Question

Let $f:R\to [2,\infty )$ be given by $f\left(x\right)=e^{e^x}+e^{-e^x}$. Then $f$ is

• A. one-one
• B. onto
• C. even
• D. bijective

Answer 1

The correct option is A one-one

$f\left(x\right)=e^{e^x}+e^{-e^x}\Rightarrow f^{\prime }\left(x\right)=e^x(e^{e^x}-e^{-e^x})\Rightarrow f^{\prime }\left(x\right)=\frac{e^x}{e^{e^x}}\left(\right(e^{e^x}{)}^2-1)$

Now, as $x\in R$
$\Rightarrow e^x\in (0,\infty )\Rightarrow e^{e^x}\in (1,\infty )\Rightarrow f^{\prime }\left(x\right)>0$
So, $f$ is strictly increasing function
$\Rightarrow f$ is one-one.

By AM-GM inequality,
$\frac{e^{e^x}+e^{-e^x}}{2}\ge \sqrt{e^{e^x}\cdot e^{-e^x}}$
$\Rightarrow e^{e^x}+e^{-e^x}\ge 2$
But $e^{e^x}+e^{-e^x}\ne 2$ for any finite value of $x$.
$\therefore$ Range $=(2,\infty )$
Range $\ne$ Co-domain $\Rightarrow f$ is into.
Clearly, $f$ is not even.