Question

Let f:[1,3]\(\rightarrow\)R be continuous and be derivable in (1,3) and f'(x)=[f(x)]²+4∀x∈(1,3). Then

• A. f(3)-f(1)=5 holds
• B. f(3)-f(1)=5 doesn't hold
• C. f(3)-f(1)=3 holds
• D. f(3)-f(1)=4 holds

Answer 1

The Correct Option is B

Step 1: Separate Variables and Integrate

We have the differential equation:

\(f'(x) = [f(x)]^2 + 4\)

\(\frac{df}{dx} = f^2 + 4\)

Separate the variables:

\(\frac{df}{f^2 + 4} = dx\)

Integrate both sides:

\(\int \frac{df}{f^2 + 4} = \int dx\)

Step 2: Evaluate the Integrals

The integral on the left is:

\(\int \frac{df}{f^2 + 2^2} = \frac{1}{2} \arctan\left(\frac{f}{2}\right) + C_1\)

The integral on the right is:

\(\int dx = x + C_2\)

So, we have:

\(\frac{1}{2} \arctan\left(\frac{f(x)}{2}\right) = x + C\), where \(C = C_2 - C_1\)

Step 3: Find f(3) - f(1)

We have the equation:

\(\frac{1}{2} \arctan\left(\frac{f(x)}{2}\right) = x + C\)

Let's evaluate this at x = 3 and x = 1:

For \(x = 3\):

\(\frac{1}{2} \arctan\left(\frac{f(3)}{2}\right) = 3 + C\) (Equation 1)

For \(x = 1\):

\(\frac{1}{2} \arctan\left(\frac{f(1)}{2}\right) = 1 + C\) (Equation 2)

Subtract Equation 2 from Equation 1:

\(\frac{1}{2} \arctan\left(\frac{f(3)}{2}\right) - \frac{1}{2} \arctan\left(\frac{f(1)}{2}\right) = 3 + C - (1 + C)\)

\(\frac{1}{2} \left[ \arctan\left(\frac{f(3)}{2}\right) - \arctan\left(\frac{f(1)}{2}\right) \right] = 2\)

\(\arctan\left(\frac{f(3)}{2}\right) - \arctan\left(\frac{f(1)}{2}\right) = 4\)

However, we can't directly find f(3) - f(1) from here. This approach appears incorrect.

Alternative approach:

We have \(\frac{df}{f^2 + 4} = dx\)

Integrate both sides from 1 to 3:

\(\int_{1}^{3} \frac{df}{dx} dx = \int_{1}^{3} (f(x)^2 + 4) dx \) Since this integral isn't directly solvable in closed form we are better looking again directly at the differential equation.

Integrate from 1 to 3 on both sides from initial equation

\(\int_{1}^{3} f'(x) dx = \int_{1}^{3} (f(x)^2 + 4) dx \)

\([f(x)]_1^3=f(3)-f(1)=\int_{1}^{3} (f(x)^2 + 4) dx \)

We know that \(f(x)^2 >=0\), hence \(f(x)^2+4 >= 4\) and \(\int_{1}^{3} (f(x)^2 + 4) dx >= \int_{1}^{3} 4 dx \) which is easily evaluated. \(\int_{1}^{3} 4 dx =4 \int_{1}^{3} dx = 4 *2 = 8 \)

So \(f(3)-f(1) >= 8\)

This implies that none of provided options is the correct answer

Conclusion:

\(f(3)-f(1)=5\) doesn't hold.

Answer 2

Given: 
Differential equation: \[ f'(x) = [f(x)]^2 + 4 \] with initial condition: \[ f(1) = 5 \] 
Step 1: Separate variables
\[ \int \frac{1}{[f(x)]^2 + 4} \, df = \int dx \] Step 2: Use standard integral
\[ \int \frac{1}{f^2 + 2^2} \, df = \frac{1}{2} \arctan\left(\frac{f}{2}\right) \] \[ \Rightarrow \frac{1}{2} \arctan\left(\frac{f(x)}{2}\right) = x + C \] \[ \Rightarrow \arctan\left(\frac{f(x)}{2}\right) = 2x + C \] Step 3: Apply initial condition \( f(1) = 5 \)
\[ \arctan\left(\frac{5}{2}\right) = 2(1) + C \Rightarrow C = \arctan\left(\frac{5}{2}\right) - 2 \] Step 4: General solution
\[ \arctan\left(\frac{f(x)}{2}\right) = 2x + \arctan\left(\frac{5}{2}\right) - 2 \] \[ \Rightarrow \frac{f(x)}{2} = \tan\left(2x + \arctan\left(\frac{5}{2}\right) - 2\right) \] \[ \Rightarrow f(x) = 2 \tan\left(2x + \arctan\left(\frac{5}{2}\right) - 2\right) \] Now compute: \[ f(3) = 2 \tan\left(6 + \arctan\left(\frac{5}{2}\right) - 2\right) \quad \text{and} \quad f(1) = 5 \] But since: \[ f(3) = 2 \tan\left(4 + \arctan\left(\frac{5}{2}\right)\right) \] this value is not guaranteed to be \(5 + f(1)\), so: 

Conclusion: f(3) - f(1) = 5 doesn't hold Correct option: (B)