If a real valued function $f$ is defined by $f(x)=$ $\frac{a x+\sqrt{a^2-x^2}}{b x}$, then $f$ is

If a real valued function $f$ is defined by $f(x)=$ $\frac{a x+\sqrt{a^2-x^2}}{b x}$, then $f$ is
  1. only one-one
  2. only onto
  3. both one-one and onto
  4. neither one-one nor onto

Solution

$f(x)=\frac{a x+\sqrt{x^2-a^2}}{b x}$ Then $f^{\prime}(x)=\frac{-a^2}{b x^2 \sqrt{x^2-a^2}}$ if $b>0 \Rightarrow f^{\prime}(x) < 0 \Rightarrow f(x)$ is decreasing $\Rightarrow f(x)$ is one-one. if $b < 0 \Rightarrow f^{\prime}(x)>0 \Rightarrow f(x)$ is increasing $\Rightarrow f(x)$ is one-one. Now, let $y=\frac{a x+\sqrt{a^2-x^2}}{b x}$ $\begin{aligned} & \Rightarrow(b x y-a x)^2=a^2-x^2 \\ & \Rightarrow x^2(b y-a)^2+x^2=a^2 \\ & \Rightarrow x^2=\frac{a^2}{1+(b y-a)^2} \\ & \Rightarrow x= \pm \frac{a}{\sqrt{1+(b y-a)^2}} \quad\left[\because 1+(b y-a)^2>0\right] \\ & \therefore y \in R \end{aligned}$ So, $\mathrm{f}(\mathrm{x})$ is onto also.

Asked in: AP EAMCET 2023 (15 May Shift 2)

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