A) 1
B) 2
C) finite but greater than 2
D) infinitely many
Correct Answer: C
Solution :
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.
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