Suppose f, g and h be three real valued function defined on R Let `f(x) =2x+|x|` `g(x) =1/3(2x-|x|)` `h(x) =f(g(x))` The range of the function `k(x) = 1 + 1/pi(cos^(-1)h(x) + cot^(-1)(h(x)))` is equal to

Options

A. `[1/4,7/4]`
B. `[5/4,11/4]`
C. `[1/4,5/4]`
D. `[7/4,11/4]`

Answer

To solve the problem step by step, we need to analyze the functions \( f(x) \), \( g(x) \), and \( h(x) \) and then find the range of the function \( k(x) \).

### Step 1: Analyze the function \( f(x) \)

The function \( f(x) \) is defined as:
\[ f(x) = 2x + |x| \]

We can break this down based on the value of \( x \):
- If \( x \geq 0 \):
\[ f(x) = 2x + x = 3x \]
- If \( x < 0 \):
\[ f(x) = 2x - x = x \]

Thus, we can summarize:
\[ f(x) = \begin{cases} 3x & \text{if } x \geq 0 \\ x & \text{if } x < 0 \end{cases} \]

### Step 2: Analyze the function \( g(x) \)

The function \( g(x) \) is defined as:
\[ g(x) = \frac{1}{3}(2x - |x|) \]

Again, we can break this down based on the value of \( x \):
- If \( x \geq 0 \):
\[ g(x) = \frac{1}{3}(2x - x) = \frac{x}{3} \]
- If \( x < 0 \):
\[ g(x) = \frac{1}{3}(2x + x) = x \]

Thus, we can summarize:
\[ g(x) = \begin{cases} \frac{x}{3} & \text{if } x \geq 0 \\ x & \text{if } x < 0 \end{cases} \]

### Step 3: Analyze the function \( h(x) \)

The function \( h(x) \) is defined as:
\[ h(x) = f(g(x)) \]

Now we will evaluate \( h(x) \) based on the cases for \( g(x) \):
- If \( x \geq 0 \):
\[ g(x) = \frac{x}{3} \quad \Rightarrow \quad h(x) = f\left(\frac{x}{3}\right) = 3\left(\frac{x}{3}\right) = x \]
- If \( x < 0 \):
\[ g(x) = x \quad \Rightarrow \quad h(x) = f(x) = x \]

Thus, we can summarize:
\[ h(x) = x \quad \text{for all } x \in \mathbb{R} \]

### Step 4: Analyze the function \( k(x) \)

The function \( k(x) \) is defined as:
\[ k(x) = 1 + \frac{1}{\pi} \left( \cos^{-1}(h(x)) + \cot^{-1}(h(x)) \right) \]

Since \( h(x) = x \), we can rewrite \( k(x) \) as:
\[ k(x) = 1 + \frac{1}{\pi} \left( \cos^{-1}(x) + \cot^{-1}(x) \right) \]

### Step 5: Determine the range of \( k(x) \)

1. **Domain of \( k(x) \)**: The functions \( \cos^{-1}(x) \) and \( \cot^{-1}(x) \) have specific domains:
- \( \cos^{-1}(x) \) is defined for \( x \in [-1, 1] \).
- \( \cot^{-1}(x) \) is defined for all real \( x \).

2. **Range of \( \cos^{-1}(x) \)**: The range is \( [0, \pi] \) for \( x \in [-1, 1] \).

3. **Range of \( \cot^{-1}(x) \)**: The range is \( (0, \pi) \).

4. **Combine the ranges**:
- For \( x = -1 \):
\[ k(-1) = 1 + \frac{1}{\pi}(\pi + \cot^{-1}(-1)) = 1 + 1 + \frac{1}{\pi}(\frac{\pi}{4}) = 2 + \frac{1}{4} \]
- For \( x = 1 \):
\[ k(1) = 1 + \frac{1}{\pi}(0 + \cot^{-1}(1)) = 1 + \frac{1}{\pi}(\frac{\pi}{4}) = 1 + \frac{1}{4} = 1.25 \]

Thus, the range of \( k(x) \) is from \( 1 \) to \( 2 \).

### Final Answer

The range of the function \( k(x) \) is:
\[ \text{Range of } k(x) = \left[ 1, 2 \right] \]