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

Question

Let $(a, b) \subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>, \theta \in(0,2 \pi)$, holds If $\alpha x^2+\beta x+\sin ^{-1}\left(x^2-6 x+10\right)+\cos ^{-1}\left(x^2-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to :

Answer 1

Step 1: Understanding the Concept:
To find the range of $ \sin^{-1}(g(x)) $, we first determine the range of the internal function $ g(x) = \frac{1}{x^2 - 2x + 2} $.
Step 2: Key Formula or Approach:
Complete the square for the denominator:
$ x^2 - 2x + 2 = (x - 1)^2 + 1 $.
Let $ D = (x - 1)^2 + 1 $. Since $ (x - 1)^2 \ge 0 $, we have $ D \ge 1 $.
Step 3: Detailed Explanation:
The range of the denominator $ D $ is $ [1, \infty) $.
As $ D \to \infty $, $ \frac{1}{D} \to 0 $.
As $ D = 1 $, $ \frac{1}{D} = 1 $.
Thus, the range of $ \frac{1}{x^2 - 2x + 2} $ is $ (0, 1] $.
Now apply $ \sin^{-1} $:
The minimum value approaches $ \sin^{-1}(0) = 0 $ (not included).
The maximum value is $ \sin^{-1}(1) = \pi/2 $ (included).
Therefore, the range is $ (0, \pi/2] $.
Step 4: Final Answer:
The range of the function is $ (0, \pi/2] $.

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

The Correct Option is C

Solution and Explanation

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