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JEE Main 2019 · 9 Jan · Shift I

JEE Main 2019 Mathematics question on Definite Integrals and Areas

If f(x)=∫5x8+7x6(x2+1+2x7)2 dx,  (x≥0)f(x) = \int \dfrac{5x^8 + 7x^6}{(x^2 + 1 + 2x^7)^2} \, dx, \; (x \geq 0), and f(0)=0f(0) = 0, then the value of f(1)f(1) is
  1. A.
    −12-\dfrac{1}{2}
  2. B.
    −14-\dfrac{1}{4}
  3. C.
    14\dfrac{1}{4}
  4. D.
    12\dfrac{1}{2}
Show answer and solution

Correct option: C

Solution

Divide numerator and denominator by x14. The denominator (x2 + 1 + 2x7)2 = x14(x-5 + x-7 + 2)2. The numerator 5x8 + 7x6 = x14(5x-6 + 7x-8).

So the integrand becomes (5x-6 + 7x-8) / (x-5 + x-7 + 2)2.

Notice that d/dx(x-5 + x-7 + 2) = -5x-6 - 7x-8, which is exactly the negative of the numerator.

Therefore the integral is -∫ d(t)/t2 where t = x-5 + x-7 + 2, giving f(x) = 1/(x-5 + x-7 + 2) + C = x7/(1 + x2 + 2x7) + C.

Applying f(0) = 0 gives C = 0, so f(1) = 1/(1 + 1 + 2) = 1/4.

[JEE Main 2019 · 9 Jan · Shift I]

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