Question

Let $f: R \rightarrow R, g: R \rightarrow R$ be differentiable

functions such that $($ fog $)(x)=x$. If $f(x)=2 x+\cos x+\sin ^{2} x$, then the value of

$\displaystyle \sum_{n=1}^{99} g(1+(2 n-1) \pi)$ is

• A. $1250 \pi$
• B. $(99)^{2} \frac{\pi}{2}$
• C. $(99)^{2} \pi$
• D. $2500 \pi$

Answer 1

Answer: (B)

$ \because \operatorname{fog}(x)=x$ and $f(x)=2 x+\cos x+\sin ^{2} x$
$g(x)$ is inverse function of $f(x)$.
So, $1+(2 n-1) \pi=2 x+\cos x+\sin ^{2} x$
$\Rightarrow 1+(2 n-1) \pi=2 x+\cos x+1-\cos ^{2} x$
$\Rightarrow 2 x+\cos x-\cos ^{2} x=(2 n-1) \pi\,...(i)$
For $ n =1, x=\frac{\pi}{2} $
$ n =2, x=\frac{3 \pi}{2} $
$ n=3, x =\frac{7 \pi}{2} $
$ \vdots \, \vdots $
$ n=n x =\frac{2 n-1}{2} \pi $
So, $ \displaystyle \sum_{n}^{99} g(1+(2 n-1) \pi) $
$=\frac{\pi}{2}+\frac{3 \pi}{2}+\frac{7 \pi}{2}+\ldots+$ upto $ 99$ terms
$=\frac{\pi}{2}[1+3+7+\ldots+$ upto $99 $ terms
$=(99)^{2} \frac{\pi}{2} $