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Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 10 Sequences and Series Ex 10(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 10 Sequences and Series Exercise 10(b)

Question 1.
Expand in ascending power of x.
(i) 2^x
Solution:

(ii) cos x
Solution:

(iii) sin x
Solution:

(iv) \(\frac{x e^{7 x}-e^{-x}}{e^{3 x}}\)
Solution:

(v) \(\boldsymbol{e}^{e^x}\) up to the term containing x⁴
Solution:

Question 2.
If x = y + \(\frac{y^2}{2 !}+\frac{y^3}{3 !}\) + ….. then show that y = x – \(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}\) +….
Solution:

Question 3.
Find the value of \(x^2-y^2+\frac{1}{2 !}\left(x^4-y^4\right)+\frac{1}{3 !}\left(x^6-y^6\right)\) + ….
Solution:

Question 4.
Show that
(i) 2\(\left(\frac{1}{3 !}+\frac{2}{5 !}+\frac{3}{7 !}+\ldots\right)=\frac{1}{e}\)
Solution:

(ii) \(\frac{9}{1 !}+\frac{19}{2 !}+\frac{35}{3 !}+\frac{57}{4 !}+\frac{85}{5 !}\) + …. = 12e – 5
Solution:

(iii) \(1+\frac{1+3}{2 !}+\frac{1+3+3^2}{3 !}+\ldots=\frac{1}{2}\left(e^3-e\right)\)
Solution:

(iv) \(\frac{1.3}{1 !}+\frac{2.4}{2 !}+\frac{3.5}{3 !}+\frac{4.6}{4 !}\) + …. = 4e
Solution:
t_n for L.H.S.
= \(\frac{n(n+2)}{n !}=\frac{n^2+2 n}{n !}\)

(v) \(\frac{1}{1.2}+\frac{1.3}{1.2 .3 .4}+\frac{1.3 .5}{1.2 .3 .4 .5 .6}\) + …. = √e – 1
Solution:

Question 5.
Prove that
(i) loge(1 + 3x + 2x²) = 3x – \(\frac{5}{2}\)x² + \(\frac{9}{3}\)x³ – \(\frac{17}{4}\)x⁴ + …..,|x| < \(\frac{1}{2}\)
Solution:

(ii) log_e(n + 1) – log_e(n – 1) = 2 \(\left[\frac{1}{n}+\frac{1}{3 n^3}+\frac{1}{5 n^5}+\ldots\right]\)
Solution:
There is a printing mistake in the question. The correct question is log_e(n + 1) – log_e(n – 1)

(iii) log_e(n + 1) – log_en = 2 \(\left[\frac{1}{2 n+1}+\frac{1}{3(2 n+1)^3}+\frac{1}{5(2 n+1)^5}+\ldots\right]\)
Solution:

(iv) log_em – log_en = \(\frac{m-n}{m}+\frac{1}{2}\left(\frac{m-n}{m}\right)^2\)
Solution:

= – [log n – log m]
= log m – log n = L.H.S.

(v) log_ea – log_eb = \(2\left[\frac{a-b}{a+b}+\frac{1}{3}\left(\frac{a-b}{a+b}\right)^3\right.\) \(\left.+\frac{1}{5}\left(\frac{a-b}{a+b}\right)^5+\ldots\right], a>b\)
Solution:

(vi) log_en = \(\frac{n-1}{n+1}+\frac{1}{2} \cdot \frac{n^2-1}{(n+1)^2}\)\(+\frac{1}{3} \cdot \frac{n^2-1}{(n+1)^3}\) + …..
Solution:

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

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

Question 1.
²ⁿC₀ + ²ⁿC₂ + …… + ²ⁿC_2n = 2^2n-1 and ²ⁿC₁ + ²ⁿC₃ + ….. + ²ⁿC_2n-1 = 2^2n-1
Solution:
We know that
(1 + x)²ⁿ = ²ⁿC₀ + ²ⁿC₁x + ²ⁿC₂x² + …..+ ²ⁿC_2nxⁿ …(1)
Putting x = 1
We get putting x = – 1  we get
∴ (²ⁿC₀ + ²ⁿC₂ + ²ⁿC₄ + ….. + ²ⁿC_2n) – (²ⁿC₁ + ²ⁿC₃ + ….. + ²ⁿC_2n-1) = 0
∴ ²ⁿC₀ + ²ⁿC₂ + ….. + ²ⁿC_2n
= ²ⁿC₁ + ²ⁿC₃ + ….. + ²ⁿC_2n-1
= \(\frac{2^{2 n}}{2}\) = 2^2n-1

