Question

The domain of the function $f(x)=\sin^{-1}(5-4x^{2})+\cos^{-1}(5-4x^{2})$ is

• A. $[1,\sqrt{\frac{3}{2}}]\cup[-\sqrt{\frac{3}{2}},-1]$
• B. $[1,\sqrt{\frac{5}{2}}]\cup[-\sqrt{\frac{3}{2}},-1]$
• C. $[1,\sqrt{\frac{3}{2}}]\cup[-\sqrt{\frac{5}{2}},-1]$
• D. $[2,\sqrt{\frac{3}{2}}]\cup[-\sqrt{\frac{3}{2}},-1]$
• E. $[1,\sqrt{\frac{3}{2}}]\cup[-\sqrt{\frac{3}{2}},-2]$

Hint:

Math Tip: When an expression features a combination of $\sin^{-1}(f(x))$ and $\cos^{-1}(f(x))$, you only need to evaluate the domain once, because their domain restrictions are identical.

Answer 1

The Correct Option is A

Concept:
Inverse Trigonometric Functions - Domain constraints.
Both $\sin^{-1}(y)$ and $\cos^{-1}(y)$ share the same domain restriction: the argument $y$ must satisfy $-1 \le y \le 1$.
Step 1: Set up the domain inequality.
Substitute the given argument $(5-4x^2)$ into the required domain condition: $$ -1 \le 5 - 4x^2 \le 1 $$
Step 2: Isolate the $x^2$ term.
Subtract $5$ from all parts of the inequality: $$ -1 - 5 \le -4x^2 \le 1 - 5 $$ $$ -6 \le -4x^2 \le -4 $$
Step 3: Divide and reverse the inequality signs.
Divide the entire inequality by $-4$. Remember to flip the direction of the inequalities when dividing by a negative number: $$ \frac{-6}{-4} \ge \frac{-4x^2}{-4} \ge \frac{-4}{-4} $$ $$ \frac{3}{2} \ge x^2 \ge 1 $$ Rearranging it to standard order gives: $$ 1 \le x^2 \le \frac{3}{2} $$
Step 4: Solve for $x$ by taking the square root.
Taking the square root of all parts yields the absolute value of $x$: $$ 1 \le |x| \le \sqrt{\frac{3}{2}} $$
Step 5: Split into positive and negative intervals.
The absolute value inequality translates to two separate regions on the number line:

• Positive region: $1 \le x \le \sqrt{\frac{3}{2}}$
• Negative region: $-\sqrt{\frac{3}{2}} \le x \le -1$

Step 6: Express the final answer in set notation.
Combine both intervals using the union symbol: $$ x \in \left[-\sqrt{\frac{3}{2}}, -1\right] \cup \left[1, \sqrt{\frac{3}{2}}\right] $$