JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Definite Integrals

f (x) = \frac{2 - x c o s x}{2 + x c o s x}

and

g (x) = lo g_{e} x, (x > 0)

then the value of the integral

\int_{- π /4}^{π /4} g (f (x)) d x

is:

Answer: (A)

Solution

Check the symmetry of the integrand. Since

f (- x) = \frac{2 + x cos x}{2 - x cos x} = \frac{1}{f ( x )}

, taking the logarithm gives

g (f (- x)) = lo g_{e} \frac{1}{f ( x )} = - g (f (x))

, so

g (f (x))

is odd. The integral of an odd function over the symmetric interval

[- \frac{π}{4}, \frac{π}{4}]

vanishes, so the value is

0 = lo g_{e} 1

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