First slide

Theory of equations

Question

If exp $\{(\sin^{2} x + \sin^{4} x + \sin^{6} x + \dots$ upto $\infty) \ln 2}$ satisfies the equation $x^{2} - 17 x + 16 = 0$ then value of $\frac{2 \cos x}{\sin x + 2 \cos x} (0 < x < π / 2)$ is

Moderate

A

$\frac{1}{2}$
B

$\frac{3}{2}$
C

$\frac{5}{2}$
D

none of these

Solution

We have $\sin^{2} x + \sin^{4} x + \sin^{6} x + \dots = \frac{\sin^{2} x}{1 - \sin^{2} x} = \tan^{2} x$
Therefore, $α = \exp \{(\sin^{2} x + \sin^{4} x + \sin^{6} x + \dots$ upto $\infty) \ln 2}$
$= \exp \{\tan^{2} x \ln 2\} = \exp \{\ln 2^{\tan^{2} x}\}$
$= 2^{\tan^{2} x}$
As $α$ satisfies the equation $x^{2} - 17 x + 16 = 0$ we get
$α = 1$ or $α = 16$
Since $0 < x < π / 2, \tan^{2} x > 0 \Rightarrow α = 2^{\tan^{2} x} > 1$ . Therefore,
$\begin{aligned} 2^{\tan^{2} x} = 16 = 2^{4} \Rightarrow \tan^{2} x = 4 \Rightarrow \tan x = 2 \\ [∵ \tan x > 0] \end{aligned}$
Thus, $\frac{2 \cos x}{\sin x + 2 \cos x} = \frac{2}{\tan x + 2} = \frac{2}{2 + 2} = \frac{1}{2}$

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