Question 2.
Find the sum of
(i) C₁ + ²C₂ + ³C₃ + ….. + ⁿC_n
Solution:
C₁ + ²C₂ + ³C₃ + ….. + ⁿC_n

(ii) C₀ + ²C₁ + ³C₂ + ….. + ⁽ⁿ⁺¹⁾C_n Hint: write (C₀ + C₁ + ….. + C_n) + (C₁ + ²C₂ + ….. + ⁿC_n) use (5) and exercise 1.

Question 3.
Compute \(\frac{(1+k)\left(1+\frac{k}{2}\right) \ldots\left(1+\frac{k}{n}\right)}{(1+n)\left(1+\frac{n}{2}\right) \ldots\left(1+\frac{n}{k}\right)}\)
Solution:

Question 4.
Show that
(i) C₀C₁ + C₁C₂ + C₂C₃ + ….. + C_n-1C_n = \(\frac{(2 n) !}{(n-1) !(n+1) !}\)
Solution:

(ii) C₀C_r + C₁C_r+1 + C₂C_r+2 + ….. + C_n-rC_n = \(\frac{(2 n) !}{(n-r) !(n+r) !}\) Hint : Proceed as in Example 13. Compare the coefficient of a^n-1 to get (i) and the coefficient of a^n-r to get (ii)
Solution:


∴ C₀C_r + C₁C_r+1 + C₂C_r+2 + ….. + C_n-rC_n = \(\frac{(2 n) !}{(n-r) !(n+r) !}\)

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

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

(v) C₀ – 2C₁ +3C₂ + ….. + (-1)ⁿ(n+1)C_n = 0
Solution:

(vi) C₀ – 3C₁ +5C₂ + ….. + (2n+1)C_n = (n+1)²ⁿ
Solution:

Question 5.
Find the sum of the following :
(i) C₁ – 2C₂ + 3C₃ + ….. + n(-1)^n-1C_n 
Solution:

(ii) 1.2C₂ + 2.3C₃ + ….. + (n-1)nC_n
Solution:

(iii) C₁ + 2²C₂ + 3²C₃ + ….. + n²C_n
Solution:

(vi) C₀ – \(\frac{1}{2}\)C₁ + \(\frac{1}{3}\)C₂ + ….. + (-1)ⁿ \(\frac{1}{n+1}\)C_n
Solution:

Question 6.
Show that
(i) C₁² + 2C₂² +3C₃² + ….. + nC_n² = \(\frac{(2 n-1) !}{\{(n-1) !\}^2}\)
Solution:

(ii) C₂ + 2C₃ + 3C₄ + ….. + (n – 1)C_n = 1 + (n – 2)2^n-1
Solution:

Question 7.
C₁ – \(\frac{1}{2}\)C₂ + \(\frac{1}{3}\)C₃ + ….. +(-1)ⁿ⁺¹ \(\frac{1}{n}\)C_n = 1 + \(\frac{1}{2}\) + …. + \(\frac{1}{n}\)
Solution:

Question 8.
C₀C₁ + C₁C₂ + ….. + C_n-1C_n = \(\frac{2^n \cdot n \cdot 1 \cdot 3 \cdot 5 \ldots(2 n-1)}{(n+1)}\)
Solution:

Question 9.
The sum \(\frac{1}{1 ! 9 !}+\frac{1}{3 ! 7 !}+\ldots+\frac{1}{7 ! 3 !}+\frac{1}{9 ! 1 !}\) can be written in the form \(\frac{2^a}{b !}\) find a and b.
Solution:

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

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

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

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

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

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

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

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

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

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

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

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

Comprehension Questions : (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ)

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

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

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

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

Session – 2 (ସୋପାନ – ୨):
SGP-2:

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

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

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

Comprehension Questions : (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

TEXT (ବିଷୟବସ୍ତୁ)

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

It was a very cool night
And there was no crab in sight.
The fox looked for one

But there was none.
“Where did they go?”
Not even one in sight!
They must be in their holes
If I’m right.”

