JEE Main 2022 — Mathematics Calculus

Here $y=\frac{9-{x}^{2}}{5-x}=5+x+\frac{16}{x-5}$y=5−x9−x2​=5+x+x−516​

$$\frac{dy}{dx}=1-\frac{16}{{(x-5)}^{2}}$$dxdy​=1−(x−5)216​

When $\frac{dy}{dx}=0\Rightarrow 1-\frac{16}{{(x-5)}^{2}}=0$dxdy​=0⇒1−(x−5)216​=0

So critical point is $x=1$x=1 in $[0,2]$[0,2]

Now we get $y(0)=\frac{9}{5},y(1)=2,y(2)=\frac{5}{3}$y(0)=59​,y(1)=2,y(2)=35​

So $\alpha =2$α=2 and $\beta =\frac{5}{3}$β=35​

$$I={\int }_{-1}^{3}\mathrm{max}(\frac{9-{x}^{2}}{5-x},x)$$I=∫−13​max(5−x9−x2​,x)

$$I={\int }_{-1}^{\frac{9}{5}}(5+x+\frac{16}{x-5})dx+{\int }_{\frac{9}{5}}^{3}xdx$$I=∫−159​​(5+x+x−516​)dx+∫59​3​xdx

$$={[5x+\frac{{x}^{2}}{2}+16\mathrm{ln}(x-5)]}_{-1}^{\frac{9}{5}}+{[\frac{{x}^{2}}{2}]}_{\frac{9}{5}}^{3}$$=[5x+2x2​+16ln(x−5)]−159​​+[2x2​]59​3​

$$I=18+16\mathrm{ln}(\frac{8}{15})$$I=18+16ln(158​)

${\alpha }_{1}=18$α1​=18 and ${\alpha }_{2}=16$α2​=16

$$\Rightarrow {\alpha }_{1}+{\alpha }_{2}=34$$⇒α1​+α2​=34