Question

Let $\left[x\right]$ denote the largest integer not exceeding $x$ and $\left\{x\right\}=x-\left[x\right]$. Then
${\int }_0^{2012}\frac{e^{\cos \left(\pi \left\{x\right\}\right)}}{e^{\cos \left(\pi \left\{x\right\}\right)}+e^{-\cos \left(\pi \left\{x\right\}\right)}}dx$ is equal to.

• A. $0$
• B. $1006$
• C. $2012$
• D. $2012\pi$

Answer 1

The correct option is B $1006$

$I={\int }_0^{2012}\frac{e^{cos\left(\pi \left\{x\right\}\right)}}{e^{cos\left(\pi \left\{x\right\}\right)}+e^{-cos\left(\pi \left\{x\right\}\right)}}$

$\left\{x\right\}$ is periodic with period $T=1$

Using the property ${\int }_0^{nT}f\left(x\right)dx=n{\int }_0^{T}f\left(x\right)dx$

$I=2012{\int }_0^1\frac{e^{cos\left(\pi \left\{x\right\}\right)}}{e^{cos\left(\pi \left\{x\right\}\right)}+e^{-cos\left(\pi \left\{x\right\}\right)}}dx$

For $x\in (0,1)\left\{x\right\}=x$

$\therefore I=2012{\int }_0^1\frac{e^{cos\left(\pi x\right)}}{e^{cos\left(\pi x\right)}+e^{-cos\left(\pi x\right)}}dx.......\left(i\right)$

$I=2012{\int }_0^1\frac{e^{cos\left(\pi \right(1-x))}}{e^{cos\left(\pi \right(1-x))}+e^{-cos\left(\pi \right(1-x))}}dx$ ........ [Using $a+b-x$ property of integral]

$I=2012{\int }_0^1\frac{e^{-cos\left(\pi x\right)}}{e^{cos\left(\pi x\right)}+e^{-cos\left(\pi x\right)}}dx...\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$

$2I=2012{\int }_0^{1}1dxI=1006\left(1\right)=1006$

Hence, option $B$ is correct.