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

To solve the given problem, we first address the equation involving x: 
\(x^2 - bx - 5 = 0\). Let the roots be \(α\) and \(β\). By Vieta's formulas, we know:

• \(α + β = b\)
• \(αβ = -5\)

Next, we proceed with solving the trigonometric and inverse trigonometric functions:

1. We have \(\cos(\sin^{-1}(x\cot(\tan^{-1}(\cos(\sin^{-1}(\text{⋅}))))) = k\).
   Denoting \(\cos(\sin^{-1}(y)) = \sqrt{1 - y^2}\), it's clear \(|x| < \frac{1}{\sqrt{2}}\), so the calculations stay within bounds.
   Given that the principal values for inverse functions are considered, simplify: \(x \cdot \text{expression} = x\) as it's multiplied by equivalent \(1\) (in this configuration).
2. It's specified \(\frac{1}{α^2} + \frac{1}{β^2}\) as a solution:
   
   • \(\frac{1}{α^2} + \frac{1}{β^2} = \frac{α^2 + β^2}{α^2β^2}\)
   • Using \(α^2 + β^2 = (α + β)^2 - 2αβ\)\)

\(k^2 x^2 = 1\), implying \(x = \frac{1}{k}\), and since \(0 < |x| < \frac{1}{\sqrt{2}}\)
Therefore, \(k = 1\).

Final Calculation:

• \(b/k^2 = b\), thus yielding \(b\)\) given \(b^2 + 10 = b\). Simplifying provides \(b = 12\), a valid integer and falls within the specified 12,12 range.

\(\therefore b/k^2 = 12\) confirming the condition.

 

Answer 2

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