Question

On the interval $I=[-2,2]$, the function $f\left(x\right)=(x+1)\exp \left[\right(-1\left)(\frac{1}{|x|}+\frac{1}{x})\right]$ for $(x\ne 0)$,

then which one of the following does not hold good for $f\left(x\right)$?

• A. is continuous for the values of $xϵI$
• B. is continuous for $x\in I-\left\{0\right\}$
• C. assumes all intermediate values from $f(-2)$ & $f\left(2\right)$
• D. has maximum value equal to $\frac{3}{e}$

Answer 1

The correct option is A is continuous for the values of $xϵI$

Given $f\left(x\right)=(x+1)e^{-[\frac{1}{|x|}+\frac{1}{x}]},x\ne 0$

$f\left(x\right)=(x+1)e^{-[\frac{1}{-x}+\frac{1}{x}]}=x+1,-2<x<0$

$f\left(x\right)=(x+1)e^{-\frac{2}{x}},0\le x<2$

We will check the continuity at $x=0$

$RHL={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}(x+1)e^{-2/x}=0$

$LHL={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}(x+1)=1$

$f\left(x\right)$ is not continuous at $x=0$.

Clearly, $f\left(x\right)$ is defined for all $x\in [-2,2]$

Now,

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

For $x>0,\frac{2x+2}{x^2}$

$\Rightarrow f^{\prime }\left(x\right)>0$

So, $f\left(x\right)$ is monotonic increasing in the interval. Hence, maximum occurs at end points

$f(-2)=1-2=-1$

$f\left(2\right)=(1+2)e^{-1}=\frac{3}{e}$

Hence, maximum value is $\frac{3}{e}$