JEE Advanced 2015Paper 1
- Exam
- JEE Advanced
- Year
- 2015
- Paper
- Paper 1
- Subject
- Mathematics
- Chapter
- Definite Integration
JEE Advanced 2015 · Paper 1 · Mathematics · Definite Integration
Let F(x) = ∫ _ x ^ x ^ 2 + (π)/(6) 2 ^ 2 tdt for all x ∈ ℝand f:...
Let for all a \in \left[0, \dfrac{1}{2}\right],{F}^{′}\left(a\right) + 2 is the area of the region bounded by then is
Correct answer
3
Step-by-step solution
\left(a\right) + 2 = {∫}_{0}^{a}f\left(x\right)dx⇒ {F}^{′′}\left(a\right) = f\left(a\right)∵ F\left(x\right) = \mathop{∫}_{x}^{{x}^{2} + \dfrac{\pi}{6}}2\cos^{2} tdt∴ {F}^{′}\left(x\right) = 2\cos^{2} \left({x}^{2} + \dfrac{\pi}{6}\right) . 2x - 2\cos^{2} x{F}^{′′}\left(x\right) = 4\cos^{2} \left({x}^{2} + \dfrac{\pi}{6}\right) - 8xcos \left({x}^{2} + \dfrac{\pi}{6}\right) . \sin \left({x}^{2} + \dfrac{\pi}{6}\right)2x + 4cos xsin xf\left(0\right) = {F}^{′′}\left(0\right) = 4 \times \dfrac{3}{4} = 3