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

Answer 1

Answer: (C)

C

Answer 2

Step-by-step Derivations:

1. Determine Set A:

Set A is defined as $A = \{x \in \mathbb{R} \mid \sin^{-1}(\sqrt{x^2+x+1}) \in [-\frac{\pi}{2}, \frac{\pi}{2}]\}$.

For $\sin^{-1}(z)$ to be defined, its argument $z$ must satisfy $-1 \le z \le 1$. Here, $z = \sqrt{x^2+x+1}$. So, we need: a. $\sqrt{x^2+x+1} \ge 0$ b. $\sqrt{x^2+x+1} \le 1$

Let's analyze the quadratic expression $x^2+x+1$. The discriminant is $\Delta = b^2 - 4ac = (1)^2 - 4(1)(1) = 1 - 4 = -3$. Since $\Delta < 0$ and the leading coefficient (1) is positive, the quadratic $x^2+x+1$ is always positive for all $x \in \mathbb{R}$. Therefore, $\sqrt{x^2+x+1}$ is always defined and positive, satisfying condition (a).

Now consider condition (b): $\sqrt{x^2+x+1} \le 1$. Since both sides are non-negative, we can square both sides without changing the inequality direction: $x^2+x+1 \le 1^2$ $x^2+x+1 \le 1$ $x^2+x \le 0$ Factor out $x$: $x(x+1) \le 0$ This inequality holds when $x$ and $(x+1)$ have opposite signs or one of them is zero. The roots are $x=0$ and $x=-1$. The expression $x(x+1)$ is a parabola opening upwards, so it is less than or equal to zero between its roots. Thus, $-1 \le x \le 0$.

The range of $\sin^{-1}(z)$ is inherently $[-\frac{\pi}{2}, \frac{\pi}{2}]$. So, if $\sin^{-1}(\sqrt{x^2+x+1})$ is defined, it will automatically fall within this range. Therefore, set A is the set of all $x$ for which the expression is defined: $A = [-1, 0]$.

2. Determine Set B:

Set B is defined as $B = \{y \in \mathbb{R} \mid y = \sin^{-1}(\sqrt{x^2+x+1}), x \in A\}$. This means B is the range of the function $f(x) = \sin^{-1}(\sqrt{x^2+x+1})$ for $x \in [-1, 0]$.

Let's find the range of the inner function $u(x) = x^2+x+1$ for $x \in [-1, 0]$. This is a parabola opening upwards. Its vertex is at $x = -\frac{b}{2a} = -\frac{1}{2(1)} = -\frac{1}{2}$. Since $x = -1/2$ is within the interval $[-1, 0]$, the minimum value of $u(x)$ occurs at the vertex: $u_{min} = u(-1/2) = (-1/2)^2 + (-1/2) + 1 = 1/4 - 1/2 + 1 = 1/4 - 2/4 + 4/4 = 3/4$.

The maximum value of $u(x)$ in the interval $[-1, 0]$ occurs at one of the endpoints: $u(-1) = (-1)^2 + (-1) + 1 = 1 - 1 + 1 = 1$. $u(0) = (0)^2 + (0) + 1 = 1$. So, for $x \in [-1, 0]$, the range of $u(x) = x^2+x+1$ is $[3/4, 1]$.

Next, consider the function $g(u) = \sqrt{u}$. This function is increasing. So, for $u \in [3/4, 1]$, the range of $\sqrt{u}$ is $[\sqrt{3/4}, \sqrt{1}] = [\frac{\sqrt{3}}{2}, 1]$.

Finally, consider the function $h(v) = \sin^{-1}(v)$. This function is also increasing. So, for $v \in [\frac{\sqrt{3}}{2}, 1]$, the range of $\sin^{-1}(v)$ is $[\sin^{-1}(\frac{\sqrt{3}}{2}), \sin^{-1}(1)]$. We know that $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{2}) = 1$. Therefore, $\sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$ and $\sin^{-1}(1) = \frac{\pi}{2}$. So, set B is $[\frac{\pi}{3}, \frac{\pi}{2}]$.

3. Evaluate the Options:

We have $A = [-1, 0]$ and $B = [\frac{\pi}{3}, \frac{\pi}{2}]$. Note that $\frac{\pi}{3} \approx 1.047$ and $\frac{\pi}{2} \approx 1.571$.

A. $A \cap B \neq \emptyset$

$A = [-1, 0]$ (contains non-positive numbers). $B = [\frac{\pi}{3}, \frac{\pi}{2}]$ (contains positive numbers greater than 1). The intervals are disjoint. $A \cap B = \emptyset$. So, option A is incorrect.

B. $A \cap B = [0, 1]$

As determined above, $A \cap B = \emptyset$. So, option B is incorrect.

C. $A^C \cap B = [\frac{\pi}{3}, \frac{\pi}{2}]$

First, find $A^C$ (complement of A in $\mathbb{R}$): $A^C = \mathbb{R} \setminus [-1, 0] = (-\infty, -1) \cup (0, \infty)$. Now, find $A^C \cap B$: $A^C \cap B = ((-\infty, -1) \cup (0, \infty)) \cap [\frac{\pi}{3}, \frac{\pi}{2}]$. Since $\frac{\pi}{3} > 0$, the entire interval $B = [\frac{\pi}{3}, \frac{\pi}{2}]$ lies within the interval $(0, \infty)$. Therefore, $A^C \cap B = [\frac{\pi}{3}, \frac{\pi}{2}]$. So, option C is correct.

D. $A \cap B = \mathbb{R} - [-1, 0] \cup [\frac{\pi}{3}, \frac{\pi}{2}]$

We know $A \cap B = \emptyset$. The set $\mathbb{R} - [-1, 0] \cup [\frac{\pi}{3}, \frac{\pi}{2}]$ is not empty. So, option D is incorrect.