Question

Let $y=e^{\left\{\right({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+\dots \left){\log }_{e}2\right\}}$ satisfy the equation $x^2-17x+16=0,$ where $0<x<\frac{\pi }{2}.$ Then match the correct value of $\text{List I}$ from $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(a)}& \frac{2\sin 2x}{1+{\cos }^{2}x}& \left(p\right)& 1\\ \text{(b)}& \frac{2\sin x}{\sin x+\cos x}& \left(q\right)& \frac{4}{9}\\ \text{(c)}& \sum _{n=1}^{\infty }(\cot x{)}^n& \left(r\right)& \frac{2}{3}\\ \text{(d)}& \sum _{n=1}^{\infty }n(\cot x{)}^{2n}& \left(s\right)& \frac{4}{3}\end{array}$

• A. $\left(a\right)\to \left(p\right),\left(b\right)\to \left(q\right)\left(c\right)\to \left(r\right),\left(d\right)\to \left(s\right)$
• B. $\left(a\right)\to \left(q\right),\left(b\right)\to \left(p\right)\left(c\right)\to \left(s\right),\left(d\right)\to \left(r\right)$
• C. $\left(a\right)\to \left(r\right),\left(b\right)\to \left(s\right)\left(c\right)\to \left(q\right),\left(d\right)\to \left(p\right)$
• D. $\left(a\right)\to \left(s\right),\left(b\right)\to \left(s\right)\left(c\right)\to \left(p\right),\left(d\right)\to \left(q\right)$

Answer 1

The correct option is D $\left(a\right)\to \left(s\right),\left(b\right)\to \left(s\right)\left(c\right)\to \left(p\right),\left(d\right)\to \left(q\right)$

$x^2-17x+16=0\Rightarrow (x-1)(x-16)=0\Rightarrow x=1,16\cdots \left(1\right)$
$y=e^{\left\{\right({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+\dots \left){\log }_{e}2\right\}}\Rightarrow y=e^{\left\{{\tan }^{2}x{\log }_{e}2\right\}}=2^{{\tan }^{2}x}$
Using equation $\left(1\right)$,
$2^{{\tan }^{2}x}=1,16$
Taking ${\log }_2$ on both sides,
${\tan }^{2}x=0,4\Rightarrow \tan x=2(\because 0<x<\frac{\pi }{2})$

Now,
$\left(a\right)$
$\frac{2\sin 2x}{1+{\cos }^{2}x}=\frac{4\sin x\cos x}{1+{\cos }^{2}x}=\frac{4\tan x}{{\sec }^{2}x+1}=\frac{4\tan x}{2+{\tan }^{2}x}=\frac{4\times 2}{2+2^2}=\frac{4}{3}$
$\left(a\right)\to \left(s\right)$

$\left(b\right)$
$\frac{2\sin x}{\sin x+\cos x}=\frac{2\tan x}{\tan x+1}=\frac{2\times 2}{2+1}=\frac{4}{3}$
$\left(b\right)\to \left(s\right)$

$\left(c\right)$
$S=\sum _{n=1}^{\infty }(\cot x{)}^n=\cot x+{\cot }^{2}x+{\cot }^{3}x+\dots$
So, the given numbers are in G.P.
As $\tan x=2\Rightarrow \cot x=\frac{1}{2}$
$S=\frac{\cot x}{1-\cot x}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$
$\left(c\right)\to \left(p\right)$

$\left(d\right)$
$\sum _{n=1}^{\infty }n(\cot x{)}^{2n}={\cot }^{2}x+2{\cot }^{4}x+3{\cot }^{6}x+\dots$
The given series is in A.G.P.
Let $S={\cot }^{2}x+2{\cot }^{4}x+3{\cot }^{6}x+\dots$
${\cot }^{2}xS={\cot }^{4}x+2{\cot }^{6}x+3{\cot }^{8}x+\dots \Rightarrow (1-{\cot }^{2}x)S={\cot }^{2}x+{\cot }^{4}x+{\cot }^{6}x+\dots \Rightarrow (1-{\cot }^{2}x)S=\frac{{\cot }^{2}x}{1-{\cot }^{2}x}\Rightarrow \sum _{n=1}^{\infty }n(\cot x{)}^{2n}=\frac{{\cot }^{2}x}{(1-{\cot }^{2}x{)}^2}\Rightarrow \sum _{n=1}^{\infty }n(\cot x{)}^{2n}=\frac{\frac{1}{4}}{{(1-\frac{1}{4})}^2}\therefore \sum _{n=1}^{\infty }n(\cot x{)}^{2n}=\frac{4}{9}$
$\left(d\right)\to \left(q\right)$