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.You need to login to perform this action.
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