Question

Let F : [3, 5] $\rightarrow$ R be a twice differentiable function on (3, 5) such that
$F(x) = e^{-x} \int_3^x (3t^2 + 2t + 4F'(t))dt$.
If $F'(4) = \frac{\alpha e^\beta - 224}{(e^\beta-4)^2}$, then $\alpha+\beta$ is equal to _________.

Hint:

In functional equations involving integrals of \(F'(t)\):
Always simplify the integral first.
Use \( \int F'(t)\,dt = F(t) \).
Convert the problem into an explicit function before differentiating.

Answer 1

Correct Answer: 16

Given: \[ F(x)=e^{-x}\int_3^x \big(3t^2+2t+4F'(t)\big)\,dt \] Multiply both sides by \(e^x\): \[ e^x F(x)=\int_3^x (3t^2+2t+4F'(t))\,dt \] Step 1: Evaluate the integral
Split the integral: \[ \int_3^x (3t^2+2t)\,dt + \int_3^x 4F'(t)\,dt \] \[ = \big[t^3+t^2\big]_3^x + 4\big[F(t)\big]_3^x \] \[ = (x^3+x^2)-(27+9)+4(F(x)-F(3)) \] From the given definition: \[ F(3)=e^{-3}\int_3^3(\cdots)\,dt=0 \] Hence, \[ e^xF(x)=x^3+x^2-36+4F(x) \] Step 2: Solve for \(F(x)\)
\[ (e^x-4)F(x)=x^3+x^2-36 \] \[ \boxed{F(x)=\frac{x^3+x^2-36}{e^x-4}} \] Step 3: Differentiate \(F(x)\)
Using the quotient rule, \[ F'(x)=\frac{(3x^2+2x)(e^x-4)-(x^3+x^2-36)e^x}{(e^x-4)^2} \] Step 4: Evaluate \(F'(4)\)
\[ F'(4)=\frac{(3\cdot16+2\cdot4)(e^4-4)-(64+16-36)e^4}{(e^4-4)^2} \] \[ =\frac{56(e^4-4)-44e^4}{(e^4-4)^2} \] \[ =\frac{12e^4-224}{(e^4-4)^2} \] --- Step 5: Compare with given form
Given: \[ F'(4)=\frac{\alpha e^\beta-224}{(e^\beta-4)^2} \] Thus, \[ \alpha=12,\quad \beta=4 \] \[ \boxed{\alpha+\beta=16} \]