If f(x) = ∫ (5 x ^ 8 + 7 x ^ 6 ) ( x ^ 2 + 1 + 2 x ^ 7 ) ^ 2 dx , (x ≥ 0), and f(0) = 0, then the value of f ( 1 ) is

Correct answer: C. (1)/(4) We have, f(x) = ∫ 5 x ^ 8 + 7 x ^ 6 ( x ^ 2 + 1 + 2 x ^ 7 ) ^ 2 dx = ∫ 5 x ^ 8 + 7 x ^ 6 x ^ 14 ( 1 x ^ 5 + 1 x ^ 7 + 2) ^ 2 dx = ∫ 5 x ^ 6 + 7 x ^ 8 ( 1 x ^ 5 + 1 x ^ 7 + 2) ^ 2 dx Put, 1 x ^ 5 + 1 x ^ 7 + 2 = t ⇒ (- 5 x ^ 6 - 7 x ^ 8 )dx = dt ∴ f(x) = - ∫ dt t ^ 2 = (1)/(t) + c , where c is the constant of integration. = x ^ 7 x ^ 2 + 1 + 2 x ^ 7 + c Now, f(0) = 0 ⇒ c = 0 ∴ f(1) = 1 1 ^ 2 + 2 + 2.1 ^ 7 = (1)/(4)

JEE Main 20199 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2019 · 9 January · Evening Shift · Mathematics · Indefinite Integration

If f(x) = ∫ (5 x ^ 8 + 7 x ^ 6 ) ( x ^ 2 + 1 + 2 x ^ 7 ) ^ 2 dx ,...

f\left(x\right) = ∫ \dfrac{\left(5{x}^{8} + 7{x}^{6}\right)}

{{\left({x}^{2} + 1 + 2{x}^{7}\right)}^{2}}dx

, \left(x \ge 0\right),

and

f\left(0\right) = 0,

then the value of

f ( 1 )

Options

A $\dfrac{- 1}{4}$
B $\dfrac{1}{2}$
C $\dfrac{1}{4}$
D $- \dfrac{1}{2}$

Correct answer

C. $\dfrac{1}{4}$

Step-by-step solution

We have,

f\left(x\right) = ∫ \dfrac{5{x}^{8} + 7{x}^{6}}{{\left({x}^{2} + 1 + 2{x}^{7}\right)}^{2}}dx

= ∫ \dfrac{5{x}^{8} + 7{x}^{6}}{{x}^{14}{\left(\dfrac{1}{{x}^{5}} + \dfrac{1}{{x}^{7}} + 2\right)}^{2}}dx

= ∫ \dfrac{\dfrac{5}{{x}^{6}} + \dfrac{7}{{x}^{8}}}{{\left(\dfrac{1}{{x}^{5}} + \dfrac{1}{{x}^{7}} + 2\right)}^{2}}dx

Put,

\dfrac{1}{{x}^{5}} + \dfrac{1}{{x}^{7}} + 2 = t

⇒ \left(- \dfrac{5}{{x}^{6}} - \dfrac{7}{{x}^{8}}\right)dx = dt

∴ f\left(x\right) = - ∫ \dfrac{dt}{{t}^{2}} = \dfrac{1}{t} + c

, where

c

is the constant of integration.

= \dfrac{{x}^{7}}{{x}^{2} + 1 + 2{x}^{7}} + c

Now,

f\left(0\right) = 0

⇒ c = 0

∴ f\left(1\right) = \dfrac{1}{{1}^{2} + 2 + {2.1}^{7}} = \dfrac{1}{4}

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