Thinking so he got ready
(The hunger too made him greedy)
To go in search of a crab hole
And he straightened his tail like a pole.
Inside the hole his bushy tail he pushed
The crab, he thought, it slightly missed.
He waited long for a sweet pull
But for long there was none at all.
Finally, he pulled out his tail
But the crab was inside he could smell.
So he changed his plan and called “Brother Crab,
Let’s some song and dance have.
The weather calls for such merrymaking
What is life without dancing and singing ?”
The crab well understood
The fox’s real mood.
Thought he to himself ‘Am I a fool ?’

And answered from his hole :
“Who is going to sing and dance
In such weather cool?
I’ll rather eat and sleep well
in my cozy little hole.”

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

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

Comprehension Questions (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ):

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 1.
The rows n = 6 and n = 7 in the pascal triangle have been kept vacant. Fill in the gaps
Solution:

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

Question 3.
Find the 3rd term in the expansion of \(\left(2 x^3-\frac{1}{x^6}\right)^4\) using rules of Pascal triangle.
Solution:

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

(b) \(\left(\frac{-9}{2} a+b\right)^7\)
Solution:

(c) \(\left(a-\frac{7}{3} c\right)^4\)
Solution:

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

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

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

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

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

Question 7.
Answer the following :
(a) If the 6th term in the expansion of (x + *)ⁿ is equal to ⁿC₅x^n-10 find *
Solution:
Let the 6th term (x + y)ⁿ is ⁿC₅x^n-10
∴ ⁿC₅x^n-5y⁵ = ⁿC₅x^n-10 = ⁿC₅x^n-5.x^-5
⇒ y⁵ = x^-5 = \(\frac{1}{x^5}\)
∴ y = \(\frac{1}{x}\) . Hence * = \(\frac{1}{x}\)

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

(c) Find the value of \(\frac{{ }^n \mathrm{C}_{r-1}}{{ }^n \mathrm{C}_r}\)
Solution:

(d) How many terms in the expansion of \(\left(\frac{3}{a}+\frac{a}{3}\right)^{10}\) have positive powers of a? How many have negative powers of a?
Solution:

Question 8.
Find the middle term(s) in the expansion of the following.
(a) \(\left(\frac{a}{b}+\frac{b}{a}\right)^6\)
Solution:
Here there is only one middle term i.e. the 4th term.
∴ 4th term i.e. (3 + 1)th term in the expansion of

(b) \(\left(x+\frac{1}{x}\right)^9\)
Solution:
Here there are two middle terms i.e. 5th and 6th terms.
∴ 5th term in the above expansion is

(c) \(\left(x^{\frac{3}{2}}-y^{\frac{3}{2}}\right)^8\)
Solution:
Here there is only one middle term i.e. 5th term.
∴ 5th term i.e. (4 + 1)th term in the expansion of

Question 9.
Find the 6th term in the expansion of \(\left(x^2+\frac{a^4}{y^2}\right)^{10}\)
Solution:
6th term i.e. (5 + 1)th term in the expansion of
(x² + \(\frac{a^4}{y^2}\))¹⁰ is ¹⁰C₅(x²)^10-5 (\(\frac{a^4}{y^2}\))⁵

Question 10.
(a) Find the fifth term in the expansion of (6x – \(\frac{a^3}{x}\))¹⁰
Solution:

(b) Is there a term independent of x? If yes find it out.
Solution:
Let (r + 1)th term in the expansion of  (6x – \(\frac{a^3}{x}\))¹⁰ is independent of x.
∴ (r + 1)th term = ¹⁰C_r(6x)^10-r(\(\frac{-a^3}{x}\))^r
= ¹⁰C_r6^10-rx^10-r(-1)^ra^3rx^-r
= (-1)^{r 10}C_r6^10-ra^3rx^10-2r
∴ x^10-2r = 1 = x⁰
or, 10 – 2r = 0 or, r = 5
∴ 6th term is term independent of x in the above expansion and the term is (-1)^{5 10}C₅6^10-5a^3.5
= – ¹⁰C₅6⁵a¹⁵ = – 252 × 6⁵a¹⁵

