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JEE Mains · Maths · STD 12 - 7.2 definite integral

माना \(\operatorname{Max}_{0 \leq x \leq 2}\left\{\frac{9-x^2}{5-x}\right\}=\alpha\) तथा \(\operatorname{Min}_{0 \leq x \leq 2}\left\{\frac{9-x^2}{5-x}\right\}=\beta\) है। यदि
\(\int \limits_{\beta-\frac{8}{3}}^{2 \alpha-1} \operatorname{Max}\left\{\frac{9-x^2}{5-x}, x\right\} d x=\alpha_1+\alpha_2 \log _e\left(\frac{8}{15}\right)\)
है, तो \(\alpha_1+\alpha_2\) बराबर है \(..........\)

  1. A \(58\)
  2. B \(23\)
  3. C \(34\)
  4. D \(38\)
Verified Solution

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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