WBJEE · Maths · Functions
If \(f: R \rightarrow R\) be defined by \(f(x)=e^{x}\) and \(g: R \rightarrow R\) be defined by \(g(x)=x^{2}\). The mapping gof : \(R \rightarrow R\) be defined by (gof) \((x)\) \(=g f(x) \mid \forall x \in R .\) Then.
- A gof is bijective but \(f\) is not injective
- B gof is injective and \(g\) is injective
- C gof is injective but \(g\) is not bijective
- D gof is surjective and \(g\) is surjective
Answer & Solution
Correct Answer
(C) gof is injective but \(g\) is not bijective
Step-by-step Solution
Detailed explanation
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 \[ \begin{aligned} (g v f)(x) &=g(f(x)) \\ &=g\left(e^{x}\right) \\ &=\left(e^{x}\right)^{2} \\ &=e^{2 x}, \forall x \in R \end{aligned} \]…
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