Question

Let $g\left(x\right)=\{\begin{array}{cc}2(x+1),& -\infty <x\le -1\\ \sqrt{1-x^2},& -1<x<1\\ |x+1|,& 1\le x<\infty \end{array}$ then

• A. $g\left(x\right)$ is discontinuous at exactly three points
• B. $g\left(x\right)$ is continuous in $(-\infty ,1]$
• C. $g\left(x\right)$ is continuous in $[1,\infty )$
• D. $g\left(x\right)$ has finite type of discontinuity at $x=1$, but continuous at $x=-1$

Answer 1

The correct option is D $g\left(x\right)$ has finite type of discontinuity at $x=1$, but continuous at $x=-1$

$g\left(x\right)=\left\{\begin{array}{c}2(x+1),-\infty <x\le -1\\ \sqrt{1-x^2},-1<x<1then\\ \mid x+1\mid ,1\le \infty \end{array}\right\}$

At $x=-1$

L.H.L(Left Hand Limit)

$\underset{x\to -1^{-}}{lim}f\left(x\right)=2(-1+1)=0$

R.H.L(Right Hand Limit)

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\sqrt{1-(-1{)}^2}=0$

$\therefore$ L.H.L=R.H.L(at $X=0$)

At $x=1$

L.H.L=$\underset{x\to 1^{+}}{lim}f\left(x\right)=\sqrt{1-1^2}=0$

R.H.L=$\underset{x\to 1^{+}}{lim}f\left(x\right)=\mid 1+1\mid =2$

$\therefore$ L.H.L$\ne$ R.H.L(at $x=2$)

But it is finite.