Question

Consider the functions $f\left(x\right)=sin(x-1)$ and $g\left(x\right)={\cot }^{-1}[x-1]$

Assertion: The function $F\left(x\right)=f\left(x\right).g\left(x\right)$ is discontinuous at $x=1$

Reason: 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 $f\left(x\right).g\left(x\right)$ is discontinuous at $x=a$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

$f\left(x\right)=sin(x-1);g\left(x\right)={\cot }^{-1}[x-1]$

$\therefore F\left(x\right)=f\left(x\right).g\left(x\right)=\{\begin{array}{c}\begin{array}{cc}-{\cot }^{-1}[x-1];& x<1\end{array}\\ \begin{array}{cc}0;& x=1\end{array}\\ \begin{array}{cc}{\cot }^{-1}[x-1];& x>1\end{array}\end{array}$

$=\{\begin{array}{c}\begin{array}{cc}-{\cot }^{-1}(-1);& x<1\end{array}\\ \begin{array}{cc}0;& x=1\end{array}\\ \begin{array}{cc}{\cot }^{-1}\left(0\right);& x>1\end{array}\end{array}=\{\begin{array}{c}\begin{array}{cc}-(\pi -{\cot }^{-1}1);& x<1\end{array}\\ \begin{array}{cc}0;& x=1\end{array}\\ \begin{array}{cc}\pi /2;& x>1\end{array}\end{array}$

$=\{\begin{array}{c}\begin{array}{cc}-3\pi /4;& x<1\end{array}\\ \begin{array}{cc}0;& x=1\end{array}\\ \begin{array}{cc}\pi /2;& x>1\end{array}\end{array}$

$\Rightarrow F\left(1^{-}\right)=-\frac{3\pi }{4};F\left(1^{+}\right)=\frac{\pi }{2}$ and $F\left(1\right)=0$

$\Rightarrow f\left(x\right)$ is discontinuous at $x=1$

Therefore assertion is correct but product of two discontinuous functions may be a continuous function.

So reason is wrong.