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

Question

Accepted Answer

To determine the range of the function $ f(x) = \sin^{-1}\left(\frac{x^2 + 1}{x^2 + 2}\right) $, we need to analyze the expression inside the inverse sine function, $\frac{x^2 + 1}{x^2 + 2}$.

### Step-by-Step Analysis:

1. **Expression Analysis**:
   - The expression $\frac{x^2 + 1}{x^2 + 2}$ is a rational function. For any real number $x$, $x^2$ is non-negative, so $x^2 + 1$ and $x^2 + 2$ are both positive.
   - The function $\frac{x^2 + 1}{x^2 + 2}$ is always less than 1 because:
     $$
     \frac{x^2 + 1}{x^2 + 2} < 1 \quad \text{since} \quad x^2 + 1 < x^2 + 2.
     $$
   - The minimum value of $\frac{x^2 + 1}{x^2 + 2}$ occurs when $x^2$ is minimized, i.e., $x = 0$. Substituting $x = 0$, we get:
     $$
     \frac{0^2 + 1}{0^2 + 2} = \frac{1}{2}.
     $$

2. **Range of the Expression**:
   - As $x$ varies over all real numbers, $\frac{x^2 + 1}{x^2 + 2}$ ranges from $\frac{1}{2}$ to values approaching 1 (but never reaching 1).
   - Therefore, the range of $\frac{x^2 + 1}{x^2 + 2}$ is $[\frac{1}{2}, 1)$.

3. **Range of the Inverse Sine Function**:
   - The function $\sin^{-1}(y)$ is defined for $y \in [-1, 1]$ and its range is $[- \frac{\pi}{2}, \frac{\pi}{2}]$.
   - For $y \in [\frac{1}{2}, 1)$, $\sin^{-1}(y)$ will range from $\sin^{-1}(\frac{1}{2})$ to $\sin^{-1}(1)$.
   - We know:
     $$
     \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} \quad \text{and} \quad \sin^{-1}(1) = \frac{\pi}{2}.
     $$
   - Therefore, the range of $f(x) = \sin^{-1}\left(\frac{x^2 + 1}{x^2 + 2}\right)$ is $[\frac{\pi}{6}, \frac{\pi}{2})$.

### Justification for Option 3:

- **Option 3: $[\frac{\pi}{6}, \frac{\pi}{2})$** is correct because it matches the calculated range of the function $f(x)$.

### Explanation for Incorrect Options:

1. **Option 1: $[0, \frac{\pi}{2}]$**:
   - This option is incorrect because the minimum value of $\sin^{-1}\left(\frac{x^2 + 1}{x^2 + 2}\right)$ is $\frac{\pi}{6}$, not 0.

2. **Option 2: $(0, \frac{\pi}{6})$**:
   - This option is incorrect because the range starts at $\frac{\pi}{6}$, not below it.

4. **Option 4: None of these**:
   - This option is incorrect because Option 3 correctly describes the range of the function.

In conclusion, the correct range of the function $f(x) = \sin^{-1}\left(\frac{x^2 + 1}{x^2 + 2}\right)$ is $[\frac{\pi}{6}, \frac{\pi}{2})$, which is why Option 3 is the correct answer.