The maximum and minimum values of `f(x)=sin^(-1)x+cos^(-1)x+tan^(-1)x` respectively is

Options

A. `(3pi)/4,pi/2`
B. `(3pi)/4,pi/4`
C. `pi/4,(-pi)/4`
D. `pi,0`

Answer

To find the maximum and minimum values of the function \( f(x) = \sin^{-1} x + \cos^{-1} x + \tan^{-1} x \), we will follow these steps:

### Step 1: Simplify the function
We know the identity:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Thus, we can rewrite the function as:
\[ f(x) = \frac{\pi}{2} + \tan^{-1} x \]

### Step 2: Determine the domain
The domains of the inverse trigonometric functions are:
- \( \sin^{-1} x \): \( x \in [-1, 1] \)
- \( \cos^{-1} x \): \( x \in [-1, 1] \)
- \( \tan^{-1} x \): \( x \in (-\infty, \infty) \)

The common domain for \( \sin^{-1} x \) and \( \cos^{-1} x \) is:
\[ x \in [-1, 1] \]

### Step 3: Analyze the behavior of \( f(x) \)
Since \( f(x) = \frac{\pi}{2} + \tan^{-1} x \), we need to analyze \( \tan^{-1} x \) within the domain \( [-1, 1] \).

### Step 4: Find the values of \( f(x) \) at the endpoints of the domain
1. **At \( x = -1 \)**:
\[ f(-1) = \frac{\pi}{2} + \tan^{-1}(-1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

2. **At \( x = 1 \)**:
\[ f(1) = \frac{\pi}{2} + \tan^{-1}(1) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4} \]

### Step 5: Determine the maximum and minimum values
From the calculations:
- The minimum value of \( f(x) \) occurs at \( x = -1 \):
\[ f(-1) = \frac{\pi}{4} \]

- The maximum value of \( f(x) \) occurs at \( x = 1 \):
\[ f(1) = \frac{3\pi}{4} \]

### Conclusion
Thus, the minimum and maximum values of \( f(x) \) are:
- Minimum value: \( \frac{\pi}{4} \)
- Maximum value: \( \frac{3\pi}{4} \)