Question

Consider $f\left(x\right)=\{\begin{array}{cc}x{\left[x\right]}^2{\log }_{(1+x)}2& \text{for}-1<x<0\\ \frac{\ln e^{x^2}+2\sqrt{\left\{x\right\}}}{\tan \sqrt{x}}& \text{for}0<x<1\end{array}$ where [*] & {*} are the greatest integer function & fractional part function respectively, then -

• A. $f\left(0\right)=ln2\Rightarrow fiscontinuousatx=0$
• B. $f\left(0\right)=2\Rightarrow fiscontinuousatx=0$
• C. $f\left(0\right)=e^2\Rightarrow fiscontinuousatx=0$
• D. f has an irremovable discontinuity at $x=0$

Answer 1

The correct option is D f has an irremovable discontinuity at $x=0$

LHL

$=\underset{h\to 0}{lim}(0-h){[0-h]}^2{\log }_{(1+0-h)}2$

$=\underset{h\to 0}{lim}\frac{-h(-1{)}^2\ln 2}{\ln (1-h)}=\ln 2$

RHL
$=\underset{h\to 0}{lim}\frac{\ln \left(e^{(0+h{)}^2}\right)+2\sqrt{\{0+h\}}}{\tan \sqrt{(0+h)}}$

$=\underset{h\to 0}{lim}\frac{\ln (e^{h^2}+2\sqrt{h})}{\tan \sqrt{\left(h\right)}}=\underset{h\to 0}{lim}\frac{\ln (e^{h^2}+2\sqrt{h})}{\sqrt{\left(h\right)}}$

$=\underset{h\to 0}{lim}[h^2+\frac{\ln (1+\frac{2\sqrt{h}}{e^{h^2}}).\frac{2\sqrt{h}}{e^{h^2}}}{\frac{2\sqrt{h}}{e^{h^2}}}]\frac{1}{\sqrt{h}}=2$

$\therefore RHL\ne LHL\Rightarrow$ f has an irremovable discontinuity.