If the range of `f(x)=tan^(1)x+2sin^(-1)x+cos^(-1)x` is `[a, b]`, then

Options

A. `a=(pi)/(4)`
B. `a=-(pi)/(2)`
C. `b=(5pi)/(4)`
D. `b=(3pi)/(2)`

Answer

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

### Step 1: Rewrite the function
We can rewrite the function by using the identity:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
Thus, we can express \( f(x) \) as:
\[ f(x) = \tan^{-1}(x) + 2 \sin^{-1}(x) = \tan^{-1}(x) + \sin^{-1}(x) + \sin^{-1}(x) + \cos^{-1}(x) \]
This simplifies to:
\[ f(x) = \tan^{-1}(x) + \frac{\pi}{2} + \sin^{-1}(x) \]

### Step 2: Determine the domain
Next, we need to find the common domain of \( \tan^{-1}(x) \) and \( \sin^{-1}(x) \):
- The domain of \( \tan^{-1}(x) \) is \( (-\infty, \infty) \).
- The domain of \( \sin^{-1}(x) \) is \( [-1, 1] \).

The common domain of both functions is:
\[ [-1, 1] \]

### Step 3: Find the minimum value
To find the minimum value of \( f(x) \) on the interval \( [-1, 1] \), we will evaluate \( f(x) \) at the endpoints:
- At \( x = -1 \):
\[ f(-1) = \tan^{-1}(-1) + \frac{\pi}{2} + \sin^{-1}(-1) \]
Calculating each term:
\[ \tan^{-1}(-1) = -\frac{\pi}{4}, \quad \sin^{-1}(-1) = -\frac{\pi}{2} \]
Thus,
\[ f(-1) = -\frac{\pi}{4} + \frac{\pi}{2} - \frac{\pi}{2} = -\frac{\pi}{4} \]

### Step 4: Find the maximum value
Now, we will evaluate \( f(x) \) at the other endpoint:
- At \( x = 1 \):
\[ f(1) = \tan^{-1}(1) + \frac{\pi}{2} + \sin^{-1}(1) \]
Calculating each term:
\[ \tan^{-1}(1) = \frac{\pi}{4}, \quad \sin^{-1}(1) = \frac{\pi}{2} \]
Thus,
\[ f(1) = \frac{\pi}{4} + \frac{\pi}{2} + \frac{\pi}{2} = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \]

### Step 5: Conclusion
The minimum value of \( f(x) \) is \( -\frac{\pi}{4} \) and the maximum value is \( \frac{5\pi}{4} \). Therefore, the range of \( f(x) \) is:
\[ \left[-\frac{\pi}{4}, \frac{5\pi}{4}\right] \]

### Final Answer
The values of \( a \) and \( b \) are:
\[ a = -\frac{\pi}{4}, \quad b = \frac{5\pi}{4} \]