JEE Main 2022 · 27 Jul · Shift 1 · Mathematics Q21

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MathematicsInteger+4 / -1Question 21 of 90

For $k\in\mathbf R$ , let the solutions of the equation $\cos\left(\sin^{-1}\left(x\cot\left(\tan^{-1}\left(\cos\left(\sin^{-1}x\right)\right)\right)\right)\right)=k$ , $0<|x|<\dfrac1{\sqrt2}$ be $\alpha$ and $\beta$ , where the inverse trigonometric functions take only principal values. If the solutions of the equation $x^2-bx-5=0$ are $\dfrac1{\alpha^2}+\dfrac1{\beta^2}$ and $\dfrac\alpha\beta$ , then $\dfrac b{k^2}$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 12

Simplify. $\cos\left(\sin^{-1}x\right)=\sqrt{1-x^2}$ , and $\cot\tan^{-1}$ of it is $\frac1{\sqrt{1-x^2}}$ . The equation becomes $\sqrt{\frac{1-2x^2}{1-x^2}}=k$ , which is even in x, so $\beta=-\alpha$ .

Roots. $\frac\alpha\beta=-1$ and $\frac2{\alpha^2}$ have product $-5$ : $\alpha^2=\frac25$ . Sum $b=5-1=4$ , and $k^2=\frac{1/5}{3/5}=\frac13$ .

$\frac b{k^2}=12$ .

Answer: 12

Topics:Inverse trigonometric functions Roots and coefficients

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JEE Main 2022 · 27 Jul · Shift 1, Mathematics question 21