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: $fog\left(x\right)=\{\begin{array}{c}\begin{array}{cc}-1,& x\in (-1,0)\cup (1,\infty )\end{array}\\ \begin{array}{cc}0,& \end{array}x\in \{-1,0,1\}\\ \begin{array}{cc}1,& x\in (-\infty ,-1)\cup (0,1)\end{array}\end{array}$ 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 but the Reason is correct.

Answer 1

The correct option is D Assertion is incorrect but the Reason is correct.

We have,

$f\left(x\right)=sgn\left(x\right)=-1,x<0$

$=0,x=0$

$=1,x>0$

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

Now, $fog\left(x\right)=-1,x(1-x^2)<0$

$=0,x(1-x^2)=0$

$=1,x(1-x^2)>0$

Solving the inequality

$x(1-x^2)<0$

i.e., $x(x-1)(x+1)>0$ i.e., $x\in (-1,0)\cup (1,\infty )$

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, $gof\left(x\right)=f(1-f^2)=-1[1-{(-1)}^2],x<0$

$=0(1-0^2),x=0$

$=1(1-1^2),x>0$

i.e., $gof\left(x\right)=0,\forall x\in R$

Which is continuous everywhere.