WBJEE · Maths · Functions
Let \(\mathrm{S}, \mathrm{T}, \mathrm{U}\) be three non-void sets and \(\mathrm{f}: \mathrm{S} \rightarrow \mathrm{T}, \mathrm{g}: \mathrm{T} \rightarrow \mathrm{U}\) and composed mapping \(\mathrm{gof}: \mathrm{S} \rightarrow U\) be defined. Let \(\mathrm{gof}\) be injective mapping. Then
- A \(f, g\) both are injective.
- B neither \(f\) nor \(g\) is injective.
- C \(\mathrm{f}\) is obviously injective.
- D \(\mathrm{g}\) is obviously injective.
Answer & Solution
Correct Answer
(C) \(\mathrm{f}\) is obviously injective.
Step-by-step Solution
Detailed explanation
Let, \(x_1, x_2 \in S\) and \(f\left(x_1\right)=f\left(x_2\right)\) \(\Rightarrow g \circ f\left(x_1\right)=g\left(f\left(x_1\right)\right)=g\left(f\left(x_2\right)\right)=g\) of \(\left(x_2\right)\) \(\Rightarrow \mathrm{x}_1=\mathrm{x}_2\) (as gof is injective)…
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