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Question

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

A

$\frac{3}{2}$

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B

$2\sqrt{3}$

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C

$\frac{1}{2}$

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D

$\sqrt{3}$

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Solution

## The correct option is C $\frac{1}{2}$Explanation for the correct option:Step 1: Simplifying the given equation:Given that,${e}^{\left({\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x+{\mathrm{cos}}^{6}x+...\infty \right){\mathrm{log}}_{e}2}$Here, ${\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x+...+\infty$ is in GP. Then$\begin{array}{rcl}{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x+...\infty & =& \frac{{\mathrm{cos}}^{2}x}{1-{\mathrm{cos}}^{2}x}\left[\because {S}_{n}=\frac{a}{1-r}\right]\\ & =& \frac{{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{2}x}\\ & =& co{t}^{2}x\end{array}$Substitute this in the given function, we get$\begin{array}{rcl}{e}^{\left({\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x+{\mathrm{cos}}^{6}x+...\infty \right){\mathrm{log}}_{e}2}& =& {e}^{co{t}^{2}x{\mathrm{log}}_{e}2}\\ & =& {e}^{{\mathrm{log}}_{e}{2}^{co{t}^{2}x}}\left[\because \mathrm{mloga}={\mathrm{loga}}^{m}\right]\\ & =& {2}^{co{t}^{2}x}\end{array}$Step 2: Find the value of $t$:$\begin{array}{rcl}{t}^{2}-9t+8& =& 0\left[\mathrm{given}\right]\\ & ⇒& {t}^{2}-8t-t+8=0\\ & ⇒& t\left(t-8\right)-1\left(t-8\right)=0\\ & ⇒& \left(t-8\right)\left(t-1\right)=0\\ & ⇒& t=8or1\end{array}$Step 3: Find the value of $x$:Since, the given equation satisfies the equation ${t}^{2}-9t+8=0,$then$\begin{array}{rcl}{2}^{co{t}^{2}x}& =& 8\\ & ⇒& {2}^{co{t}^{2}x}={2}^{3}\\ & ⇒& co{t}^{2}x=3\left[\mathrm{by}\mathrm{comparing}\right]\\ & ⇒& cotx=\sqrt{3}\\ & ⇒& x=\frac{\mathrm{\pi }}{6}\end{array}$ or $\begin{array}{rcl}{2}^{co{t}^{2}x}& =& 1\\ & ⇒& {2}^{co{t}^{2}x}={2}^{0}\\ & ⇒& co{t}^{2}x=0\left[\mathrm{by}\mathrm{comparing}\right]\\ & ⇒& cotx=cot\left(\frac{\mathrm{\pi }}{2}\right)\\ & ⇒& x=\frac{\mathrm{\pi }}{2}\end{array}$But it is given $0so, 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.$\begin{array}{rcl}\frac{2\mathrm{sin}x}{\mathrm{sin}x+\sqrt{3}\mathrm{cos}x}& =& \frac{2\mathrm{sin}\left(\frac{\mathrm{\pi }}{6}\right)}{\mathrm{sin}\left(\frac{\mathrm{\pi }}{6}\right)+\sqrt{3}\mathrm{cos}\left(\frac{\mathrm{\pi }}{6}\right)}\\ & =& \frac{2×\frac{1}{2}}{\frac{1}{2}+\sqrt{3}×\frac{\sqrt{3}}{2}}\\ & =& \frac{2}{4}\\ & =& \frac{1}{2}\end{array}$Hence, the correct option is C.

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