If $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x)=e^{x}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $g(x)=x^{2}$. The mapping $g \circ f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $(g \circ f)(x)=g[f(x)] \forall x \in \mathbb{R}$, Then

Question

$$\begin{aligned} &\text { If } f: \mathbb{R} \rightarrow \mathbb{R} \text { be defined by } f(x)=e^{x} \text { and } g: \mathbb{R} \rightarrow \mathbb{R} \text { be defined by } g(x)=x^{2} \text {. The mapping } g \circ f: \mathbb{R} \rightarrow \mathbb{R} \text { be defined by } \\ &\begin{array}{ll} (g \circ f)(x)=g[f(x)] \forall x \in \mathbb{R}, \text { Then } & \text { (B) } g \circ f \text { is injective and } g \text { is injective } \\ \text { (A) } g \circ f \text { is bijective but } f \text { is not injective } & \text { (D) } g \circ f \text { is surjective and } g \text { is surjective } \\ \text { (C) } g \text { of is injective but } g \text { is not bijective } & \end{array} \\ &\text {  }(\text { ) } \end{aligned}$$

Answer 1

Ans: (C)
Hint: $g$ is neither injective nor surjective
$g \circ f(x)=e^{2 x}, x \in \mathbb{R} \therefore g \circ f$ is injective

Answer 2

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Answer 3

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