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JEE Main 2019 · 8 Apr · Shift II

JEE Main 2019 Mathematics question on Definite Integrals and Areas

If f(x)=2−xcos⁡x2+xcos⁡xf(x) = \dfrac{2 - x\cos x}{2 + x\cos x} and g(x)=log⁡exg(x) = \log_e x, (x>0)(x > 0) then the value of the integral ∫−π/4π/4g(f(x)) dx\int_{-\pi/4}^{\pi/4} g(f(x))\,dx is
  1. A.
    log⁡e3\log_e 3
  2. B.
    log⁡ee\log_e e
  3. C.
    log⁡e2\log_e 2
  4. D.
    log⁡e1\log_e 1
Show answer and solution

Correct option: D

Solution

g(f(x)) = ln((2 - x·cosx)/(2 + x·cosx)). Notice that replacing x with -x gives (2 + x·cosx)/(2 - x·cosx), so g(f(-x)) = -g(f(x)), meaning the integrand is an odd function. The integral of any odd function over a symmetric interval [-a, a] is 0 = ln 1, so the answer is loge 1.

[JEE Main 2019 · 8 Apr · Shift II]

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