The function f : R→R defined by f(x)={x}{{1+x²}} is

Correct Answer is : injective but not surjective

neither injective nor surjective

Correct Answer is : injective but not surjective

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TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 1 of 160

Marks: +1, -0

The function

f

f(x)=\frac{x}{\sqrt{1+x^{2}}}

Solution:

We have,

f(x)=\;\frac{x}{\sqrt{1+x^{2}}}

Injective

Let

x_{1}, x_{2} \in R

such that

\Rightarrow \;\; \;\frac{x_{1}}{\sqrt{1+x_{1}^{2}}}=\;\frac{x_{2}}{\sqrt{1+x_{2}^{2}}}

\Rightarrow\;\frac{x_{1}^{2}}{1+x_{1}^{2}}=\;\frac{x_{2}^{2}}{1+x_{2}^{2}}

\Rightarrow \;\; x_{1}^{2}=x_{2}^{2}

\Rightarrow \;\; x_{1}=x_{2}

\therefore f(x)

is injective

Surjective

Let

y=\;\frac{x}{\sqrt{1+x^{2}}}

\Rightarrow \;\; y^{2}(1+x^{2})=x^{2} \Rightarrow y^{2}+y^{2}x^{2}=x^{2}

\Rightarrow \;\; x^{2}(1-y^{2})=y^{2} \Rightarrow x=\sqrt{\;\frac{y^{2}}{1-y^{2}}}

\Rightarrow \;\; \;\frac{y^{2}}{1-y^{2}} \ge 0

\therefore y \in (-\infty,-1) \cup [0,1)

\therefore

Range of

f(x)=(-\infty,-1) \cup [0,1)

So,

f(x)

is not surjective.

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