Question

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

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

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

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

Answer 1

Answer: (B)

$\because \mathrm{fog}(x)=x$ and $f(x)=2x+\cos x+{\sin }^{2}x$
$g(x)$ is inverse function of $f(x)$.
So, $1+(2n-1)\pi =2x+\cos x+{\sin }^{2}x$
$\Rightarrow 1+(2n-1)\pi =2x+\cos x+1-{\cos }^{2}x$
$\Rightarrow 2x+\cos x-{\cos }^{2}x=(2n-1)\pi ...(i)$
For $n=1,x=\frac{\pi }{2}$
$n=2,x=\frac{3\pi }{2}$
$n=3,x=\frac{7\pi }{2}$
$⋮⋮$
$n=nx=\frac{2n-1}{2}\pi$
So, $\sum _n^{99}g(1+(2n-1)\pi )$
$=\frac{\pi }{2}+\frac{3\pi }{2}+\frac{7\pi }{2}+\dots +$ upto $99$ terms
$=\frac{\pi }{2}[1+3+7+\dots +$ upto $99$ terms
$=(99{)}^2\frac{\pi }{2}$