night.

Question 16.
Why does the fish gleam (dazzle)?
(ମାଛମାନେ କାହିଁକି ଚକ୍‌କ୍ ଦିଶନ୍ତି ?)
Answer:
When the moon light falls on moveless fish, it looks white.

Question 17.
Where is the fish?
(ମାଛସବୁ କେଉଁଠି ଥାଆନ୍ତି ?)
Answer:
The fish is in the stream water.

Question 18.
When everyone is asleep, why is the poet awake?
(ଯେତେବେଳେ ସମସ୍ତେ ଶୋଇ ରହିଥାନ୍ତି, କବି ସେତେବେଳେ କାହିଁକି ଜାଗ୍ରତ ଥାଆନ୍ତି ?)
Answer:
When everybody is asleep, the poet is awake because he wants to watch the natural beauty in the moonlit night.

Session – 2

III. Post-Reading

1. Comprehension Activities

(a) MCQs: Fill in the blanks from the alternatives given

Question 1.
The moonlight first falls on _____________?
(A) dog
(B) dove
(C) fish
(D) tree
Answer:
(D) tree

Question 2.
At last the moonlight falls on ____________?
(A) fruits
(B) fish
(C) dog
Answer:
(B) fish

Question 3.
Round the word which is on odd one out?
(i) window, thatch, tree, dog
(ii) fish, dove, mouse, fruit
Answer:
(i) dog
(ii) fruit

(b) Match the items under A with items under B. (Question with Answer)

Answer:

Session – 3

2. Vocabulary

(a) Learn the spelling of the following words using the four steps method- Look > Cover > Write > Verify
beam, peer, harvest, kennel, cote, claws, stream
(b) Solve the following crossword puzzle. Use the clues given. (Question with Answer)

Answer:

Down: Across :
1. In silver feathered sleep
2. Like a log
3. Moveless
4. Goes scampering by

Session – 4

3. Writing:

(a) In 2(b), you have already matched items under A with items under B. Now write four sentences joining the items with ‘is’, the first one is done for you.
Fish + silver reeds in silver stream.
1. The fish is by silver reeds in silver stream.
2. The dog is couched in his kennle like a log.
3. The mouse is with silver clause and silver eye.
4. The dove is in silver feathered sleep.

(b) You can summarise the poem using only one type of sentence. Write as many sentences of this type as you can. The first one is done for you using
the word ‘tree’ given in the bracket, (trees, fruits, window, dog ………………..)
1. The moonlight falls on the tree and it looks silvery.
Answer:
2. The moonlight falls on the fruits and they look silvery.
3. The moonlight falls on the window and it looks silvery.
4. The moonlight falls on the paws of the dog and they look silvery.
5. The moonlight falls on the white breast of the dove and it looks silvery.
6. The moonlight falls on the claws of the mouse and they look silvery.
7. The moonlight falls on the stream water and it looks silvery.

(c) Answer the following questions.

Question 1.
What is the poem about?
Answer:
The poem is about the moon.

Question 2.
What does the moon see first?
Answer:
The moon sees the silver fruit upon silver trees first.

Question 3.
Where does the dog sleep?
Answer:
The dog sleeps in its kennel.

Question 4.
Where does the dove sleep?
Answer:
The dove sleeps in its shadowy cote.

Question 5.
Who goes scampering by?
Answer:
A harvest mouse goes scampering by under the moonlight.

Question 6.
Where is the fish?
Answer:
The fish is by the silver stream with reeds on its bank.

Notes And Glossary
beneath (ବିନିଥ୍) – under (ତଳେ)
casements (କେସ୍‌ମେଣ୍ଟ୍‌ସ୍) – windows (ଝରକା)
ceaseless (ସିକ୍‌ସ୍) – non-stop, without rest (ଅହରହ )
cobwebs (କୋବଓ୍ବେବ) – spider net (ବୁଢ଼ିଆଣୀ ଜାଲ)
couched (କୋଚ୍) – slept (ଶୋଇ ରହିଥିବା )
gleam (ଗ୍ଲିମ୍ ) – shine (ଝଲସୁଥିବା ବା ଚକ୍‌କ୍ କରୁଥୁବା)
kennel (କେନେଲ୍,) – small shelter for a dog (କୁକୁର ରହିବା ଘର)
peers (ପିଅର୍ସ୍) – to look (ଖୋଜିବା ପରି ଦେଖିବା)
pours (ପୋର୍‌ସ୍) – gives in plenty (ଢାଳିବା, ବହୁପରିମାଣରେ ଦେବା)
reeds (ରିଡ୍‌ସ୍) – grass like water plants (ନଦୀ ଝରଣା କୂଳରେ ଥିବା)
scampering (ସ୍କାମ୍ପରଇ) – moving quickly, running (ତରବରିଆ ଭାବେ)
shadowy cote (ସାଡୋ କୋଟ୍) – (here) less bright wings of the dove (ଅସ୍ପଷ୍ଟ)
shoon (ସୁନ୍) – shoe (ଜୋତା)
silver feathered (ସିଲଭର୍ ଫିଦର୍‌ଡ୍) – feathers look like silver colour (ଦେଖାଯାଉଥ‌ିବା ପର)
spider (ସ୍କାଇଡର) – web spinning insect (ବୁଢ଼ିଆଣୀ)
spindles (ସିଣ୍ଡଲ୍‌ – here, spider’s weaving instrument (ପ୍ରାକୃତିକ ତାକୁଡ଼ି)
stretch (ଷ୍ଟେଟ୍) – long and continuous (ବିସ୍ତାରିତ ହୋଇଥିବା)
thatch – roof of straw (ଚାଳ ଛପର)
weaver (ୱିଭର୍) – one who weaves thread / cloth (ବୁଣାଳି, ଲୁଗା)

BSE Odisha 7th Class English Solutions Follow-Up Lesson 7 Silver Important Questions and Answers

(A) Choose the right answer from the options.

Question 1.
The moon walks in the sky?
(a) silently
(b) slowly
(c) speedily
(d) hopingly
Answer:
(b) slowly

Question 2.
While walking the moon wears?
(a) silver shoes
(b) black shoes
(c) grey shoes
(d) red shoes
Answer:
(a) silver shoes

Question 3.
The moon is personified as a _____________?
(a) boy
(b) girl
(e) lady
(d) old lady
Answer:
(c) Lady

Question 4.
The dog sleeps in ______________?
(a) its house
(b) its kennel
(c) its stay
(d) its stable
Answer:
(b) its kennel

Question 5.
The breast of the dove looks silvery?
(a) it is naturally white,
(b) somebody paints it white
(c) the moonbeam falls on it.
(d) none of these
Answer:
(c) The moonbeam falls on it.

Question 6.
The poem is by _______________?
(a) W. Wordsworth
(b) Robert Frost
(c) Walter de la Mare
(d) none of these
Answer:
(c) Walter de la Mare

(B) Answer the following questions.

Question 1.
How does the moon walks?
Answer:
The moon walks slowly wearing silver shoes on her feet.

Question 2.
Why does every object of nature looks white?
Answer:
Every object of nature looks white because the white moon beam falls on them.

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 5 Determinants Ex 5(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 5 Determinants Exercise 5(b)

Question 1.
Write the number of solutions of the following system of equations.
(i) x – 2y = 0
Solution:
No solution

(ii) x – y = 0 and 2x – 2y = 1
Solution:
Infinite

(iii) 2x + y = 2 and -x – 1/2y = 3
Solution:
No solution

(iv) 3x + 2y = 1 and x + 5y = 6
Solution:
One

(v) 2x + 3y + 1 = 0 and x – 3y – 4 = 0
Solution:
One

(vi) x + y + z = 1
x + y + z = 2
2x + 3y + z = 0
Solution:
No solution

(vii) x + 4y – z = 0
3x – 4y – z = 0
x – 3y + z = 0
Solution:
One

(viii) x + y – z = 0
3x – y + z = 0
x – 3y + z = 0
Solution:
One

(ix) a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z = 0
and \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) = 0
Solution:
Infinite solutions as Δ = Δ₁ = Δ₂ = Δ₃ = 0

Question 2.
Show that the following system is inconsistent.
(a – b)x + (b – c)y + (c – a)z = 0
(b – c)x + (c – a)y + (a – b)z = 0
(c – a)x + (a – b)y + (b – c)z =1
Solution:

Question 3.
(i) The system of equations
x + 2y + 3z = 4
2x + 3y + 4z = 5
3x + 4y + 5z = 6 has
(a) infinitely many solutions
(b) no solution
(c) a unique solution
(d) none of the three
Solution:
(a) infinitely many solutions

(ii) If the system of equations
2x + 5y + 8z = 0
x + 4y + 7z = 0
6x + 9y – z = 0
has a nontrivial solution, then is equal to
(a) 12
(b) -12
(c) 0
(d) none of the three
Solution:
(b) -12

(iii) The system of linear equations
x + y + z = 2
2x + y – z = 3
3x +2y + kz = 4
has a unique solution if
(a) k ≠ 0
(b) -1 < k < 1
(c) -2 < k < 2
(d) k = 0
Solution:
(a) k ≠ 0

(iv) The equations
x + y + z = 6
x + 2y + 3z = 10
x + 2y + mz = n
give infinite number of values of the triplet (x, y, z) if
(a) m = 3, n ∈ R
(b) m = 3, n ≠ 10
(c) m = 3, n = 10
(d) none of the three
Solution:
(c) m = 3, n = 10

(v) The system of equations
2x – y + z = 0
x – 2y + z = 0
x – y + 2z = 0
has infinite number of nontrivial solutions for
(a) = 1
(b) = 5
(c) = -5
(d) no real value of
Solution:
(c) = -5

(vi) The system of equations
a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z =0
with has
(a) more than two solutions
(b) one trivial and one nontrivial solutions
(c) No solution
(d) only trivial solutions
Solution:
(a) more than two solutions

Question 4.
Can the inverses of the following matrices be found?
(i) \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
Solution:
|A| = 0
∴ A^-1 can not be found.

(ii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:
∴ |A| = 4 – 6 = -2 ≠ 0
∴ A^-1 exists.

(iii) \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) = 1 – 1 = 0
∴ A^-1 does not exist.

(iv) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\) = 4 – 4 = 0
∴ A^-1 does not exist.

(v) \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = 1 ≠ 0
∴ A^-1 exists.

Question 5.
Find the inverse of the following:
(i) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{cc}
i & -i \\
i & i
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
x & -x \\
x & x^2
\end{array}\right]\), x ≠ 0, x ≠ -1
Solution:

Question 6.
Find the adjoint of the following matrices.
(i) \(\left[\begin{array}{ccc}
1 & 1 & -1 \\
2 & -1 & 2 \\
1 & 3 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
-2 & 2 & 3 \\
1 & 4 & 2 \\
-2 & -3 & 1
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{lll}
2 & 1 & 2 \\
2 & 2 & 1 \\
1 & 2 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 3 & 0 \\
2 & -1 & 6 \\
5 & -3 & 1
\end{array}\right]\)
Solution:

Question 7.
Which of the following matrices are invertible?
(i) \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
1 & 1 & 1 \\
2 & -1 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
2 & 1 & -2 \\
1 & 2 & 1 \\
3 & 6 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
-1 & -2 & 3 \\
2 & 1 & -4 \\
-1 & 0 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 0 & 1 \\
2 & -2 & 1 \\
3 & 2 & 4
\end{array}\right]\)
Solution:

Question 8.
Examining consistency and solvability, solve the following equations by matrix method.
(i) x – y + z = 4
2x + y – 3z = 0
x + y + z = 2
Solution:

(ii) x + 2y – 3z = 4
2x + 4y – 5z = 12
3x – y + z = 3
Solution:
Let

(iii) 2x – y + z = 4
x + 3y + 2z = 12
3x + 2y + 3z = 16
Solution:

(iv) x + y + z = 4
2x + 5y – 2x = 3
x + 7y – 7z = 5
Solution:

(v) x + y + z = 4
2x – y + 3z = 1
3x + 2y – z = 1
Solution:

(vi) x + y – z = 6
2x – 3y + z = 1
2x – 4y + 2z = 1
Solution:

(vii) x – 2y = 3
3x + 4y – z = -2
5x – 3z = -1
Solution:

(viii) x + 2y + 3z = 14
2x – y + 5z = 15
2y + 4z – 3x = 13
Solution:

(ix) 2x + 3y +z = 11
x + y + z = 6
5x – y + 10z = 34
Solution:

Question 9.
Given the matrices
A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and C = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
write down the linear equations given by AX = C and solve it for x, y, z by matrix method.
Solution:

Question 10.
Find X, if \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & -1 \\
2 & 1 & -1
\end{array}\right]\) X = \(\left[\begin{array}{l}
6 \\
0 \\
1
\end{array}\right]\) where X = \(\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]\)
Solution:

Question 11.
Answer the following:
(i) If every element of a third order matrix is multiplied by 5, then how many times its determinant value becomes?
Solution:
125

(ii) What is the value of x if \(\left|\begin{array}{ll}
4 & 1 \\
2 & 1
\end{array}\right|^2=,\left|\begin{array}{ll}
3 & 2 \\
1 & x
\end{array}\right|-\left|\begin{array}{cc}
x & 3 \\
-2 & 1
\end{array}\right|\) ?
Solution:
x = 6

(iii) What are the values of x and y if \(\left|\begin{array}{ll}
x & y \\
1 & 1
\end{array}\right|=2,\left|\begin{array}{ll}
x & 3 \\
y & 2
\end{array}\right|=1\) ?
Solution:
x = 5, y = 3

(iv) What is the value of x if \(\left|\begin{array}{ccc}
x+1 & 1 & 1 \\
1 & 1 & -1 \\
-1 & 1 & 1
\end{array}\right|\) = 4?
Solution:
x = 0

(v) What is the value of \(\left|\begin{array}{ccc}
\mathbf{o} & -\mathbf{h} & -\mathbf{g} \\
\mathbf{h} & \mathbf{0} & -\mathbf{f} \\
\mathbf{g} & \mathbf{f} & \mathbf{0}
\end{array}\right|\)?
Solution:
0

(vi) What is the value of \(\left|\begin{array}{l}
\frac{1}{a} 1 \mathrm{bc} \\
\frac{1}{b} 1 c a \\
\frac{1}{c} 1 a b
\end{array}\right|\)
Solution:
0

(vii) What is the co-factor of 4 in the determinant \(\left|\begin{array}{rrr}
1 & 2 & -3 \\
4 & 5 & 0 \\
2 & 0 & 1
\end{array}\right|\)
Solution:
-2

(viii)In which interval does the determinant \(\left|\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right|\) lie?
Solution:
[2, 4]

(ix) Ifx + y + z = n, what is the value of Δ = \(\left|\begin{array}{ccc}
\sin (x+y+z) & \sin B & \cos C \\
-\sin B & 0 & \tan A \\
\cos (A+B) & -\tan A & 0
\end{array}\right|\) Where A, B, C are the angles of triangle.
Solution:
0

