# Solve this following

Question:

Let $\mathrm{F}:[3,5] \rightarrow \mathbf{R}$ be a twice differentiable function on $(3,5)$ such that

$F(x)=e^{-x} \int_{3}^{x}\left(3 t^{2}+2 t+4 F^{\prime}(t)\right) d t$

If $\mathrm{F}^{\prime}(4)=\frac{\alpha \mathrm{e}^{\beta}-224}{\left(\mathrm{e}^{\beta}-4\right)^{2}}$, then $\alpha+\beta$ is equal to

Solution:

$\mathrm{F}(3)=0$

$e^{x} F(x)=\int_{3}^{x}\left(3 t^{2}+2 t+4 F^{\prime}(t)\right) d t$

$e^{x} F(x)+e^{x} F^{\prime}(x)=3 x^{2}+2 x+4 F^{\prime}(x)$

$\left(e^{x}-4\right) \frac{d y}{d x}+e^{x} y=\left(3 x^{2}+2 x\right)$

$\frac{d y}{d x}+\frac{e^{x}}{\left(e^{x}-4\right)} y=\frac{\left(3 x^{2}+2 x\right)}{\left(e^{x}-4\right)}$

$y e^{\int \frac{e^{x}}{\left(e^{x}-4\right)} d x}=\int \frac{\left(3 x^{2}+2 x\right)}{\left(e^{x}-4\right)} e^{\int \frac{e^{x}}{e^{x}-4} d x} d x$

$y \cdot\left(e^{x}-4\right)=\int\left(3 x^{2}+2 x\right) d x+c$

$y\left(e^{x}-4\right)=x^{3}+x^{2}+c$

Put $x=3 \Rightarrow c=-36$

$F(x)=\frac{\left(x^{3}+x^{2}-36\right)}{\left(e^{x}-4\right)}$

$F^{\prime}(x)=\frac{\left(3 x^{2}+2 x\right)\left(e^{x}-4\right)-\left(x^{3}+x^{2}-36\right) e^{x}}{\left(e^{x}-4\right)^{2}}$

Now put value of $x=4$ we will get $\alpha=12 \& \beta=4$