Question

Graph of f(x) = ${\log }_a(x+\sqrt{x^2+1})(a>0,\ne 1)$ is symmetric about

• A. x-axis
• B. origin
• C. y-axis
• D. y = x

Answer 1

The correct option is A x-axis

$f\left(x\right)={\log }_a(x+\sqrt{x^2+1})(-a>0,\ne 1)$

Checking symmetry about X-axis,

Let $y={\log }_a(x+\sqrt{x^2+1})$

$\Rightarrow x+\sqrt{x^2+1}=e^y$

$\Rightarrow x^2+1=e^{2y}+x^2-e^{y}x$

$\Rightarrow 1=e^{2y}-e^{y}x$

$\Rightarrow x=\frac{e^{2y}-1}{e^y}$

Now $f\left(f\right)=\frac{e^{-2y}-1}{e^{-y}}$

If $f\left(y\right)$ is symmetri about $x,-f\left(y\right)=f(-y)$

$\Rightarrow f(-y)=\frac{e^{-2y}-1}{e^{-y}}$

$=\frac{1-e^{2y}}{e^y}$

$=\frac{-(e^{2y}-i)}{e^y}$

$=-f\left(y\right)$

So, $f\left(x\right)$ is symmetric about X-axis.