Question

Function whose jump (non-negative difference of LHL and RHL) of discontinuity is greater than or equal to one, is/are-

• A. $f\left(x\right)=\{\begin{array}{c}\frac{(e^{1/x}+1)}{(e^{1/x}-1)},x<0\\ \frac{(1-\cos x)}{x},x>0\end{array}$
• B. $g\left(x\right)=\{\begin{array}{c}\frac{x^{1/3}-1}{x^{1/2}-1};x>1\\ \frac{\ln x}{(x-1)};\frac{1}{2}<x<1\end{array}$
• C. $u\left(x\right)=\{\begin{array}{c}\frac{{\sin }^{-1}2x}{{\tan }^{-1}3x};x\in (0,\frac{1}{2}]\\ \frac{\left|\sin x\right|}{x};x<0\end{array}$
• D. $w\left(x\right)=\{\begin{array}{c}{\log }_3(x+2);x>2\\ {\log }_{1/2}(x^2+5);x<2\end{array}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $f\left(x\right)=\{\begin{array}{c}\frac{(e^{1/x}+1)}{(e^{1/x}-1)},x<0\\ \frac{(1-\cos x)}{x},x>0\end{array}$
C $u\left(x\right)=\{\begin{array}{c}\frac{{\sin }^{-1}2x}{{\tan }^{-1}3x};x\in (0,\frac{1}{2}]\\ \frac{\left|\sin x\right|}{x};x<0\end{array}$
D $w\left(x\right)=\{\begin{array}{c}{\log }_3(x+2);x>2\\ {\log }_{1/2}(x^2+5);x<2\end{array}$

We will check by options
Option A:
$f\left(x\right)=\{\begin{array}{c}\frac{(e^{1/x}+1)}{(e^{1/x}-1)},x<0\\ \frac{(1-\cos x)}{x},x>0\end{array}$
$LHL={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{h\to 0}f(0-h)$
$={lim}_{h\to 0}\frac{(e^{-1/h}+1)}{(e^{-1/h}-1)}$
$=\frac{0+1}{0-1}=-1$
$RHL={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{h\to 0}f(0+h)$
$={lim}_{h\to 0}\frac{(1-\cos h)}{h}$
$={lim}_{h\to 0}\frac{2{\sin }^{2}h/2}{h}$
$={lim}_{h\to 0}\frac{h}{4}\frac{2{\sin }^{2}h/2}{\frac{h}{2}\frac{h}{2}}$
$=0\times 1=0$
Difference between LHL and RHL $=0-(-1)=1$
Hence, jump of discontinuity is 1.
Option B:
$f\left(x\right)=\{\begin{array}{c}\frac{x^{1/3}-1}{x^{1/2}-1};x>1\\ \frac{\ln x}{(x-1)};\frac{1}{2}<x<1\end{array}$
$RHL={lim}_{x\to 1^{+}}f\left(x\right)={lim}_{h\to 0}f(1+h)$
$={lim}_{h\to 0}\frac{{(1+h)}^{1/3}-1}{{(1+h)}^{1/2}-1}$
$={lim}_{h\to 0}\frac{(1+\frac{h}{3}-\frac{1}{9}{h}^2+....)-1}{(1+\frac{1}{2}h-\frac{1}{8}{h}^2.....)-1}$
$={lim}_{h\to 0}\frac{h(\frac{1}{3}-\frac{1}{9}h+....)}{h(\frac{1}{2}-\frac{1}{8}h.....)}$
$RHL=\frac{2}{3}$
$LHL={lim}_{x\to 1^{-}}f\left(x\right)={lim}_{h\to 0}f(1-h)$
$={lim}_{h\to 0}\frac{\ln (1-h)}{1-h-1}$
${lim}_{h\to 0}\frac{\ln (1+(-h\left)\right)}{-h}$
$=1$ ($\because {lim}_{x\to 0}\frac{\ln (1+x)}{x}=1$)
Difference between LHL and RHL $1-\frac{2}{3}=\frac{1}{3}<1$
Here, jump of discontinuity for f(x) is not 1 or greater than 1.
Option C:
$f\left(x\right)=\{\begin{array}{c}\frac{{\sin }^{-1}2x}{{\tan }^{-1}3x};x\in (0,\frac{1}{2}]\\ \frac{\left|\sin x\right|}{x};x<0\end{array}$
$LHL={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{h\to 0}f(0-h)$
$={lim}_{h\to 0}\frac{\sin |0-h|}{-h}$
$={lim}_{h\to 0}\frac{|-\sin h|}{-h}$
$={lim}_{h\to 0}\frac{\sin h}{-h}=-1$
$RHL={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{h\to 0}f(0+h)$
$={lim}_{h\to 0}\frac{{\sin }^{-1}2h}{{\tan }^{-1}3h}$
$={lim}_{h\to 0}\frac{2h\frac{{\sin }^{-1}2h}{2h}}{3h\frac{{\tan }^{-1}3h}{3h}}$
$RHL=\frac{2}{3}$
Difference between LHL and RHL $\frac{2}{3}-\left(1\right)=\frac{5}{3}>1$
Hence, jump of discontinuity for f(x) is greater than 1.
Option D:
$f\left(x\right)=\{\begin{array}{c}{\log }_3(x+2);x>2\\ {\log }_{1/2}(x^2+5);x<2\end{array}$
$RHL={lim}_{x\to 2^{+}}f\left(x\right)={lim}_{h\to 0}f(2+h)$
$={lim}_{h\to 0}{\log }_3(h+4)$
$RHL={\log }_{3}4=1.26$
$LHL={lim}_{x\to 2^{-}}f\left(x\right)={lim}_{h\to 0}f(2-h)$
$={lim}_{h\to 0}{\log }_{1/2}(h^2-4h+9)$
$LHL={\log }_{1/2}9=-0.653$
Difference between LHL and RHL $1.26-(-0.653)>1$
Hence, jump of discontinuity for f(x) is greater than 1.