Question

If $f:R\to R$ be defined by $f\left(x\right)=e^x$ and $g:R\to R$ be defined by $g\left(x\right)=x^2$. The mapping $g\circ f:R\to R$ be defined by $(g\circ f(x\left)\right)=g\left(f\right(x\left)\right)\forall x\in R$. Then

• A. $g\circ f$ is injective but $f$ is not injective
• B. $g\circ f$ is injective and $g$ is injective
• C. $g\circ f$ is injective but $g$ is not injective
• D. $g\circ f$ is surjective and $g$ is surjective

Answer 1

The correct option is C $g\circ f$ is injective but $g$ is not injective

$f\left(x\right)=e^x:R\to R$
$g\left(x\right)=x^2:R\to R$ as (-x and +x will be having the same outcome , so it is not injective)
$g\left(f\right(x\left)\right)=g\left(e^x\right)=e^{2x}$ for all $x$ belonging to $R$ (it will have distinct outcomes for different values of $x$)
clearly $g\left(f\right(x\left)\right)$ is injective and g(x) is not injective
Therefore Answer is $C$