Question

If $e^{\left({\cos }^{2}x+{\cos }^{4}x+{\cos }^{6}x+...\infty \right){\log }_{e}2}$ satisfies the equation $t^2-9t+8=0,$ then the value of $\frac{2\sin x}{\sin x+\sqrt{3}\cos x}$ ,$0<x<\frac{\mathrm{\pi }}{2}$ is

• A. $\frac{3}{2}$
• B. $2\sqrt{3}$
• C. $\frac{1}{2}$
• D. $\sqrt{3}$

Answer 1

The correct option is C

$\frac{1}{2}$

Explanation for the correct option:

Step 1: Simplifying the given equation:

Given that,

$e^{\left({\cos }^{2}x+{\cos }^{4}x+{\cos }^{6}x+...\infty \right){\log }_{e}2}$

Here, ${\cos }^{2}x+{\cos }^{4}x+...+\infty$ is in GP. Then

$\begin{array}{rcl}{\cos }^{2}x+{\cos }^{4}x+...\infty & =& \frac{{\cos }^{2}x}{1-{\cos }^{2}x}\left[\because S_n=\frac{a}{1-r}\right]\\ & =& \frac{{\cos }^{2}x}{{\sin }^{2}x}\\ & =& co{t}^{2}x\end{array}$

Substitute this in the given function, we get

$\begin{array}{rcl}{e}^{\left({\cos }^{2}x+{\cos }^{4}x+{\cos }^{6}x+...\infty \right){\log }_{e}2}& =& e^{co{t}^{2}x{\log }_{e}2}\\ & =& e^{{\log }_e{2}^{co{t}^{2}x}}\left[\because \mathrm{mloga}={\mathrm{loga}}^m\right]\\ & =& 2^{co{t}^{2}x}\end{array}$

Step 2: Find the value of $t$:

$\begin{array}{rcl}{t}^2-9t+8& =& 0\left[\mathrm{given}\right]\\ & \Rightarrow & t^2-8t-t+8=0\\ & \Rightarrow & t\left(t-8\right)-1\left(t-8\right)=0\\ & \Rightarrow & \left(t-8\right)\left(t-1\right)=0\\ & \Rightarrow & t=8or1\end{array}$

Step 3: Find the value of $x$:

Since, the given equation satisfies the equation $t^2-9t+8=0,$then

$\begin{array}{rcl}{2}^{co{t}^{2}x}& =& 8\\ & \Rightarrow & 2^{co{t}^{2}x}=2^3\\ & \Rightarrow & co{t}^{2}x=3\left[\mathrm{by}\mathrm{comparing}\right]\\ & \Rightarrow & cotx=\sqrt{3}\\ & \Rightarrow & x=\frac{\mathrm{\pi }}{6}\end{array}$ or $\begin{array}{rcl}{2}^{co{t}^{2}x}& =& 1\\ & \Rightarrow & 2^{co{t}^{2}x}=2^0\\ & \Rightarrow & co{t}^{2}x=0\left[\mathrm{by}\mathrm{comparing}\right]\\ & \Rightarrow & cotx=cot\left(\frac{\mathrm{\pi }}{2}\right)\\ & \Rightarrow & x=\frac{\mathrm{\pi }}{2}\end{array}$

But it is given $0<x<\frac{\mathrm{\pi }}{2}$so, take only $x=\frac{\mathrm{\pi }}{6}$.

Step 4: Find the value of the expression:

By substituting $x=\frac{\mathrm{\pi }}{6}$ in the expression.

$\begin{array}{rcl}\frac{2\sin x}{\sin x+\sqrt{3}\cos x}& =& \frac{2\sin \left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)+\sqrt{3}\cos \left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)}\\ & =& \frac{2\times {\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}+\sqrt{3}\times {\displaystyle \frac{\sqrt{3}}{2}}}\\ & =& \frac{2}{4}\\ & =& \frac{1}{2}\end{array}$

Hence, the correct option is C.