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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

f(x)=\displaystyle\int 2e^{x}\cos^{2}x\,(-\tan^{2}x+\tan x+1)\,dx

and

y=f(x)

passes through

(\pi,0)

, then

f(0)+f'(0)

equals

2e

3e

2

correct

3

Solution

Step 1: Expand

\cos^{2}x(-\tan^{2}x+\tan x+1)

term by term:

\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^{2}x-\sin^{2}x)+\sin x\cos x=\cos 2x+\dfrac{1}{2}\sin 2x

Step 2: Substitute into the integral:

f(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

\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)=e^{x}\sin 2x+C

. Step 4: Apply the point

(\pi,0)

f(\pi)=e^{\pi}\sin 2\pi+C=0+C=0\Rightarrow C=0

. Hence

f(x)=e^{x}\sin 2x

. