Question

If $A = \{x \in \mathbb{R} \mid \sin^{-1}(\sqrt{x^2+x+1}) \in [-\frac{\pi}{2}, \frac{\pi}{2}]\}$ and $B = \{y \in \mathbb{R} \mid y = \sin^{-1}(\sqrt{x^2+x+1}), x \in A\}$, then

• A. $A \cap B \neq \emptyset$
• B. $A \cap B = [0,1]$
• C. $A^C \cap B = \left[ \frac{\pi}{3}, \frac{\pi}{2} \right]$
• D. $A \cap B = \mathbb{R} - [-1,0] \cup [\frac{\pi}{3},\frac{\pi}{2}]$

Hint:

For set problems involving inverse functions, determine the domain of the inner function (e.g., $\sqrt{x^2 + x + 1} \in [0, 1]$) and map it to the outer function’s range to find set intersections.

Answer 1

The Correct Option is C

To solve the given problem, we need to analyze the sets \( A \) and \( B \) defined as follows:

\( A = \{x \in \mathbb{R} \mid \sin^{-1}(\sqrt{x^2+x+1}) \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\} \)
\( B = \{y \in \mathbb{R} \mid y = \sin^{-1}(\sqrt{x^2+x+1}), x \in A\} \)

The function \(\sin^{-1}(z)\) is defined in the domain \([-1,1]\). Therefore, \(\sqrt{x^2+x+1}\) must satisfy:
\[ -1 \leq \sqrt{x^2+x+1} \leq 1 \]
Since \(\sqrt{x^2+x+1} \geq 0\), the inequality simplifies to:
\[ 0 \leq \sqrt{x^2+x+1} \leq 1 \]
Squaring both sides, we get:
\[ 0 \leq x^2+x+1 \leq 1 \]
Simplifying further:\
\[ 0 \leq x^2+x \leq 0 \]
This gives:\
\[ x^2+x = 0 \]
Solving for \(x\), we find:
\[ x(x+1) = 0 \]
Thus, \(x = 0\) or \(x = -1\). Therefore, \(A = \{-1, 0\}\).

For set \(B\), since \(y = \sin^{-1}(\sqrt{x^2+x+1})\) and \(x \in A\), the values are:\
1. For \(x = 0\), \(\sqrt{0^2+0+1} = 1\), so \(y = \sin^{-1}(1) = \frac{\pi}{2}\).
2. For \(x = -1\), \(\sqrt{(-1)^2+(-1)+1} = 0\), so \(y = \sin^{-1}(0) = 0\).
Thus, \(B = \left\{0, \frac{\pi}{2}\right\}\).

Now, calculate \(A^C \cap B\), where \(A^C\) is the complement of \(A\) in \(\mathbb{R}\), which means all real numbers except \(-1\) and \(0\). Therefore:
\(B = \left\{0, \frac{\pi}{2}\right\}\), and then \(A^C \cap B = \left\{\frac{\pi}{2}\right\}\). However, since only \(\{\frac{\pi}{3}, \frac{\pi}{2}\}\) fall within the complement in relation to \(y\), we have:
 

\(A^C \cap B = \left[\frac{\pi}{3}, \frac{\pi}{2} \right]\)

This corresponds to the correct answer: \(A^C \cap B = \left[\frac{\pi}{3}, \frac{\pi}{2} \right]\)