Question

$\text{List I}$ has four entries and $\text{List II}$ has five entries. Each entry of $\text{List I}$ is to be correctly matched with one or more than one entries of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(P)}& {(\frac{1}{y^2}{\left(\frac{\cos \left({\tan }^{-1}y\right)+y\sin ({\tan }^{-1}y{)}^2}{\cot \left({\sin }^{-1}y\right)+\tan \left({\sin }^{-1}y\right)}\right)}^2+y^4)}^{1/2}& \text{(1)}& \frac{1}{2}\sqrt{\frac{5}{3}}\\ & \text{takes value}& & \\ \text{(Q)}& \text{If}\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z& \text{(2)}& \sqrt{2}\\ & \text{then possible value of}\cos \frac{x-y}{2}\text{is}& & \\ \text{(R)}& \text{If}\cos (\frac{\pi }{4}-x)\cos 2x+\sin x\sin 2x\sec x& \text{(3)}& \frac{1}{2}\\ & =\cos x\sin 2x\sec x+\cos (\frac{\pi }{4}+x)\cos 2x& & \\ & \text{then possible value of}\sec x\text{is}& & \\ \text{(S)}& \text{If}\cot \left({\sin }^{-1}\sqrt{1-x^2}\right)=\sin {\tan }^{-1}\left(x\sqrt{6}\right)),x\ne 0& \text{(4)}& 1\\ & \text{then possible value of}x\text{is}& & \end{array}$

Which of the following is the only CORRECT combination?

• A. $\left(P\right)\to \left(4\right),\left(Q\right)\to \left(3\right),\left(R\right)\to \left(1\right),\left(S\right)\to \left(2\right)$
• B. $\left(P\right)\to \left(4\right),\left(Q\right)\to \left(3\right),\left(R\right)\to \left(2\right),\left(S\right)\to \left(1\right)$
• C. $\left(P\right)\to \left(3\right),\left(Q\right)\to \left(4\right),\left(R\right)\to \left(2\right),\left(S\right)\to \left(1\right)$
• D. $\left(P\right)\to \left(3\right),\left(Q\right)\to \left(4\right),\left(R\right)\to \left(1\right),\left(S\right)\to \left(2\right)$

Answer 1

The correct option is B $\left(P\right)\to \left(4\right),\left(Q\right)\to \left(3\right),\left(R\right)\to \left(2\right),\left(S\right)\to \left(1\right)$

$\left(P\right)$
${[\frac{1}{y^2}{\left[\frac{\frac{1}{\sqrt{1+y^2}}+\frac{y\cdot y}{\sqrt{1+y^2}}}{\frac{\sqrt{1+y^2}}{y}+\frac{y}{\sqrt{1-y^2}}}\right]}^2+y^4]}^{\frac{1}{2}}$
$\Rightarrow [1-y^4+y^4{]}^{\frac{1}{2}}=1$

$\left(Q\right)$
$\cos x+\cos y=-\cos z\cdots \left(1\right)$
$\sin x+\sin y=-\sin z\cdots \left(2\right)$
Squaring and adding $\left(1\right)$ and $\left(2\right)$
$\cos (x-y)=\frac{-1}{2}$
$\Rightarrow {\cos }^2\frac{(x-y)}{2}=\frac{1}{4}$
$\Rightarrow \cos \frac{x-y}{2}=\pm \frac{1}{2}$

$\left(R\right)$
$(\cos (\frac{\pi }{4}-x)-\cos (\frac{\pi }{4}+x))\cos 2x=\sin 2x[\cot x-\tan x]$
$\Rightarrow \sin \frac{\pi }{4}\sin x\cos 2x=\sin 2x2\cot 2x$
$\Rightarrow \frac{\sin x}{\sqrt{2}}=1$ (Not possible)
and $\cos 2x=0$
$\Rightarrow {\cos }^{2}x=\frac{1}{2}$
$\Rightarrow \cos x=\frac{1}{\sqrt{2}}$

$\left(S\right)$
$\cot \left({\sin }^{-1}\sqrt{1-x^2}\right)=\sin {\tan }^{-1}\left(x\sqrt{6}\right))$


$\Rightarrow \frac{x}{\sqrt{1-x^2}}=\frac{x\sqrt{6}}{\sqrt{1+6x^2}}$
$x=0$ (not possible)
$\therefore \frac{1}{\sqrt{1-x^2}}=\frac{6}{\sqrt{1+6x^2}}$
$\Rightarrow x=\frac{1}{2}\sqrt{\frac{5}{3}}$