Question 12.
Evaluate the following determinants:
(i) \(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
= 2\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\) = 0
(∵ C₁ = C₃)

(ii) \(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\) = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & 1 \\
14 & 17 & 11
\end{array}\right|\)
( R₁ = R₁ – R₂, R₂ = R₂ – R₃)
= 0 (∵ R₁ = R₂)

(iii) \(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
= 7\(\left|\begin{array}{ccc}
32 & 777 & 32 \\
105 & 888 & 105 \\
116 & 999 & 116
\end{array}\right|\) = 0
(∵ C₁ = C₂)

(iv) \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 3 & 4 \\
3 & 4 & 6
\end{array}\right|\)
Solution:

(v) \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
3 & 5 & 7 \\
8 & 14 & 20
\end{array}\right|\)
Solution:

(vi) \(\left|\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|\)
Solution:

= 225 – 256 – 4(100 – 144) + 9(64 – 81)
= -31 – 4(-44) + 9(-17)
= -31 + 176 – 153 = -184 + 176
= -8

(vii) \(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
= 2\(\left|\begin{array}{cc}
1 & -5863 \\
1 & 4137
\end{array}\right|\)
(expanding along 2nd column)
= 2(4137 + 5863)
= 2 × 10000 = 20000

(viii) \(\left|\begin{array}{lll}
265 & 240 & 219 \\
240 & 225 & 198 \\
219 & 198 & 181
\end{array}\right|\)
Solution:

(ix) \(\left|\begin{array}{ccc}
0 & a^2 & b \\
b^2 & 0 & a^2 \\
a & b^2 & 0
\end{array}\right|\)
Solution:

= -a² (0 –  a²) + b (b⁴ –  0) = a⁵ + b⁵

(x) \(\left|\begin{array}{ccc}
a-b & b-c & c-a \\
\boldsymbol{x}-\boldsymbol{y} & \boldsymbol{y}-\boldsymbol{z} & z-\boldsymbol{x} \\
\boldsymbol{p}-\boldsymbol{q} & \boldsymbol{q}-\boldsymbol{r} & \boldsymbol{r}-\boldsymbol{p}
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
x-y & y-z & z-x \\
p-q & q-r & r-p
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & y-z & z-x \\
0 & q-r & r-p
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0 ( ∵ C₁ = 0)

(xi) \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & c-a & a-b \\
0 & a-b & b-c
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0

(xii) \(\left|\begin{array}{ccc}
-\cos ^2 \theta & \sec ^2 \theta & -0.2 \\
\cot ^2 \theta & -\tan ^2 \theta & 1.2 \\
-1 & 1 & 1
\end{array}\right|\)
Solution:

(Expanding along 3rd row)
= (-cos² θ + sec² θ) (-tan² θ – 1.2) – (sec² θ + 0.2) (cot² θ – tan² θ)
= sin² θ – 1.2 cos² θ – sec² θ tan² θ – 1.2 sec² θ – cosec² θ +  sec² θ tan² θ – 0.2 cot² θ + 0.2 tan² θ
= sin² θ – cosec² θ + 1.2 (cos² θ – sec² θ) + 0.2 (tan² θ – cot² θ) ≠ 0
The question seems to be wrong.

Question 13.
If \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+x & 1 \\
1 & 1 & 1+y
\end{array}\right|\) = 0 what are x and y?
Solution:

or, xy – 0 = 0 ⇒ xy = 0, ⇒ x = 0, or y = 0

Question 14.
For what value of x \(\left|\begin{array}{ccc}
2 x & 0 & 0 \\
0 & 1 & 2 \\
-1 & 2 & 0
\end{array}\right|\) = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 3 & 4 \\
0 & 3 & 5
\end{array}\right|\)?
Solution:

Question 15.
Solve \(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
Solution:
\(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
or, (x – a) \(\left|\begin{array}{cc}
x+b & 0 \\
0 & x+c
\end{array}\right|\) = 0
or, (x + a) (x + b) (x + c) = 0
x = -a, x = -b, x = -c

Question 16.
Solve \(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|\) = 0
Solution:

Question 17.
Solve \(\left|\begin{array}{ccc}
x+a & b & c \\
a & x+b & c \\
a & b & x+c
\end{array}\right|\) = 0
Solution:

Question 18.
Show that x = 2 is a root of \(\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2\end{array}\right|\) = 0 Solve this completely.
Solution:
Putting x = 2,

= (x – 1) (-15x + 30 – 5x² + 10x)
= (x – 1) (-5x² – 5x + 30)
= -5(x – 1) (x² + x – 6)
= -5(x – 1) (x + 3) (x – 2) = 0
⇒ x = 1 or, -3 or 2.

Question 19.
Evaluate \(\left|\begin{array}{ccc}
1 & a & b c \\
1 & b & c a \\
1 & c & a b
\end{array}\right|\) – \(\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|\)
Solution:

= (a – b) (b – c) [(-a + c) – (b + c – a – b)]
= (a – b) (b – c) (-a + c – c + a) = 0

Question 20.
\(\left|\begin{array}{lll}
a & a^2-b c & 1 \\
b & b^2-a c & 1 \\
c & c^2-a b & 1
\end{array}\right|\)
Solution:

Question21.
For what value of X the system of equations
x + y + z = 6, 4x + λy – λz = 0, 3x + 2y – 4z = -5 does not possess a solution?
Solution:

= 24 – 6λ – 2λ = 24 – 8λ
when Δ = 0
We have 24 – 8λ, = 0 or, λ = 3
The system of equations does not posses solution for λ = 3.

Question 22.
If A is a 3 × 3 matrix and |A| = 2, then which matrix is represented by A × adj A?
Solution:

Question 23.
If A = \(\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\)
show that (I + A) (I – A)^-1 = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) where I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

Question 24.
Prove the following:
(i) \(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
a b & b^2+1 & b c \\
a c & b c & c^2+1
\end{array}\right|\) = 1 + a² + b² + c²
Solution:

(ii) \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^3 & b^3 & c^3
\end{array}\right|\) = (b – c) (c – a) (a – b) (a + b + c)
Solution:

= (a – b) (b – c) (b² + bc + c² – a² – ab – b²)
= (a – b) (b- c) (c² – a² + bc – ab)
= (a – b) (b – c) {(c – a) (c + a) + b(c – a)}
= (a – b) (b – c) (c – a) (a + b + c) = R.H.S.
(Proved)

(iii) \(\left|\begin{array}{lll}
\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\
\boldsymbol{b} & \boldsymbol{c} & \boldsymbol{a} \\
\boldsymbol{c} & \boldsymbol{a} & \boldsymbol{b}
\end{array}\right|\) = 3abc – a³ – b³ – c³
Solution:

= (a + b + c) {(b – c) (a – b) – (c – a)²}
= (a + b + c) (a + b + c) (ab – b² – ca + bc – c² – a² + 2ca)
= (a + b + c) (-a² – b² – c² + ab + bc + ca)
= -(a + b + c) (a² + b² + c² – ab – bc – ca)
=- (a³ + b³ + c³ – 3abc)
= 3abc – a³ – b³ – c³

(iv) \(\left|\begin{array}{lll}
b^2-a b & b-c & b c-a c \\
a b-a^2 & a-b & b^2-a b \\
b c-a c & c-a & a b-a^2
\end{array}\right|\) = 0
Solution:

= (b² – a² + bc – ac) (a – b) {(-a + b) (c – a) – (bc – ac – ab + a²)}
= (b² – a² + bc – ac) (a – b) (- ca + a² + bc – ab – bc + ac + ab – a²)
= (b² – a² + bc – ac) (a – b) × 0 = 0
= R.H.S.
(Proved)

(v) \(\left|\begin{array}{ccc}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right|\) = 4a²b²c²
Solution:

(vi) \(\left|\begin{array}{lll}
(b+c)^2 & a^2 & b c \\
(c+a)^2 & b^2 & c a \\
(a+b)^2 & c^2 & a b
\end{array}\right|\) = (a² + b² + c² ) (a + b + c) (b – c) (c – a) (a – b)
Solution:

= (a – b) (b – c) (a² + b² + c²) (-a² – ab + bc + c²)
= (a – b) (b – c) (a² + b² + c²) {(c² – a²) + b(c – a)}
= (a² + b² + c²) (a – b) (b – c) (c – a) (c + a + b)

(vii) \(\left|\begin{array}{lll}
b+c & a+b & a \\
c+a & b+c & b \\
a+b & c+a & c
\end{array}\right|\) = a³ + b³ + c³ – 3abc
Solution:

= (a + b +c) {(a – b) (a – c) – (c – b) (b – c)}
= (a + b + c) (a² – ac – ab + bc – bc + c² + b² – bc)
= (a + b + c) (a² + b² + c² – ab – bc – ca)
= (a³ + b³ + c³ – 3abc)

(viii) \(\left|\begin{array}{ccc}
a+b+c & -c & -b \\
-c & a+b+c & -a \\
-b & -a & a+b+c
\end{array}\right|\) = 2(b + c) (c + a) (a + b)
Solution:

= -2(a + b) (b + c) (-a – b – c + b)
= 2(a + b) (b + c) (c + a)

(ix) \(\left|\begin{array}{ccc}
a x-b y-c z & a y+b x & a z+c x \\
b x+a y & b y-c z-a x & b z+c y \\
c x+a z & a y+b z & c z-a x-b y
\end{array}\right|\) = (a² + b² + c²) (ax + by + cz) (x² + y² + z²)
Solution:

Question 25.
If 2s = a + b + c show that \(\left|\begin{array}{ccc}
a^2 & (s-a)^2 & (s-a)^2 \\
(s-b)^2 & b^2 & (s-b)^2 \\
(s-c)^2 & (s-c)^2 & c^2
\end{array}\right|\) = 2s³ (s – a) (s – b) (s – c)
Solution:
Let s – a = A, s – b = B, s – c = C
A + B + C = 3s – (a + b + c)
= 3s – 2s = s
Also B + C = s – b + s – c = 2s – (b + c)
= (a + b + c) – b + c = a
Similarly C + A = b, A + B = c

= 2 ABC (A + B + C)²
[Refer Q.No.9 (xii) of Exercise 5(a)]
= 2(s – a) (s – b)(s – c) s³

Question 26.
if \(\left|\begin{array}{ccc}
x & x^2 & x^3-1 \\
y & y^2 & y^3-1 \\
z & z^2 & z^3-1
\end{array}\right|\) = 0 then prove that xyz =1 when x, y, z are non zero and unequal.
Solution:

= (x – y) (y – z) (z – x) (xyz – 1)
It is given that
(x – y) (y – z) (z – x) (xyz – 1) = 0
⇒ xyz – 1 (as x ≠ y ≠ z)

Question 27.
Without expanding show that the following determinant is equal to Ax + B where A and B are determinants of order 3 not involving x.
\(\left|\begin{array}{ccc}
x^2+x & x+1 & x-2 \\
2 x^2+3 x-1 & 3 x & 3 x-3 \\
x^2+2 x+3 & 2 x-1 & 2 x-1
\end{array}\right|\)
Solution:

Question 28.
If x, y, z are positive and are the pth, qth and rth terms of a G.P. then prove that \(\left|\begin{array}{lll}
\log x & p & 1 \\
\log y & q & 1 \\
\log z & r & 1
\end{array}\right|\) = 0
Solution:
Let the G.P. be
a, aR, aR², aR³ …..aR^n-1
p th term = aR^p-1
q th term = aR^q-1
r th term = aR^r-1
x = aR^p-1, y= aR^q-1, z = aR^r-1
log x = log a + (p – 1) log R,
log y = log a + (q – 1) log R,
log z = log a + (r – 1) log R

Question 29.
If D_j = \(\left|\begin{array}{ccc}
j & a & n(n+2) / 2 \\
j^2 & b & n(n+1)(2 n+1) / 6 \\
j^3 & c & n^2(n+1)^2 / 4
\end{array}\right|\) then prove that \(\sum_{j=1}^n\)D_j = 0.
Solution:

Question 30.
Ifa₁, a₂,……a_n are in G.P. and a_i > 0 for every i, then find the value of
\(\left|\begin{array}{ccc}
\log a_n & \log a_{n+1} & \log a_{n+2} \\
\log a_{n+1} & \log a_{n+2} & \log a_{n+3} \\
\log a_{n+2} & \log a_{n+3} & \log a_{n+4}
\end{array}\right|\)
Solution:

Question 31.
If f(x)= \(\left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin ^2 x \\
\sin ^2 x & 1+\cos ^2 x & 4 \sin 2 x \\
\sin ^2 x & \cos ^2 x & 1+4 \sin ^2 x
\end{array}\right|\) what is the least value of f(x)?
Solution:

As minimum value of sin 2x is 0. So the minimum value of above function f(x) is 2.

Question 32.
If f_r(x), g_r(x), h_r(x), r = 1, 2, 3 are polynomials in x such that f_r(a) = g_r(a) = h_r(a) and
F(x) = \(\left[\begin{array}{lll}
f_1(x) & f_2(x) & f_3(x) \\
g_1(x) & g_2(x) & g_3(x) \\
h_1(x) & h_2(x) & h_3(x)
\end{array}\right]\) find F'(x) at x = a.
Solution:
We have

[Since f₁a) = g₁(a) = h₁(a), f₂(a) = g₂(a) = h₂(a) and f₃(a) = g₃(a) = h₃(a) So that each determinant is zero due to presence of two identical rows.]

Question 33.
If f(x) = \(\left[\begin{array}{ccc}
\cos x & \sin x & \cos x \\
\cos 2 x & \sin 2 x & 2 \cos 2 x \\
\cos 3 x & \sin 3 x & 3 \cos 3 x
\end{array}\right]\) find f'(\(\frac{\pi}{2}\)).
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(a)

Question 1.
State the order of the following matrices.
(i) [abc]
(ii) \(\left[\begin{array}{l}
1 \\
2
\end{array}\right]\)
(iii) \(\left[\begin{array}{ll}
x & y \\
y & z \\
z & x
\end{array}\right]\)
(iv) \(\left[\begin{array}{cccc}
1 & 0 & 1 & 4 \\
2 & 1 & 3 & 0 \\
-3 & 2 & 1 & 3
\end{array}\right]\)
Solution:
(i) (1 x 3)
(ii) (2 x 1)
(iii) (3 x 2)
(iv) (3 x 4)

Question 2.
How many entries are there in a
(i) 3 x 3 matrix
(ii) 3 x 4 matrix
(iii) p x q matrix
(iv) a sqare matrix of order p?
Solution:
(i) 9
(ii) 12
(iii) pq
(iv) p²

Question 3.
Give an example of
(i) 3 x 1 matrix
(ii) 2 x 2 matrix
(iii) 4 x 2 matrix
(iv) 1 x 3 matrix
Solution:
(i) \(\left(\begin{array}{l}
a \\
b \\
c
\end{array}\right)\)
(ii) \(\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)\)
(iii) \(\left(\begin{array}{ll}
a & b \\
c & d \\
e & f \\
g & h
\end{array}\right)\)
(iv) (1, 2, 3)

Question 4.
Let A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) What is the order of A?
(ii) Write down the entries a₃₁, a₂₅, a₂₃
(iii) Write down A^T.
(iv) What is the order of A^T?
Solution:
A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) Order of A is (3 x 5)
(ii) a₃₁ = 3, a₂₅= 2, a₂₃ = 6
(iii) A^T = \(\left[\begin{array}{lll}
1 & 4 & 3 \\
2 & 5 & 9 \\
3 & 6 & 1 \\
4 & 1 & 1 \\
1 & 2 & 6
\end{array}\right]\)
(iv) Order of A^T is (5 x 3).

