Let `f(0)=0,f((pi)/(2))=1,f((3pi)/(2))=-1` be a continuos and twice differentiable function.
Statement I `|f''(x)|le1` for atleast one `x in (0,(3pi)/(2))` because
Statement II According to Rolle's theorem, if y=g(x) is
continuos and differentiable, `AAx in [a,b]andg(a)=g(b),`
then there exists atleast one such that g'(c)=0.

Options

A. Statement I is true, Statement II is also true, Statement II is the correct explanation of statement I.
B. Statement I is true, Statement II is also true, Statement II is not the correct explanation of Statement I.
C. Statement I is true, Statement II is false
D. Statement I is false, Statement II is true

Answer

To solve the given problem, we need to analyze the two statements provided about the function \( f \).

### Step 1: Analyze Statement I
We are given:
- \( f(0) = 0 \)
- \( f\left(\frac{\pi}{2}\right) = 1 \)
- \( f\left(\frac{3\pi}{2}\right) = -1 \)

We need to check if \( |f''(x)| \leq 1 \) for at least one \( x \) in \( \left(0, \frac{3\pi}{2}\right) \).

Since \( f(0) \neq f\left(\frac{3\pi}{2}\right) \), we can apply the Mean Value Theorem (MVT). According to MVT, there exists at least one \( c \) in \( \left(0, \frac{3\pi}{2}\right) \) such that:

\[ f'(c) = \frac{f\left(\frac{3\pi}{2}\right) - f(0)}{\frac{3\pi}{2} - 0} = \frac{-1 - 0}{\frac{3\pi}{2}} = \frac{-2}{3\pi} \]

### Step 2: Find \( f''(x) \)
Now, we will apply the Mean Value Theorem again on the interval \( \left(0, \frac{\pi}{2}\right) \) and \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \).

1. For the interval \( \left(0, \frac{\pi}{2}\right) \):
- \( f(0) = 0 \) and \( f\left(\frac{\pi}{2}\right) = 1 \)
- By MVT, there exists \( d \in \left(0, \frac{\pi}{2}\right) \) such that:
\[ f'(d) = \frac{f\left(\frac{\pi}{2}\right) - f(0)}{\frac{\pi}{2} - 0} = \frac{1 - 0}{\frac{\pi}{2}} = \frac{2}{\pi} \]

2. For the interval \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \):
- \( f\left(\frac{\pi}{2}\right) = 1 \) and \( f\left(\frac{3\pi}{2}\right) = -1 \)
- By MVT, there exists \( e \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \) such that:
\[ f'(e) = \frac{f\left(\frac{3\pi}{2}\right) - f\left(\frac{\pi}{2}\right)}{\frac{3\pi}{2} - \frac{\pi}{2}} = \frac{-1 - 1}{\pi} = \frac{-2}{\pi} \]

### Step 3: Apply the Mean Value Theorem Again
Now we have two values of \( f' \):
- \( f'(d) = \frac{2}{\pi} \)
- \( f'(e) = \frac{-2}{\pi} \)

By applying the Mean Value Theorem again on the interval \( \left(d, e\right) \), there exists \( x \in (d, e) \) such that:

\[ f''(x) = \frac{f'(e) - f'(d)}{e - d} \]

Calculating \( f''(x) \):
\[ f''(x) = \frac{\frac{-2}{\pi} - \frac{2}{\pi}}{e - d} = \frac{-4/\pi}{e - d} \]

### Step 4: Conclusion for Statement I
Since \( |f''(x)| \) can potentially be greater than 1 depending on the values of \( e \) and \( d \), we cannot conclude that \( |f''(x)| \leq 1 \) for at least one \( x \) in \( \left(0, \frac{3\pi}{2}\right) \). Thus, Statement I is **False**.

### Step 5: Analyze Statement II
According to Rolle's Theorem, if \( f(a) = f(b) \) for a continuous and differentiable function \( f \) on the interval \([a, b]\), then there exists at least one \( c \) in \( (a, b) \) such that \( f'(c) = 0 \).

In our case, since \( f(0) \neq f\left(\frac{3\pi}{2}\right) \), we cannot apply Rolle's theorem directly. However, since we have established that there are points where \( f' \) takes both positive and negative values, it implies that there is at least one point in between where \( f' \) must equal zero.

Thus, Statement II is **True**.

### Final Conclusion
- Statement I: **False**
- Statement II: **True**