Let `f(x) = sin x and " g(x)" = {{:(max {f(t)","0 le x le pi},"for", 0 le x le pi),((1-cos x)/(2)",","for",x gt pi):}` Then, g(x) is

Options

A. A. differentiable for all `x in R`
B. B. differentiable for all `x in R - {pi}`
C. C. differentiable for all `x in (0, oo)`
D. D. differentiable for all `x in (0, oo) - {pi}`

Answer

To solve the problem, we need to analyze the function \( g(x) \) defined in two parts based on the value of \( x \):

1. For \( 0 \leq x \leq \pi \), \( g(x) = \max(f(t)) \) where \( f(x) = \sin x \).
2. For \( x > \pi \), \( g(x) = \frac{1 - \cos x}{2} \).

### Step 1: Determine \( g(x) \) for \( 0 \leq x \leq \pi \)

Since \( f(x) = \sin x \) is continuous and differentiable in the interval \( [0, \pi] \), we find the maximum value of \( f(x) \) in this interval.

- The maximum value of \( \sin x \) occurs at \( x = \frac{\pi}{2} \), where \( \sin \frac{\pi}{2} = 1 \).
- Therefore, for \( 0 \leq x \leq \pi \), \( g(x) = 1 \).

### Step 2: Determine \( g(x) \) for \( x > \pi \)

For \( x > \pi \), we have:

\[ g(x) = \frac{1 - \cos x}{2} \]

### Step 3: Check continuity at \( x = \pi \)

To check if \( g(x) \) is continuous at \( x = \pi \), we need to find the left-hand limit (LHL) and right-hand limit (RHL) at \( x = \pi \):

- **LHL as \( x \to \pi^- \)**:
\[ g(\pi) = 1 \]

- **RHL as \( x \to \pi^+ \)**:
\[ g(\pi) = \frac{1 - \cos(\pi)}{2} = \frac{1 - (-1)}{2} = \frac{2}{2} = 1 \]

Since LHL = RHL = 1, \( g(x) \) is continuous at \( x = \pi \).

### Step 4: Check differentiability at \( x = \pi \)

To check differentiability, we need to find the derivative of \( g(x) \) in both intervals and check if the derivatives match at \( x = \pi \).

- For \( 0 \leq x < \pi \):
\[ g'(x) = 0 \]

- For \( x > \pi \):
\[ g(x) = \frac{1 - \cos x}{2} \]
Using the chain rule:
\[ g'(x) = \frac{1}{2} \cdot \sin x \]

Now, we find the limit of \( g'(x) \) as \( x \to \pi \):

- **LHL as \( x \to \pi^- \)**:
\[ g'(\pi) = 0 \]

- **RHL as \( x \to \pi^+ \)**:
\[ g'(\pi) = \frac{1}{2} \cdot \sin(\pi) = \frac{1}{2} \cdot 0 = 0 \]

Since LHL = RHL = 0, \( g(x) \) is differentiable at \( x = \pi \).

### Conclusion

Since \( g(x) \) is continuous and differentiable for all \( x \) in \( \mathbb{R} \), the correct option is:

**g(x) is differentiable for all \( x \in \mathbb{R}**.