Question

Consider $f(x)={\tan }^{-1}\left(\frac{(\sqrt{12}-2)x^2}{x^4+2x^2+3}\right)$ and $m$ and $M$ are respectively minimum and maximum values of $f(x)$ and $x=a(a>0)$ is the point in the domain of $f(x)$ where $f(x)$ attains its maximum value.

Column I		Column II
A	If ${\sin }^{-1}2\sqrt{x}=3{\tan }^{-1}(\tan (m+M))$ then $8x$ equals	P	0
B	If ${\cos }^{-1}x+{\cos }^{-1}y=3\left\{{\tan }^{-1}\left(\tan \frac{7M}{2}\right)+{\tan }^{-1}\left(m+\tan \frac{3\pi }{8}\right)\right\}$ then $(x+y)$ equals	Q	2
C	The value of $\tan \left({\sec }^{-1}\left(\frac{2}{a^2}\right)+M\right)$ equals(	R	- 2
D	If $\alpha$ and $\beta$ are roots of the equation $x^2-\left(\tan \left(3{\sin }^{-1}(\sin M)\right)\right)x+a^4=0$, then $\alpha \beta -(\alpha +\beta )$ equals	S	1
		T	-1

• A. (A) S; (B) R; (C) S ; (D) P
• B. (A) S; (B) R; (C) S ; (D) Q
• C. (A) S; (B) R; (C) S ; (D) R
• D. (A) S; (B) R; (C) S ; (D) S

Answer 1

Answer: (B)

We have $f(x)={\tan }^{-1}\left(\frac{2(\sqrt{3}-1)}{x^2+\frac{3}{x^2}+2}\right)$
As $x^2+\frac{3}{x^2}\ge 2\sqrt{3}$ (UsingA.M.-GM. inequality)
$\Rightarrow x^2+\frac{3}{x^2}+2\ge 2+2\sqrt{3}$
${\therefore f(x)|}_{\max }={\tan }^{-1}\left(\frac{2(\sqrt{3}-1)}{2(\sqrt{3}+1)}\right)=\frac{\pi }{12}=M$,
which occurs at $x^2=\frac{3}{x^2}\Rightarrow x=3^{\frac{1}{4}}=a$ ${f(x)|}_{\text{min }}=0=m$, which occurs at $x=0$
(A)${\sin }^{-1}(2\sqrt{x})=3{\tan }^{-1}\left(\tan \frac{\pi }{12}\right)=\frac{\pi }{4}$
$2\sqrt{x}=\frac{1}{\sqrt{2}}\Rightarrow 8x=1$
(B) ${\cos }^{-1}x+{\cos }^{-1}y=3\left[{\tan }^{-1}\left(\tan \frac{7\pi }{24}\right)+{\tan }^{-1}\left(0+\tan \frac{3\pi }{8}\right)\right]=3\left[\frac{7\pi }{24}+\frac{3\pi }{8}\right]=3\left(\frac{16\pi }{24}\right)=2\pi$
$\therefore x=y=-1\Rightarrow x+y=-2$
(C) $\tan \left({\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)+\frac{\pi }{12}\right)=\tan \left(\frac{\pi }{6}+\frac{\pi }{12}\right)=1$
(D) $x^2-\tan \left(3{\sin }^{-1}\left(\sin \frac{\pi }{12}\right)\right)x+3=0$