question_answer

The function \[f:R\to \left[ -\frac{1}{2},\frac{1}{2} \right]\]defined as \[f(x)=\frac{x}{1+{{x}^{2}}},\]is:      JEE Main Solved Paper-2017

A)  neither injective nor surjective.

B)  invertible.

C)  injective but not surjective.

D)  surjective but not injective

Correct Answer: D

Solution :

 \[f:R\to \left[ -\frac{1}{2},\frac{1}{2} \right],\]                 \[f(x)=\frac{x}{1+{{x}^{2}}}\forall x\in R\]                 \[\Rightarrow \]\[f'(x)=\frac{(1+{{x}^{2}}).1-x.2x}{{{(1+{{x}^{2}})}^{2}}}=\frac{-(x+1)(x-1)}{{{(1+{{x}^{2}})}^{2}}}\]                                 \[\therefore \]From above diagram of \[f(x),f(x)\]is surjective but not injective.