Let operatorname{Max} limits _{0 leq x leq 2} | vedclass

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(C) Let $f(x) = \frac{9-x^2}{5-x} = \frac{16-(x-5)^2}{5-x} = 5+x+\frac{16}{x-5}$.$f'(x) = 1 - \frac{16}{(x-5)^2}$. Setting $f'(x)=0$,we get $(x-5)^2 =

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(C) Let $f(x) = \frac{9-x^2}{5-x} = \frac{16-(x-5)^2}{5-x} = 5+x+\frac{16}{x-5}$.$f'(x) = 1 - \frac{16}{(x-5)^2}$. Setting $f'(x)=0$,we get $(x-5)^2 = 16$,so $x-5 = \pm 4$,$x=1$ or $x=9$. In $[0,2]$,the critical point is $x=1$.$f(0) = 9/5 = 1.8$,$f(1) = 8/4 = 2$,$f(2) = 5/3 \approx 1.67$.Thus,$\alpha = 2$ and $\beta = 5/3$.The integral is $I = \int_{\beta-8/3}^{2\alpha-1} \max\{f(x), x\} dx = \int_{5/3-8/3}^{4-1} \max\{f(x), x\} dx = \int_{-1}^{3} \max\{f(x), x\} dx$.We find the intersection of $f(x) = x \implies \frac{9-x^2}{5-x} = x \implies 9-x^2 = 5x-x^2 \implies 5x=9 \implies x=9/5$.For $x \in [-1, 9/5]$,$f(x) \geq x$. For $x \in [9/5, 3]$,$x \geq f(x)$.$I = \int_{-1}^{9/5} (5+x+\frac{16}{x-5}) dx + \int_{9/5}^{3} x dx$.$I = [5x + \frac{x^2}{2} + 16 \ln|x-5|]_{-1}^{9/5} + [\frac{x^2}{2}]_{9/5}^{3}$.$I = (9 + 81/50 + 16 \ln(16/25)) - (-5 + 1/2 + 16 \ln(6)) + (9/2 - 81/50) = 14 + 16 \ln(8/15) + 4 = 18 + 16 \ln(8/15)$.Thus $\alpha_1 = 18, \alpha_2 = 16$. $\alpha_1 + \alpha_2 = 34$.

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