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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 F(x)=∫xx2+π62cos⁡2tdtF\left(x\right) = \mathop{∫}_{x}^{{x}^{2} + \dfrac{\pi}{6}}2\cos^{2} tdt for all x∈Randf:[0,12]→[0,∞)beacontinuousfunction.Forx \in \mathbb{R}and f: \left[0, \dfrac{1}{2}\right] \to \left[0, \infty\right) be a continuous function. Fora \in \left[0, \dfrac{1}{2}\right],ifif{F}^{′}\left(a\right) + 2 is the area of the region bounded by x=0,y=0,y=f(x)andx=a,x = 0, y = 0, y = f\left(x\right)andx = a, then f(0)f ( 0 ) is

Correct answer

3

Step-by-step solution

F′{F}^{′}\left(a\right) + 2 = {∫}_{0}^{a}f\left(x\right)dx<br/><br />⇒ {F}^{′′}\left(a\right) = f\left(a\right)...(i)<br/>...(i)<br />∵ F\left(x\right) = \mathop{∫}_{x}^{{x}^{2} + \dfrac{\pi}{6}}2\cos^{2} tdt<br/><br />∴ {F}^{′}\left(x\right) = 2\cos^{2} \left({x}^{2} + \dfrac{\pi}{6}\right) . 2x - 2\cos^{2} x<br/><br />{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 x<br/><br />f\left(0\right) = {F}^{′′}\left(0\right) = 4 \times \dfrac{3}{4} = 3
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