Question

The domain of the function \(f(x) = \sin^{-1} \left( \frac{3x^2 + x - 1}{(x - 1)^2} \right) + \cos^{-1} \left( \frac{x - 1}{x + 1} \right)\) is :

• A. \([0, \frac{1}{4}]\)
• B. \([0, \frac{1}{2}]\)
• C. \([\frac{1}{4}, \frac{1}{2}] \cup \{0\}\)
• D. \([-2, 0] \cup [\frac{1}{4}, \frac{1}{2}]\)

Hint:

When dealing with \(\{0\}\) as part of an answer choice, check if \(x=0\) specifically satisfies all individual function constraints. Here at \(x=0\), the arguments are \(-1\) and \(-1\), both of which are valid for inverse trig functions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The domain of \(f(x) = g(x) + h(x)\) is the intersection of the domains of \(g(x)\) and \(h(x)\). For inverse sine and cosine, the argument must lie in the interval \([-1, 1]\).
Step 2: Detailed Explanation:
Let \(D_1\) be the domain of \(\sin^{-1} \left( \frac{3x^2 + x - 1}{(x - 1)^2} \right)\): \[ -1 \le \frac{3x^2 + x - 1}{(x - 1)^2} \le 1 \] Since \((x-1)^2>0\) for \(x \ne 1\), we have: (i) \(3x^2 + x - 1 \le (x - 1)^2 = x^2 - 2x + 1 \implies 2x^2 + 3x - 2 \le 0\) \[ (2x - 1)(x + 2) \le 0 \implies x \in [-2, \frac{1}{2}] \] (ii) \(3x^2 + x - 1 \ge -(x - 1)^2 = -x^2 + 2x - 1 \implies 4x^2 - x \ge 0\) \[ x(4x - 1) \ge 0 \implies x \in (-\infty, 0] \cup [\frac{1}{4}, \infty) \] Intersection for \(D_1\): \([-2, 0] \cup [\frac{1}{4}, \frac{1}{2}]\).
Let \(D_2\) be the domain of \(\cos^{-1} \left( \frac{x - 1}{x + 1} \right)\): \[ -1 \le \frac{x - 1}{x + 1} \le 1 \] This simplifies to \(\frac{x}{x+1} \ge 0\) and \(\frac{-2}{x+1} \le 0\), which gives \(x \ge 0\) (with \(x \ne -1\)).
Intersection \(D_1 \cap D_2\): \[ ([-2, 0] \cup [\frac{1}{4}, \frac{1}{2}]) \cap [0, \infty) = \{0\} \cup [\frac{1}{4}, \frac{1}{2}] \]
Step 3: Final Answer:
The domain is \([\frac{1}{4}, \frac{1}{2}] \cup \{0\}\).