Check whether following pairs of function are identical or not?
`f(x)=sqrt((1+cos2x)/2)` and `g(x)=cosx`

Explanation

To determine if the functions \( f(x) = \sqrt{\frac{1 + \cos(2x)}{2}} \) and \( g(x) = \cos(x) \) are identical, we need to check if they have the same domain and range.

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

The function \( f(x) \) is defined as \( f(x) = \sqrt{\frac{1 + \cos(2x)}{2}} \).

1. The expression inside the square root, \( \frac{1 + \cos(2x)}{2} \), must be non-negative for \( f(x) \) to be defined.
2. The cosine function, \( \cos(2x) \), ranges from -1 to 1. Thus, \( 1 + \cos(2x) \) ranges from \( 0 \) to \( 2 \).
3. Therefore, \( \frac{1 + \cos(2x)}{2} \) ranges from \( 0 \) to \( 1 \), which is non-negative.

Thus, the domain of \( f(x) \) is all real numbers, \( \mathbb{R} \).

### Step 2: Determine the range of \( f(x) \)

1. Since \( f(x) = \sqrt{\frac{1 + \cos(2x)}{2}} \), the minimum value occurs when \( \cos(2x) = -1 \), giving \( f(x) = \sqrt{0} = 0 \).
2. The maximum value occurs when \( \cos(2x) = 1 \), giving \( f(x) = \sqrt{1} = 1 \).

Thus, the range of \( f(x) \) is \( [0, 1] \).

### Step 3: Determine the domain of \( g(x) \)

The function \( g(x) = \cos(x) \) is defined for all real numbers.

Thus, the domain of \( g(x) \) is also \( \mathbb{R} \).

### Step 4: Determine the range of \( g(x) \)

1. The cosine function oscillates between -1 and 1.
2. Therefore, the range of \( g(x) \) is \( [-1, 1] \).

### Step 5: Compare the ranges of \( f(x) \) and \( g(x) \)

- The range of \( f(x) \) is \( [0, 1] \).
- The range of \( g(x) \) is \( [-1, 1] \).

Since the ranges are not equal (the range of \( f(x) \) does not include negative values), the functions \( f(x) \) and \( g(x) \) are not identical.

### Conclusion

The functions \( f(x) \) and \( g(x) \) are not identical because their ranges are different.

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