JEE Main 2021 — Mathematics Calculus

Given,

$$F(x)={e}^{-x}{\int }_{3}^{x}(3{t}^{2}+2t+4{F}^{'}(t))dt.$$F(x)=e−x∫3x​(3t2+2t+4F′(t))dt.

So, $\displaystyle F(3)={e}^{-3}{\int }_{3}^{3}(3{t}^{2}+2t+4{F}^{'}(t))dt.$F(3)=e−3∫33​(3t2+2t+4F′(t))dt.

$$\Rightarrow F(3)=0$$⇒F(3)=0

Again, $\displaystyle {e}^{x}F(x)={\int }_{3}^{x}(3{t}^{2}+2t+4{F}^{'}(t))dt$exF(x)=∫3x​(3t2+2t+4F′(t))dt

Differentiate both sides

$\Rightarrow {e}^{x}F(x)+{e}^{x}{F}^{'}(x)=3{x}^{2}+2x+4{F}^{'}(x)$⇒exF(x)+exF′(x)=3x2+2x+4F′(x) (Newton Leibnitz Theorem for Definite Integral)

$\Rightarrow ({e}^{x}-4)\frac{dy}{dx}+{e}^{x}y=(3{x}^{2}+2x)$⇒(ex−4)dxdy​+exy=(3x2+2x) (Use $y=F(x)$y=F(x))

$$\Rightarrow \frac{dy}{dx}+\frac{{e}^{x}}{({e}^{x}-4)}y=\frac{(3{x}^{2}+2x)}{({e}^{x}-4)}$$⇒dxdy​+(ex−4)ex​y=(ex−4)(3x2+2x)​

Here, integrating factor is $\displaystyle {e}^{\int \frac{{e}^{x}}{{e}^{x}-4}dx}={e}^{\mathrm{ln}({e}^{x}-4)}={e}^{x}-4$e∫ex−4ex​dx=eln(ex−4)=ex−4

So, the solution is

$$y\cdot ({e}^{x}-4)=\int (3{x}^{2}+2x)dx+c$$y⋅(ex−4)=∫(3x2+2x)dx+c

$$\Rightarrow y\cdot ({e}^{x}-4)=({x}^{3}+{x}^{2})+c$$⇒y⋅(ex−4)=(x3+x2)+c

Put $x=3,y=0\Rightarrow c=-36$x=3,y=0⇒c=−36

So, $F(x)=\frac{({x}^{3}+{x}^{2}-36)}{({e}^{x}-4)}$F(x)=(ex−4)(x3+x2−36)​

$$\Rightarrow {F}^{'}(x)=\frac{(3{x}^{2}+2x)({e}^{x}-4)-({x}^{3}+{x}^{2}-36){e}^{x}}{{({e}^{x}-4)}^{2}}$$⇒F′(x)=(ex−4)2(3x2+2x)(ex−4)−(x3+x2−36)ex​

Now put value of $x=4$x=4 we will get

$${F}^{'}(4)=\frac{(3{(4)}^{2}+2(4))({e}^{4}-4)-({(4)}^{3}+{(4)}^{2}-36){e}^{4}}{{({e}^{4}-4)}^{2}}=\frac{12{e}^{4}-224}{{({e}^{4}-4)}^{2}}$$F′(4)=(e4−4)2(3(4)2+2(4))(e4−4)−((4)3+(4)2−36)e4​=(e4−4)212e4−224​

So, \alpha =12&\beta =4

Hence, $\alpha +\beta =16$α+β=16