Mathematics · 2019
JEE Main · 8 April 2019, Shift 1 · Q75
If f(x)=(2-x cos x)/(2+x cos x) and g(x)= log_e x, (x>0) then the value of the integral ∫_-π / 4^π / 4 g(f(x)) d x is:
If $\displaystyle f(x)=\frac{2-x \cos x}{2+x \cos x}$ and $\displaystyle g(x)=\log _{\mathrm{e}} x$, $\displaystyle (x>0)$ then the value of the integral
$$\int_{-\pi / 4}^{\pi / 4} g(f(x)) \mathrm{d} x \text { is : }
$$
Official answer
(3)
$\displaystyle \log _{\mathrm{e}} 1$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.