Question

Let $f\left(x\right)=\frac{e^x+1}{e^x-1}$ and ${\int }_0^1\frac{e^x+1}{e^x-1}.$ $xdx=\lambda .$ Then ${\int }_{-1}^{1}tf\left(t\right)$ is equal to

• A. $0$
• B. $2\lambda$
• C. $\lambda$
• D. none of these

Answer 1

Answer: (B)

$2\lambda$

$f\left(x\right)=\frac{e^x+1}{e^x-1}$
$f(-x)=\frac{e^{-x}+1}{e^{-x}-1}=\frac{1+e^x}{1-e^x}$
$=-\frac{e^x+1}{e^x-1}$
$\Rightarrow f(-x)=-f\left(x\right)$
Hence, f(x) is an odd function.
Consider, ${\int }_{-1}^{1}tf\left(t\right)dt=2{\int }_0^{1}tf\left(t\right)dt(\because tf(t)$ is even)
$=2{\int }_0^{1}xf\left(x\right)dx$ (Integration is independent of change of variable)
$=2{\int }_0^{1}x\frac{e^x+1}{e^x-1}dx$
$=2\lambda (\because {\int }_0^1\frac{e^x+1}{e^x-1}.xdx=\lambda$ given)
$\Rightarrow {\int }_{-1}^{1}tf\left(t\right)dt=2\lambda$