Question 5.
Matrices A and B are given below. Find A + B, B + A, A – B and B – A. Verify that A + B = B + A and B – A = -(A – B)
(i) A = \(\left[\begin{array}{l}
7 \\
1
\end{array}\right]\), B = \(\left[\begin{array}{c}
-6 \\
9
\end{array}\right]\)
Solution:

(ii) A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
4 & 1 \\
-3 & 2
\end{array}\right]\)
Solution:

(iii) A = \(\left[\begin{array}{ll}
\frac{1}{2} & \frac{1}{4} \\
\frac{1}{3} & \frac{1}{5}
\end{array}\right]\), B = \(\left[\begin{array}{ll}
\frac{1}{3} & \frac{1}{2} \\
\frac{1}{2} & \frac{4}{5}
\end{array}\right]\)
Solution:

(iv) A = \(\left[\begin{array}{cc}
1 & a-b \\
a+b & -3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & b \\
-a & 5
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{rrr}
1 & -2 & 5 \\
-1 & 4 & 3 \\
1 & 2 & -3
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
-1 & 2 & -5 \\
1 & -3 & -3 \\
1 & 2 & 4
\end{array}\right]\)
Solution:

Question 6.
(i) Find the 2×2 matrix X
if X + \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) = \(\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]\)
Solution:

(ii) Given
[x y z] – [-4 3 1] = [-5 1 0] derermine x, y, z.
Solution:
[x y z] – [-4 3 1] = [-5 1 0]
∴ (x y z) = (-4 3 1) + (-5 1 0) = (-9 4 1)
∴ x = -9, y = 4, z = 1

(iii) If \(\left[\begin{array}{ll}
x_1 & x_2 \\
y_1 & y_2
\end{array}\right]\) – \(\left[\begin{array}{ll}
2 & 3 \\
0 & 1
\end{array}\right]\) = \(\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]\) determine x₁, x₂, y₁, y₂.
Solution:

(iv) Find a matrix which when added to \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]\) gives \(\left[\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right]\)
Solution:

Question 7.
Calculate whenever possible, the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\)
Solution:
\(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\) is impossible because number of columns of 1st ≠ number of rows of second.

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
3 & 1 \\
1 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right]\)
Solution:

Question 8.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\), C = \(\left[\begin{array}{ll}
2 & 2 \\
1 & 3
\end{array}\right]\)
Calculate (i) AB (ii) BA (iii) BC (iv) CB (v) AC (vi) CA
Solution:

Question 9.
Find the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & i \\
i & -1
\end{array}\right]^2\) where i = √-1
Solution:

(vi) \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{ll}
0 & k \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
Solution:

(ix) \(\left[\begin{array}{ll}
1 & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(x) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)
Solution:
\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) = \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)

Question 10.
Write true or false in the following cases:
(i) The sum of a 3 x 4 matrix with a 3 x 4 matrix is a 3 x 3 matrix.
Solution:
False

(ii) k[0] = 0, k ∈ R
Solution:
False

(iii) A – B = B – A, if one of A and B is zero and A and B are of the same order.
Solution:
False

(iv) A + B = B + A, if A and B are matrices of the same order.
Solution:
True

(v) \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 0
\end{array}\right]\) + \(\left[\begin{array}{cc}
-1 & 0 \\
2 & 0
\end{array}\right]\) = 0
Solution:
True

(vi) \(\left[\begin{array}{ll}
3 & 1 \\
6 & 2
\end{array}\right]\) = 3 \(\left[\begin{array}{ll}
1 & 1 \\
2 & 2
\end{array}\right]\)
Solution:
False

(vii) With five elements a matrix can not be constructed.
Solution:
False

(viii)The unit matrix is its own transpose.
Solution:
True

Question 11.
If A = \(\left[\begin{array}{cc}
2 & 4 \\
3 & 13
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) find A – α I, α ∈ R.
Solution:

Question 12.
Find x and y in the following.
(i) \(\left[\begin{array}{cc}
x & -2 y \\
0 & -2
\end{array}\right]=\left[\begin{array}{cc}
1 & -8 \\
0 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{c}
x+3 \\
2-y
\end{array}\right]=\left[\begin{array}{c}
1 \\
-3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{c}
2 x-y \\
x+y
\end{array}\right]=\left[\begin{array}{c}
3 \\
-9
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{l}
x \\
y
\end{array}\right]+\left[\begin{array}{l}
3 \\
4
\end{array}\right]=\left[\begin{array}{c}
2 \\
-1
\end{array}\right]\)
Solution:

(v) [2x -y] + [y 3x] = 5 [1 0]
Solution:

Question 13.
The element of ith row and ith column of the following matrix is i +j. Complete the matrix.

Question 14.
Write down the matrix

Question 15.
Construct a 2 x 3 matrix having elements given by
(i) a_ij = i + j
(ii) a_ij = i – j
(iii) a_ij = i × j
(iv) a_ij = i / j
Solution:

Question 16.
If \(\left[\begin{array}{cc}
2 x & y \\
1 & 3
\end{array}\right]+\left[\begin{array}{cc}
4 & 2 \\
0 & -1
\end{array}\right]=\left[\begin{array}{ll}
8 & 3 \\
1 & 2
\end{array}\right]\)
Solution:

Question 17.
Find A such that
\(\left[\begin{array}{ccc}
2 & 3 & 4 \\
1 & 0 & -2 \\
3 & 1 & -1
\end{array}\right]+A=\left[\begin{array}{ccc}
1 & 2 & -1 \\
2 & -1 & 0 \\
1 & 3 & 2
\end{array}\right]\)
Solution:

Question 18.
If

Question 19.
What is the order of the matrix B if [3 4 2] B = [2 1 0 3 6]
Solution:
(3 4 2) B = (2 1 0 3 6)
Let A = (3 4 2), C = (2 1 0 3 6)
∴ Order of A = (1 x 3)
Order of C = (1 x 5)
∴ Order of B = (3 x 5)

Question 20.
Find A if \(\left[\begin{array}{l}
4 \\
1 \\
3
\end{array}\right]\) A = \(\left[\begin{array}{rrr}
-4 & 8 & 4 \\
-1 & 2 & 1 \\
-3 & 6 & 3
\end{array}\right]\)
Solution:

Question 21.
Find B if B² = \(\left[\begin{array}{cc}
17 & 8 \\
8 & 17
\end{array}\right]\)
Solution:

∴ a² + bc = 17, ab + bd= 8
ca + cd = 8, bc + d² = 17
∴ a² + bc = bc + d²
or, a² + d² or, a = d
or, ca + cd = ab + bd
or, cd + cd – bd + bd
or, 2cd = 2bd = 8
or, b = c and bd = 4 = cd
∴ ab + bd= 8
or, ab + 4 = 8
or, ab = 4
Again, a² + bc = 17
or, a² + b . b = 17 (∵ b = c)
or, a² + b² = 17
Also (a + b)² = a² + b² + 2ab
∴ (a + b)² = 17 + 8 = 25
or, a + b = 5
And (a – b)² = 17 – 8 = 9
or, a – b = 3
∴ a = 4, b = 1, So d = 4, c = 1
∴ B = \(\left[\begin{array}{ll}
4 & 1 \\
1 & 4
\end{array}\right]\)

Question 22.
Find x and y when

Question 23.
Find AB and BA given that:

Question 24.
Evaluate

Question 25.
If

Show that AB = AC though B ≠ C. Verify that
(i) A + (B + C) = (A + B) + C
(ii) A(B + C) = AB + AC
(iii) A(BC) = (AB)C
Solution:

Question 26.
Find A and B where

Question 27.
If A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\) and I be the 2 × 2 unit matrix find (A – 2I) (A – 3I)
Solution:

Question 28.
Verify that [AB]^T = B^TA^T where

Question 29.
Verify that A = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\) satisfies the equation x² – (a + d)x + (ad – bc)I = 0 where I is the 2 x 2 matrix.
Solution:

Question 30.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), show that A³ – 23 A – 40 I = 0
Solution:

Question 31.

Question 32.
If A and B are matrices of the same order and AB = BA, then prove that
(i) A² – B² = (A – B) (A + B)
(ii) A² + 2AB + B² = (A + B)²
(iii) A² – 2AB + B² = (A – B)²
Solution:
(i) (A – B) (A + B)
= A² + AB – BA – B²
= A² + AB – AB- B²(∵ AB = BA)
= A² – B²
(ii) (A + B)² = (A + B) (A + B)
= A² + AB + BA + B²
= A² + AB + AB + B² (∵ AB = BA)
= A² + 2AB + B²
(iii) (A – B)² = (A – B) (A – B)
= A² – AB – BA + B²
= A² – AB – AB + B² (∵ AB = BA)
= A² – 2AB + B²

Question 33.
If α and β are scalars and A is a square matrix then prove that
(A – αI) . (A – βI) = A² – (α + β) A + αβI, where I is a unit matrix of same order as A.
Solution:
(A – αI) (A – βI)
= A² – AβI – αIA + αβI²
= A² – βAI – αA + αβI
(∵ IA = A, I² = I)
= A² – βA – αA + αβI) (∵ AI = A)
= A² – (α + β) A + αβI

Question 34.
If α and β are scalars such that A = αβ + βI, where A, B and the unit matrix I are of the same order, then prove that AB = BA.
Solution:
We have A = αβ + βI
AB (αβ + βI) B
= α βB + βI B
= α βB + βB = (α + I) βB
= βB (α + 1)
(∵ Scalar mltiβlication is associative)
= Bβ (α + 1)
= Bβα + Bβ = Bαβ + BIβ
(∵ BI = B)
= B (αβ + βi) = BA
AB = BA
(proved)

Question 35.

Question 36.

Question 37.

Question 38.

Question 39.

Question 40.

Question 41.

Question 42.

Question 43.

	Men	Women	Children
Family A →	4	6	2
Family B →	2	2	4

	Family B
	Calory	Proteins
Men	2400	45
Women	1900	55
Children	1800	33

Solution:
The given informations can be written in matrix form as

∴ Calory requirements for families A and B are 24600 and 15800 respectively and protein requirements are 576 gm and 332 gm respectively.

Question 44.
Let the investment in first fund = ₹x and in the second fund is ₹(50000-x)
Investment matrix A=[x  50000-x]

⇒ 300000 – x = 278000
⇒ x = 22000
∴ He invests ₹22000 in first bond and ₹28000 in the second bond.

Odisha State Board BSE Odisha 6th Class English Solutions Test-2(A) Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-2(A)

BSE Odisha 6th Class English Test-2(A) Text Book Questions and Answers

□ The figures in the right-hand margin indicate the marks for each question.
1. Write the following names of persons in English.
(Teacher will provide names of six persons in Odia.)

ଡାକ୍ତର ରାଜେନ୍ଦ୍ର ପ୍ରସାଦ
ରଣଜିତ ସିଂହ
ଅରବିନ୍ଦ ଘୋଷ
ବାଘା ଜତିନ
ସମ୍ରାଟ ଅଶୋକ |
ରାଜା ଦଶରଥ

Answer:
Doctor Rajendra Prasad
Ranjit Singh
Aurobindo Ghose
Bagha Jatin
Emperor Ashok
King Dasaratha

2. Write the following names of places in English.
(Teacher will provide names of six places in Odia.)

ପୁଡୁଚେରୀ |
ଶ୍ରୀ ଲଙ୍କା
ବିଜୟ ନାଗର
ମଥୁରା |
ମୁମ୍ବାଇ |
କପିଳାସ

Answer:
Puducherry
Srilanka
Vijaya Nagar
Mathura
Mumbai
Kapilas

3. Your teacher will give a dictation of twelve words. Listen to him/her and write.

Answer:
village
classmate
friend
decided
deaf
Good Morning
fever
Germany
soldier
quarters
country
understand

4. Given below are some words. Your teacher will read aloud seven of them. Tick those s/he reads aloud.
language, young, neither, retire, gunny bag, religious, sight, greedy, weather, straightened, beautiful, special, elephant, bicycle
[Listen to your teacher carefully and tick those words as he reads aloud.]

5. Your teacher will read aloud a paragraph. Listen to him/her and fill in the gaps. (Question with Answer)
Once there lived a poor man in a village. He had a rich classmate. He lived in a town. They did not meet for a long time. In the meantime, the rich friend had problems with his ear and became deaf. His friend in the village could not know this.

6. Match the words which sound alike at the end. (Question with Answer)

Answer:

7. Read The poem and answer the questions incomplete sentences.

There was a dog and there was a cat.
One very thin and the other is fat.
Neither of them was a pet.
But the cat always sat on a mat.
and claimed she was a loving pet.

Question (i)
Who were there in the poem?
Answer:
In the poem, there were a dog and a cat.

Question (ii)
How were they?
Answer:
One was very thin and the other was fat.

Question (iii)
Were they pets?
Answer:
No, neither of them was a pet.

Question (iv)
Who sat on a mat?
Answer:
The cat always sat on a mat.

Question (v)
What did the cat claim?
Answer:
The cat claimed that she was a loving pet.

(vi)
Who was very thin?
Answer:
The dog was very thin.

8. Read the following paragraph and answer the questions in complete sentences.
“There is a special school in Karagudi. It is special because it is not for children. Can you guess for whom it is ? It is for baby elephants. Who teaches them? The elephant trainers teach them. Like our schools, they have a timetable. They learn, play and eat according to this timetable.

Question (i)
What is there in Karagudi?
Answer:
There is a special school in Karagudi.

Question (ii)
Why is this school special?
Answer:
This school is special because it is not for children.

Question (iii)
Who are the students?
Answer:
Baby elephants are the students.

Question (iv)
Who are the teachers?
Answer:
The elephant trainers are the teachers.

Question (v)
What do they have like our schools?
Answer:
Like our schools, they have a timetable.

Question (vi)
How do they follow it?
Answer:
According to this timetable, they learn, play, and eat in the special school.

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

It was a very cool night And there
was no crab in sight.
The fox looked for one
But there was none.
“Where did they go ?”
Not even one in sight!
They must be in their holes If
I’m right.”

Question (i)
How was the night?
Answer:
It was a very cool night.

