Q49 / 160 · AP EAMCET 2020 (17 Sep Shift 2) · Mathematics

The function \(f: \mathbf{R} \rightarrow \mathbf{R}\) defined by \(f(x)=\frac{x}{\sqrt{1+x^2}}\) is ........

  • A
    Surjective but not injective
  • B
    Bijective
  • Correct answer
    C
    Injective but not surjective
  • D
    Neither injective nor surjective

Solution

\(\begin{aligned} & f(x)=\frac{x}{\sqrt{1+x^2}} \\ & f\left(x_1\right)=f(x)_2 \Rightarrow \frac{x_1}{\sqrt{1+x_1^2}}=\frac{x_2}{\sqrt{1+x_2^2}} \end{aligned}\) \(\begin{array}{rlrl} \Rightarrow & & x_1^2\left(1+x_2^2\right) & =x_2^2\left(1+x_1^2\right) \\ \Rightarrow & & x_1^2 & =x_2^2 \\ x_1 & =x_2 \end{array}\) \(\therefore F\) is injective in nature Also, Let \(f(x)=y=\frac{x}{\sqrt{1+x^2}}\) \(\begin{aligned} & \Rightarrow \quad y^2\left(1+x^2\right)=x^2 \\ & \Rightarrow \quad \frac{y^2}{1-y^2}=x^2 \Rightarrow x=\frac{y}{\sqrt{1-y^2}} \end{aligned}\) As \(y^2 \leq 1 \Rightarrow-1 \leq|y| \leq 1\) So, \(f\) is non surjective.

Hence, the correct answer is option: 'C'.

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