Question

Let $\underset{0\le x\le 2}{\mathrm{Max}}\left\{\frac{9-x^2}{5-x}\right\}=\alpha$ and $\underset{0\le x\le 2}{\mathrm{Min}}\left\{\frac{9-x^2}{5-x}\right\}=\beta$ If $\underset{\beta -\frac{8}{3}}{\overset{2\alpha -1}{\int }}\text{Max}\left\{\frac{9-x^2}{5-x},x\right\}dx={\alpha }_1+{\alpha }_2{\log }_e\left(\frac{8}{15}\right)$ then ${\alpha }_1+{\alpha }_2$ is equal to ____

Answer 1

Answer: 34

$y=\frac{9-x^2}{5-x}=5+x+\frac{16}{x-5}$
$\frac{dy}{dx}=1-\frac{16}{(x-5{)}^2}$
So critical point is $x=1$ in $[0,2]$
$y(0)=\frac{9}{5},y(1)=2,y(2)=\frac{5}{3}$
So $\alpha =2$ and $\beta =\frac{5}{3}$
$I=\underset{-1}{\overset{3}{\int }}\max \left(\frac{9-x^2}{5-x},x\right)$
$I=\underset{-1}{\overset{9/5}{\int }}\frac{9-x^2}{5-x}dx+\underset{9/5}{\overset{3}{\int }}xdx$
$I=\underset{-1}{\overset{9/5}{\int }}5+x+\frac{16}{x-5}dx+\underset{9/5}{\overset{3}{\int }}xdx$
After solving
$I=14+\frac{28}{25}+16\ln \left(\frac{8}{15}\right)+\frac{72}{25}$
${\alpha }_1=18\text{ and }{\alpha }_2=16$