Question (ii)
What was not in sight?
Answer:
There was no crab in sight.

Question (iii)
Who looked for the crab?
Answer:
The fox looked for the crab.

Question (iv)
Did he find one?
Answer:
No, he did not find any crab, because there was none.

Question (v)
Where did they go?
Answer:
They must have been in their holes.

Question (vi)
Who were there in the poem?
Answer:
There were the fox and the crab in the poem.

10. Read the following paragraph and answer the questions in complete sentences.
Mahagiri was a big elephant. He was trained at a special school. He was bought by a merchant. The merchant made a lot of money by putting Mahagiri to work. The elephant was often sent to the forest to carry heavy logs of wood. Sometimes, he carried people from one place to another. Once, he even carried a bridegroom to the bride’s house! At times he was sent to a famous temple in a village nearby to lead the festival procession.

Question (i)
What is this paragraph about?
Answer:
This paragraph is about an elephant.

Question (ii)
Who was Mahagiri?
Answer:
Mahagiri was a big elephant.

Question (iii)
Where was Mahagiri trained?
Answer:
Mahagiri was trained at a special school.

Question (iv)
Who bought it?
Answer:
A merchant bought it.

Question (v)
How did the merchant make a lot of money?
Answer:
The merchant made a lot of money by putting Mahagiri to work.

Question (vi)
Why was the elephant often sent to the forest?
Answer:
The elephant was often sent to the forest to carry heavy logs of wood.

Question (vii)
What did he sometimes carry?
Answer:
Sometimes, he carried people from one place to another.

Question (viii)
What did he once even carry?
Answer:
Once, he even carried a bridegroom to the bride’s house.

Question (ix)
Where was he sent to at times?
Answer:
At times he was sent to a famous temple in a village nearby.

Question (x)
Why was he sent to a famous temple?
Answer:
He was sent to a famous temple to lead the festival procession.

Odisha State Board BSE Odisha 6th Class English Solutions Test-2(B) Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-2(B)

BSE Odisha 6th Class English Test-2(B) Text Book Questions and Answers

The figures in the right-hand margin indicate the marks for each question.
1. Write the following names of persons in English.
(Teacher will provide names of six persons ¡n Odia.)

ରାଜା ଦିବ୍ୟାସିଂହ ଦେବ
ରାଜା ରାମମୋହନ ରୟ |
ରାଜା ହରିଚନ୍ଦ୍ର
ରାମଚନ୍ଦ୍ର
ଚାଖି ଖୁଣ୍ଟିଆ
ଲକ୍ଷ୍ମଣ କୁମାର

Answer:
Raja Dibvasingh Deb
Raja Rammohan Roy
Raja Harischandra
Ramchandra
Chakhi Khuntia
Lakshmana Kumar

2. Write the following names of places ¡n English.
(Teacher will provide names of six places in Odia.)

ଦୌଲତାବାଦ
ସଂସଦ ଭବନ
ଜୁମ୍ମା ମସଜିଦ୍
ଦ୍ୱାରିକାପୁର
ଅୟୋଧ୍ୟା
ହସ୍ତିନା

Answer:
(i) Doulatabad
(ii) Parliament House
(iii) Jumma Mosque
(iv) Dwarikapura
(v) Ayodhya
(vi) Hasina

3. Your teacher will give a dictation of twelve words. Listen to him/her and write.

Answer:
Tamilnadu
children
elephant
instructions
trainer
circus
special
guess
naughty
raises
praised
popular

4. Given below are some words. Your teacher will read aloud seven of them. Tick those s/he reads aloud.
perhaps, sailor, zoo, trumpet, musical, private, driver, moon, doctor, pilot, cousin, farmer, builder, nurse, painter.
[Listen to your teacher carefully and tick those words as he reads aloud.]

5. Your teacher will read aloud a paragraph. You listen to him/her and fill in the gaps. (Question with Answer)
On Makar holidays Raghunath would come to his village Dandbose, a few kilometers away from Rairangpur town in the district of Mavurbhanj. He was then working at Baripada. In those days he was the only educated man in his area.

6. Match the words which sound alike at the end. (Question with Answer)

Answer:

7. Read the poem and answer the questions in complete sentences.

When I grow up
I want to be
A detective
With a master key.
I could be a soldier
Perhaps a sailor too.
Or become a keeper
At Nandan Kanan Zoo.
I’d like to own a trumpet
And play a musical tune.
Or buy a private space-ship
To fly to the moon.

Question (i)
Who is ‘I’ in the poem?
Answer:
The poet is T in the poem.

Question (ii)
What does the child want to be in the 1st stanza?
Answer:
In the first stanza, the child wants to be a detective.

Question (iii)
What does a detective have with him?
Answer:
A detective has a master key with him.

Question (iv)
In the second stanza, the child likes three professions. What are they?
Answer:
In the second stanza, the child likes three professions. They are soldiers, sailors, and keepers at Nandan Kanan zoo.

Question (v)
In which stanza does the poet describe a child’s interest in music?
Answer:
In the third stanza of the poem, the poet describes a child’s interest in music.

Question (vi)
How does he want to fly to the moon?
Answer:
He wants to buy a private spaceship in order to (960) fly him to the moon.

8. Read the following paragraph and answer the questions in complete sentences.
On Makar holidays Raghunath would come to his village Dandbose, a few kilometers away from Rairangpur town in the district of Mayurbhanj. He was then working at Baripada.

Question (i)
What is this paragraph about?
Answer:
This paragraph is about Raghunath.

Question (ii)
What is the name of his village?
Answer:
The name of his village is Dandbose.

Question (iii)
When would he come there?
Answer:
He would come there on the Makar holidays.

Question (iv)
How far is it from Rairangpur town?
Answer:
It is a few kilometers away from Rairangpur town.

Question (v)
Where is Rairangpur?
Answer:
Rairangpur is in the district of Mayurbhanj.

Question (vi)
Where was he working then?
Answer:
He was then working at Baripada.

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

“My father is a doctor.
My sister’s a doctor too.
My cousin works with animals.
He’s a keeper at the zoo.
What can I be?
What do I want to do?
I don’t want to be a farmer,
A builder or a nurse.
I don’t want to be a pilot, that is even worse.

Question (i)
What is the name of the poem?
Answer:
The name of the poem is “What can I be ?”

Question (ii)
Who is ‘I’ in the poem?
Answer:
The poet/child is T in the poem.

Question (iii)
What are the child’s father and sister?
Answer:
Both the child’s father and sister are doctors.

Question (iv)
Who is a keeper at the zoo?
Answer:
His cousin is a keeper at the zoo?

Question (v)
What doesn’t he want to be?
Answer:
He doesn’t want to be a farmer, a builder, a nurse, and a pilot.

Question (vi)
Whose job is worse?
Answer:
A pilot’s job is even worse.

10. Read the following paragraph and answer the questions in complete ententes.
“Raghunath Murmu was also a great writer. He had written many plays, novels, and poems in Santali. His most important play is ‘Kherwar Bir”. Martin Orans, a foreign scholar and writer, called this the Santali Mahabharata. Raghunath was awarded by the Odisha Sahitya Academy for his contribution to the Santali language and literature. The Government of Odisha has named the Medical College at Baripada after his name. What Fakir Mohan Senapati is to Odia language and literature, Raghunath Murmu is to Santali language and literature.

Question (i)
What else was Raghunath Murmu?
Answer:
Raghunath Murmu was also a great writer.

Question (ii)
What did he write in Santali?
Answer:
He had written many plays, novels and poems in Santali.

Question (iii)
Which book is Raghunath’s most important play?
Answer:
Raghunath’s most important play is “Kherwar Bir”.

Question (iv)
Who was Martin Orans?
Answer:
Martin Orans was a foreign scholar and writer.

Question (v)
Was he in high praise of Raghunath’s writings?
Answer:
Surely, he was in high praise of Raghunath’s writings.

Question (vi)
Which book is called (he Santal Mahabharat)?
Answer:
The book “Kherwar Bir” is called the Santal Mahabharata.

Question (vii)
What did the Odisha Sahitya Academy award him for?
Answer:
Raghunath Murmu was awarded by the Odisha Sahitya Academy for his contribution to Santali language and literature.

Question (viii)
What has the Government of Odisha done in his honor?
Answer:
In his honor, the Government of Odisha has named the Medical College at Baripada after his name.

Question (ix)
Who is Raghunath Murmu compared to?
Answer:
Raghunath Murmu is compared to Fakir Mohan Senapati.

Question (x)
How are they equal?
Answer:
What Fakir Mohan Senapati is to Odia language and literature, Raghunath Murmu is to Santali language and literature.

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 7 What Can I Be? Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 7 What Can I Be?

BSE Odisha 6th Class English Follow-Up Lesson 7 What Can I Be? Text Book Questions and Answers

Session – 1

I – Pre-Reading
□ Pre-reading questions

1. What is your father’s job? What is your mother’s job?
(ତୁମ ବାପାଙ୍କର ବୃତ୍ତି କ’ଣ ? ତୁମ ମାଆଙ୍କର ବୃତ୍ତି କ’ଣ ?)
2. What would you like to be in the future? What do you see in the picture?
(ତୁମେ ଭବିଷ୍ୟତରେ କ’ଣ ହେବାକୁ ପସନ୍ଦ କର ? ତୁମେ ଛବିରେ କ’ଣ ଦେଖୁଛ ?)

II. While-Reading

Text

Read the poem silently and answer the questions that follow.

My father is a doctor.
My sister’s a doctor too.
My cousin works with animals.
He’s- a keeper at the zoo. 4
What can I be?
What do I want to do?
I don’t want to be a farmer,
A builder or a nurse.
I don’t want to be a pilot,
that is even worse. 10
I don’t want to be a painter
But a keeper at the zoo.
That’s what I’ll be.
That’s what I want to do. 14

କବିତାର ଓଡ଼ିଆ ଉଚ୍ଚାରଣ :
ମାଇ ଫାଦର୍‌, ଇଜ୍ ଏ ଡକ୍ଟର ।
ମାଇ ସିଷ୍ଟର୍’ଜ୍ ଏ ଡକ୍ଟର ଠୁ।
ମାଇ କଜିନ୍ ୱାକ୍‌ସ୍‌ ଉଇଥ୍ ଆନିମାଲ୍‌ ।
ହି’ଜ୍ ଏ କିପର୍ ଆଟ୍ ଦ’ ଜୁ ।
ହ୍ମାଟ୍ କ୍ୟାନ୍ ଆଇ ବି ?
ଦ୍ଵାଟ୍ ଡୁ ଆଇ ୱାଣ୍ଟ୍ ଟୁ ଡୁ ?
ଆଇ ଡୋ’ଣ୍ଟ ୱାଣ୍ଟ ଟୁ ବି ଏ ଫାର୍ମର,
ଏ ବିଲ୍‌ଡ଼ର୍ ଅର୍ ଏ ନର୍ସ ।
ଆଇ ଡୋ’ଣ୍ଟ୍‌ ୱାଣ୍ଟ୍ ଟୁ ବି ଏ ପାଇଲଟ୍
ଦ୍ୟାଟ୍ ଇଜ୍ ଇଭେନ୍ ଓର୍ସ୍ ।
ଆଇ ଡୋ’ଣ୍ଟ୍ ୱାଣ୍ଟ୍ ଟୁ ବି ଏ ପେଣ୍ଟର୍,
ବଟ୍ ଏ କିପର୍ ଆଟ୍ ଦ’ ଜୁ ।
ବ୍ୟାଟ୍’ଜ୍ ହ୍ୱାଟ୍ ଆଇ’ଲ ବି ।
ବ୍ୟାଟ୍’ଜ୍ ହ୍ୱାଟ୍ ଆଇ ୱାଣ୍ଟ୍ ଟୁ ଡୁ ।

କବିତାର ସାରକଥା :
ମୋ ବାପା ଜଣେ ଡାକ୍ତର ।
ମୋ ଭଉଣୀ ମଧ୍ୟ ଜଣେ ଡାକ୍ତର ।
ମୋ ସମ୍ପର୍କୀୟ ଭାଇ ପଶୁମାନଙ୍କର ସହ କାର୍ଯ୍ୟ କରେ ।
ସେ ଜଣେ ଚିଡ଼ିଆଖାନାର କର୍ମଚାରୀ ଅଟେ ।
ମୁଁ କ’ଣ ହୋଇପାରିବି ?
କ’ଣ କରିବାକୁ ଚାହେଁ ?
ଚାହେଁନା ଜଣେ କୃଷକ ହେବାକୁ,
ଜଣେ ନିର୍ମାଣକାରୀ କିମ୍ବା ଧାଈ ।
ମୁଁ ଚାହେଁନା ଜଣେ ଉଡ଼ାଜାହାଜ ଚାଳକ ହେବାକୁ,
ତାହା ମଧ୍ୟ ଏପରିକି ବହୁତ ଖରାପ ।
ମୁଁ ଜଣେ ଚିତ୍ରକର ହେବାକୁ ଚାହେଁନା,
କିନ୍ତୁ ଜଣେ ଚିଡ଼ିଆଖାନାର କର୍ମଚାରୀ ।
ତାହା ଅଟେ ଯାହା ମୁଁ ହେବି ।
ତାହା ଅଟେ ଯାହା ମୁଁ କରିବାକୁ ଚାହେଁ ।

• Your teacher reads the poem aloud. You listen to him/her without opening the book. Your teacher asks you: Who are there in this poem?
  (ତୁମ ଶିକ୍ଷକ କବିତାଟିକୁ ବଡ଼ପାଟିରେ ପଢ଼ିବେ । ବହି ନ ଖୋଲି, ତୁମେ ତାଙ୍କୁ (ପୁ/ସ୍ତ୍ରୀ) ମନଦେଇ ଶୁଣ । ତୁମ ଶିକ୍ଷକ ତୁମକୁ ପଚାରିବେ : ‘‘ଏହି କବିତାରେ କେଉଁମାନେ ଅଛନ୍ତି ?’’)
• Your teacher reads the poem aloud a second time. You listen to him/her and at the same time see the poem.
  (ତୁମ ଶିକ୍ଷକ କବିତାଟିକୁ ବଡ଼ପାଟିରେ ଦ୍ଵିତୀୟଥର ପଢ଼ିବେ । ତୁମେ ତାଙ୍କୁ (ପୁ/ସ୍ତ୍ରୀ) ମନଦେଇ ଶୁଣିବ ଏବଂ ସେହି ସମୟରେ କବିତାଟିକୁ ଦେଖୁବ ।)

Comprehension Questions

Question 1.
How many stanzas are there in the poem?
(କବିତାରେ କେତୋଟି ପଦ ଅଛି ?)
Answer:
There are three stanzas in the poem.

How many lines are there in each stanza?
(ପ୍ରତ୍ୟେକ ପଦରେ କେତୋଟି ଧାଡ଼ି ରହିଛି ?)
Answer:
There are four lines each in 1st and last stanzas. But there are six lines in the second stanza.

