If `f''(x) gt forall in R, f(3)=0 and g(x) =f(tan^(2)x-2tanx+4y)0ltxlt(pi)/(2)`,then g(x) is increasing in

Options

A. `(0,(pi)/(4))`
B. `((pi)/(6),(pi)/(3))`
C. `(0,(pi)/(3))`
D. `((pi)/(4),(pi)/(2))`

Explanation

To solve the problem, we need to analyze the function \( g(x) = f(\tan^2 x - 2\tan x + 4) \) and determine the intervals where it is increasing.

### Step 1: Understand the given information
We know that \( f'(x) > 0 \) for all \( x \in \mathbb{R} \), which means that \( f(x) \) is a strictly increasing function. We also have \( f(3) = 0 \).

### Step 2: Differentiate \( g(x) \)
To find where \( g(x) \) is increasing, we need to compute its derivative \( g'(x) \):
\[ g'(x) = f'(\tan^2 x - 2\tan x + 4) \cdot \frac{d}{dx}(\tan^2 x - 2\tan x + 4) \]

### Step 3: Differentiate the inner function
Now, we differentiate the inner function \( \tan^2 x - 2\tan x + 4 \):
\[ \frac{d}{dx}(\tan^2 x) = 2\tan x \sec^2 x \]
\[ \frac{d}{dx}(-2\tan x) = -2\sec^2 x \]
Thus, the derivative of the inner function is:
\[ \frac{d}{dx}(\tan^2 x - 2\tan x + 4) = 2\tan x \sec^2 x - 2\sec^2 x = 2\sec^2 x (\tan x - 1) \]

### Step 4: Substitute back into \( g'(x) \)
Now we substitute this back into the expression for \( g'(x) \):
\[ g'(x) = f'(\tan^2 x - 2\tan x + 4) \cdot 2\sec^2 x (\tan x - 1) \]

### Step 5: Determine when \( g'(x) > 0 \)
Since \( f'(x) > 0 \) for all \( x \), the sign of \( g'(x) \) depends on the term \( 2\sec^2 x (\tan x - 1) \). The term \( 2\sec^2 x \) is always positive for \( x \in (0, \frac{\pi}{2}) \). Therefore, we need to analyze when \( \tan x - 1 > 0 \):
\[ \tan x - 1 > 0 \implies \tan x > 1 \]
This occurs when:
\[ x > \frac{\pi}{4} \]

### Step 6: Conclusion about the intervals
Thus, \( g'(x) > 0 \) when \( x > \frac{\pi}{4} \). Therefore, \( g(x) \) is increasing in the interval:
\[ \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \]

### Final Answer
The function \( g(x) \) is increasing in the interval \( \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \).