JEE Main 2022 · 30 Jun · Shift 1 · Mathematics Q19

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MathematicsSingle correct+4 / -1Question 19 of 90

Let $\alpha=\tan\left(\dfrac{5\pi}{16}\sin\left(2\cos^{-1}\left(\dfrac1{\sqrt5}\right)\right)\right)$ and $\beta=\cos\left(\sin^{-1}\left(\dfrac45\right)+\sec^{-1}\left(\dfrac53\right)\right)$ where the inverse trigonometric functions take principal values. Then, the equation whose roots are $\alpha$ and $\beta$ is :

15x^2-8x-7=0

5x^2-12x+7=0

25x^2-18x-7=0

25x^2-32x+7=0

Answer and worked solution

Official answer: C

α. $\sin\left(2\cos^{-1}\frac1{\sqrt5}\right)=\frac45$ , so $\alpha=\tan\frac\pi4=1$ .

β. $\sin^{-1}\frac45=\sec^{-1}\frac53=\phi$ , $\cos2\phi=1-\frac{32}{25}=-\frac7{25}$ .

Equation $25x^2-18x-7=0$ .

(C)

Topics:Inverse trigonometric functions Roots and coefficients

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JEE Main 2022 · 30 Jun · Shift 1, Mathematics question 19