Q. Given the functions `f(x)=e^(cos^-1(sin(x+pi/3)))`, `g(x)=cosec^-1((4-2cos x)/3)` & the function `h(x)=f(x)` defined only for those values of x, which are common to the domains of the functions `f(x)` and `g(x)`, Calculate the range of the function `h(x)` .

Options

A. `[e^((pi)/6),e^(pi)]`
B. `[e^(-(pi)/6),e^(pi)]`
C. `(e^((pi)/6),e^(pi))`
D. `[e^(-(pi)/6),e^((pi)/6)]`

Explanation

To solve the problem step-by-step, we need to find the range of the function \( h(x) \) which is defined as \( h(x) = f(x) \) for values of \( x \) that are common to the domains of the functions \( f(x) \) and \( g(x) \).

### Step 1: Determine the Domain of \( f(x) \)

The function \( f(x) \) is given by:
\[ f(x) = e^{\cos^{-1}(\sin(x + \frac{\pi}{3}))} \]

The \( \cos^{-1}(y) \) function is defined for \( y \) in the interval \([-1, 1]\). Therefore, we need:
\[ \sin(x + \frac{\pi}{3}) \in [-1, 1] \]
Since the sine function always lies within this interval for all \( x \), the domain of \( f(x) \) is:
\[ \text{Domain of } f(x) = \mathbb{R} \]

### Step 2: Determine the Domain of \( g(x) \)

The function \( g(x) \) is given by:
\[ g(x) = \csc^{-1}\left(\frac{4 - 2\cos x}{3}\right) \]

The \( \csc^{-1}(y) \) function is defined for \( y \leq -1 \) or \( y \geq 1 \). Thus, we need to consider both cases:

1. **Case 1:**
\[ \frac{4 - 2\cos x}{3} \leq -1 \]
This simplifies to:
\[ 4 - 2\cos x \leq -3 \implies 7 \leq 2\cos x \implies \cos x \geq \frac{7}{2} \]
Since \( \cos x \) can only take values in \([-1, 1]\), this case yields no valid solutions.

2. **Case 2:**
\[ \frac{4 - 2\cos x}{3} \geq 1 \]
This simplifies to:
\[ 4 - 2\cos x \geq 3 \implies 1 \geq 2\cos x \implies \cos x \leq \frac{1}{2} \]

The values of \( x \) for which \( \cos x \leq \frac{1}{2} \) occur in the intervals:
\[ x \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right] \cup \left[\frac{2\pi}{3}, \frac{4\pi}{3}\right] \]

### Step 3: Find the Intersection of the Domains

The domain of \( f(x) \) is \( \mathbb{R} \) and the domain of \( g(x) \) is:
\[ \left[-\frac{\pi}{3}, \frac{\pi}{3}\right] \cup \left[\frac{2\pi}{3}, \frac{4\pi}{3}\right] \]

Thus, the domain of \( h(x) \) is:
\[ \text{Domain of } h(x) = \left[-\frac{\pi}{3}, \frac{\pi}{3}\right] \cup \left[\frac{2\pi}{3}, \frac{4\pi}{3}\right] \]

### Step 4: Calculate the Range of \( h(x) \)

Now we need to find the range of \( h(x) = f(x) = e^{\cos^{-1}(\sin(x + \frac{\pi}{3}))} \).

1. **Find the range of \( \sin(x + \frac{\pi}{3}) \)**:
- The transformation \( x + \frac{\pi}{3} \) shifts the sine function.
- The range of \( \sin(x + \frac{\pi}{3}) \) for \( x \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right] \) is:
\[ \sin\left(-\frac{\pi}{3} + \frac{\pi}{3}\right) = \sin(0) = 0 \quad \text{to} \quad \sin\left(\frac{\pi}{3} + \frac{\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
- Thus, the range of \( \sin(x + \frac{\pi}{3}) \) is \( [0, \frac{\sqrt{3}}{2}] \).

2. **Find the range of \( \cos^{-1}(y) \)**:
- The range of \( \cos^{-1}(y) \) for \( y \in [0, \frac{\sqrt{3}}{2}] \) is:
\[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \quad \text{to} \quad \cos^{-1}(0) = \frac{\pi}{2} \]

3. **Find the range of \( h(x) = e^{\cos^{-1}(\sin(x + \frac{\pi}{3}))} \)**:
- The minimum value occurs at \( \cos^{-1}(0) = \frac{\pi}{2} \):
\[ h(x)_{\text{min}} = e^{\frac{\pi}{2}} \]
- The maximum value occurs at \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \):
\[ h(x)_{\text{max}} = e^{\frac{\pi}{6}} \]

Thus, the range of \( h(x) \) is:
\[ \left[e^{\frac{\pi}{6}}, e^{\frac{\pi}{2}}\right] \]

### Final Answer
The range of the function \( h(x) \) is:
\[ \boxed{\left[e^{\frac{\pi}{6}}, e^{\frac{\pi}{2}}\right]} \]