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

Answer: (B)

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

Answer 2

The given differential equation is: f'(x) = [f(x)]^2 + 4

We'll attempt to solve this differential equation with the initial condition f(1) = 5.

Separating variables and integrating both sides, we get: ∫ (1 / [f(x)^2 + 4]) df = ∫ dx

The left side can be integrated using the arctangent function:

(1/2) * arctan(f(x) / 2) = x + C

Solving for f(x):

arctan(f(x) / 2) = 2x + C

Now, applying the initial condition f(1) = 5:

arctan(5/2) = 2 * 1 + C C = arctan(5/2) - 2

So, we have:

arctan(f(x) / 2) = 2x + arctan(5/2) - 2

Solving for f(x):

f(x) / 2 = tan(2x + arctan(5/2) - 2)

Finally:

f(x) = 2 * tan(2x + arctan(5/2) - 2)

This is the correct solution for the given problem, satisfying the differential equation f'(x) = [f(x)]^2 + 4 and the initial condition f(1) = 5.

The correct answer is option (B): f(3)-f(1)=5 doesn't hold.