Let F:[3,5]R be a twice differentiable function on (3,5) such that F(x)=e^-x_3^x(3t^2+2t+4…

Question

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

Answer: 16

Accepted Answer

$F(3)=0$. From $e^xF(x)=\displaystyle\int_3^x(3t^2+2t+4F'(t))dt$, differentiating:
$$e^xF(x)+e^xF'(x)=3x^2+2x+4F'(x)\Rightarrow(e^x-4)y'+e^xy=3x^2+2x\quad(y=F(x))$$

This is linear in $y$; the integrating factor is $e^x-4$ (verified via $\frac{d}{dx}[(e^x-4)y]=(e^x-4)y'+e^xy$), so
$$y(e^x-4)=\int(3x^2+2x)\,dx+c=x^3+x^2+c$$

Using $F(3)=0$: $0\cdot(e^3-4)=27+9+c\Rightarrow c=-36$. So
$$F(x)=\frac{x^3+x^2-36}{e^x-4}\ \Rightarrow\ F'(x)=\frac{(3x^2+2x)(e^x-4)-(x^3+x^2-36)e^x}{(e^x-4)^2}$$

At $x=4$: comparing with the given form gives $\alpha=12$, $\beta=4$, so $\alpha+\beta=16$.

Solutions · 1