The range of the function `f(x)=tan^(- 1)((x^2+1)/(x^2+sqrt(3)))` `x in R` is

Options

A. `[(pi)/(6),(pi)/(2))`
B. `[(pi)/(6),(pi)/(3))`
C. `[(pi)/(6),(pi)/(4))`
D. none of these

Explanation

To find the range of the function \( f(x) = \tan^{-1} \left( \frac{x^2 + 1}{x^2 + \sqrt{3}} \right) \), we will follow these steps:

### Step 1: Define the function
Let \( y = \frac{x^2 + 1}{x^2 + \sqrt{3}} \).

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ y (x^2 + \sqrt{3}) = x^2 + 1 \]
This simplifies to:
\[ yx^2 + \sqrt{3}y = x^2 + 1 \]

### Step 3: Rearrange the equation
Rearranging the terms, we get:
\[ yx^2 - x^2 = 1 - \sqrt{3}y \]
Factoring out \( x^2 \):
\[ x^2 (y - 1) = 1 - \sqrt{3}y \]
Thus, we can express \( x^2 \) as:
\[ x^2 = \frac{1 - \sqrt{3}y}{y - 1} \]

### Step 4: Determine the conditions for \( x^2 \geq 0 \)
Since \( x^2 \) must be non-negative, we require:
\[ \frac{1 - \sqrt{3}y}{y - 1} \geq 0 \]
This inequality holds when both the numerator and denominator are either both positive or both negative.

### Step 5: Analyze the numerator and denominator
1. **Numerator**: \( 1 - \sqrt{3}y \geq 0 \) implies \( y \leq \frac{1}{\sqrt{3}} \).
2. **Denominator**: \( y - 1 > 0 \) implies \( y > 1 \).

Combining these inequalities, we find:
- From the numerator, \( y \leq \frac{1}{\sqrt{3}} \).
- From the denominator, \( y > 1 \).

These two conditions cannot be satisfied simultaneously, so we need to analyze the critical points.

### Step 6: Find critical points
To find where the expression changes sign, we can set the numerator and denominator to zero:
1. \( 1 - \sqrt{3}y = 0 \) gives \( y = \frac{1}{\sqrt{3}} \).
2. \( y - 1 = 0 \) gives \( y = 1 \).

### Step 7: Test intervals
We test the intervals determined by these critical points:
- For \( y < \frac{1}{\sqrt{3}} \): both numerator and denominator are positive.
- For \( \frac{1}{\sqrt{3}} < y < 1 \): numerator is negative, denominator is positive.
- For \( y > 1 \): numerator is negative, denominator is negative.

From this analysis, we conclude that \( y \) must be in the interval \( \left( \frac{1}{\sqrt{3}}, 1 \right) \).

### Step 8: Find the range of \( f(x) \)
Now, we need to find the range of \( f(x) = \tan^{-1}(y) \) where \( y \) is in the interval \( \left( \frac{1}{\sqrt{3}}, 1 \right) \).

Calculating the values:
- \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \)
- \( \tan^{-1}(1) = \frac{\pi}{4} \)

Thus, the range of \( f(x) \) is:
\[ \left( \frac{\pi}{6}, \frac{\pi}{4} \right) \]

### Final Answer
The range of the function \( f(x) = \tan^{-1} \left( \frac{x^2 + 1}{x^2 + \sqrt{3}} \right) \) is \( \left( \frac{\pi}{6}, \frac{\pi}{4} \right) \).

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