Question

$\begin{array}{cccc}& ListI& & ListII\\ P.& \text{Let}y\left(x\right)=\cos \left(3{\cos }^{-1}x\right),x\in [-1,1],x\ne \pm \frac{\sqrt{3}}{2}.\text{Then}\frac{1}{y\left(x\right)}\left\{\right(x^2-1)\frac{d^{2}y\left(x\right)}{d{x}^2}+x\frac{dy\left(x\right)}{dx}\}\text{equals}& 1.& 1\\ Q.& \text{Let}{A}_1,A_2,\cdots ,A_n(n>2)\text{be the vertices of a regular polygon of}n\text{sides with its centre at the origin. Let}\overrightarrow{a_k}\text{be the position vector of the point}{A}_k,k=1,2,\cdots ,n.\text{If}\left|\sum _{k=1}^{n-1}(\overrightarrow{a_k}\times \overrightarrow{a_{k+1}})\right|=\left|\sum _{k=1}^{n-1}(\overrightarrow{a_k}\cdot \overrightarrow{a_{k+1}})\right|,\text{then}\text{the minimum value of}n\text{is}& 2.& 2\\ R.& \text{If the normal from the point}P(h,1)\text{on the ellipse}\frac{x^2}{6}+\frac{y^2}{3}=1\text{is perpendicular to the line}x+y=8,\text{then the value of}h\text{is}& 3.& 8\\ S.& \text{Number of positive solutions satisfying the equation}{\tan }^{-1}\left(\frac{1}{2x+1}\right)+{\tan }^{-1}\left(\frac{1}{4x+1}\right)={\tan }^{-1}\left(\frac{2}{x^2}\right)\text{is}& 4.& 9\end{array}$

Which of the following option is correct?

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

Answer 1

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

$\left(P\right)$
$y\left(x\right)=\cos \left(3{\cos }^{-1}x\right)\Rightarrow y\left(x\right)=4x^3-3x\Rightarrow y^{\prime }=12x^2-3\Rightarrow y^{\prime \prime }=24x\frac{1}{y\left(x\right)}\left\{\right(x^2-1)\frac{d^{2}y\left(x\right)}{d{x}^2}+x\frac{dy\left(x\right)}{dx}\}=\frac{1}{4x^3-3x}\left\{\right(x^2-1)24x+x(12x^2-3\left)\right\}=\frac{1}{4x^3-3x}\{36x^3-27x\}=9$

$\left(Q\right)$
$\left|\sum _{k=1}^{n-1}\right(\overrightarrow{a_k}\times \overrightarrow{a_{k+1}}\left)\right|=\left|\sum _{k=1}^{n-1}\right(\overrightarrow{a_k}\cdot \overrightarrow{a_{k+1}}\left)\right|\Rightarrow \sin \left(\frac{2\pi }{n}\right)=\cos \left(\frac{2\pi }{n}\right)\Rightarrow \frac{2\pi }{n}+\frac{2\pi }{n}=\frac{(2k+1)\pi }{2}\Rightarrow n=\frac{8}{(2k+1)}$
For $n$ to be integer,$k=0$
$\therefore n=8$

$\left(R\right)$
Slope of the line $x+y=8$
$m=-1$
Slope of the tangent,
$m_t=-1$
Ellipse given:
$\frac{x^2}{6}+\frac{y^2}{3}=1\Rightarrow \frac{2x}{6}+\frac{2y{y}^{\prime }}{3}=0$
Putting $(h,1),y^{\prime }=-1$
$\Rightarrow \frac{2h}{6}-\frac{2}{3}=0\Rightarrow h=2$

$\left(S\right)$
${\tan }^{-1}\left(\frac{1}{2x+1}\right)+{\tan }^{-1}\left(\frac{1}{4x+1}\right)={\tan }^{-1}\left(\frac{2}{x^2}\right)\Rightarrow \frac{\frac{1}{2x+1}+\frac{1}{4x+1}}{1-\frac{1}{2x+1}\times \frac{1}{4x+1}}=\frac{2}{x^2}\Rightarrow \frac{3x+1}{4x^2+3x}=\frac{2}{x^2}\Rightarrow (3x+2)(x-3)=0$
Number of positive solution is one $(x=3)$