JEE PYQ: Trigonometric Equations And Inequations Question 4
Question 4 - 2021 (24 Feb Shift 1)
If $e^{(\cos^2 x + \cos^4 x + \cos^6 x + \cdots \infty) \ln 2}$ satisfies the equation $t^2 - 9t + 8 = 0$, then the value of $\frac{2 \sin x}{\sin x + \sqrt{3} \cos x}$ $\left(0 < x < \frac{\pi}{2}\right)$ is:
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(1) $\frac{3}{2}$
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(2) $2\sqrt{3}$
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(3) $\frac{1}{2}$
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(4) $\sqrt{3}$
Show Answer
Answer:
Solution
$2^{\cot^2 x} = 1, 8 \Rightarrow \cot^2 x = 0, 3$
$\frac{2\sin x}{\sin x + \sqrt{3}\cos x} = \frac{2}{1 + \sqrt{3}\cot x} = \frac{2}{4} = \frac{1}{2}$
BETA
