Let `f:R rarr R` be a function defined by `f(x)=log_(m) {sqrt(2)(sin x-cos x)+m-2}`,for some `m`,such that the range of `f` is `[0,2]`.Then the value of `m` is

Options

A. `3`
B. `5`
C. `4`
D. `2`

Answer

To solve the problem, we need to find the value of \( m \) such that the function \( f(x) = \log_{\sqrt{m}} \left( \sqrt{2} (\sin x - \cos x) + m - 2 \right) \) has a range of \([0, 2]\).

### Step-by-Step Solution:

1. **Understanding the Range of the Logarithmic Function**:
The logarithmic function \( f(x) \) will have a range of \([0, 2]\) if the argument of the logarithm, \( \sqrt{2}(\sin x - \cos x) + m - 2 \), lies within the interval \([1, m]\). This is because:
- \( f(x) = 0 \) when the argument is \( 1 \).
- \( f(x) = 2 \) when the argument is \( m \).

2. **Finding the Range of \( \sqrt{2}(\sin x - \cos x) \)**:
The expression \( \sin x - \cos x \) can be rewritten using the identity:
\[ \sin x - \cos x = \sqrt{2} \sin \left( x - \frac{\pi}{4} \right) \]
The maximum value of \( \sin \left( x - \frac{\pi}{4} \right) \) is \( 1 \) and the minimum value is \( -1 \). Therefore:
\[ \sin x - \cos x \in [-\sqrt{2}, \sqrt{2}] \]
Consequently:
\[ \sqrt{2}(\sin x - \cos x) \in [-2, 2] \]

3. **Setting Up the Inequalities**:
We need to ensure that:
\[ -2 + m - 2 \geq 1 \quad \text{(minimum condition)} \]
\[ 2 + m - 2 \leq m \quad \text{(maximum condition)} \]

4. **Solving the Minimum Condition**:
From the minimum condition:
\[ m - 4 \geq 1 \implies m \geq 5 \]

5. **Solving the Maximum Condition**:
From the maximum condition:
\[ 2 \leq m \implies m \geq 2 \]
However, since \( m \geq 5 \) is stricter, we only need to consider \( m \geq 5 \).

6. **Finding the Exact Value of \( m \)**:
To ensure that the maximum value of the argument equals \( m \):
\[ 2 + m - 2 = m \implies 2 = 0 \]
This condition is satisfied for any \( m \geq 5 \).

7. **Conclusion**:
The smallest value of \( m \) that satisfies both conditions is \( m = 5 \).

### Final Answer:
Thus, the value of \( m \) is \( \boxed{5} \).