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JEE MAIN - Mathematics (2022 - 27th July Morning Shift - No. 16)

For $$k \in \mathbb{R}$$, let the solutions of the equation $$\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k, 0<|x|<\frac{1}{\sqrt{2}}$$ be $$\alpha$$ and $$\beta$$, where the inverse trigonometric functions take only principal values. If the solutions of the equation $$x^{2}-b x-5=0$$ are $$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$$ and $$\frac{\alpha}{\beta}$$, then $$\frac{b}{k^{2}}$$ is equal to ____________.
Answer
12

Explanation

$$\cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}} \right)} \right)} \right)} \right)} \right) = k$$

$$ \Rightarrow \cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\sqrt {1 - {x^2}} } \right)} \right)} \right) = k$$

$$ \Rightarrow \cos \left( {{{\sin }^{ - 1}}\left( {{x \over {\sqrt {1 - {x^2}} }}} \right)} \right) = k$$

$$ \Rightarrow {{\sqrt {1 - 2{x^2}} } \over {\sqrt {1 - {x^2}} }} = k$$

$$ \Rightarrow {{1 - 2{x^2}} \over {1 - {x^2}}} = {k^2}$$

$$ \Rightarrow 1 - 2{x^2} = {k^2} - {k^2}{x^2}$$

$$\therefore$$JEE Main 2022 (Online) 27th July Morning Shift Mathematics - Inverse Trigonometric Functions Question 32 English Explanation

$${1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} = 2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right)$$ ...... (1)

and $${\alpha \over \beta } = - 1$$ ...... (2)

$$\therefore$$ $$2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right)( - 1) = - 5$$

$$ \Rightarrow {k^2} = {1 \over 3}$$

and $$b = S.R = 2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right) - 1 = 4$$

$$\therefore$$ $${b \over {{k^2}}} = {4 \over {{1 \over 3}}} = 12$$

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