JEE Main 202224 Jun, Morning shiftMathematicsHardIntegral Calculus369× since 2002Let Max0 ≤x ≤2{9−x25−x}=α\mathop {Max}\limits_{0\, \le x\, \le 2} \left\{ {{{9 - {x^2}} \over {5 - x}}} \right\} = \alpha0≤x≤2Max{5−x9−x2}=α and Min0 ≤x ≤2{9−x25−x}=β\mathop {Min}\limits_{0\, \le x\, \le 2} \left\{ {{{9 - {x^2}} \over {5 - x}}} \right\} = \beta0≤x≤2Min{5−x9−x2}=β. If ∫β−832α−1Max{9−x25−x,x}dx=α1+α2loge(815)\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)}β−38∫2α−1Max{5−x9−x2,x}dx=α1+α2loge(158) then α1+α2{\alpha _1} + {\alpha _2}α1+α2 is equal to _____________.Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution
