Question

If $f : R \to R$ be defined by $f(x) = e^x$ and $g : R \to R$ be defined by $g(x) = x^2$. The mapping $g of : R \to R$ be defined by $(g o f ) (x) = g[f(x)] \forall x \in R$ , Then

• A. $g \, of$ is bijective but f is not injective
• B. $g o f$ is injective and g is injective
• C. $g o f$ is injective but g is not bijective
• D. $gof$ is surjective and g is surjective

Answer 1

Answer: (C)

$g o f$ is injective but g is not bijective

Answer 2

We have, $f: R \rightarrow R$, defined by $f(x)=e^{x}$
and $g: R \rightarrow R$ defined by $g(x)=x^{2}$
Now, We have
$(g o f)(x) =g(f(x))$
$=g\left(e^{x}\right)$
$=\left(e^{x}\right)^{2}$
$=e^{2 x}, \forall x \in R$
$\Rightarrow gof$ is injective and $g$ is neither injective nor surjective.
$\Rightarrow gof$ is injective but $g(x)$ is not bijective.