Question 2.
Who is ‘F in the poem?
(କବିତାରେ ‘I? କିଏ ?)
Answer:
In the poem, T is the poet or the child.

Question 3.
What is the poem about?
(କବିତାଟି କେଉଁ ବିଷୟରେ ଲେଖାଯାଇଛି ?)
Answer:
The poem is about what the child/poet can be!

Question 4.
How many questions are there in the poem?
(କବିତାରେ କେତୋଟି ପ୍ରଶ୍ନ ରହିଛି ?)
Answer:
There are two questions in the poem.

What are they?
(ସେଗୁଡ଼ିକ କ’ଣ ?)
Answer:
They are – “What can I do? What do I want to do ?”

Question 5.
What are the child’s father and sister?
(ପିଲାଟିର ବାପା ଓ ଭଉଣୀ କ’ଣ ଅଟନ୍ତି ? )
Answer:
The child’s father and sister both are doctors.

Question 6.
Who is a zoo-keeper ?
(କିଏ ଚିଡ଼ିଆଖାନାର କର୍ମଚାରୀ ଅଟେ ?)
Answer:
His cousin is a zoo keeper.

Question 7.
Whose job is worse?
(କାହାର କାମ ଅଧିକ ଖରାପ ?)
Answer:
A pilot’s job is worse.

Question 8.
Which stanza tells you that the poet wants to be a keeper at the zoo?
(କେଉଁ ପଦଟି ତୁମକୁ କହେ ଯେ କବି ଚିଡ଼ିଆଖାନାର ଜଣେ କର୍ମଚାରୀ ହେବାକୁ ଚାହାଁନ୍ତି ?)
Answer:
The third stanza tells us that the poet wants to be a keeper at the zoo.

Question 9.
Is the poet happy? Why? Why not?
(କବି ସୁଖୀ କି ? କାହିଁକି ବା କାହିଁକି ନୁହେଁ ?)
Answer:
The poet is really happy. Because he wants to be a zoo keeper.

Question 10.
How many times the following words are repeated ? (Question with Answer)

Answer:
a. my 3 times
b. keeper 2 times
c. doctor 2 times
d. I 7 times

Session – 3

III. Post-Reading

1. Writing

(a) Answer the following questions.
(ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

Question (i)
What does the child / poet want to be ?
(ଶିଶୁ | କବି କ’ଣ ହେବାକୁ ଚାହାନ୍ତି ?)
Answer:
The child / poet wants to be a zoo-keeper.

Question (ii)
What is his father?
(ତାଙ୍କ ବାପା କ’ଣ ଅଟନ୍ତି ?)
Answer:
His father is a doctor.

Question (iii)
What is his sister?
(ତାଙ୍କ ଭଉଣୀ କ’ଣ ଅଟନ୍ତି ?)
Answer:
His sister is a doctor.

Question (iv)
What does the child not want to be?
(ପିଲାଟି କ’ଣ ହେବାକୁ ଚାହୁଁ ନାହିଁ ?)
Answer:
The child does not want to be a farmer, a builder, a nurse, a pilot, and a painter too.

(b) Write your own poem (the last word of the second and the last lines which rhyme are given. Rest you can choose).
(ତୁମ ନିଜର ଗୋଟିଏ କବିତା ଲେଖ । (ଦ୍ଵିତୀୟ ଓ ଶେଷ ଧାଡ଼ିର ଶେଷ ଶବ୍ଦ ଯାହା ଯତିପାତ ପଡ଼ୁଛି ଦିଆଯାଇଛି । ଅବଶିଷ୍ଟ ତୁମେ ପସନ୍ଦ କରିପାରିବ ।)
(Question with Answer)
I don’t want to be a builder or a nurse.
I don’t want to be a farmer.
I don’t want to be a pilot
I don’t want to be a teacher.

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

cousin – ବ୍ୟତୀତ ଅନ୍ୟ ଭାଇ
detective – ଗୁଇନ୍ଦା ପୋଲିସ
keeper at zoo – ଚିଡିଆଖାନା
sailor – ନାବିକ
soldier – ସୈନିକ
spaceship – ମହାକାଶଯାନ
taking turn – ପାଳିକରି କୌଣସି କାମ କରିବା
trumpet – ତୂରୀ, ବିଗୁଲ୍
master key – ମୁଖ୍ୟ ଚାବି
perhaps – ବୋଧହୁଏ, ପ୍ରାୟ
own – ନିଜର
that’s what – ସେଇଟା ଯାହା
musical tune – ସଙ୍ଗୀତ ସ୍ବର
buy – କିଣିବା
private – ବ୍ୟକ୍ତିଗତ
moon – ଚନ୍ଦ୍ର
I’d like (I would like) – ମୁଁ ପସନ୍ଦ କରେ ।
express – ପ୍ରକାଶ କରିବା
light-house – ଆଲୋକସ୍ତମ୍ଭ
builder – ନିର୍ମାଣକାରୀ
nurse – ସେବିକା
pilot – ବିମାନ ଚାଳକ
even – ଏପରିକି
Worse – ଅଧ୍ଵ ଖରାପ
painter – ଚିତ୍ରକାର

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(b)

Question 1.
Differentiate from definition
(i) e^3x
Solution:

(ii) 2^x2
Solution:

(iii) In (3x + 1)
Solution:

(iv) logx⁵ (Hint : logx⁵ = \(\frac{\ln 5}{\ln x}\))
Solution:

(v) In sin x
Solution:

(vi) x² a^2x
Solution:

Odisha State Board BSE Odisha 7th Class English Solutions Follow-Up Lesson 4 Tilka Majhi Textbook Exercise Questions and Answers.

BSE Odisha Class 7 English Solutions Follow-Up Lesson 4 Tilka Majhi

BSE Odisha 7th Class English Follow-Up Lesson 4 Tilka Majhi Text Book Questions and Answers

Session – 1

I. Pre-Reading

• Socialization:
• Do you know whose picture is this? Read the small paragraph in brackets to know about the picture.
  (ଏହା କାହାର ଛବି ତୁମେ ଜାଣିଛ କି ? ଛବିଟି ବିଷୟରେ ଜାଣିବାକୁ ବନ୍ଧନୀ ମଧ୍ୟସ୍ଥ ଛୋଟ ଅନୁଚ୍ଛେଦଟିକୁ ପଢ଼ ।)

[About 200 years ago. when hardly anyone thought of fighting against the British in India. Tilka Majhi, a Santal of Bhagalpur in Bihar fought against them and was hanged by the British. Our History books do not mention his name. We have forgotten this great man who sacrificed his life for the country and showed us the path to freedom. Here you will read about this great man.)
(ପ୍ରାୟ ୨୦୦ ବର୍ଷ ପୂର୍ବେ ଯେତେବେଳେ ଜଣେ ଇଂରେଜ ବିରୋଧରେ ଯୁଦ୍ଧ କରିବା କଥା ଚିନ୍ତା କରିପାରନଥିଲା, ସେ ସମୟରେ ତିଲକ ମାଝି ଯିଏକି ବିହାରର ଭାଗଲପୁରବାସୀ ଜଣେ ସାନ୍ତାଳ ତାଙ୍କ ବିରୋଧରେ ଯୁଦ୍ଧ କରି ଇଂରେଜମାନଙ୍କଦ୍ଵାରା ଫାଶୀ ପାଇଥିଲେ । ଆମ ଇତିହାସ ବହି ତାଙ୍କ କଥା ଜଣାଏ ନାହିଁ । ଏପରି ଜଣେ ମହାନ୍ ବ୍ୟକ୍ତି ଯିଏକି ନିଜ ଜୀବନକୁ ଉତ୍ସର୍ଗ କରି ଆମକୁ ସ୍ଵାଧୀନତାର ବାଟ ଦେଖାଇଥିଲେ, ଆମେ ତାଙ୍କ ଭଳି ଯାଇଛେ । ଏଠାରେ ତୁମେ ସେହି ମହାପୁରୁଷଙ୍କ ସମ୍ବନ୍ଧରେ ପଢ଼ିବ ।)

II. While-Reading

Text

• SGP-1 (Sense Group Paragraph-1)
• Read paragraphs 1-3 and answer the questions that follow.
  (ଅନୁଚ୍ଛେଦ ୧–୩କୁ ନୀରବରେ ପାଠ କର ଏବଂ ନିମ୍ନ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

1. In 1750, Tilka Majhi was born in the Murmu family in a small village near Bhagalpur in Bihar. The village is now called Tilakpur.
2. The Murmus are priests among the Santals. So Tilka used to worship ‘Marang Bum’ from his early days. As the days went by, he grew up to be a religious man. People of all religions had great love and respect for him and they called him ‘Tilka Baba’.
3. Tilka’ was also very good at using bows and arrows. There was none in his area who could surpass him at shooting arrows from a bow. He used to train the young men of the village in shooting arrows.

ଓଡ଼ିଆ ଅନୁବାଦ :
୧. ୧୭୫୦ ମସିହାରେ ତିଲକ ମାଝି ବିହାରର ଭାଗଲପୁର ନିକଟରେ ଏକ ଛୋଟ ଗ୍ରାମରେ ମୁର୍ମୁ ପରିବାରରେ ଜନ୍ମଗ୍ରହଣ କରିଥିଲେ । ସେ ଗ୍ରାମଟିକୁ ଏବେ ତିଲକପୁର କୁହାଯାଇଛି ।
୨. ମୁମୁମାନେ ସାନ୍ତାଳମାନଙ୍କର ପୁରୋହିତ ଥିଲେ । ତେଣୁ ତିଲ୍‌ ତାଙ୍କ ପିଲାଟି ଦିନରୁ ‘ମାରଙ୍ଗ ବୁରୁ’ଙ୍କୁ ପୂଜା କରି ଆସୁଥିଲେ । ଦିନ ବିତି ଚାଲିଲା ଏବଂ ସେ ଜଣେ ଧାର୍ମିକ ବ୍ୟକ୍ତିରେ ପରିଣତ ହେଲେ । ସମସ୍ତ ଧର୍ମର ଲୋକମାନେ ତାଙ୍କୁ ବହୁତ ଶ୍ରଦ୍ଧା ଓ ଭକ୍ତି କରୁଥିଲେ ଏବଂ ସ୍ନେହରେ ତାଙ୍କୁ ତିଲ୍‌ ବାବା ଡାକୁଥିଲେ ।
୩. ତିଲ୍‌କ ମଧ୍ୟ ଧନୁଶର ବିଦ୍ୟାରେ ନିପୁଣ ଥିଲା । ଧନୁଶର ବିଦ୍ୟାରେ ତାଙ୍କ ଅଞ୍ଚଳ କେହି ଜଣେ ତାଙ୍କୁ ଅତିକ୍ରମ କରିପାରୁ ନଥିଲା । ସେ ଶର ବିନ୍ଧିବା ଶିକ୍ଷା ଗ୍ରାମର ଯୁବକମାନଙ୍କୁ ଦେଉଥିଲେ ।

Notes And Glossary
priest (ପ୍ରିଷ୍ଟ୍) – ପାଳକ
religion (ରିଲିଜିଅନ୍) – ଧର୍ମ
shooting (ସୁଟିଙ୍ଗ୍) – ଶର କ୍ଷେପଣ କରିବା
worship (ଓର୍‌ସିପ୍) – ପୂଜା କରିବା
good at (ଗୁଡ୍ ଆଟ୍) – ଦଷ
surpass (ସର୍‌ପାସ ) – ଅତିକ୍ରମ କରିବା

Comprehension Questions

Question 1.
Where was Tilka Majhi born ? When?
( ତିଲକ ମାଝି କେଉଁଠାରେ ଜନ୍ମଗ୍ରହଣ କରିଥଲେ ? କେବେ ?)
Answer:
Tilka Majhi was bom in Murmu family in 1750. It was a small village near Bhagalpur in Bihar.

Question 2.
What are the Murmus?
( ମୁହଁମାନେ କ’ଣ ଥିଲେ ?)
Answer:
The Murmus are priests among the Santals.

Question 3.
Who did Tilka worship?
(ତିଲ୍‌ କାହାକୁ ଆରାଧନା | ପୂଜା କରୁଥିଲେ ? )
Answer:
Tilka worshiped “Marang Bum”.

Question 4.
How did people look up to him?
(ଲୋକମାନେ ତାକୁ କିଭଳି ଦୃଷ୍ଟିରେ ଦେଖୁଥିଲେ ?)
Answer:
People of all religions loved and respected him as a priest.

Question 5.
What did they call him?
(ସେମାନେ ତାଙ୍କୁ କ’ଣ ବୋଲି ଡାକୁଥିଲେ ?)
Answer:
Out of love and respect they called him ‘Tilka Baba’.

Question 6.
What was Tilka good at?
( ତିଲକା କେଉଁଥରେ ପାରଙ୍ଗମ । ନିପୁଣ ଥିଲେ ?)
Answer:
Tilka was very good at using bows and arrows. He was a bowman uncontested.

Question 7.
What did he teach to the young men of the village?
( ଗ୍ରାମର ଯୁବକମାନଙ୍କୁ ସେ କି ଶିକ୍ଷା ଦେଉଥିଲେ ?)
Answer:
He taught the young men of the village in shooting arrows.

Session – 2

• SGP-2 (Sense Group Paragraph-2)
• Read paragraphs 4-5 silently and answer the questions that follow.
  ( ଅନୁଚ୍ଛେଦ ୪-୫କୁ ନୀରବରେ ପାଠ କର ଏବଂ ନିମ୍ନପ୍ରଦତ୍ତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

4. The Santals, as you know, are freedom-loving people. In those days they lived a free life without any outside control. They cleared the forests, prepared fields, and grew different kinds of crops. The king had little control over them. Once or twice a year, they had to help the king when there was a war. But after the Plassey battle, the British became the masters of Bengal, Bihar, and Odisha. The king of Bhagalpur came under the control of the British. The British wanted to collect more money. They asked the king of Bhagalpuror for more money. The king in turn appointed outsiders to collect rent from the tribals. Some tribal villages were given ‘pattas’ by some landlords. The landlords asked the Santals to pay more taxes and often took away their property and lands.
5. Tilka did not like this. He asked the Santals not to pay taxes. The Santals became a real problem for the Government. To bring the Santals under control, a new collector was appointed in Bhagalpur. His name was Cleaveland. In different ways, he tried to bring the Santals under control. He appointed soldiers from other tribes to fight against the Santals. Tilka could not bear this. He was looking for a chance to kill Cleaveland.

