Question

A function whose graph is symmetrical about the $y-$axis is given by

• A. $f\left(x\right)={\log }_e(x+\sqrt{x^2+1})$
• B. $f(x+y)=f\left(x\right)+f\left(y\right)\forall x,y\in R$
• C. $f\left(x\right)=\cos x+\sin x$
• D. None of these

Answer 1

Answer: (D)

None of these

A function which is even in nature has a graph symmetric about $y$ axis.

Thus the function has to satisfy

$f\left(x\right)=f(-x)$.
Consider option A

$f\left(x\right)=\ln (x+\sqrt{x^2+1})$

Therefore

$f(-x)=\ln (-x+\sqrt{x^2+1})$
$f\left(x\right)+f(-x)$
$=\log (\sqrt{x^2+1}+x)+\log (\sqrt{x^2+1}-x)$
$=\log \left(\right[\sqrt{x^2+1}+x]\times [\sqrt{x^2+1}-x\left]\right)$
$=\log (x^2+1-x^2)$
$=\log 1$

$=0$

Hence, $f\left(x\right)+f(-x)=0$.

This is an odd function and hence it is not symmetric about $y$ axis.
Consider option B

$f(x+y)=f\left(x\right)+f\left(y\right)$ is a linear function where $f\left(x\right)=\lambda x$.

Since $f\left(x\right)\ne f(-x)$ it is not symmetric about $y$ axis.
Consider option C

$f\left(x\right)=\sin x+\cos x$
$f(-x)=-\sin x+\cos x$
$f\left(x\right)+f(-x)$
$=2\cos x$
Hence

$f\left(x\right)\ne f(-x)$

Hence this too is not symmetric about $y$ axis.
Hence the answer is none of these.