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 $x^{2}-bx-5=0$ are $\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$ and $\frac{\alpha}{\beta}$, then $\frac{b}{k^{2}}$ is equal to ____.

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Inverse Trigonometric Functions

For

k \in R

, let the solutions of the equation

cos (sin^{- 1} (x cot (tan^{- 1} (cos (sin^{- 1} x))))) = k

0 < ∣ x ∣ < \frac{1}{2}

α

and

β

, where the inverse trigonometric functions take only principal values. If the solutions of the equation

x^{2} - b x - 5 = 0

are

\frac{1}{α ^{2}} + \frac{1}{β ^{2}}

and

\frac{α}{β}

, then

\frac{b}{k ^{2}}

is equal to ____.

Answer: 12

Simplifying the nested inverse trigonometric expression:

cos (sin^{- 1} x) = 1 - x^{2}

cot (tan^{- 1} 1 - x^{2}) = \frac{1}{1 - x ^{2}}

sin^{- 1} (\frac{x}{1 - x ^{2}})

leads to

cos (sin^{- 1} (\frac{x}{1 - x ^{2}})) = \frac{1 - 2 x ^{2}}{1 - x ^{2}} = k

Squaring:

1 - 2 x^{2} = k^{2} (1 - x^{2})

, giving

x^{2} = \frac{k ^{2} - 1}{k ^{2} - 2}

α^{2} = β^{2} = \frac{k ^{2} - 1}{k ^{2} - 2}

and

\frac{α}{β} = - 1

\frac{1}{α ^{2}} + \frac{1}{β ^{2}} = \frac{2 ( k ^{2} - 2 )}{k ^{2} - 1}

Product of roots:

\frac{2 ( k ^{2} - 2 )}{k ^{2} - 1} \times (- 1) = - 5

, giving

k^{2} = \frac{1}{3}

.

Sum of roots:

b = \frac{2 ( k ^{2} - 2 )}{k ^{2} - 1} - 1 = 5 - 1 = 4

.

Therefore,

\frac{b}{k ^{2}} = \frac{4}{1/3} = 12

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer