A) Injective
B) Surjective
C) Bijective
D) None of these
Correct Answer: D
Solution :
We have \[f(x)=x+\sqrt{{{x}^{2}}}=x+|x|\] Clearly f is not one-one as \[f(-1)=f(-2)=0\] but \[-1\ne 2\]. Also f is not onto as \[f(x)\ge 0,\,\forall x\in R,\] Also, range of \[f=(0,\infty )\subset R\].You need to login to perform this action.
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