JEE Main 2021 · 24 Feb · Shift 1 · Mathematics Q65

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MathematicsSingle correct+4 / -1Question 65 of 90

If $e^{\left(\cos^2x+\cos^4x+\cos^6x+\dots\infty\right)\log_e2}$ satisfies the equation $t^2-9t+8=0$ , then the value of $\dfrac{2\sin x}{\sin x+\sqrt3\cos x}\left(0<x<\dfrac\pi2\right)$ is :

\dfrac12

\sqrt3

\dfrac32

2\sqrt3

Answer and worked solution

Official answer: A

Series $\cos^2x+\cos^4x+\dots=\cot^2x$ , so the number is $2^{\cot^2x}$ , a root 1 or 8. In $(0,\frac\pi2)$ , $\cot^2x=3$ , $x=\frac\pi6$ .

Value $\dfrac{2\cdot\frac12}{\frac12+\frac32}=\dfrac12$ .

(A)

Topics:Trigonometric equations Geometric progression

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JEE Main 2021 · 24 Feb · Shift 1, Mathematics question 65