Question

Find the domain of the real valued function: \[ f(x) \;=\; \sqrt{\cos(\sin x)} + \cos^{-1}\left(\frac{1+x^2}{2x}\right). \]

• A. \((-1, 1)\)
• B. \([-1, 1]\)
• C. \(\mathbb{R} \setminus (-1, 1)\)
• D. \(\{-1, 1\}\)

Hint:

Check the boundaries of functions involving inverse trigonometric functions since their domains are restricted to \([-1, 1]\).

Answer 1

The Correct Option is D

Step 1: Analyzing \( \cos^{-1}\left(\frac{1+x^2}{2x}\right) \).
The expression inside the \( \cos^{-1} \) function needs to be within \([-1, 1]\) for real values: \[ \left|\frac{1+x^2}{2x}\right| \leq 1 \quad \Rightarrow \quad -2x \leq 1 + x^2 \leq 2x \] This simplifies to \(x^2 - 2x - 1 \leq 0\) and \(x^2 + 2x - 1 \geq 0\), which holds only for \(x = -1\) and \(x = 1\). Step 2: Verifying the domain.
Checking at \(x = -1\) and \(x = 1\): \[ \cos(\sin(-1)) + \cos^{-1}\left(\frac{1+(-1)^2}{2(-1)}\right) = \cos(\sin(-1)) + \cos^{-1}(1) = \cos(\sin(-1)) + 0, \] \[ \cos(\sin(1)) + \cos^{-1}\left(\frac{1+1^2}{2 \cdot 1}\right) = \cos(\sin(1)) + \cos^{-1}(1) = \cos(\sin(1)) + 0, \] both of which are valid since \(\cos(\sin(\pm 1))\) is defined.