Question

$\text{List I}$ contains functions and $\text{List II}$ contains behaviours of functions in their respective domain. Each entry of $\text{List I}$ is to be matched with one or more than one entries of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& f\left(x\right)={\left(\right(\text{sgn}x{)}^{\text{sgn}x})}^n;x\ne 0& \text{(P)}& \text{odd function}\\ & \text{where}n\text{is an odd integer and}& & \\ & \text{sgn}x\text{denotes the signum function}& & \\ \text{(B)}& f\left(x\right)=\frac{x}{e^x-1}+\frac{x}{2}+1& \text{(Q)}& \text{even function}\\ \text{(C)}& f\left(x\right)=\{\begin{array}{cc}0,& \text{if}x\text{is rational}\\ 1,& \text{if}x\text{is irrational}\end{array}& \text{(R)}& \text{neither odd nor}\\ & & & \text{even function}\\ \text{(D)}& f\left(x\right)=max\{\tan x,\cot x\}& \text{(S)}& \text{periodic}\end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(A)}\to \text{(P)},\text{(S)}$
• B. $\text{(B)}\to \text{(Q)}$
• C. $\text{(A)}\to \text{(Q)},\text{(S)}$
• D. $\text{(B)}\to \text{(R)}$

Answer 1

The correct option is B $\text{(B)}\to \text{(Q)}$

$\text{(A)}$
$f\left(x\right)={\left(\right(\text{sgn}x{)}^{\text{sgn}x})}^n=\{\begin{array}{cc}\left[\right(1{)}^1{]}^n,& x>0\\ \left[\right(-1{)}^{-1}{]}^n,& x<0\end{array}=\{\begin{array}{cc}1,& x>0\\ -1,& x<0\end{array}$
Hence, $f\left(x\right)$ is an odd function.

$\text{(B)}$
$f\left(x\right)=\frac{x}{e^x-1}+\frac{x}{2}+1f(-x)=\frac{-x}{e^{-x}-1}-\frac{x}{2}+1=\frac{x{e}^x}{e^x-1}-\frac{x}{2}+1=\frac{x{e}^x-x+x}{e^x-1}-\frac{x}{2}+1=x+\frac{x}{e^x-1}-\frac{x}{2}+1=\frac{x}{e^x-1}+\frac{x}{2}+1=f\left(x\right)$
Hence, $f\left(x\right)$ is an even function.