Q41 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics
The range of the real valued function $f(x)=\sin ^{-1}\left(\frac{1+x^2}{2 x}\right)+\cos ^{-1}\left(\frac{2 x}{1+x^2}\right)$ is
- Correct answerA$\left\{\frac{\pi}{2}\right\}$
- B$\mathbf{R}$
- C$\mathbf{Q}$
- D$\left\{-\frac{\pi}{2}, \frac{\pi}{2}\right\}$
Solution
$f(x)=\sin ^{-1}\left(\frac{1+x^2}{2 x}\right)+\cos ^{-1}\left(\frac{2 x}{1+x^2}\right)$
We know, domain of $\sin ^{-1}(x)$ is $[-1,1]$
$\therefore-1 \leq \frac{1+x^2}{2 x} \leq 1 \text { and }-1 \leq \frac{2 x}{1+x^2} \leq 1$
When, $-1 \leq \frac{1+x^2}{2 x} \leq 1 \Rightarrow \frac{1+x^2}{2 x} \leq 1$ and $\frac{1+x^2}{2 x} \geq-1$
$\begin{aligned} & \Rightarrow \frac{(1-x)^2}{2 x} \leq 0 \text { and } \frac{(1+x)^2}{2 x} \geq 0 \\ & \Rightarrow x \lt 0, x=1 \text { and } x\gt0, x=-1 \Rightarrow x=-1,1...(i) \end{aligned}$
$\begin{aligned} & \text { When, }-1 \leq \frac{2 x}{1+x^2} \leq 1 \\ & \Rightarrow \frac{2 x}{1+x^2} \leq 1 \text { and } \frac{2 x}{1+x^2} \geq-1 \\ & \Rightarrow \frac{-(1-x)^2}{1+x^2} \leq 0 \text { and } \frac{(1+x)^2}{1+x^2} \geq 0 \Rightarrow x \in \mathrm{R}...(ii) \end{aligned}$
From equations (i) \& (ii), $x=-1,1$; Domain $=\{-1,1\}$ and $f(1)=\sin ^{-1}\left(\frac{1+1}{2}\right)+\cos ^{-1}\left(\frac{2}{1+1}\right)=\frac{\pi}{2}$ $f(-1)=\sin \left(\frac{1+1}{-2}\right)+\cos ^{-1}\left(\frac{-2}{1+1}\right)=-\frac{\pi}{2}+\pi=\frac{\pi}{2}$ $\therefore \quad$ Range $(f)=\left\{\frac{\pi}{2}\right\}$
We know, domain of $\sin ^{-1}(x)$ is $[-1,1]$
$\therefore-1 \leq \frac{1+x^2}{2 x} \leq 1 \text { and }-1 \leq \frac{2 x}{1+x^2} \leq 1$
When, $-1 \leq \frac{1+x^2}{2 x} \leq 1 \Rightarrow \frac{1+x^2}{2 x} \leq 1$ and $\frac{1+x^2}{2 x} \geq-1$
$\begin{aligned} & \Rightarrow \frac{(1-x)^2}{2 x} \leq 0 \text { and } \frac{(1+x)^2}{2 x} \geq 0 \\ & \Rightarrow x \lt 0, x=1 \text { and } x\gt0, x=-1 \Rightarrow x=-1,1...(i) \end{aligned}$
$\begin{aligned} & \text { When, }-1 \leq \frac{2 x}{1+x^2} \leq 1 \\ & \Rightarrow \frac{2 x}{1+x^2} \leq 1 \text { and } \frac{2 x}{1+x^2} \geq-1 \\ & \Rightarrow \frac{-(1-x)^2}{1+x^2} \leq 0 \text { and } \frac{(1+x)^2}{1+x^2} \geq 0 \Rightarrow x \in \mathrm{R}...(ii) \end{aligned}$
From equations (i) \& (ii), $x=-1,1$; Domain $=\{-1,1\}$ and $f(1)=\sin ^{-1}\left(\frac{1+1}{2}\right)+\cos ^{-1}\left(\frac{2}{1+1}\right)=\frac{\pi}{2}$ $f(-1)=\sin \left(\frac{1+1}{-2}\right)+\cos ^{-1}\left(\frac{-2}{1+1}\right)=-\frac{\pi}{2}+\pi=\frac{\pi}{2}$ $\therefore \quad$ Range $(f)=\left\{\frac{\pi}{2}\right\}$
Hence, the correct answer is option: 'A'.