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


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

The correct option is D Assertion is incorrect and Reason is correct
We have, $$\displaystyle f\left( x \right)=sgn\left( x \right) =\begin{cases} \begin{matrix} -1, & x<0 \end{matrix} \\ \begin{matrix} 0, & x=0 \end{matrix} \\ \begin{matrix} 1, & x>0 \end{matrix} \end{cases}$$
and $$g\left( x \right)=x\left( 1-{ x }^{ 2 } \right) $$
Now, $$\displaystyle fog\left( x \right) =\begin{cases} \begin{matrix} -1, & x\left( 1-{ x }^{ 2 } \right) <0 \end{matrix} \\ \begin{matrix} 0, & x\left( 1-{ x }^{ 2 } \right) =0 \\ 1, & x\left( 1-{ x }^{ 2 } \right) >0 \end{matrix} \end{cases}$$
Solving the inequality 
$$x\left( 1-{ x }^{ 2 } \right) <\Rightarrow x\left( x-1 \right) \left( x+1 \right) >0\Rightarrow x\in \left( -1,0 \right) \cup \left( 1,\infty  \right) $$
Thus, we have
$$\displaystyle fog\left( x \right) =\begin{cases} \begin{matrix} -1, & x\in \left( -1,0 \right) \cup \left( 1,\infty  \right)  \end{matrix} \\ \begin{matrix} 0, & x\in \left\{ -1,0,1 \right\}  \end{matrix} \\ \begin{matrix} 1, & x\in \left( -\infty ,-1 \right) \cup \left( 1,0 \right)  \end{matrix} \end{cases}$$
which is continuous everywhere except at $$x\in \left\{ -1,0,1 \right\} $$
Also, $$\displaystyle gof\left( x \right) =f\left( 1-{ f }^{ 2 } \right) =\begin{cases} \begin{matrix} -1\left[ 1-{ \left( -1 \right)  }^{ 2 } \right] , & x<0 \end{matrix} \\ \begin{matrix} 0\left( 1-{ 0 }^{ 2 } \right) , & x=0 \end{matrix} \\ \begin{matrix} 1\left( 1-{ 1 }^{ 2 } \right) , & x>0 \end{matrix} \end{cases}$$
$$\Rightarrow gof\left( x \right) =0,\quad \forall \quad x\in R$$ which is continuous everywhere.

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