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 x² – bx – 5 = 0 are
\(\frac{1}{α^2}+\frac{1}{β^2} \)and \(\frac{α}{β}\)
, then b/k2 is equal to_____.

Answer 1

Correct Answer: 12

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