The range of the function : `f(x)=tan^(- 1)x+1/2sin^(- 1)x`

Options

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

Answer

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

### Step 1: Determine the Domain of the Function
The function consists of two parts: \( \tan^{-1} x \) and \( \frac{1}{2} \sin^{-1} x \).

- The domain of \( \tan^{-1} x \) is all real numbers, \( x \in \mathbb{R} \).
- The domain of \( \sin^{-1} x \) is restricted to \( x \in [-1, 1] \).

Thus, the overall domain of \( f(x) \) is the intersection of these two domains, which is:
\[ x \in [-1, 1] \]

### Step 2: Evaluate the Function at the Endpoints of the Domain
Next, we will evaluate \( f(x) \) at the endpoints of the domain to find the minimum and maximum values.

#### Evaluate at \( x = -1 \):
\[ f(-1) = \tan^{-1}(-1) + \frac{1}{2} \sin^{-1}(-1) \]
- \( \tan^{-1}(-1) = -\frac{\pi}{4} \)
- \( \sin^{-1}(-1) = -\frac{\pi}{2} \)

Thus,
\[ f(-1) = -\frac{\pi}{4} + \frac{1}{2} \left(-\frac{\pi}{2}\right) = -\frac{\pi}{4} - \frac{\pi}{4} = -\frac{\pi}{2} \]

#### Evaluate at \( x = 1 \):
\[ f(1) = \tan^{-1}(1) + \frac{1}{2} \sin^{-1}(1) \]
- \( \tan^{-1}(1) = \frac{\pi}{4} \)
- \( \sin^{-1}(1) = \frac{\pi}{2} \)

Thus,
\[ f(1) = \frac{\pi}{4} + \frac{1}{2} \left(\frac{\pi}{2}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \]

### Step 3: Determine the Range of the Function
From our evaluations:
- The minimum value of \( f(x) \) is \( -\frac{\pi}{2} \) at \( x = -1 \).
- The maximum value of \( f(x) \) is \( \frac{\pi}{2} \) at \( x = 1 \).

Therefore, the range of the function \( f(x) \) is:
\[ \text{Range of } f(x) = \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]

### Final Answer
The range of the function \( f(x) = \tan^{-1} x + \frac{1}{2} \sin^{-1} x \) is:
\[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]

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