Question

Let $\underset{0 \leq x \leq 2}{\text{Max}} \left\{\frac{9-x^{2}}{5-x}\right\}=\alpha$ and $\underset{0 \leq x \leq 2}{\text{Min}} \left\{\frac{9-x^{2}}{5-x}\right\}=\beta$ If $\int\limits_{\beta-\frac{8}{3}}^{2 \alpha-1} \text{Max}\left\{\frac{9- x ^{2}}{5- x }, x \right\} dx =\alpha_{1}+\alpha_{2} \log _{ e }\left(\frac{8}{15}\right)$ then $\alpha_{1}+\alpha_{2}$ is equal to ____

Answer 1

$y=\frac{9-x^{2}}{5-x}=5+x+\frac{16}{x-5}$
$\frac{d y}{d x}=1-\frac{16}{(x-5)^{2}}$
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 $\beta=\frac{5}{3}$
$I=\int\limits_{-1}^{3} \max \left(\frac{9-x^{2}}{5-x}, x\right)$
$I=\int\limits_{-1}^{9 / 5} \frac{9-x^{2}}{5-x} d x+\int\limits_{9 / 5}^{3} x d x$
$I=\int\limits_{-1}^{9 / 5} 5+x+\frac{16}{x-5} d x+\int\limits_{9 / 5}^{3} x d x$
After solving
$I=14+\frac{28}{25}+16 \ln \left(\frac{8}{15}\right)+\frac{72}{25} $
$\alpha_{1}=18 \text { and } \alpha_{2}=16$