Question

Jee revision Question

Answer 1

Topic :-

Definite Integration

Given :-

$f(x)=\dfrac{2-x\cos x}{2+x\cos x}$

$g(x)=log_ex,\:\:(x>0)$

To Find :-

$\displaystyle \int_{-\pi/4}^{\pi/4}g(f(x))\:dx$

Concept Used :-

$\displaystyle \int_{-a}^{a}f(x)dx=\displaystyle \int_{0}^{a}[f(x)+f(-x)]dx=\left\{\begin{matrix}0;\:\:if\:f(x)\:is\:an\:odd\:function\\2\displaystyle \int_{0}^{a}f(x)\:dx;\:\:if\:f(x)\:is\:an\:even\:function\end{matrix}\right.$

Solution :-

Calculate g(f(x)),

$g(f(x)) = g\left (\dfrac{2-x\cos x}{2+x\cos x}\right )$

$g\left (\dfrac{2-x\cos x}{2+x\cos x}\right )=log_e\left(\dfrac{2-x\cos x}{2+x\cos x}\right )$

Check for 'Odd' or 'Even' function,

$g(f(x))=log_e\left(\dfrac{2-x\cos x}{2+x\cos x}\right )$

Replace 'x' with '-x',

$g(f(-x))=log_e\left(\dfrac{2-(-x)\cos (-x)}{2+(-x)\cos (-x)}\right )$

$\because \cos(-x)=\cos x$

$g(f(-x))=log_e\left(\dfrac{2+x\cos x}{2-x\cos x}\right )\:or$

$g(f(-x))=log_e\left(\dfrac{2-x\cos x}{2+x\cos x}\right )^{-1}$

$\because log_ex^a=a.log_ex$

$g(f(-x))=-log_e\left(\dfrac{2-x\cos x}{2+x\cos x}\right )$

We can clearly observe that g(f(-x)) = -g(f(x)), which means g(f(x)) is an odd function.

Solving the integral,

$\displaystyle \int_{-\pi/4}^{\pi/4}g(f(x))\:dx$

$\displaystyle \int_{0}^{\pi/4}[g(f(x))+g(f(-x))]\:dx$

$\displaystyle \int_{0}^{\pi/4}[g(f(x))-g(f(x))]\:dx$

$\displaystyle \int_{0}^{\pi/4}0\:dx$

0

$0=log_e1$

Note : We could have directly mentioned that value of integral is Zero as g(f(x)) is an odd function.

Answer :-

$So,\:the\:value\:of\:integral\:is\:\bold{log_e1}\:which\:is\:\bold{option\:4}.$