Question

Consider $f\left(x\right)={\sin }^{-1}(1-2\sqrt{x})+{\sec }^{-1}\left(\frac{1}{2\sqrt{\sqrt{x}-x}}\right)+{\tan }^{-1}\left(\frac{\sqrt{2}-1-\sqrt{x}}{1+\sqrt{2x}-\sqrt{x}}\right).$
Which of the following statements is (are) CORRECT?

• A. $f\left(x\right)$ is a decreasing function
• B. Minimum value of $f\left(x\right)$ is $\frac{-\pi }{8}$
• C. $f^{\prime }\left(\frac{1^{-}}{4}\right)=\frac{-24}{5}$
• D. $f^{\prime }\left(\frac{1^{+}}{4}\right)=\frac{-4}{5}$

Answer 1

Answer: (A), (C), (D)

$f^{\prime }\left(\frac{1^{+}}{4}\right)=\frac{-4}{5}$

$f\left(x\right)={\sin }^{-1}(1-2\sqrt{x})+{\cos }^{-1}\left(2\sqrt{\sqrt{x}-x}\right)+{\tan }^{-1}\left(\frac{\sqrt{2}-1-\sqrt{x}}{1+\sqrt{2x}-\sqrt{x}}\right)$
Domain : $x\in (0,1)$
We know ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\frac{x-y}{1+xy}$ for $x,y>0$
$\therefore f\left(x\right)={\sin }^{-1}(1-2\sqrt{x})+{\sin }^{-1}(|1-2\sqrt{x}|)+{\tan }^{-1}(\sqrt{2}-1)-{\tan }^{-1}\left(\sqrt{x}\right)$

$f\left(x\right)=\{\begin{array}{cc}2{\sin }^{-1}(1-2\sqrt{x})+\frac{\pi }{8}-{\tan }^{-1}\sqrt{x},& x\in (0,\frac{1}{4}]\\ \frac{\pi }{8}-{\tan }^{-1}\sqrt{x},& x\in (\frac{1}{4},1)\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}\frac{2}{\sqrt{1-(1-2\sqrt{x}{)}^2}}\left(\frac{-2}{2\sqrt{x}}\right)-\frac{1}{1+x}\cdot \frac{1}{2\sqrt{x}},& x\in (0,\frac{1}{4}]\\ \frac{-1}{1+x}\cdot \frac{1}{2\sqrt{x}},& x\in (\frac{1}{4},1)\end{array}$

$\therefore f^{\prime }\left(\frac{1^{+}}{4}\right)=\frac{-1}{1+\frac{1}{4}}\cdot \frac{1}{2\cdot \frac{1}{2}}=\frac{-4}{5}$

and $f^{\prime }\left(\frac{1^{-}}{4}\right)=\frac{2}{\sqrt{1-0}}(-2)-\frac{4}{5}=\frac{-24}{5}$
Also, $f^{\prime }\left(x\right)<0$ for all $x\in (0,1)$
$\therefore f\left(x\right)$ is decreasing in $(0,1)$
Also, minimum value of $f\left(x\right)$ does not exist.