Question

Let $F:[3,5]\to R$ be a twice differentiable function on (3, 5) such that $F\left(x\right)=e^{-x}\underset{3}{\overset{x}{\int }}(3t^2+2t+4F^{\prime }(t\left)\right)dt$. If $F^{\prime }\left(4\right)=\frac{\alpha e^{\beta }-224}{(e^{\beta }-4{)}^2}$, then $\alpha +\beta$ is equal to

Answer 1

Given: $F\left(x\right)=e^{-x}\underset{3}{\overset{x}{\int }}(3t^2+2t+4F^{\prime }(t\left)\right)dt$
$\Rightarrow e^{x}F\left(x\right)=\underset{3}{\overset{x}{\int }}(3t^2+2t+4F^{\prime }(t\left)\right)dt$
Differentiating w.r.t. $x,$ we get
$e^x{F}^{\prime }\left(x\right)+e^{x}F\left(x\right)=3x^2+2x+4F^{\prime }\left(x\right)$
$\Rightarrow (e^x-4)F^{\prime }\left(x\right)+e^{x}F\left(x\right)=3x^2+2x$
$\Rightarrow F^{\prime }\left(x\right)+\left(\frac{e^x}{e^x-4}\right)F\left(x\right)=\frac{3x^2+2x}{e^x-4}$
$I.F.=e^x-4$
Now, the general solution
$F\left(x\right)(e^x-4)=\int \frac{3x^2+2x}{e^x-4}(e^x-4)dx$
$\Rightarrow F\left(x\right)(e^x-4)=\int (3x^2+2x)dx$
$\Rightarrow F\left(x\right)(e^x-4)=x^3+x^2+C$
Since $F\left(3\right)=0,$ therefore $C=-36$
$\therefore F\left(x\right)=\frac{x^3+x^2-36}{(e^x-4)}$
Differentiating w.r.t. $x,$ we get
$F^{\prime }\left(x\right)=\frac{(e^x-4)(3x^2+2x)-(x^3+x^2-36)e^x}{(e^x-4{)}^2}$
$\Rightarrow F^{\prime }\left(x\right)=\frac{e^x(-x^3+2x^2+2x+36)-4(3x^2+2x)}{(e^x-4{)}^2}$
$\Rightarrow F^{\prime }\left(4\right)=\frac{e^4(-4^3+2\cdot 4^2+2\cdot 4+36)-4(3\cdot 4^2+2\cdot 4)}{(e^4-4{)}^2}$
$\Rightarrow F^{\prime }\left(4\right)=\frac{12e^4-224}{(e^4-4{)}^2}$
$\therefore \alpha =12,\beta =4$
Hence, $\alpha +\beta =16$