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f(0)+f'(0) from ∫2eˣcos²x(−tan²x+tanx+1) dx | JEE

JEE Maths question with a full step-by-step solution.

Question
If f(x)=∫2excos⁡2x (−tan⁡2x+tan⁡x+1) dxf(x)=\displaystyle\int 2e^{x}\cos^{2}x\,(-\tan^{2}x+\tan x+1)\,dx and y=f(x)y=f(x) passes through (π,0)(\pi,0), then f(0)+f′(0)f(0)+f'(0) equals
A2e2e
B3e3e
C22correct
D33
Solution
Step 1: Expand cos⁡2x(−tan⁡2x+tan⁡x+1)\cos^{2}x(-\tan^{2}x+\tan x+1) term by term:
cos⁡2x(−tan⁡2x)=−sin⁡2x,cos⁡2xtan⁡x=sin⁡xcos⁡x,cos⁡2x⋅1=cos⁡2x\cos^{2}x(-\tan^{2}x)=-\sin^{2}x,\quad \cos^{2}x\tan x=\sin x\cos x,\quad \cos^{2}x\cdot 1=\cos^{2}x
 Summing,
=(cos⁡2x−sin⁡2x)+sin⁡xcos⁡x=cos⁡2x+12sin⁡2x=(\cos^{2}x-\sin^{2}x)+\sin x\cos x=\cos 2x+\dfrac{1}{2}\sin 2x
 Step 2: Substitute into the integral:
f(x)=∫2ex ⁣(cos⁡2x+12sin⁡2x)dx=∫ex(2cos⁡2x+sin⁡2x) dxf(x)=\int 2e^{x}\!\left(\cos 2x+\dfrac{1}{2}\sin 2x\right)dx=\int e^{x}\big(2\cos 2x+\sin 2x\big)\,dx
 Step 3: Recognise the derivative. Since
ddx(exsin⁡2x)=exsin⁡2x+2excos⁡2x=ex(sin⁡2x+2cos⁡2x),\dfrac{d}{dx}\big(e^{x}\sin 2x\big)=e^{x}\sin 2x+2e^{x}\cos 2x=e^{x}\big(\sin 2x+2\cos 2x\big),
 we get f(x)=exsin⁡2x+Cf(x)=e^{x}\sin 2x+C. Step 4: Apply the point (π,0)(\pi,0): f(π)=eπsin⁡2π+C=0+C=0⇒C=0f(\pi)=e^{\pi}\sin 2\pi+C=0+C=0\Rightarrow C=0. Hence f(x)=exsin⁡2xf(x)=e^{x}\sin 2x. Step 5: Compute the required values. f(0)=e0sin⁡0=0f(0)=e^{0}\sin 0=0. Differentiating, f′(x)=ex(sin⁡2x+2cos⁡2x)f'(x)=e^{x}(\sin 2x+2\cos 2x), so f′(0)=1⋅(0+2)=2f'(0)=1\cdot(0+2)=2.
∴f(0)+f′(0)=0+2=2\therefore f(0)+f'(0)=0+2=2
 Correct answer: (3)
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