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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