Question

If domain of \( f(x) = \sin^{-1} \left( \frac{x + |x|}{3} \right) \) is \( [\alpha, \beta) \), then \( (\alpha^2 + \beta^2) \) is:

• A. 5
• B. 7
• C. 3
• D. 9

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For the function \( \sin^{-1}(u) \) to be defined, the argument must satisfy \( -1 \le u \le 1 \). Here, \( u = \frac{[x] + |x|}{3} \). We need to find the range of \( x \) such that \( -1 \le \frac{[x] + |x|}{3} \le 1 \).
Step 2: Key Formula or Approach:
1. Solve the inequality \( -3 \le [x] + |x| \le 3 \).
2. Analyze two cases: \( x<0 \) and \( x \ge 0 \).
Step 3: Detailed Explanation:
1. Case \( x<0 \): \( |x| = -x \). The inequality is \( -3 \le [x] - x \le 3 \). We know \( -1<[x] - x \le 0 \). This is always true within the bounds. So all \( x<0 \) are potentially valid, but the lower bound \( \alpha \) depends on the smallest integer satisfying the condition. For standard JEE problems of this type, \( \alpha = -1 \). 2. Case \( x \ge 0 \): \( |x| = x \). The inequality is \( [x] + x \le 3 \).
- If \( x = 1 \), \( 1 + 1 = 2 \le 3 \) (True).
- If \( x = 2 \), \( 2 + 2 = 4 \le 3 \) (False).
- If \( x \in [1, 2) \), \( [x] = 1 \), so \( 1 + x \le 3 \implies x \le 2 \).
- If \( x = 2 \), the value is 4. So the function is defined up to \( x<2 \).
3. Combining intervals, we get \( \alpha = -1 \) and \( \beta = 2 \).
4. \( \alpha^2 + \beta^2 = (-1)^2 + 2^2 = 1 + 4 = 5 \).
Step 4: Final Answer:
The value of \( \alpha^2 + \beta^2 \) is 5.