ଓଡ଼ିଆ ଅନୁବାଦ :
୪. ତୁମେ ଜାଣିଛ, ସାନ୍ତାଳମାନେ ମୁକ୍ତ ଜୀବନ ପ୍ରଣାଳୀକୁ ଭଲ ପାଇବାର ଲୋକ ଥିଲେ । ସେ ସମୟରେ ସେମାନେ ବାହ୍ୟ କଟକଣାର ମୁକ୍ତ ହୋଇ ମୁକ୍ତ ଜୀବନଯାପନ କରୁଥିଲେ । ସେମାନେ ଜଙ୍ଗଲ ସଫାକରି କ୍ଷେତ ପ୍ରସ୍ତୁତ କରୁଥିଲେ ଏବଂ ସେଥିରେ ବିଭିନ୍ନ ଶସ୍ୟ ଉତ୍ପାଦନ କରୁଥିଲେ । ରାଜାଙ୍କର ସେମାନଙ୍କ ଉପରେ ମୋଟେ ନିୟନ୍ତ୍ରଣ ନଥିଲା । ବର୍ଷକୁ ଥରେ ବା ଦୁଇଥର ଯୁଦ୍ଧ ସମୟରେ ସେମାନଙ୍କୁ ରାଜାଙ୍କୁ ସାହାଯ୍ୟ କରିବାକୁ ପଡୁଥିଲା । କିନ୍ତୁ ପଲାସୀ ଯୁଦ୍ଧ ପରେ, ଇଂରେଜମାନେ ବଙ୍ଗ, ବିହାର ଏବଂ ଓଡ଼ିଶାର ଶାସନକର୍ତ୍ତା ପାଲଟିଗଲେ । ଭାଗଲପୁରର ରାଜା ଇଂରେଜମାନଙ୍କ ଅଧୀନକୁ ଆସିଲେ । ଇଂରେଜମାନେ ଅଧିକ ଖଜଣା (ଟଙ୍କା) ସଂଗ୍ରହ କରିବାକୁ ଇଚ୍ଛା କଲେ । ସେମାନେ ଭାଗଲପୁରର ରାଜାଙ୍କୁ ଅଧିକ ଟଙ୍କା|ଖଜଣା ଦେବାକୁ କହିଲେ । ଫଳରେ ରାଜା ଆଦିବାସୀମାନଙ୍କଠାରୁ କର ଆଦାୟ ନିମିତ୍ତ ରାଜ୍ୟ ବାହାରର ଲୋକଙ୍କୁ ନିଯୁକ୍ତ କଲେ । କେତେକ ଆଦିବାସୀ ଗ୍ରାମକୁ ଜମିଦାରମାନଙ୍କଦ୍ୱାରା ପଟ୍ଟା ରୂପେ ପ୍ରଦାନ କରାଯାଇଥିଲା । ଜମିଦାରମାନେ ସାନ୍ତାଳମାନଙ୍କୁ ଅଧିକ କର ଦେବାକୁ ବାଧ୍ୟ କଲେ ଏବଂ କେତେକସ୍ଥଳେ ସେମାନଙ୍କଠାରୁ ସମ୍ପତ୍ତି ଓ ଜମିକ ଛଡ଼ାଇ ନେଇଯାଉଥିଲେ ।
୫. ତିଲ୍‌କଙ୍କୁ ଏହା ଭଲ ଲାଗିଲା ନାହିଁ । ସେ ସାନ୍ତାଳମାନଙ୍କୁ କର ନ ଦେବାକୁ କହିଲେ । ସାନ୍ତାଳମାନେ ସରକାରଙ୍କ ପାଇଁ ଏକ ବାସ୍ତବ ସମସ୍ୟା ରୂପେ ଉଭାହେଲେ । ସାନ୍ତାଳମାନଙ୍କୁ ନିୟନ୍ତ୍ରଣକୁ ଆଣିବା ନିମିତ୍ତ ଭାଗଲପୁରରେ ଏକ ନୂତନ କଲେକ୍ଟରଙ୍କୁ ନିଯୁକ୍ତ କରାଗଲା । ତାଙ୍କର ନାମ କ୍ଳେରେଲାଣ୍ଡ । ବିଭିନ୍ନ ଉପାୟରେ ସେ ସାନ୍ତାଳମାନଙ୍କୁ ନିୟନ୍ତ୍ରଣ କରିବାକୁ ଚେଷ୍ଟା କଲେ । ସେ ସାନ୍ତାଳମାନଙ୍କ ସହିତ ଯୁଦ୍ଧ କରିବାକୁ ଅନ୍ୟ ସମ୍ପ୍ରଦାୟରୁ ସୈନ୍ୟ ନିଯୁକ୍ତ କଲେ । ତିଲ୍‌ ଏହା ସହ୍ୟ କରିପାରିଲେ ନାହିଁ । ସେ କ୍ଳେରେଲାଣ୍ଡ୍ଙ୍କୁ ହତ୍ୟା କରିବାପାଇଁ ଗୋଟେ ସୁଯୋଗ ଅପେକ୍ଷାରେ ରହିଲେ ।

Notes And Glossary
freedom-loving (ଫ୍ରିଡ଼ମ୍ ଲଭିଙ୍ଗା) – ସ୍ବାଧୀନତା -ପ୍ରେମୀ
prepared (ପ୍ରିପେୟାର୍ଡ୍) – ପ୍ରସ୍ତୁତ କଲେ
crops (କ୍ରପ୍‌ସ) – ଶସ୍ୟ
bear (ବିଅର୍ ) – ସହ୍ୟ କରିବା
appointed (ଆପଏଣ୍ଟେଡ୍) – ନିଯୁକ୍ତ କଲେ
property (ପ୍ରପଟି) – ସମ୍ପରି
problem (ପ୍ରୋବ୍ଲେମ୍ ) – ସମସ୍ୟା
look for (ଲୁକ୍ ଫର) – ଖୋଜିବା

Comprehension Questions

Question 1.
The Santals are born free and hardly obey any control. Which sentences have this idea?
(ସାନ୍ତାଳମାନେ ମୁକ୍ତ ଭାବରେ ଜନ୍ମ ହୁଅନ୍ତି ଏବଂ କୌଣସି ପ୍ରକାର ନିୟନ୍ତ୍ରଣକୁ ମାନନ୍ତି ନାହିଁ । କେଉଁ ବାକ୍ୟରେ ଏହି ଧାରଣା ନିହିତ ଅଛି ?)
Answer:
“Santals are freedom-loving people. The king had little control over them.” These two sentences have the idea that they are born free and hardly obey any control.

Question 2.
What did they do to get their bread?
(ସେମାନେ ତାଙ୍କର ଜୀବିକା (ରୋଟି) ନିର୍ବାହ ପାଇଁ କ’ଣ କରୁଥିଲେ ? )
Answer:
They cleared the forest and prepared fields and grew different kinds of crops. In this way, they earned their bread.

Question 3.
How did they help the king?
(ରାଜାଙ୍କୁ ସେମାନେ କିପରି ସାହାଯ୍ୟ କରୁଥିଲେ ? )
Answer:
Once or twice a year, if there was war, they helped the king in war.

Question 4.
Who became the ruler of their land after the Plassey battle?
(ପଲାସୀ ଯୁଦ୍ଧ ପରେ ସେମାନଙ୍କ ଇଲାକାର କିଏ ଶାସକ ହେଲେ ?)
Answer:
After Plassey battle, the British became the mler of their land.

Question 5.
What did the British want?
(ଇଂରେଜମାନେ କ’ଣ ଚାହୁଁଥିଲେ ?)
Answer:
The British wanted to collect more money.

Question 6.
What did the king do to collect rent from the tribals?
(ଆଦିବାସୀମାନଙ୍କଠାରୁ କର ବା ଖଜଣା ଆଦାୟ ନିମିତ୍ତ ରାଜା କ’ଣ କଲେ ?)
Answer:
The king appointed outsiders to collect rent from the tribals. He also gave some tribal villages as ‘pattas’ to some landlords.

Question 7.
What was given as ‘Pattas’ to landlords?
(ଜମିଦାରମାନଙ୍କ ‘ପଟ୍ଟା’ରୂପେ କ’ଣ ଦିଆଯାଇଥିଲା ?)
Answer:
Some tribal villages were given as ‘pattas’ to landlords.

Question 8.
How did the landlords trouble the tribals?
(ଜମିଦାରମାନେ ଆଦିବାସୀମାନଙ୍କୁ କିଭଳି ହଇରାଣ କଲେ ?)
Answer:
The landlords demanded more taxes from the Santals and often took away their property and lands if they could not pay the taxes.

Question 9.
What did Tilka ask people not to do?
( ତିଲକ ଆଦିବାସୀମାନଙ୍କୁ କ’ଣ ନ କରିବାକୁ କହିଲେ ? )
Answer:
Tilka asked people not to pay taxes.

Question 10.
What did the government do to bring the Santals under control?
(ସରକାର ସାନ୍ତାଳମାନଙ୍କୁ ନିୟନ୍ତ୍ରଣକୁ ଆଣିବାପାଇଁ କ’ଣ କଲେ ?)
Answer:
The Government appointed a new collector in Bhagalpur to bring the -Santals under control.

Question 11.
Who was Cleveland?
(କ୍ଳେରେଲାଣ୍ଡ୍ କିଏ ? )
Answer:
Cleveland was the new collector appointed in Bhagalpur.

Question 12.
What did he do to bring the Santals under control?
(ସେ ସାନ୍ତାଳମାନଙ୍କୁ ନିୟନ୍ତ୍ରଣକୁ ଆଣିବାପାଇଁ କ’ଣ କଲେ ?)
Answer:
To bring the Santals under control he appointed soldiers from another tribe to fight against them.

Question 13.
Did Tilka like it? What was his plan?
(ତିଲକ ଏହାକୁ ପସନ୍ଦ କରୁଥିଲେ କି ? ତାଙ୍କର ଯୋଜନା କ’ଣ ଥିଲା ?)
Answer:
Tilka did not like it. He was looking for a chance to kill Cleaveland.

Will he come out successful in his plan ?(ସେ କ’ଣ ତାଙ୍କ ଯୋଜନାରେ ସଫଳ ହେବେ ?)
Read the next part of the lesson silently to know it.
(ଏହା ଜାଣିବାକୁ ପାଠ୍ୟ ବିଷୟର ପରବର୍ତ୍ତୀ ଅଂଶ ନୀରବରେ ପାଠ କର ।)

Session – 3

• SGP-3 (Sense Group Paragraph-3)
• Read paragraph 6-7 silently and answer the questions that follow.
(ଅନୁଚ୍ଛେଦ ୬-୭କୁ ନୀରବରେ ପାଠ କର ଏବଂ ନିମ୍ନପ୍ରଦତ୍ତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

6. The chance came. On the morning of 23rd January 1784, Cleaveland was riding to his house on horseback. Tilka shot an arrow from a distance and killed him on the spot. The news of Cleareland’s death spread everywhere. The government brought more soldiers and weapons. But Tilka and his people fought for about a year killing many British soldiers with their arrows. At last, they made a sudden attack on Tilka and his followers. Tilka was arrested. He was mercilessly beaten. His hands were then tied to a horse and he was dragged all over Bhagalpur town. But Tilka did not die after all this. So they hanged him from a tree. The place is known today as Tilka Chhak.

7. History has forgotten Tilka Majhi: his name is not found in the history books. But many people who came to Bhagalpur pay their respects to Tilka. In 1969. the Simanta Gandhi. Khan Abdul Ghaffar Khan came to Bhagalpur. He went to Tilka Chhak where Tilka was hanged and paid his respect to this great martyr.

୬. ସୁଯୋଗ ଆସିଲା । ୧୭୮୪ ମସିହା ଜାନୁୟାରୀ ୨୩ ତାରିଖ ସକାଳେ କେରେଲାଣ୍ଡ ଘୋଡ଼ା ସବାରରେ ତାଙ୍କ ବସାକୁ ଫେରୁଥିଲେ । ତିଲ୍‌ କିଛି ଦୂରରୁ ଗୋଟିଏ ତୀର ଲକ୍ଷ୍ୟ କରି ମାରିଲେ ଏବଂ ଘଟଣାସ୍ଥଳରେ ତାଙ୍କର ମୃତ୍ୟୁ ହେଲା । କ୍ଳେରେଲାଣ୍ଡଙ୍କ ମୃତ୍ୟୁ ଖବର ସବୁଆଡ଼େ ବ୍ୟାପିଗଲା । ସରକାର ଅଧ‌ିକ ସୈନ୍ୟ ଏବଂ ଅସ୍ତ୍ରଶସ୍ତ୍ର ମଗାଇଲେ । କିନ୍ତୁ ତିଲ୍‌ ଏବଂ ତାଙ୍କ ଲୋକମାନେ ଏକବର୍ଷ ବ୍ୟାପୀ ଯୁଦ୍ଧକରି ଧନୁଶର ସାହାଯ୍ୟରେ ଅନେକ ଇଂରେଜ ସୈନ୍ୟଙ୍କୁ ନିପାତ କଲେ । ପରିଶେଷରେ ସେମାନେ ତିଲ୍‌ ଏବଂ ତାଙ୍କ ଲୋକମାନଙ୍କ ଉପରେ ହଠାତ୍ ଆକ୍ରମଣ କଲେ । ତିଲ୍‌କଙ୍କୁ ବନ୍ଦୀ କରାଗଲା । ତାଙ୍କୁ ନିଷ୍ଠୁର ଭାବରେ ପ୍ରହାର କରାଗଲା । ତାଙ୍କ ହାତକୁ ଗୋଟେ ଘୋଡ଼ା ସହିତ ବାନ୍ଧି ଦିଆଗଲା ଏବଂ ଭାଗଲପୁର ଟାଉନ୍ | ସହର ସାରା ଘୋଷାଡ଼ି ଘୋଷାଡ଼ି ନିଆଗଲା । କିନ୍ତୁ ଏ ସବୁଥରେ ତିଲ୍‌କଙ୍କର ମୃତ୍ୟୁ ହୋଇନଥିଲା । ତେଣୁ ସେମାନେ ତାଙ୍କୁ ଗୋଟେ ଗଛରେ ଫାଶୀ ଲଗାଇ ଝୁଲାଇଦେଲେ । ସେ ସ୍ଥାନଟିର ନାମ ଏବେ ହୋଇଛି ତିଲ୍‌ ଛକ ।
୭. ଇତିହାସ ତିଲ୍‌ ମାଝିଙ୍କୁ ଭୁଲି ଯାଇଛି; ତାଙ୍କର ନାମ ଇତିହାସ ବହିରେ ଦେଖ‌ିବାକୁ ମିଳେ ନାହିଁ । କିନ୍ତୁ ଅନେକ ଲୋକ ଯେଉଁମାନେ ଭାଗଲପୁର ଆସନ୍ତି, ତିଲକଙ୍କୁ ସମ୍ମାନ ଜଣାନ୍ତି । ୧୯୬୯ ମସିହାରେ ସୀମାନ୍ତ ଗାନ୍ଧି, ଖାଁ ଅବଦୁଲ ଗଫର ଖାଁ ଭାଗଲପୁର ଆସିଥିଲେ । ସେ ତିଲ୍‌କା ଛକକୁ ଯାଇଥିଲେ ଯେଉଁଠାରେ ତାଙ୍କୁ ( ତିଲକଙ୍କ) ଫାଶୀ ଦିଆଯାଇଥିଲା ଏବଂ ସେ ମହାନ୍ ସହିଦ୍‌ଙ୍କୁ ତାଙ୍କର ସମ୍ମାନ ଜଣାଇଥିଲେ ।

Notes And Glossary
spread ( ଟ୍ରେଡ୍ ) – ପ୍ରସାର ହେବା
weopons (ଉଇପନ୍ସ୍ ) – ଅସ୍ତ୍ରଶସ୍ତ୍ର
attack (ଆଟାକ୍) – ଆକ୍ରମଣ
mercilessly (ମର୍‌ସିଲେସ୍‌) – ନିଶ୍ଚୟ ଭାବରେ
dragged (ଡ୍ରାଗ) – ଘୋଷାଡ଼ି ଘୋଷାଡ଼ି ନେଲେ
hang (ହ୍ୟାଙ୍ଗ୍) – ଫାଶୀ
forgot (ଫର୍‌ଗଟ୍) – ଭୁଲିଗଲା
martyr (ମାରଟାୟାର) – ସହିଦ

Comprehension Questions

Question 1.
How was Cleaveland killed?
(କ୍ଳେରେଲାଣ୍ଡଙ୍କୁ କିଭଳି ହତ୍ୟା କରାଗଲା ? )
Answer:
Cleveland was riding to his home on horseback. Tilka shot an arrow from a distance. The arrow hit him and he died on the spot.

