NCERT Grade 12-Integrals-Answers

NCERT Solutions for Class 12 Maths

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Comparing coefficients x, of 2A = 6 ⇒ A = 3

Comparing constants, -9A + B = 7

On solving, we get A = 3, B = 34

Putting the values of A and B in eq. (ii),

6x + 7 = 3(2x – 9) + 34

Putting this value of 6x + 7 in eq. (i),

105. 

Comparing coefficients of x, –2A = 1 ⇒ A =

Comparing constants, 4A + B = 2

On solving, we get A = , B = 4

Putting the values of A and B in eq. (ii),

106.  

Comparing coefficients of x, 2A = 1 ⇒ A = 

Comparing constants, 2A + B = 2

On solving, we get A = , B = 1

Putting the values of A and B in eq. (ii),

107.  

Comparing coefficients of x, 2A = 1 ⇒ A = 

Comparing constants, -2A + B = 3

On solving, we get ⇒ A = , B = 4

Putting the values of A and B in eq.(ii), x + 3 = (2x – 2) + 4

Putting this value of x + 3 in eq.(i),

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Comparing coefficients of x on both sides A + B = 1 …….(ii)

Comparing constants 2A + B = 0 …….(iii)

Solving eq. (ii) and (iii), we get A = -1 and B = 2

Putting these values of A and B in eq. (i),

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Comparing coefficients of x2 : A + B + C = 0 ….(ii)

Comparing coefficients of x : -5A – 4B – 3C = 3

⇒ 5A + 4B + 3C = -3    …(iii)

Comparing constants: 6A + 3B + 2C = –1 ……….(iv)

On solving eq. (i), (ii) and (iii), we get A = 1, B = –5, C = 4

Putting the values of A, B and C in eq. (i),

114.  

Comparing coefficients of x2 : A + B + C = 0    ….(ii)

Comparing coefficients of x : -5A – 4B – 3C = 1

⇒ 5A + 4B + 3C = -1       ….(iii)

Comparing constants: 6A + 3B + 2C = 0 ……….(iv)

On solving eq. (i), (ii) and (iii), we get 

Putting the values of A, B and C in eq. (i),

115. 

Comparing coefficients of x on both sides A + B = 2 ……….(ii)

Comparing constants 2A + B = 0 …….(iii)

Solving eq. (ii) and (iii), we get A = -2 and B = 4

Putting these values of A and B in eq. (i),

116. 

Comparing coefficients of on both sides 

Comparing constants A = 1 ……….(iv)

Solving eq. (ii) and (iii), we get

Putting these values of A and B in eq. (ii),

117. 

Comparing coefficients of A + C = 0 ……….(ii)

Comparing coefficients of , –A + B = 1 ……….(iii)

Comparing constant terms, –B + C = 0 ……….(iv)

Solving eq. (ii), (iii) and (iv), we get 

Putting the values of A, B and C in eq.(i),

118. 

Comparing coefficients of x2 : A + C = 0 ……….(ii)

Comparing coefficients of x : A + B – 2C= 1 ……….(iii)

Comparing constants: –2A + 2B + C = 0 ……….(iv)

On solving eq. (i), (ii) and (iii), we get 

Putting the values of A, B and C in eq. (i), 

119. 

Comparing coefficients of x2 : A + C = 0 ……….(ii)

Comparing coefficients of x : B – 2C= 3 ……….(iii)

Comparing constants: –2A + B + C = 5 ……….(iv)

On solving eq. (i), (ii) and (iii), we get 

Putting the values of A, B and C in eq. (i),

120. 

Comparing coefficients of x2 : 2A + 2B + C = 0 ……….(ii)

Comparing coefficients of x : 5A + B = 2 ……….(iii)

Comparing constants: 3A – 3B – C = –3 ……….(iv)

On solving eq. (i), (ii) and (iii), we get 

Putting the values of A, B and C in eq. (i),

121. 

Comparing coefficients of x2 : A + +B + C = 0 ……….(ii)

Comparing coefficients of x : –B + 3C = 5 ……….(iii)

Comparing constants: –4A – 2B + 2C = 0 ……….(iv)

On solving eq. (i), (ii) and (iii), we get 

Putting the values of A, B and C in eq. (i),

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Comparing the coefficients of x2 A – B = 0 ……….(ii)

Comparing the coefficients of x B – C = 0 ……….(iii)

Comparing constants A + C = 2 ……….(iv)

On solving eq. (ii), (iii) and (iv), we get A = 1, B = 1, C = 1

Putting these values of A, B and C in eq. (i),

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Comparing the coefficients of y A + B = 0 ………(ii)

Comparing constants A – B = 1 ……….(iii)

On solving the eq. (ii) and (iii), we get 

Putting the values of A, B and in eq. (i),

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Comparing coefficients of y A + B = –4 ……….(v)

Comparing constants 4A + 3B = –10 ……….(vi)

On solving eq. (v) and (vi), we get A = 2, B = –6

Putting the values of A, B and y in eq. (iii),

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Comparing coefficients of x A + B = 1 ……….(ii)

Comparing constants –2A – B = 0 ……….(iii)

On solving eq. (ii) and (iii), we get A = –1, B = 2

Putting these values of A and B in eq. (i),

Therefore, option (B) is correct.

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To change the limits of integration from x to t

when x = 0, t = x2 + 1 = 0 + 1 = 1

when x = 1, t = x2 + 1 = 1 + 1 = 2

∴ From eq. (i),

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Comparing the coefficients of x2 –A + B – C = 0 ……….(ii)

Comparing the coefficients of x B + C = 0 ……….(iii)

Comparing constants A = 1 ……….(iv)

On solving eq. (ii), (iii) and (iv), we get 

Putting these values in eq. (i),

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Comparing coefficients of x2 A + B = 0 ……….(iii)

Comparing coefficients of x B + C = 5 ……….(iv)

Comparing constants 9A + C = 0 ……….(v)

On solving eq. (iii), (iv) and (v), we get 

Putting these values of A, B and C in eq. (ii),

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Comparing coefficients of x2 A + C = 1 ……….(iii)

Comparing coefficients of x 3A + B + 2C = 1 ……….(iv)

Comparing constants 2A + 2B + C = 1 ……….(v)

On solving eq. (iii), (iv) and (v), we get A = –2, B = 1, C = 3

Putting these values of A, B and C in eq. (ii),

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Putting sin x – cos x = t

⇒ (cos x + sin x) dx = dt

Again (sin x – cos x)2 = t2

⇒ sin2 x + cos2 x – 2 sin x cos x = t2

⇒ sin 2x = 1 – t2

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Putting sin x – cos x = t

⇒ (cos x + sin x) dx = dt

Again (sin x – cos x)2 = t2

⇒ sin2 x + cos2 x – 2 sin x cos x = t2

⇒ sin 2x = 1 – t2

Limits of integration when x = 0, t = 0 – 1 = -1 and

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261. Taking RHS

Comparing coefficients of x A + C + 0 ……….(iii)

Comparing coefficients of x A + B = 0 ……….(iv)

Comparing constants B = 1

On solving eq. (iii), (iv) and (v), we get A = -1, B = 1, C = 1

Putting these values of A, B and C in eq. (ii),

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