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.