`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 solve the problem, we need to find \( f(g(x)) \) for the given piecewise functions \( f(x) \) and \( g(x) \).

### Step 1: Define the Functions
The functions are defined as follows:

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

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

### Step 2: Determine the Range of \( g(x) \)
We need to evaluate \( g(x) \) in both intervals to find the range.

1. For \( 0 \leq x < \frac{\pi}{4} \):
- At \( x = 0 \): \( g(0) = 1 + \tan(0) = 1 \)
- At \( x = \frac{\pi}{4} \): \( g\left(\frac{\pi}{4}\right) = 1 + \tan\left(\frac{\pi}{4}\right) = 2 \)
- Therefore, the range of \( g(x) \) in this interval is \( [1, 2) \).

2. For \( \frac{\pi}{4} \leq x < \pi \):
- At \( x = \frac{\pi}{4} \): \( g\left(\frac{\pi}{4}\right) = 3 - \cot\left(\frac{\pi}{4}\right) = 2 \)
- At \( x = \pi \): \( g(\pi) = 3 - \cot(\pi) = 3 - 0 = 3 \)
- Therefore, the range of \( g(x) \) in this interval is \( (2, 3) \).

Combining both ranges, we find that \( g(x) \) maps to \( [1, 3) \).

### Step 3: Evaluate \( f(g(x)) \)
Now we can evaluate \( f(g(x)) \) based on the ranges we found.

1. For \( 0 \leq x < \frac{\pi}{4} \):
- Since \( g(x) \) is in \( [1, 2) \), we use the first case of \( f(x) \):
\[ f(g(x)) = g(x) + 2 = (1 + \tan x) + 2 = 3 + \tan x \]

2. For \( \frac{\pi}{4} \leq x < \pi \):
- Since \( g(x) \) is in \( (2, 3) \), we use the second case of \( f(x) \):
\[ f(g(x)) = 6 - g(x) = 6 - (3 - \cot x) = 3 + \cot x \]

### Step 4: Combine Results
Thus, we have:

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

### Step 5: Check Continuity and Differentiability at \( x = \frac{\pi}{4} \)
To check continuity at \( x = \frac{\pi}{4} \):

- Left-hand limit (LHL):
\[ \lim_{x \to \frac{\pi}{4}^-} f(g(x)) = 3 + \tan\left(\frac{\pi}{4}\right) = 3 + 1 = 4 \]

- Right-hand limit (RHL):
\[ \lim_{x \to \frac{\pi}{4}^+} f(g(x)) = 3 + \cot\left(\frac{\pi}{4}\right) = 3 + 1 = 4 \]

- Value at \( x = \frac{\pi}{4} \):
\[ f(g(\frac{\pi}{4})) = 3 + \cot\left(\frac{\pi}{4}\right) = 4 \]

Since LHL = RHL = value at \( x = \frac{\pi}{4} = 4 \), \( f(g(x)) \) is continuous at \( x = \frac{\pi}{4} \).

### Step 6: Check Differentiability at \( x = \frac{\pi}{4} \)
- Left-hand derivative (LHD):
\[ \lim_{x \to \frac{\pi}{4}^-} \frac{f(g(x)) - 4}{x - \frac{\pi}{4}} = \lim_{x \to \frac{\pi}{4}^-} \frac{(3 + \tan x) - 4}{x - \frac{\pi}{4}} = \lim_{x \to \frac{\pi}{4}^-} \frac{\tan x - 1}{x - \frac{\pi}{4}} = \sec^2\left(\frac{\pi}{4}\right) = 2 \]

- Right-hand derivative (RHD):
\[ \lim_{x \to \frac{\pi}{4}^+} \frac{f(g(x)) - 4}{x - \frac{\pi}{4}} = \lim_{x \to \frac{\pi}{4}^+} \frac{(3 + \cot x) - 4}{x - \frac{\pi}{4}} = \lim_{x \to \frac{\pi}{4}^+} \frac{\cot x - 1}{x - \frac{\pi}{4}} = -\csc^2\left(\frac{\pi}{4}\right) = -2 \]

Since LHD ≠ RHD, \( f(g(x)) \) is not differentiable at \( x = \frac{\pi}{4} \).

### Final Result
Thus, we conclude that \( f(g(x)) \) is continuous but not differentiable at \( x = \frac{\pi}{4} \).