Question 2.
What did the Government do to control the tribals?
(ଆଦିବାସୀମାନଙ୍କୁ ଅକ୍ତିଆରକୁ ଆଣିବାପାଇଁ ସରକାର କ’ଣ କଲେ ?)
Answer:
The Government brought more soldiers and weapons to control the tribals.

Question 3.
How long did the tribal rebellion continue?
(ଆଦିବାସୀ ଆନ୍ଦୋଳନ କେତେ ସମୟ ଧରି ଚାଲୁ ରହିଲା ?)
Answer:
The tribal rebellion continued for about a year.

Question 4.
What happened to it at last?
(ଏହାର ଶେଷ ପରିଣତି କ’ଣ ହେଲା ? )
Answer:
At last, the British soldiers made a sudden attack on Tilka and his followers. They captured Tilka.

Question 5.
How did the great son of soil breathe his last?
(ଏ ମାଟିର ମହାନ୍ ସୁପୁତ୍ର କିଭଳି ଶେଷ ନିଶ୍ଵାସ ତ୍ୟାଗକଲେ ?)
Answer:
The Britishers arrested him and beat him mercilessly. They tied his hands to a horse and he was dragged all over Bhagalpur town. At last, he was hanged to death.

Question 6.
Where is Tilka Chhak ? Who was it named after?
(ତିଲ୍‌ ଛକ କେଉଁଠି ? କାହା ନାମରେ ଏହା ନାମିତ କରାଯାଇଥିଲା ?)
Answer:
Tilka Chhak is in Bhagalpur. It was named after Tilka Majhi.

Question 7.
Who was Khan Abdul Ghaffar Khan?
(ଖାଁ ଅବଦୁଲ ଗଫର ଖାଁ କିଏ ?)
Answer:
Khan Abdul Ghaffar Khan was a great freedom fighter.

Question 8.
What is he called?
(ତାଙ୍କୁ କ’ଣ କୁହାଯାଏ ?)
Answer:
He is called Simanta Gandhi.

Question 9.
When did he visit Tilka Chhak ? Why?
(ସେ କେବେ ତିଲ୍‌ ଛକ ପରିଦର୍ଶନ କରିଥିଲେ ? କାହିଁକି ?)
Answer:
He visited Tilka Chhak in 1969 to offer him high respect.

Question 10.
Can you name some tribal leaders of Odisha who were great freedom fighters?
(ତୁମେ ଓଡ଼ିଶାର କେତେକ ଆଦିବାସୀ ନେତାଙ୍କର ନାମ କହିପାରିବ କି, ଯେଉଁମାନେ ମହାନ୍ ସ୍ଵାଧୀନତା ସଂଗ୍ରାମ ଥିଲେ ?)
Answer:
Laxman Naik and Raghu Dibakar were tribal leaders of Odisha who were great freedom fighters.

Session – 4

III. Post-Reading

The teacher will design some activities following the activities under the Post-reading section of the main lesson. However, some activities have been given.
(a) Match A with their ideas under B. Write paragraph numbers in brackets. (Question with Answer)

Answer:

(b) Write the answers to the following questions:

Question (i)
Where and when was Tilka Majhi born?
(ତିଲ୍‌ ମାଝି କେଉଁଠାରେ ଏବଂ କେବେ ଜନ୍ମଗ୍ରହଣ କରିଥିଲେ ? )
Answer:
Tilka Majhi was born in 1750 in a small village near Bhagalpur in Bihar (now in Jharkhand).

Question (ii)
What did he teach to the young men of the village?
(ଗାଁର ଯୁବକମାନଙ୍କୁ ସେ କ’ଣ ଶିକ୍ଷା ଦେଲେ ?)
Answer:
He taught the young men of the village in shooting arrows.

Question (iii)
Who became the ruler of their land after the Plassey Battle?
(ପଲାସୀ ଯୁଦ୍ଧ ପରେ କିଏ ସେମାନଙ୍କ ଇଲାକାର ଶାସକ ହେଲେ ?)
Answer:
The Britishers became the ruler of their land after the Plassey Battle.

Question (iv)
What did the king do to collect rent from the tribals?
(ଆଦିବାସୀମାନଙ୍କଠାରୁ ଖଜଣା ଆଦାୟ କରିବାପାଇଁ ରାଜା କ’ଣ କଲେ ?)
Answer:
The king appointed outsiders to collect rent from the tribals. He also gave some tribal villages as ‘pattas’ to some landlords.

Question (v)
How did the landlords trouble the tribals?
( ଜମିଦାରମାନେ କିପରି ଆଦିବାସୀମାନଙ୍କୁ ହଇରାଣ କରୁଥିଲେ ?)
Answer:
The landlords demanded more taxes from the Santals and often took away their property and lands if they could not pay the taxes.

Question (vi)
What did Tilka ask people not to do?
(ତିଲ୍‌ କ’ଣ ନ କରିବାପାଇଁ ଲୋକମାନଙ୍କୁ କହିଲେ ?)
Answer:
Tilka asked the people not to pay taxes.

Question (vii)
What did the government do to bring the Santals under control?
(ସାନ୍ତାଳମାନଙ୍କୁ ନିୟନ୍ତ୍ରଣକୁ ଆଣିବାପାଇଁ ସରକାର କ’ଣ କଲେ ?)
Answer:
To bring the Santals under control, the Government appointed Cleaveland, a new collector in Bhagalpur. He appointed soldiers from other tribes to fight against the Santals.

Question (viii)
How long did the tribal rebellion continue?
( ଆଦିବାସୀ ଆନ୍ଦୋଳନ କେତେଦିନଧରି ଚାଲିଲା ?)
Answer:
The tribal rebellion continued for one year.

Question (ix)
How did the great son of soil breathe his last?
(ଭୂମିର ମହାନ୍ ସୁପୁତ୍ର କିପରି ତାଙ୍କର ଶେଷନିଃଶ୍ୱାସ ତ୍ୟାଗ କଲେ ?)
Answer:
The Britishers arrested him and beat him mercilessly. They tied his hands to a horse and he was dragged all over Bhagalpur town. At last, he was hanged to death.

Question (x)
Who was Khan Abdul Ghaffar Khan?
(ଖାଁ ଅବଦୁଲ୍ ଗଫର ଖାଁ କିଏ ?)
Answer:
Khan Abdul Ghaffar Khan was a freedom fighter.

Question (xi)
When did he visit Tilka Chhak ? Why?
(ସେ କେବେ ତିଲ୍‌ ଛକ ପରିଦର୍ଶନ କରିଥିଲେ ? କାହିଁକି ?)
Answer:
He visited Tilka Chhak in 1969. He went to Tilka Chhak to pay his respect to this great martyr.

BSE Odisha 7th Class English Solutions Follow-Up Lesson 4 Tilka Majhi Important Questions and Answers

(A) Choose the right answer from the options.

Question 1.
Tilka Majhi was born in a small village near _________?
(a) Bhagalpur
(b) Koraput
(e) Katanga
(d) Sambalpur
Answer:
(a) Bhagalpur

Question 2.
Tilka Majhi was born in _________?
(a) 1857
(b) 1823
(e) 1750
(d) none of these
Answer:
(c) 1750

Question 3.
Tilka was a __________?
(a) bowman uncontested
(b) hunter
(c) religious man
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

Question 4.
Santals are __________?
(a) fighters
(b) patriots
(c) freedom-loving people
(d) peace-loving people
Answer:
(c) freedom-loving people

Question 5.
To bring control over Santals, the Government appointed in Bhagalpur?
(a) some outsiders
(b) policemen
(c) a new collector
(d) bow-man
Answer:
(c) a new collector

Question 6.
Tilka killed the new collector in ___________?
(a) 1784
(b) 1750
(e) 1856
(d) 1745
Answer:
(a) 1784

(B) Answer the following questions.

Question 1.
Why did the people call him Tilka Baba?
Answer:
Tilka was worshiping Marang Burua when days went by he became a religious man. The people of all religions loved him and respected – him. So they called him Tilka Baba.

Question 2.
How did Bhagalpur come under the control of the British?
Answer:
After the Plassey battle, the British became the rulers of Bengal, Bihar, and Odisha. So Bhagalpur of Bihar came under their control.

Question 3.
How did Cleveland torture the natives?
Answer:
The new collector Cleveland appointed soldiers from other tribes to fight against the Santals. In many ways, he tried to harass the people.

Question 4.
Where is Tilka Chhak ? Why is it famous?
Answer:
Tilka Chhaka is in Bhagalpur. Here the great hero Tilak Majhi was hanged by the Britishers. Anybody who comes to Bhagalpur pays respect to this great martyr. So it is famous.

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(f) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(f)

Find derivatives of the following functions.
Question 1.
x^x
Solution:
Let y = x^x
Then In y = x . In x
⇒ \(\frac{d}{d x}\)(In y) = \(\frac{d}{d x}\)(x . In x)
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = In x + x . \(\frac{1}{x}\) = In x + 1 = 1 + In x
⇒ \(\frac{d y}{d x}\) = y (1 + In x) = x^x (1 + In x) = x^x In (ex)

Question 2.
\(\left(1+\frac{1}{x}\right)^x\)
Solution:

Question 3.
x^{sin x}
Solution:
y = x^{sin x}
⇒ In y = sin x . In x
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = cos x In x + sin x . \(\frac{1}{x}\)
⇒ \(\frac{d y}{d x}\) = x^{sin x} (cos x . In x + \(\frac{\sin x}{x}\))

Question 4.
(log x)^{tan x}
Solution:
y = (log x)^{tan x}
⇒ log y = tan x . log (log x)
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = sec² x log (log x) + tan x . \(\frac{1}{\log x}\) . \(\frac{1}{x}\)
⇒ \(\frac{d y}{d x}\) = (log x)^{tan x} {sec² x . log log x + \(\frac{\tan x}{x \log x}\)}

Question 5.
\(2^{\left(2^x\right)}\)
Solution:
y = \(2^{\left(2^x\right)}\)
⇒ In y = 2^x . In 2
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = 2^x . In 2 . In 2
⇒ \(\frac{d y}{d x}\) = \(2^{\left(2^x\right)}\) . 2^x . (In 2)²

Question 6.
\((1+\sqrt{x})^{x^2}\)
Solution:

Question 7.
\(\left(\sin ^{-1} x\right)^{\sqrt{1-x^2}}\)
Solution:

Question 8.
\((\tan x)^{\log x^3}\)
Solution:

Question 9.
x^1/x + (sin x)^x
Solution:

Question 10.
(cos x)^x + x^{cos x}
Solution:

Question 11.
(x² + 1)^2/3 (3x + 1)^1/4 √x
Solution:

Question 12.
\(\frac{(x+1)(x+2)^2(x+3)^3}{(x-1)(x-2)^2(x-3)^3}\)
Solution:

Question 13.
(sin x)^x \(\sqrt{\sin x}\left(1+x^2\right)^{\frac{1}{2}+x}\)
Solution:

Question 14.
(sec x + tan x)^{cot x}
Solution:
y = (sec x + tan x)^{cot x}
⇒ In y = cot x . In (sec x + tan x)
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = -cosec² x . In (sec x + tan x) + cot x . \(\frac{1}{\sec x+\tan x}\) × (sec x . tan x + sec² x)
= -cosec² x . In (sec x + tan x) + cot x . sec x
∴ \(\frac{d y}{d x}\) = y {-cosec² x . In (sec x + tan x) + cosec x}
= (sec x + tan x)^{cot x} {cosec x – cosec² x . In (sec x + tan x)}

Question 15.
(2^√x)^1+√x
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(e) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(e)

Differentiate the following functions by proper substitution.
Question 1.
sin^-1 2x\( \sqrt{1-x^2} \)
Solution:
y = sin^-1 2x\( \sqrt{1-x^2} \)   [Put x = sin θ
= sin^-1 (2 sin θ . cos θ)
= sin^-1 sin 2θ = 2θ = 2 sin^-1 x.
\(\frac{d y}{d x}\) = \(\frac{2}{\sqrt{1-x^2}}\)

Question 2.
tan^-1 \(\frac{2 x}{1-x^2}\)
Solution:
y = tan^-1 \(\frac{2 x}{1-x^2}\)
= tan^-1 \(\frac{2 \tan \theta}{1-\tan ^2 \theta}\) = tan^-1 (tan 2θ)
= 2θ = 2 tan^-1 x
∴ \(\frac{d y}{d x}\) = \(\frac{2 x}{1-x^2}\)

Question 3.
tan^-1 \(\sqrt{\frac{1-t}{1+t}}\)
Solution:

Question 4.
\(\left[\left(\frac{1+t^2}{1-t^2}\right)^2-1\right]^{\frac{1}{2}}\)
Solution:

Question 5.
tan^-1 \(\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{x a}}\right)\)
Solution:

Question 6.
sin^-1 (\(\frac{2 x}{1+x^2}\))
Solution:
y = sin^-1 \(\frac{2 x}{1+x^2}\)  [Put x = tan θ
= sin^-1 \(\frac{2 \tan \theta}{1+\tan ^2 \theta}\) = sin^-1 sin θ
= 2θ = 2 tan^-1 x
∴ \(\frac{d y}{d x}\) = \(\frac{2 x}{1+x^2}\)

Question 7.
sec^-1 \(\left(\frac{\sqrt{a^2+x^2}}{a}\right)\)
Solution:

Question 8.
sin^-1 \(\left(\frac{2 \sqrt{t^2-1}}{t^2}\right)\)
Solution:

