The range of `f(x)=sin^(-1)((x^2+1)/(x^2+2))` is `(a)[0,pi/2]` (b) `(0,pi/6)` (c) `[pi/6,pi/2]` (d) none of these

Options

A. `[0,pi//2]`
B. `(0,pi//6)`
C. `[pi//6,pi//2)`
D. None of these

Explanation

To find the range of the function \( f(x) = \sin^{-1}\left(\frac{x^2 + 1}{x^2 + 2}\right) \), we will follow these steps:

### Step 1: Analyze the function
The function can be rewritten as:
\[ f(x) = \sin^{-1}\left(1 - \frac{1}{x^2 + 2}\right) \]
This transformation helps us understand how the term \( \frac{x^2 + 1}{x^2 + 2} \) behaves.

### Step 2: Determine the range of \( \frac{x^2 + 1}{x^2 + 2} \)
We know that \( x^2 \) is always non-negative for real \( x \). Therefore, \( x^2 + 2 \) is always greater than or equal to 2.

Now, let's find the minimum and maximum values of \( \frac{x^2 + 1}{x^2 + 2} \):
- As \( x^2 \to 0 \), \( \frac{x^2 + 1}{x^2 + 2} \to \frac{1}{2} \).
- As \( x^2 \to \infty \), \( \frac{x^2 + 1}{x^2 + 2} \to 1 \).

Thus, the range of \( \frac{x^2 + 1}{x^2 + 2} \) is \( \left[\frac{1}{2}, 1\right) \).

### Step 3: Apply the arcsine function
Now we apply the arcsine function to the range we just found:
\[ \text{Range of } f(x) = \sin^{-1}\left(\left[\frac{1}{2}, 1\right)\right] \]
Calculating the arcsine:
- \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \)
- \( \sin^{-1}(1) = \frac{\pi}{2} \)

Thus, the range of \( f(x) \) is:
\[ \left[\frac{\pi}{6}, \frac{\pi}{2}\right) \]

### Step 4: Conclusion
The range of the function \( f(x) \) is \( \left[\frac{\pi}{6}, \frac{\pi}{2}\right) \).

### Step 5: Check options
From the given options:
- (a) \([0, \frac{\pi}{2}]\)
- (b) \((0, \frac{\pi}{6})\)
- (c) \([\frac{\pi}{6}, \frac{\pi}{2}]\)
- (d) none of these

The correct answer is option (c) \([\frac{\pi}{6}, \frac{\pi}{2}]\).