Let `f(x)= cos^(-1)((x^(2))/(x^(2)+1))`. Then , the range of the f , is

Options

A. `(0, pi//2]`
B. `[-pi//2, pi//2]`
C. `[-pi//2, 0]`
D. none of these

Explanation

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

### Step 1: Determine the range of the inner function
We start by analyzing the expression inside the cosine inverse function, which is \( \frac{x^2}{x^2 + 1} \).

**Hint:** Consider the behavior of the function as \( x \) varies over all real numbers.

### Step 2: Analyze the limits of \( \frac{x^2}{x^2 + 1} \)
As \( x \) approaches \( 0 \):
\[ \frac{x^2}{x^2 + 1} = \frac{0}{0 + 1} = 0 \]

As \( x \) approaches \( \infty \):
\[ \frac{x^2}{x^2 + 1} \approx \frac{x^2}{x^2} = 1 \]

Since \( x^2 \) is always non-negative, \( \frac{x^2}{x^2 + 1} \) will always yield values between \( 0 \) and \( 1 \).

**Hint:** Check the continuity and monotonicity of the function \( \frac{x^2}{x^2 + 1} \).

### Step 3: Establish the range of \( \frac{x^2}{x^2 + 1} \)
The function \( \frac{x^2}{x^2 + 1} \) is continuous and increases from \( 0 \) to \( 1 \) as \( x \) moves from \( 0 \) to \( \infty \) (and symmetrically from \( 0 \) to \( -\infty \)). Therefore, the range of \( \frac{x^2}{x^2 + 1} \) is:
\[ [0, 1) \]

**Hint:** Remember that the function approaches \( 1 \) but never actually reaches it.

### Step 4: Apply the cosine inverse function
Now, we apply \( \cos^{-1} \) to the interval \( [0, 1) \):
\[ f(x) = \cos^{-1}\left(y\right) \quad \text{where } y \in [0, 1) \]

The cosine inverse function \( \cos^{-1}(y) \) is defined for \( y \in [0, 1] \) and its range is:
\[ [0, \frac{\pi}{2}] \]

**Hint:** Consider the endpoints of the interval when applying the inverse cosine function.

### Step 5: Determine the final range of \( f(x) \)
Since \( y \) can take values from \( 0 \) to just below \( 1 \), we find:
\[ f(x) \in [0, \frac{\pi}{2}) \]

Thus, the range of the function \( f(x) \) is:
\[ \boxed{[0, \frac{\pi}{2})} \]