Question

The Range of $f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x + 2} \right)$ is:

• A. $(0, \frac{\pi}{2})$
• B. $[0, \frac{\pi}{2}]$
• C. $(0, \frac{\pi}{2}]$
• D. $[0, \frac{\pi}{2})$

Answer 1

The Correct Option is C

Step 1: Analyze the quadratic expression in the denominator.
Let $g(x) = x^2 - 2x + 2$.
By completing the square: $g(x) = (x-1)^2 + 1$.
Since $(x-1)^2 \ge 0$, the minimum value of $g(x)$ is $1$ (at $x=1$).
Thus, $g(x) \in [1, \infty)$.


Step 2: Find the range of the inner fraction.
Let $u = \frac{1}{g(x)} = \frac{1}{(x-1)^2 + 1}$.
As $g(x) \to \infty$, $u \to 0$ (from the positive side).
As $g(x) = 1$, $u = 1$.
So, $u \in (0, 1]$.


Step 3: Find the range of the function $f(x) = \sin^{-1}(u)$.
The function $\sin^{-1}(u)$ is strictly increasing for $u \in (0, 1]$.
Minimum value (limit as $u \to 0^+$): $\lim_{u \to 0} \sin^{-1}(u) = 0$. Since $u$ never actually reaches 0, the lower bound is open: $(0, ...$.
Maximum value (at $u = 1$): $\sin^{-1}(1) = \frac{\pi}{2}$.
So the range is $(0, \frac{\pi}{2}]$.

Conclusion: The range is $(0, \frac{\pi}{2}]$, which corresponds to option (3).