Let $$\mathop {Max}\limits_{0\, \le x\, \le 2} \left\{ {{{9 - {x^2}} \over {5 - x}}} \right\} = \alpha $$ and $$\mathop {Min}\limits_{0\, \le x\, \le 2} \left\{ {{{9 - {x^2}} \over {5 - x}}} \right\} = \beta $$.
If $$\int\limits_{\beta - {8 \over 3}}^{2\alpha - 1} {Max\left\{ {{{9 - {x^2}} \over {5 - x}},x} \right\}dx = {\alpha _1} + {\alpha _2}{{\log }_e}\left( {{8 \over {15}}} \right)} $$ then ${\alpha _1} + {\alpha _2}$ is equal to _____________.
Answer (integer)
34
Solution
Let $f(x)=\frac{x^{2}-9}{x-5} \Rightarrow f^{\prime}(x)=\frac{(x-1)(x-9)}{(x-5)^{2}}$
<br/><br/>
So, $\alpha=f(1)=2$ and $\beta=\min (f(0), f(2))=\frac{5}{3}$
<br/><br/>
Now, $\int_{-1}^{3} \max \left\{\frac{x^{2}-9}{x-5}, x\right\} d x=\int_{-1}^{9 / 5} \frac{x^{2}-9}{x-5} d x+\int_{9 / 5}^{3} x d x$
<br/><br/>
$$
=\int_{-1}^{9 / 5}\left(x+5+\frac{16}{x-5}\right) d x+\left.\frac{x^{2}}{2}\right|_{9 / 5} ^{3}
$$
<br/><br/>
$$
=\frac{28}{25}+14+16 \ln \left(\frac{8}{15}\right)+\frac{72}{25}=18+16 \ln \left(\frac{8}{15}\right)
$$
<br/><br/>
Clearly $\alpha_{1}=18$ and $\alpha_{2}=16$, so $\alpha_{1}+\alpha_{2}=34$.
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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