Let `f(x)` be a continuous function for all `x`, which is not identically zero such that `{f(x)}^2=int_0^x\ f(t)\ (2sec^2t)/(4+tant)\ dt and f(0)=ln4,` then

Options

A. `f((pi)/(4))=log(5)`
B. `f((pi)/(4))=(3)/(4)`
C. `f((pi)/(2))=2`
D. None of these

Answer

To solve the problem, we start with the given equation:

\[ f(x)^2 = \int_0^x f(t) \frac{2 \sec^2 t}{4 + \tan t} \, dt \]

and the initial condition \( f(0) = \ln 4 \).

### Step 1: Differentiate both sides
Using the Fundamental Theorem of Calculus and the chain rule, we differentiate both sides with respect to \( x \):

\[ \frac{d}{dx}[f(x)^2] = \frac{d}{dx}\left[\int_0^x f(t) \frac{2 \sec^2 t}{4 + \tan t} \, dt\right] \]

This gives us:

\[ 2 f(x) f'(x) = f(x) \frac{2 \sec^2 x}{4 + \tan x} \]

### Step 2: Simplify the equation
Assuming \( f(x) \neq 0 \) (since \( f(x) \) is not identically zero), we can divide both sides by \( f(x) \):

\[ 2 f'(x) = \frac{2 \sec^2 x}{4 + \tan x} \]

### Step 3: Solve for \( f'(x) \)
Now, we can simplify this to find \( f'(x) \):

\[ f'(x) = \frac{\sec^2 x}{4 + \tan x} \]

### Step 4: Integrate to find \( f(x) \)
Next, we integrate both sides to find \( f(x) \):

\[ f(x) = \int \frac{\sec^2 x}{4 + \tan x} \, dx \]

Using the substitution \( u = \tan x \), we have \( du = \sec^2 x \, dx \), thus:

\[ f(x) = \int \frac{1}{4 + u} \, du = \ln |4 + u| + C = \ln |4 + \tan x| + C \]

### Step 5: Apply the initial condition
We know that \( f(0) = \ln 4 \). Substituting \( x = 0 \):

\[ f(0) = \ln |4 + \tan(0)| + C = \ln 4 + C \]

Setting this equal to \( \ln 4 \):

\[ \ln 4 + C = \ln 4 \implies C = 0 \]

Thus, we have:

\[ f(x) = \ln(4 + \tan x) \]

### Step 6: Conclusion
The function \( f(x) \) is:

\[ f(x) = \ln(4 + \tan x) \]