JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(\operatorname{Max} \limits _{0 \leq x \leq 2}\left\{\frac{9-x^{2}}{5-x}\right\}=\alpha\) and \(\operatorname{Min} \limits _ {0 \leq x \leq 2}\left\{\frac{9-x^{2}}{5-x}\right\}=\beta\) If \(\int\limits_{\beta-\frac{8}{3}}^{2 a-1} \operatorname{Max}\left\{\frac{9- x ^{2}}{5- x }, x \right\} dx =\alpha_{1}+\alpha_{2} \log _{e}\left(\frac{8}{15}\right)\) then \(\alpha_{1}+\alpha_{2}\) is equal to
- A \(58\)
- B \(23\)
- C \(34\)
- D \(38\)
Answer & Solution
Correct Answer
(C) \(34\)
Step-by-step Solution
Detailed explanation
\(y=\frac{9-x^{2}}{5-x}=5+x+\frac{16}{x-5}\) \(\frac{ dy }{ dx }=1-\frac{16}{( x -5)^{2}}\) 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 \(\beta=\frac{5}{3}\)…
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