Question

Let $f(x)=x^2,g(x)={\log }_{e}x$ . The number of values of for which $(fog)(x)=(gof)(x),$ is

• A. 1
• B. 2
• C. finite but greater than 2
• D. infinitely many

Answer 1

Answer: (C)

Given, $f(x)=x^2$ $g(x)={\log }_{e}x$ and $(fog)(x)=(gof)(x)$ $(fog)(x)=f(g(x))$
$=f({\log }_{e}x)={({\log }_{e}x)}^2$ ?.(i) and $gof(x)=g(f(x))$
$=g(x^2)={\log }_e{x}^2$ ..(ii) From Eqs. (i) and (ii), we get ${({\log }_{e}x)}^2={\log }_e{x}^2$
$\Rightarrow$ ${({\log }_{e}x)}^2=2{\log }_{e}x$
$\Rightarrow$ ${\log }_{e}x=2\Rightarrow e^2=x$
Hence, the value of x is finite, but greater than 2.