JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 10)
If $$f(x) = {{2 - x\cos x} \over {2 + x\cos x}}$$ and g(x) = logex, (x > 0) then
the value of integral
$$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {g\left( {f\left( x \right)} \right)dx{\rm{ }}} $$ is
$$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {g\left( {f\left( x \right)} \right)dx{\rm{ }}} $$ is
loge3
loge2
loge1
logee
Explanation
$$g\left( {f\left( x \right)} \right)$$ = $$\ln \left( {f\left( x \right)} \right)$$ = $$\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)$$
$$ \therefore $$ I = $$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)dx} $$
( Using property $$\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left( {f\left( x \right) + f\left( { - x} \right)} \right)} dx$$ )
I = $$\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right) + \ln \left( {{{2 + x\cos x} \over {2 - x\cos x}}} \right)} \right)dx} $$
= $$\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( 1 \right)} \right)dx} $$
= 0 = loge1
$$ \therefore $$ I = $$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)dx} $$
( Using property $$\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left( {f\left( x \right) + f\left( { - x} \right)} \right)} dx$$ )
I = $$\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right) + \ln \left( {{{2 + x\cos x} \over {2 - x\cos x}}} \right)} \right)dx} $$
= $$\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( 1 \right)} \right)dx} $$
= 0 = loge1
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