Question 11.
(a) Find the coefficient of \(\frac{1}{y^{10}}\) in the expansion of \(\left(y^3+\frac{a^7}{y^5}\right)^{10}\)
Solution:

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

Question 12.
(a) Find the coefficient of x⁴ in the expansion of (1 + 3x + 10x²)(x + \(\frac{1}{x}\))¹⁰
Solution:

(b) Find the term independent of x in the above expansion.
Solution:

Question 13.
Show that the coefficient of a^m and aⁿ in expansion of (1 + a)^m+n are equal.
Solution:
(m + 1)th and (n + 1)th terms in the expansion of (1 + a)^m+n are ^m+nC_ma^m and ^m+nC_naⁿ
∴ The coefficient of a^m and aⁿ are ^m+nC_m and ^m+nC_n which are equal.

Question 14.
An expression of the form (a + b + c + d + …. ) consisting of a sum of many distinct symbols called a multinomial. Show that (a + b + c)ⁿ is
the sum of all terms of the form \(\frac{\boldsymbol{n} !}{\boldsymbol{p} ! \boldsymbol{q} ! \boldsymbol{r} !}\) a^pb^qc^r where p, q and r range over all positive triples of non-negative integers such that p + q + r = n.
Solution:

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

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

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

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

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

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

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

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

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

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

(iv) \(\frac{1}{8 !}+\frac{1}{9 !}+\frac{1}{10 !}\)
Solution:

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

Question 4.
Show that 2.6.10 ……. to n factors = \(\frac{(2 n) !}{n !}\)
Solution:
2.6.10 ……. to n factors = \(\frac{(2 n) !}{n !}\)

Question 5.
Find r if P(20, r) = 13. P (20, r – 1).
Solution:

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

Question 7.
If P (n – 1, 3) : P (n + 1, 3) = 5: 12, Find n.
Solution:

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

Question 9.
Show that
(i) P(n, n) = P(n, n – 1) for all positive integers.
Solution:

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

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

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

1st

2nd

3rd

4th

5th

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

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

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

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

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

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

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

Question 18.
Find the numbers of all 4-lettered words (not necessarily having meaning) that can be formed using the letters of the word BOOKLET.
Solution:
We have to form 4-lettered words using the letters of the word BOOKLET. The word contains 7 letters out of which there are 20’s. So there are 6 letters.
∴ The number of 4-lettered words = ⁷P₄ – ⁶P₄ = 7.6.5.4 – 6.5.4.3 = 480

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

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

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

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

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

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

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

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

X	0	1
y	0	2

x – 2y = 0

X	0	2
y	0	1

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

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

X	1	0
y	0	-1

y – x = 1 ⇒ y = x + 1

X	0	-1
y	1	0

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

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

X	-2	0
y	0	1

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

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

X	0	-1
y	1	0

3x + 4y = 12

X	4	0
y	0	3

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

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

X	1	0
y	0	1

3x – y = 3
⇒ y = 3x – 3

X	1	0
y	0	-3

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

X	-3	0
y	0	1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(iii) – 47 + 8 √-1
Solution:

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

(iv) – 8 + √-1
Solution:

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

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

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

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

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

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

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

(ii) z³ = i
Solution:

(iii) z⁶ = – i
Solution:

(iv) z³ = 1 + i
Solution:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

BSE Odisha Class 7 English Solutions Test-2

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

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

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

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

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

4. Your teacher will dictate ten names of persons. Listen to him/her and write in the space provided. [10]

Answer:
ବିରାକିଶୋର ଦାସ |  – Birakishore Das
ନିଲମାନି ମାର୍ଗ        – Nilamani Routray
ମଧୁସୂଦନ ଦାସ       – Madhusudan Das
ଗୋପାଳକୃଷ୍ଣ ଗୋଖଲେ | – Gopalkrushna Gokhle
ତିଲ୍କା ମାଜି             – Tilka Majhi
ଲକ୍ଷ୍ମଣ ନାୟକ |       – Laxman Nayak
ସୁରେନ୍ଦ୍ର ସାଇ          – Surendra Sai
ପଦ୍ମଲୋଚନ ବେହେରା | – Padmalochan Behera
ଲିଙ୍ଗରାଜ ନନ୍ଦା |      – Lingaraj Nanda
ରାଧନାଥ ରୟ        – Radhanath Roy

