Question

$f:R\to R,f\left(x\right)=x^2,g:R\to R,g\left(x\right)=2^x$, then $x|\left(fog\right)\left(x\right)=\left(gof\right)\left)\right(x)=.........$

Answer 1

Given that ,

$f\left(x\right)=x^{2}andg\left(x\right)=2^x$

$\Rightarrow \left(fog\right)\left(x\right)=f\left(g\right(x\left)\right)$

$\Rightarrow \left(fog\right)\left(x\right)=f\left(2^x\right)$

$\Rightarrow \left(fog\right)\left(x\right)=f\left(2^x\right)$

$\Rightarrow \left(fog\right)\left(x\right)=(2^x{)}^2=2^{2x}$

Therefore, $\Rightarrow \left(fog\right)\left(x\right)=2^{2x}$. ....$\left(1\right)$

$\Rightarrow \left(gof\right)\left(x\right)=g\left(f\right(x\left)\right)$

$\Rightarrow \left(gof\right)\left(x\right)=g\left(x^2\right)$

$\Rightarrow \left(gof\right)\left(x\right)=2^{x^2}$ ...$\left(2\right)$

Given that $\left(fog\right)\left(x\right)=\left(gof\right)\left(x\right)$

Equating $\left(1\right)$ and $\left(2\right)$, we get

$\Rightarrow 2^{2x}=2^{x^2}$

$\Rightarrow 2x=x^2$

$\Rightarrow x^2-2x=0$

$\Rightarrow x(x-2)=0$

Therefore, $x=2andx=0$