Let `f:X rarr Y` be a function defined by
`f(x)=2sin(x+pi/4)-sqrt(2)cosx+c.`
`bb"Statement I"` For set `X,x in [0,pi/2] cup [pi,(3pi)/2]`, f(x) is one-one function.
`bb"Statement II" f'(x) ge 0,x in [0,pi/2]`

Answer

To solve the problem, we need to analyze the function \( f(x) = 2\sin\left(x + \frac{\pi}{4}\right) - \sqrt{2}\cos x + c \) and evaluate the two statements provided.

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

We start with the function:
\[ f(x) = 2\sin\left(x + \frac{\pi}{4}\right) - \sqrt{2}\cos x + c \]
Using the sine addition formula, we can rewrite \( \sin\left(x + \frac{\pi}{4}\right) \):
\[ \sin\left(x + \frac{\pi}{4}\right) = \sin x \cos\frac{\pi}{4} + \cos x \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} \sin x + \frac{\sqrt{2}}{2} \cos x \]
Thus, we have:
\[ f(x) = 2\left(\frac{\sqrt{2}}{2} \sin x + \frac{\sqrt{2}}{2} \cos x\right) - \sqrt{2}\cos x + c \]
This simplifies to:
\[ f(x) = \sqrt{2} \sin x + \sqrt{2} \cos x - \sqrt{2} \cos x + c \]
\[ f(x) = \sqrt{2} \sin x + c \]

### Step 2: Analyze Statement I

**Statement I**: For the set \( X = [0, \frac{\pi}{2}] \cup [\pi, \frac{3\pi}{2}] \), \( f(x) \) is a one-one function.

To determine if \( f(x) \) is one-one, we need to check if \( f(a) = f(b) \) implies \( a = b \).

From our simplification:
\[ f(x) = \sqrt{2} \sin x + c \]
The sine function is not one-one over the intervals \( [0, \frac{\pi}{2}] \) and \( [\pi, \frac{3\pi}{2}] \) because:
- In the interval \( [0, \frac{\pi}{2}] \), \( \sin x \) is increasing.
- In the interval \( [\pi, \frac{3\pi}{2}] \), \( \sin x \) is decreasing.

However, since \( \sin x \) takes the same value at \( x = 0 \) and \( x = \pi \) (both yield \( \sin 0 = 0 \)), we have:
\[ f(0) = f(\pi) = c \]
This means \( f(x) \) is not one-one.

Thus, **Statement I is false**.

### Step 3: Analyze Statement II

**Statement II**: \( f'(x) \geq 0 \) for \( x \in [0, \frac{\pi}{2}] \).

Now, we compute the derivative:
\[ f'(x) = \frac{d}{dx}(\sqrt{2} \sin x + c) = \sqrt{2} \cos x \]
In the interval \( [0, \frac{\pi}{2}] \), \( \cos x \) is non-negative:
\[ \cos x \geq 0 \quad \text{for} \quad x \in [0, \frac{\pi}{2}] \]
Thus, \( f'(x) \geq 0 \) in this interval.

Therefore, **Statement II is true**.

### Conclusion

- Statement I is **false**.
- Statement II is **true**.