Question

The domain of the function $f(x) = \sin^{-1}\{\log_2(\frac{1}{2}x^2)\}$ is

• A. [-2, -1) $\cup$ [1, 2]
• B. (-2, -1] $\cup$ [1, 2]
• C. [-2, -1] $\cup$ [1, 2]
• D. (-2, -1) $\cup$ (1, 2)

Answer 1

Answer: (C)

[-2, -1] $\cup$ [1, 2]

Answer 2

For the function $f(x) = \sin^{-1}\{\log_2(\frac{1}{2}x^2)\}$ to be defined, two conditions must be met:

1. Domain of $\sin^{-1}(u)$: The argument $u$ must lie in the interval $[-1, 1]$. Here, $u = \log_2(\frac{1}{2}x^2)$. So, we must have: $-1 \le \log_2\left(\frac{1}{2}x^2\right) \le 1$

2. Domain of $\log_b(v)$: The argument $v$ must be strictly positive. Here, $v = \frac{1}{2}x^2$. So, we must have: $\frac{1}{2}x^2 > 0 \implies x^2 > 0 \implies x \ne 0$

Now, let's solve the inequality from condition 1: $-1 \le \log_2\left(\frac{1}{2}x^2\right) \le 1$ Since the base of the logarithm is $2$ (which is greater than 1), we can convert the logarithmic inequalities to exponential inequalities without changing the direction of the inequality signs: $2^{-1} \le \frac{1}{2}x^2 \le 2^1$ $\frac{1}{2} \le \frac{1}{2}x^2 \le 2$ This can be split into two separate inequalities:

• Inequality 1: $\frac{1}{2} \le \frac{1}{2}x^2$ Multiply both sides by 2: $1 \le x^2$ $x^2 \ge 1$ This implies $x \le -1$ or $x \ge 1$. In interval notation, $x \in (-\infty, -1] \cup [1, \infty)$.

• Inequality 2: $\frac{1}{2}x^2 \le 2$ Multiply both sides by 2: $x^2 \le 4$ This implies $-2 \le x \le 2$. In interval notation, $x \in [-2, 2]$.

To satisfy both inequalities simultaneously, we need to find the intersection of their solution sets: $((-\infty, -1] \cup [1, \infty)) \cap [-2, 2]$ This intersection is $[-2, -1] \cup [1, 2]$.

Finally, we must also satisfy condition 2, which states $x \ne 0$. The interval $[-2, -1] \cup [1, 2]$ does not include $0$, so this condition is already satisfied.

Therefore, the domain of the function $f(x)$ is $[-2, -1] \cup [1, 2]$.