5. Your teacher will dictate ten names of persons. Listen to him/her and write in the space provided. [10]

Answer:
ବରିପଡା |  – Baripada
କେନ୍ଦୁଝର   – Keonjhar
କାକଟପୁର  – Kakatpur
ମୁମ୍ବାଇ |     – Mumbai
ରାୟାଗଡା | – Raygada
ବୋରିଗୁମା  – Boriguma
ସମ୍ବଲପୁର   – Sambalpur
କୋଲକାତା – Kolkata
ବେଙ୍ଗାଲୁରୁ  – Bengaluru
ଆହ୍ଲାବାଦ  – Allahabad

6. Match the pair of words that sound alike at the end.

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 1.

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

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

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

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

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

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

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

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

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

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

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

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

(ii) √3 sin θ – cos θ = 2
Solution:
√3 sin θ – cos θ = 2
or, \(\frac{\sqrt{3}}{2}\) sin θ – 1/2 cos θ = 1

which is the only principal solution.

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

(iv) sin 4x + sin 2x = 0
Solution:

(v) sin x + cos x = \(\frac{1}{\sqrt{2}}\)
Solution:

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

(ii) sin(x° + 40°) = \(\frac{1}{\sqrt{2}}\)
Solution:

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

(iv) tan ax = cot bx
Solution:

(v) tan² 3θ = 3
Solution:

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

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

(iii) 3 sin x + 4 cos x = 5
Solution:

(iv) 3 tan x + cot x = 5 cosec x
Solution:

(v) cos x + √3 sin x = √2
Solution:

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

(vii) sec θ + tan θ = √3
Solution:

(viii) cos 2θ – cos θ = sin θ – sin 20
Solution:

(ix) sin θ + sin 2θ + sin 3θ + sin 4θ = 0
Solution:

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

(xi) tan θ + tan 2θ = tan 3θ
Solution:

(xii) tan θ + tan (\(\theta+\frac{\pi}{3}\)) + tan (\(\theta+\frac{2\pi}{3}\)) = 3
Solution:

(xiii) cot² θ – tan² θ = 4 cot 2θ
Solution:

(xiv) cos 2θ = \((\sqrt{2}+1)\left(\cos \theta-\frac{1}{\sqrt{2}}\right)\)
Solution:

(xv) sec θ – 1 = \((\sqrt{2}-1)\) tan θ
Solution:

⇒ θ = 2nπ + \(\frac{\pi}{4}\)

(xvi) 3cot² θ – 2 sin θ = 0
Solution:

(xvii) 4 cos x. cos 2x . cos 3x = 1
Solution:
4 cos x cos 2x cos 3x = 1
⇒ 2 cos x cos 2x. 2 cos 3x = 1
⇒ (cos 3x + cos x) 2 cos 3x = 1
⇒ 2 cos² 3x + 2 cos 3x cos x = 1
⇒ 2 cos² 3x – 1 + cos 4x + cos 2x = 0
⇒ cos 6x + cos 4x + cos 2x = 0
⇒ cos 6x + cos 2x + cos 4x = 0
⇒ 2 cos 4x cos 2x + cos 4x = 0
⇒ cos 4x (2 cos 2x + 1) = 0
⇒ cos 4x = 0, cos 2x = – 1/2
cos 4x = 0 ⇒ 4x = (2n + 1) \(\frac{\pi}{2}\)

(xviii) cos 3x – cos 2x = sin 3x
Solution:

⇒ 1 – 2 sin x cos x = y²
∴ Equation (1) reduces to
1 – 2(1 – y²) + y = 0
⇒ 2y² + y – 1 = 0
⇒ (2y- 1) (y + 1) = 0

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

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

(xxi) 2(sec² θ + sin² θ) = 5
Solution:

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

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

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

Long Questions With Answers

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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