Verified Solution • 2025–26 Session moderate

QJEEM22Mabb20: JEE Main 2022 Mathematics - Class 12 JEEM22Mabb20 Solution: Let M a x 0 ≤ x ≤ 2 ⁡ { 9 − x 2 5 − x } = α and M i n 0 ≤ x ≤ 2 …

Let Max0≤x≤2⁡{9−x25−x}=α and Min0≤x≤2⁡{9−x25−x}=β.

If ∫β−832α−1Max{9−x25−x,x}dx=α1+α2loge(815) then α1+α2 is equal to _____________.

Step-by-Step Verified Solution

Let f(x)=x2−9x−5⇒f′(x)=(x−1)(x−9)(x−5)2f(x)=\frac{x^{2}-9}{x-5} \Rightarrow f^{\prime}(x)=\frac{(x-1)(x-9)}{(x-5)^{2}}

So, α=f(1)=2\alpha=f(1)=2 and β=min⁡(f(0),f(2))=53\beta=\min (f(0), f(2))=\frac{5}{3}

Now, ∫−13max⁡{x2−9x−5,x}dx=∫−19/5x2−9x−5dx+∫9/53xdx\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

=∫−19/5(x+5+16x−5)dx+x22|9/53

=2825+14+16ln⁡(815)+7225=18+16ln⁡(815)

Clearly α1=18\alpha_{1}=18 and α2=16\alpha_{2}=16 , so α1+α2=34\alpha_{1}+\alpha_{2}=34 .

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring