Question

Consider the function $f\left(x\right)=\text{sgn}(x-1)$ and $g\left(x\right)={\cot }^{-1}[x-1],$ where $[.]$ denotes the greatest integer function.

Statement $1:$ The function $F\left(x\right)=f\left(x\right)\cdot g\left(x\right)$ is discontinuous at $x=1.$
Statement $2:$ If $f\left(x\right)$ is discontinuous at $x=a$ and $g\left(x\right)$ is also discontinuous at $x=a,$ then the product function $f\left(x\right)\cdot g\left(x\right)$ is discontinuous at $x=a.$

• A. Both the statements are true and Statement $2$ is the correct explanation of Statement $1.$
• B. Both the statements are true and Statement $2$ is not the correct explanation of Statement $1.$
• C. Statement $1$ is true and Statement $2$ is false.
• D. Statement $1$ is false and Statement $2$ is true.

Answer 1

The correct option is C Statement $1$ is true and Statement $2$ is false.

$f\left(x\right)=\text{sgn}(x-1);g\left(x\right)={\cot }^{-1}[x-1]\therefore F\left(x\right)=f\left(x\right)\cdot g\left(x\right)=\{\begin{array}{c}-{\cot }^{-1}[x-1],x\to 1^{-}\\ 0,x=1\\ {\cot }^{-1}[x-1],x\to 1^{+}\end{array}=\{\begin{array}{c}-\frac{3\pi }{4},x\to 1^{-}\\ 0,x=1\\ \frac{\pi }{2},x\to 1^{+}\end{array}\Rightarrow F\left(1^{-}\right)=-\frac{3\pi }{4};F\left(1^{+}\right)=\frac{\pi }{2}$ and $F\left(1\right)=0$

$\therefore F\left(x\right)$ is discontinuous at $x=1$
So, statement $1$ is true but product of two discontinuous functions may be a continuous function.
So, statement $2$ is false.