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 \(\int_{\beta - \frac83}^{2\alpha - 1} \mathrm{Max}\left\{\frac{9 - x^2}{5 - x},\ x\right\}dx = \alpha_1 + \alpha_2\log_e\left(\frac{8}{15}\right)\), then \(\alpha_1 + \alpha_2\) is equal to ____________.

Hint:

Find alpha = 2 and beta = 5/3 from the derivative of g(x) = (9 - x^2)/(5 - x); the limits become -1 and 3, and g(x) >= x exactly for x <= 9/5.

Answer 1

Correct Answer: 34

Let \(g(x) = \frac{9 - x^2}{5 - x}\). \(g'(x) = \frac{(x^2 - 10x + 9)}{(5 - x)^2} = \frac{(x - 1)(x - 9)}{(5 - x)^2}\), which is positive on [0, 1) and negative on (1, 2].
So the maximum is \(g(1) = 2 = \alpha\); the minimum is the smaller of \(g(0) = \frac95\) and \(g(2) = \frac53\), so \(\beta = \frac53\).
Limits: lower \(\beta - \frac83 = -1\), upper \(2\alpha - 1 = 3\).
\(g(x) - x = \frac{9 - 5x}{5 - x}\), which is positive for \(x \lt \frac95\) and negative for \(\frac95 \lt x \le 3\). So Max{g, x} = g on [-1, 9/5] and = x on [9/5, 3].
\(\frac{9 - x^2}{5 - x} = x + 5 + \frac{16}{x - 5}\) (since \(x^2 - 9 = (x - 5)(x + 5) + 16\)).
\(\int_{-1}^{9/5}\left(x + 5 + \frac{16}{x - 5}\right)dx = \left[\frac{x^2}{2} + 5x + 16\ln|x - 5|\right]_{-1}^{9/5}\)
\(= \left(\frac{81}{50} + 9 + 16\ln\frac{16}{5}\right) - \left(\frac12 - 5 + 16\ln 6\right) = \frac{378}{25} + 16\ln\frac{8}{15}\).
\(\int_{9/5}^{3} x\,dx = \frac92 - \frac{81}{50} = \frac{72}{25}\).
Total \(= \frac{450}{25} + 16\ln\frac{8}{15} = 18 + 16\ln\frac{8}{15}\).
So \(\alpha_1 = 18\), \(\alpha_2 = 16\), and \(\alpha_1 + \alpha_2 = 34\).
Answer: 34