Verify Rolle's Theorem for the functions :
`f(x)=sinxcosx " in "[0,pi/2]`

Answer

To verify Rolle's Theorem for the function \( f(x) = \sin x \cos x \) on the interval \([0, \frac{\pi}{2}]\), we will follow these steps:

### Step 1: Check Continuity
We need to check if the function \( f(x) \) is continuous on the closed interval \([0, \frac{\pi}{2}]\).

**Solution:**
The function \( f(x) = \sin x \cos x \) is a product of two continuous functions, \( \sin x \) and \( \cos x \). Since both sine and cosine functions are continuous everywhere, their product is also continuous. Therefore, \( f(x) \) is continuous on \([0, \frac{\pi}{2}]\).

### Step 2: Check Differentiability
Next, we need to check if \( f(x) \) is differentiable on the open interval \((0, \frac{\pi}{2})\).

**Solution:**
The function \( f(x) = \sin x \cos x \) is differentiable everywhere since it is composed of sine and cosine functions, which are differentiable everywhere. Thus, \( f(x) \) is differentiable on \((0, \frac{\pi}{2})\).

### Step 3: Check the Endpoints
Now we need to verify that \( f(a) = f(b) \) where \( a = 0 \) and \( b = \frac{\pi}{2} \).

**Solution:**
Calculate \( f(0) \) and \( f\left(\frac{\pi}{2}\right) \):
- \( f(0) = \sin(0) \cos(0) = 0 \cdot 1 = 0 \)
- \( f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) = 1 \cdot 0 = 0 \)

Since \( f(0) = f\left(\frac{\pi}{2}\right) = 0 \), we have \( f(a) = f(b) \).

### Step 4: Find \( c \) such that \( f'(c) = 0 \)
According to Rolle's Theorem, there exists at least one \( c \) in \( (0, \frac{\pi}{2}) \) such that \( f'(c) = 0 \).

**Solution:**
First, we need to find the derivative \( f'(x) \):
Using the product rule:
\[ f'(x) = \sin x \cdot \frac{d}{dx}(\cos x) + \cos x \cdot \frac{d}{dx}(\sin x) = \sin x (-\sin x) + \cos x (\cos x) = \cos^2 x - \sin^2 x \]

Now we set \( f'(x) = 0 \):
\[ \cos^2 x - \sin^2 x = 0 \]
This implies:
\[ \cos^2 x = \sin^2 x \implies \tan^2 x = 1 \implies \tan x = 1 \]
Thus, \( x = \frac{\pi}{4} \) is a solution in the interval \( (0, \frac{\pi}{2}) \).

### Conclusion
Since all conditions of Rolle's Theorem are satisfied, we conclude that there exists at least one \( c \) in \( (0, \frac{\pi}{2}) \) such that \( f'(c) = 0 \).

---