Given the inverse trigonometric function assumes principal values only. Let $ x, y $ be any two real numbers in $ [-1, 1] $ such that \[ \cos^{-1}x - \sin^{-1}y = \alpha, \, -\frac{\pi}{2} \leq \alpha \leq \pi. \] Then, the minimum value of $ x^2 + y^2 + 2xy \sin \alpha $ is:

Question

If the domain of the function $f(x)=\sin^{-1}\!\left(\dfrac{1}{x^2-2x-2}\right)$ is $(-\infty,\alpha)\cup[\beta,\gamma]\cup[\delta,\infty)$, then $\alpha+\beta+\gamma+\delta$ is equal to

Answer 1

The expression $ x^2 + y^2 + 2xy \sin \alpha $ is equivalent to $ (x + y \sin \alpha)^2 $, which is always non-negative.

The condition $ \cos^{-1} x - \sin^{-1} y = \alpha $ constrains $ x $ and $ y $ to the interval $[-1, 1]$, the principal range for these inverse trigonometric functions.

Rewriting the expression yields:

\[ x^2 + y^2 + 2xy \sin \alpha = (x + y \sin \alpha)^2. \]

The term $ (x + y \sin \alpha)^2 $ achieves its minimum value of 0 when $ x + y \sin \alpha = 0 $.

Therefore, the minimum value of $ x^2 + y^2 + 2xy \sin \alpha $ is 0.

Given the inverse trigonometric function assumes principal values only. Let \( x, y \) be any two real numbers in \( [-1, 1] \) such that \[ \cos^{-1}x - \sin^{-1}y = \alpha, \, -\frac{\pi}{2} \leq \alpha \leq \pi. \] Then, the minimum value of \( x^2 + y^2 + 2xy \sin \alpha \) is:

The Correct Option is B

Solution and Explanation

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