`Let f(x) ={{:( x+2, 0 le x lt 2),( 6-x, x ge 2):}, g(x) ={{:( 1+ tan x, 0le x lt (pi) /(4)),( 3-cotx,(pi)/(4) le x lt pi ):}`
`f(g(x))` is

Options

A. discontinuous at `x=pi//4`
B. differentiable at `x=pi //4`
C. continuous but non - differentiable `x= pi//4`
D. differentiable at` x=pi//4` , but derivative is not continuous

Answer

To find \( f(g(x)) \), we need to evaluate the composite function using the definitions of \( f(x) \) and \( g(x) \).

### Step 1: Define the functions

The functions are defined as follows:

1. \( f(x) = \begin{cases} x + 2 & \text{for } 0 \leq x < 2 \\ 6 - x & \text{for } x \geq 2 \end{cases} \)

2. \( g(x) = \begin{cases} 1 + \tan x & \text{for } 0 \leq x < \frac{\pi}{4} \\ 3 - \cot x & \text{for } \frac{\pi}{4} \leq x < \pi \end{cases} \)

### Step 2: Evaluate \( g(x) \)

We will evaluate \( g(x) \) for two intervals:

- For \( 0 \leq x < \frac{\pi}{4} \):
\[ g(x) = 1 + \tan x \]

- For \( \frac{\pi}{4} \leq x < \pi \):
\[ g(x) = 3 - \cot x \]

### Step 3: Determine the range of \( g(x) \)

1. For \( 0 \leq x < \frac{\pi}{4} \):
- As \( x \) approaches \( 0 \), \( g(0) = 1 + \tan(0) = 1 \).
- As \( x \) approaches \( \frac{\pi}{4} \), \( g\left(\frac{\pi}{4}\right) = 1 + \tan\left(\frac{\pi}{4}\right) = 1 + 1 = 2 \).
- Therefore, \( g(x) \) ranges from \( 1 \) to \( 2 \).

2. For \( \frac{\pi}{4} \leq x < \pi \):
- At \( x = \frac{\pi}{4} \), \( g\left(\frac{\pi}{4}\right) = 3 - \cot\left(\frac{\pi}{4}\right) = 3 - 1 = 2 \).
- As \( x \) approaches \( \pi \), \( g(x) \) approaches \( 3 - 0 = 3 \).
- Therefore, \( g(x) \) ranges from \( 2 \) to \( 3 \).

Combining both intervals, we find that \( g(x) \) takes values in the interval \( [1, 3) \).

### Step 4: Evaluate \( f(g(x)) \)

Now we will evaluate \( f(g(x)) \) based on the ranges of \( g(x) \):

1. For \( 1 \leq g(x) < 2 \):
- Here, \( f(g(x)) = g(x) + 2 \).
- Therefore, \( f(g(x)) = (1 + \tan x) + 2 = 3 + \tan x \) for \( 0 \leq x < \frac{\pi}{4} \).

2. For \( 2 \leq g(x) < 3 \):
- Here, \( f(g(x)) = 6 - g(x) \).
- Therefore, \( f(g(x)) = 6 - (3 - \cot x) = 3 + \cot x \) for \( \frac{\pi}{4} \leq x < \pi \).

### Final Result

Thus, we can summarize the result as follows:

\[ f(g(x)) = \begin{cases} 3 + \tan x & \text{for } 0 \leq x < \frac{\pi}{4} \\ 3 + \cot x & \text{for } \frac{\pi}{4} \leq x < \pi \end{cases} \]