Question

Assertion :If $f\left(x\right)=sgn\left(x\right)$ and $g\left(x\right)=x(1-x^2),$ then $fog\left(x\right)$ and $gof\left(x\right)$ are continuous everywhere Reason: $fog=\{\begin{array}{c}\begin{array}{cc}-1,& x\in (-1,0)\cup (1,\infty )\end{array}\\ \begin{array}{cc}0,& x\in \{-1,0,1\}\end{array}\\ \begin{array}{cc}1,& x\in (-\infty ,-1)\cup (0,1)\end{array}\end{array}$ and $gof\left(x\right)=0,\forall x\in R$

• 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. Assertion is incorrect and Reason is correct

Answer 1

The correct option is D Assertion is incorrect and Reason is correct

We have, $f\left(x\right)=sgn\left(x\right)=\{\begin{array}{c}\begin{array}{cc}-1,& x<0\end{array}\\ \begin{array}{cc}0,& x=0\end{array}\\ \begin{array}{cc}1,& x>0\end{array}\end{array}$

and $g\left(x\right)=x(1-x^2)$

Now, $fog\left(x\right)=\{\begin{array}{c}\begin{array}{cc}-1,& x(1-x^2)<0\end{array}\\ \begin{array}{cc}0,& x(1-x^2)=0\\ 1,& x(1-x^2)>0\end{array}\end{array}$

Solving the inequality

$x(1-x^2)<\Rightarrow x(x-1)(x+1)>0\Rightarrow x\in (-1,0)\cup (1,\infty )$

Thus, we have

$fog\left(x\right)=\{\begin{array}{c}\begin{array}{cc}-1,& x\in (-1,0)\cup (1,\infty )\end{array}\\ \begin{array}{cc}0,& x\in \{-1,0,1\}\end{array}\\ \begin{array}{cc}1,& x\in (-\infty ,-1)\cup (1,0)\end{array}\end{array}$

which is continuous everywhere except at $x\in \{-1,0,1\}$

Also, $gof\left(x\right)=f(1-f^2)=\{\begin{array}{c}\begin{array}{cc}-1[1-{(-1)}^2],& x<0\end{array}\\ \begin{array}{cc}0(1-0^2),& x=0\end{array}\\ \begin{array}{cc}1(1-1^2),& x>0\end{array}\end{array}$

$\Rightarrow gof\left(x\right)=0,\forall x\in R$ which is continuous everywhere.