f is a function defined on the interval [-1,1] such that f(sin2x)=sinx+cosx. `bb"Statement I"` If `x in [-pi/4,pi/4]`, then `f(tan^(2)x)=secx` `bb "Statement II" f(x)=sqrt(1+x), forall x in [-1,1]`

Answer

To solve the problem, we need to analyze the two statements given about the function \( f \) defined on the interval \([-1, 1]\) such that \( f(\sin 2x) = \sin x + \cos x \).

### Step 1: Understanding the Function
We start with the equation:
\[ f(\sin 2x) = \sin x + \cos x \]
We know that \(\sin 2x = 2 \sin x \cos x\). Therefore, we can express \(f\) in terms of \(x\).

### Step 2: Analyzing Statement II
Let's assume Statement II is true:
\[ f(x) = \sqrt{1+x} \quad \text{for all } x \in [-1, 1] \]
We will check if this assumption satisfies the initial condition.

### Step 3: Substitute \( \sin 2x \) into \( f \)
If \( f(x) = \sqrt{1+x} \), then:
\[ f(\sin 2x) = f(2 \sin x \cos x) = \sqrt{1 + 2 \sin x \cos x} \]
Now, we need to check if this equals \( \sin x + \cos x \).

### Step 4: Simplifying \( \sqrt{1 + 2 \sin x \cos x} \)
We can rewrite \( \sin x + \cos x \):
\[ \sin x + \cos x = \sqrt{(\sin x + \cos x)^2} = \sqrt{\sin^2 x + \cos^2 x + 2 \sin x \cos x} = \sqrt{1 + 2 \sin x \cos x} \]
Thus, we have:
\[ f(\sin 2x) = \sqrt{1 + 2 \sin x \cos x} = \sin x + \cos x \]
This confirms that Statement II is consistent with the initial condition.

### Step 5: Verifying Statement I
Now we need to verify Statement I:
\[ f(\tan^2 x) = \sec x \quad \text{for } x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \]
Using our assumption from Statement II:
\[ f(\tan^2 x) = \sqrt{1 + \tan^2 x} = \sqrt{\sec^2 x} = \sec x \]
This shows that Statement I is also true under the assumption that Statement II is true.

### Conclusion
Since we have verified both statements, we conclude that:
- Statement I is true if Statement II is true.
- Statement II is indeed a valid explanation for Statement I.

Thus, both statements are true.