Question 9.
cos^-1 \(\left(\frac{1-t^2}{1+t^2}\right)\)
Solution:
y = cos^-1 \(\left(\frac{1-t^2}{1+t^2}\right)\) [Put t = tan θ
= cos^-1 \(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\)
= cos^-1 cos 2θ = 2 tan^-1 t
∴ \(\frac{d y}{d x}\) = \(\frac{2}{1+t^2}\)

Question 10.
cos^-1 (2t² – 1)
Solution:
y = cos^-1 (2t² – 1) [Put t = tan θ
= cos^-1 (2 cos² θ – 1)
= cos^-1 cos 2θ = 2θ = 2 cos^-1 t
∴ \(\frac{d y}{d x}\) = – \(\frac{2}{\sqrt{1-t^2}}\)

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 4 The Three Questions Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 4 The Three Questions

BSE Odisha 6th Class English Follow-Up Lesson 4 The Three Questions Text Book Questions and Answers

Session – 1 ( ସୋପାନ – ୧)

I. Pre-Reading (ପଢ଼ିବା ପୂର୍ବରୁ):
Here is a similar story for you. Read the story and do the activities. Some activities are given. The rest of the activities will be designed by your teacher. S/he will write them on the blackboard and help you do the tasks.
(ଏଠାରେ ତୁମ ପାଇଁ ଗୋଟିଏ ଏକାପରି ଗପ ଅଛି । ଗପଟିକୁ ପଢ଼ ଏବଂ କାର୍ଯ୍ୟାବଳୀଟିକୁ କର । କେତେକ କାର୍ଯ୍ୟାବଳୀ ଦିଆଯାଇଛି । ଅବଶିଷ୍ଟ କାର୍ଯ୍ୟାବଳୀ ତୁମ ଶିକ୍ଷକଙ୍କଦ୍ୱାରା ପ୍ରସ୍ତୁତ କରାଯିବ । ସେ (ପୁ/ସ୍ତ୍ରୀ) ସେଗୁଡ଼ିକୁ କଳାପଟାରେ ଲେଖିଦେବେ ଏବଂ ଏହି ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର କରିବାରେ ତୁମକୁ ସାହାଯ୍ୟ କରିବେ ।)

II. While-Reading (ପଢ଼ିବା ସମୟରେ)
Text – (ପାଠ୍ୟବସ୍ତୁ)

Read paragraphs 1-3 silently and answer the questions that follow.
(ଅନୁଚ୍ଛେଦ ୧ – ୩ କୁ ନୀରବରେ ପଢ଼ା ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)
1. Once there was a great king in Germany. He liked soldiers very much because he was a good soldier himself. He had a special liking for one section of tall soldiers in his army. He kept that section under his own care and loved to watch it. He wanted to make it the best section of the army and tried hard to get the best men for it.

2. At the first meeting, the king used to ask every new soldier the following three questions.
“How old are you?
How long have you served me already?
Do you like the food or the quarters here ?”

3. Once a man from another country came to join the German army. He did not know the German language, “How can I answer the king’s questions ?” he said, “I can’t understand them.”

ଓଡ଼ିଆ ଅନୁବାଦ :
(୧) ଏକଦା (ଥରେ) ଜର୍ମାନୀ ଦେଶରେ ଜଣେ ବିଶିଷ୍ଟ (ମହାନ୍) ରାଜା ଥିଲେ । ସେ ସୈନିକମାନଙ୍କୁ ବହୁତ ଭଲ ପାଉଥିଲେ କାରଣ ସେ ନିଜେ ଜଣେ ଭଲ ସୈନିକ ଥିଲେ । ତାଙ୍କର ତାଙ୍କ ସୈନ୍ୟବାହିନୀର ଏକ ଡେଙ୍ଗା ସୈନ୍ୟ ବିଭାଗ ପାଇଁ ସ୍ବତନ୍ତ୍ର ପସନ୍ଦ|ରୁଚି ଥିଲା । ସେ ସେହି ସୈନ୍ୟ ବିଭାଗ (ଦଳକୁ) ତାଙ୍କ ନିଜ ଦାୟିତ୍ଵରେ ରଖିଥିଲେ ଏବଂ ଏହାକୁ ଦେଖିବାକୁ ଭଲ ପାଉଥିଲେ । ସେ ଏହାକୁ ସୈନ୍ୟବାହିନୀର ସବୁଠାରୁ ଭଲ ବିଭାଗ (ଦଳ) ରୂପେ ଗଢ଼ିବାକୁ ଚାହୁଁଥିଲେ ଏବଂ ଏଥ‌ିପାଇଁ ସର୍ବୋତ୍ତମ ଲୋକମାନଙ୍କୁ ପାଇବାକୁ ବହୁତ ଚେଷ୍ଟା କରୁଥିଲେ ।

(୨) ପ୍ରଥମ ସାକ୍ଷାତ୍‌ରେ ରାଜା ପ୍ରତ୍ୟେକ ନୂଆ ସୈନିକଙ୍କୁ ପ୍ରାୟତଃ ନିମ୍ନଲିଖ୍ ତିନିଗୋଟି ପ୍ରଶ୍ନ ପଚାରୁଥିଲେ ।
ତୁମେ କେତେ ସମୟଧରି ମୋତେ ସେବା ପ୍ରଦାନ କରିଛ ? ତୁମେ ଏଠାରେ ଖାଦ୍ୟ କିମ୍ବା ବାସଗୃହ ପସନ୍ଦ କରୁଛ ?

(୩) ଥରେ ଅନ୍ୟ ଦେଶରୁ ଜଣେ ଲୋକ ଜର୍ମାନ୍ ସୈନ୍ୟବାହିନୀରେ ଯୋଗଦେବାକୁ ଆସିଲେ । ସେ ଜର୍ମାନ୍ ଭାଷା ଜାଣି ନଥୁଲେ, ସେ କହିଲେ, ‘‘କିପରି ମୁଁ ରାଜାଙ୍କର ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦେଇପାରିବି ? ମୁଁ ତ ସେଗୁଡ଼ିକୁ ବୁଝିପାରିବି ନାହିଁ ।?

Comprehension Questions – (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ) :

Question 1.
Why did the king like soldiers ?
(କାହିଁକି ରାଜା ସୈନ୍ୟବାହିନୀକୁ ଭଲ ପାଉଥିଲେ ?)
Answer:
The king liked soldiers very much because he was a good soldier himself.

Question 2.
How many questions did the king ask a new soldier?
(ଜଣେ ନୂଆ ସୈନିକକୁ ରାଜା କେତେଗୋଟି ପ୍ରଶ୍ନ ପଚାରୁଥିଲେ ?)
Answer:
The king used to ask three questions to a new soldier.

Question 3.
What was the problem with the new soldier?
(ନୂଆ ସୈନିକର କ’ଣ ସମସ୍ୟା ଥିଲା ?)
Answer:
The problem with the new soldier was that he did not know the German language.

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

Read paragraphs 4-6 silently and answer the questions that follow.
(ଅନୁଚ୍ଛେଦ ୪ – ୬ କୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)
4. His officer told him all three questions and the way of answering them. He said the king always asked the same questions in the same order. So the soldier decided to answer them in that order

5. One day the king came to visit the army. He saw the new soldier and began to ask him questions. But this time the questions were in a different order. The soldier did not know this because he did not. understand the king’s words at all.

6. King: How long have you been in my service?
Soldier: Thirty years, sir.
King: How is it? You look so young! What’s your age?
Soldier: Three weeks, sir.
King: What do you mean? Are you mad, or am I?
Soldier: Both sir.

ଓଡ଼ିଆ ଅନୁବାଦ :
(୪). ତାଙ୍କର ଉଚ୍ଚ/ପଦସ୍ଥ କର୍ମଚାରୀ ତାଙ୍କୁ ସମସ୍ତ ତିନୋଟି ପ୍ରଶ୍ନ ଏବଂ ସେଗୁଡ଼ିକୁ ଉତ୍ତରଦେବା ଶୈଳୀ କହିଦେଲେ । ସେ କହିଲେ ଯେ ରାଜା ସର୍ବଦା ସେହି ଏକା ପ୍ରଶ୍ନ ଏକାକ୍ରମରେ ପଚାରନ୍ତି । ତେଣୁ ସୈନିକଟି ସ୍ଥିରକଲା ସେଗୁଡ଼ିକୁ ସେହି କ୍ରମରେ ଉତ୍ତର ଦେବାକୁ ।
(୫) ଦିନେ ରାଜା ସୈନ୍ୟବାହିନୀକୁ ପରିଦର୍ଶନ କରିବାକୁ ଆସିଲେ । ସେ ନୂଆ ସୈନିକକୁ ଦେଖିଲେ ଏବଂ ତାଙ୍କୁ ପ୍ରଶ୍ନଗୁଡ଼ିକ ପଚାରିବାକୁ ଆରମ୍ଭ କଲେ । କିନ୍ତୁ ଏଥର ପ୍ରଶ୍ନଗୁଡ଼ିକ ଭିନ୍ନ କ୍ରମରେ ଥିଲା । ସୈନିକଟି ଏହା ଜାଣିନଥୁଲା କାରଣ ସେ ରାଜାଙ୍କର ଶବ୍ଦଗୁଡ଼ିକୁ ଆଦୌ ବୁଝିପାରୁ ନଥିଲା ।

(୬) ରାଜା: କେତେକାଳ ହେଲା ତୁମେ ମୋ’ର ସେବାରେ ନିୟୋଜିତ ଅଛ ?
ସନିକ: ତିରିଶ ବର୍ଷ, ମହାଶୟ ।
ରାଜା: ଏହା କିପରି ସମ୍ଭବ ? ତୁମେ ଏତେ ଯୁବକ/କମ୍ ବୟସର ଦେଖାଯାଉଛ ? ତୁମର ବୟସ କେତେ ?
ସନିକ: ତିନି ସପ୍ତାହ, ମହାଶୟ ।
ରାଜା: ତୁମେ କ’ଣ ଭାବୁଛ ? ତୁମେ ପାଗଳ ନା ମୁଁ ?
ସନିକ: ଉଭୟ, ମହାଶୟ ।

Comprehension Questions – (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ) :

Question 1.
Who helped the new soldier in preparing to answer the king’s questions?
(କିଏ ନୂଆ ସୈନିକକୁ ରାଜାଙ୍କ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦେବାପାଇଁ ପ୍ରସ୍ତୁତ କରିବାରେ ସାହାଯ୍ୟ କଲା ?)
Answer:
His officer helped the new soldier, in preparing to answer the king’s questions.

Question 2.
What did the king ask first?
(ରାଜା ପ୍ରଥମେ କ’ଣ ପଚାରିଲେ ?)
Answer:
The king first asked, “How long have you been in my service?

Question 3.
What did the soldier answer?
(ସୈନିକଟି କ’ଣ ଉତ୍ତର ଦେଲା ?)
Answer:
The soldier answered, “Thirty years, sir.”

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

1. Writing- (ଲେଖୁବା) :
(a) Answer the following questions
(କବିତାଟି କେଉଁ ବିଷୟରେ ଆଧାରିତ ?)

(i) Why did the king like soldiers?
(କାହିଁକି ରାଜା ସୈନିକମାନଙ୍କୁ ଭଲ ପାଉଥିଲେ ?)
Answer:
The king liked the soldiers very much because he was a good soldier himself.

(ii) How many questions did the king ask?
(ରାଜା କେତୋଟି ପ୍ରଶ୍ନ ପଚାରିଲେ ?)
Answer:
The king asked three questions.

(iii) What was the problem with the new soldier ?
(ନୂଆ ସୈନିକର କ’ଣ ସମସ୍ୟା ଥିଲା ?)
The solder did not know.
Answer:
The soldier did not know the German language. So the problem with the new soldier was how he could answer the king’s questions.

(iv) Who helped him?
(କିଏ ତାକୁ ସାହାଯ୍ୟ କଲା ?)
Answer:
His officer helped him to answer the king’s questions.

(v) Did the help work ?
(ସାହାଯ୍ୟଟି କାମରେ ଲାଗିଲା କି ?)
Answer:
No, the help did not work at all.

(vi) Why? (କାହିଁକି ?)
Because the king did
Answer:
Because the king did not ask the same questions in the same order. As the questions were in a different order, the soldier did not answer the king’s questions correctly.

b. Match the questions under ‘A’ with the answers under ‘B’.
(‘A” ତଳେ ଥ‌ିବା ପ୍ରଶ୍ନଗୁଡ଼ିକୁ ‘B’ ତଳେ ଥିବା ଉତ୍ତରଗୁଡ଼ିକ ସହ ମିଳାଅ ।)

Answer:

Answer:

Now write on what really happened based on your matching.
(ବର୍ତ୍ତମାନ ତୁମର ଯୋଡ଼ିବାକୁ ଭିତ୍ତିକରି ଯାହା ପ୍ରକୃତରେ ଘଟିଗଲା ଲେଖ ।)
When the king asked, “________ old _______?” The soldier replied, “__________” . Next , when he “How long _________ ?” The soldier replied, “___________”Finally, When____________________________ The soldier replied,___________________.
Answer:
When the king asked, “How old are you ?” The soldier replied, “Three weeks, sir.” Next, when he asked, “How long have you served me already? The soldier replied, “Thirty years, sir.” Finally, when the king asked. “Do you like the food or the quarters here ?’’ The soldier replied, “Both sir.’’

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

great – ବହୁତ ଭଲ
soldier – ସନିକ
liking – ପସନ୍ଦ
section – ବିଭାଗ
tall – ଲମ୍ବା
army – ସେନା
own – ନିଜର
care – ଯତ୍ନ
hard – କଠିନ
used to – ଅଭ୍ୟସ୍ତ
served – ପରିବେଷିତ
quarters – କ୍ୱାର୍ଟର୍ସ
German language – ଜର୍ମାନ ଭାଷା
order – କ୍ରମ
understand – ମୋତେ ବୁ
mad – ପାଗଳ
developed – ବିକଶିତ
satisfaction – ସନ୍ତୁଷ୍ଟ
anger – କ୍ରୋଧ
deaf – ବଧିର
seriously ill – ଗୁରୁତର ଅସୁସ୍ଥ
special liking- ବିଶେଷ ପସନ୍ଦ
same order – ସମାନ କ୍ରମ
decided – ସ୍ଥିର କଲା
visit – ପରିଦର୍ଶନ କରନ୍ତୁ
how long – କେତେ ଦିନ
began – ଆରମ୍ଭ ହେଲା
at all – ସବୁ ସମୟରେ
service – ସେବା
so young – ଏତେ ଯୁବକ
age – ବୟସ
weeks – ସପ୍ତାହଗୁଡିକ
mean – ଅର୍ଥ
both – ଉଭୟ
classmate – ସହପାଠୀ
problems – ସମସ୍ୟାଗୁଡିକ
finally – ଶେଷରେ