Question

For k ∈ R , let the solution of the equation
$\cos\left(\sin^{-1}\left(x \cot\left(\tan^{-1}\left(\cos\left(\sin^{-1}\right)\right)\right)\right)\right) = k, \quad 0 < |x| < \frac{1}{\sqrt{2}}$
Inverse trigonometric functions take only principal values. If the solutions of the equation x 2 – bx – 5 = 0 are
$\frac{1}{α^2}+\frac{1}{β^2}$and $\frac{α}{β}$
, then b/k2 is equal to_____.

Answer 1

The correct answer is 12
$\cos\left(\sin^{-1}\left(x \cot\left(\tan^{-1}\left(\cos\left(\sin^{-1}\right)\right)\right)\right)\right) = k$
⇒ $$\cos\left(\sin^{-1}\left(x \cot\left(\tan^{-1}\left(\sqrt{1 - x^2}\right)\right)\right)\right) = k
⇒ $$\cos\left(\sin^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right)\right) = k
$⇒$ $\frac{\sqrt{1 - 2x^2}}{\sqrt{1 - x^2}} = k$
$⇒$ $\frac{1 - 2x^2}{1 - x^2} = k^2$
$⇒ 1-2x^2$
$= k^2-k^2x^2$
∴ α,β be the roots of x2-(k2-1)/(k2-2) = 0
$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 2\left(\frac{k^2 - 2}{k^2 - 1}\right) \dots (1)$
and $\frac{α}{β} = -1....(2)$
$∴$ $2\left(\frac{k^2 - 2}{k^2 - 1}\right)(-1) = -5$
$⇒ k^2 = \frac{1}{3}$
and b = S.R
$2\left(\frac{k^2 - 2}{k^2 - 1}\right)( - 1) = 4$
$\therefore \frac{b}{k^2} = \frac{4}{\frac{1}{3}}$
= 12