Question

Let $f,g$ and $h$ be three functions defined as
$f\left(x\right)=\{\begin{array}{cc}x& \text{for}\left|x\right|\le 1\\ 1& \text{for}\left|x\right|>1\end{array},$

$g\left(x\right)=\{\begin{array}{cc}\cos \left(\frac{\pi }{2}x\right)& \text{for}\left|x\right|\le 1\\ |x-1|& \text{for}\left|x\right|>1\end{array}$

and $h\left(x\right)=\{\begin{array}{cc}\frac{\left|x\right|-1}{{\log }_a\left|x\right|}& \text{for}\left|x\right|\ne 1\\ \ln a& \text{for}\left|x\right|=1\end{array}$ for $a>0$ and $a\ne 1$

If $l,m,n$ denote the number of points of discontinuity of the functions $f,g$ and $h$ in their domain respectively, over $R$, then $(l,m,n)$ is

• A. $(0,0,0)$
• B. $(1,1,1)$
• C. $(2,2,2)$
• D. $(1,1,0)$

Answer 1

The correct option is D $(1,1,0)$

$f\left(x\right)=\{\begin{array}{cc}x& \text{for}-1\le x\le 1\\ 1& \text{for}x>1\text{or}x<-1\end{array}$

Clearly, $f\left(x\right)$ is discontinuous at $x=-1$

$g\left(x\right)=\{\begin{array}{cc}\cos \frac{\pi x}{2}\text{for}-1\le x\le 1& \\ x-1\text{for}x>1& \\ 1-x\text{for}x<-1& \end{array}$

Clearly, $g\left(x\right)$ is discontinuous at $x=-1$

$h\left(x\right)=\{\begin{array}{cc}\frac{x-1}{{\log }_{a}x}& \text{for}x>0,x\ne 1\\ \frac{-x-1}{{\log }_a(-x)}& \text{for}x<0,x\ne -1\\ \ln a& \text{for}x=-1\text{or}x=1\end{array}$

$x=0$ is not in the domain of $h\left(x\right)$
$\underset{x\to 1^{\pm }}{lim}h\left(x\right)=\left(\frac{x-1}{{\log }_{a}x}\right)$ $\frac{0}{0}$ form
Applying L'Hospital rule
$\underset{x\to 1^{\pm }}{lim}h\left(x\right)=\underset{x\to 1^{\pm }}{lim}x\ln a=\ln a=h\left(1\right)$
$\therefore h\left(x\right)$ is continuous at $x=1$

$\underset{x\to -1^{\pm }}{lim}h\left(x\right)=\left(\frac{-x-1}{{\log }_a(-x)}\right)$ $\frac{0}{0}$ form
Applying L'Hospital rule
$\underset{x\to -1^{\pm }}{lim}h\left(x\right)=\underset{x\to -1^{\pm }}{lim}(-x\ln a)=\ln a=h(-1)$
$\therefore h\left(x\right)$ is continuous at $x=-1$
Clearly, $h\left(x\right)$ is continuous for all $x$ in its domain.
Hence, there is no point of discontinuity.

Now, $(l,m,n)=(1,1,0)$