Consider `f(x)=[[2(sinx-sin^3x)+|sinx-sin^3 x|)/(2(sinx-sin^3 x)-|sinx-sin^3x|]], x != pi/2` for `x in (0,pi), f(pi/2) = 3` where [ ] denotes the greatest integer function then,

Options

A. f is continuous and differentiable at `x = pi//2`
B. f is continuous but not differentiable at `x = pi//2`
C. f is neither continuous nor differentiable at `x = pi//2`
D. None of the above

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as:

\[ f(x) = \left\lfloor \frac{2(\sin x - \sin^3 x) + |\sin x - \sin^3 x|}{2(\sin x - \sin^3 x) - |\sin x - \sin^3 x|} \right\rfloor \]

for \( x \neq \frac{\pi}{2} \) and \( f\left(\frac{\pi}{2}\right) = 3 \).

### Step 1: Analyze the expression inside the function

First, we simplify the expression \( \sin x - \sin^3 x \).

\[ \sin x - \sin^3 x = \sin x(1 - \sin^2 x) = \sin x \cos^2 x \]

### Step 2: Determine the sign of \( \sin x - \sin^3 x \)

Since \( x \) is in the interval \( (0, \pi) \), \( \sin x \) is positive for \( x \in (0, \pi) \), and thus \( \sin x - \sin^3 x \) is also positive for \( x \neq \frac{\pi}{2} \).

### Step 3: Simplify the absolute value

Given that \( \sin x - \sin^3 x > 0 \), we have:

\[ |\sin x - \sin^3 x| = \sin x - \sin^3 x \]

### Step 4: Substitute back into the function

Now substituting back into \( f(x) \):

\[ f(x) = \left\lfloor \frac{2(\sin x - \sin^3 x) + (\sin x - \sin^3 x)}{2(\sin x - \sin^3 x) - (\sin x - \sin^3 x)} \right\rfloor \]

This simplifies to:

\[ f(x) = \left\lfloor \frac{3(\sin x - \sin^3 x)}{\sin x - \sin^3 x} \right\rfloor \]

### Step 5: Cancel the common terms

Since \( \sin x - \sin^3 x \neq 0 \) for \( x \neq \frac{\pi}{2} \):

\[ f(x) = \left\lfloor 3 \right\rfloor = 3 \]

### Step 6: Conclusion for \( x \neq \frac{\pi}{2} \)

Thus, for all \( x \in (0, \pi) \) except \( x = \frac{\pi}{2} \):

\[ f(x) = 3 \]

### Step 7: Check the value at \( x = \frac{\pi}{2} \)

We are given that \( f\left(\frac{\pi}{2}\right) = 3 \). Therefore, \( f(x) \) is continuous at \( x = \frac{\pi}{2} \).

### Step 8: Check differentiability

Since \( f(x) = 3 \) for all \( x \) in the interval and is constant, it is differentiable everywhere in that interval, including at \( x = \frac{\pi}{2} \).

### Final Answer

Thus, we conclude that \( f(x) \) is continuous and differentiable at \( x = \frac{\pi}{2} \).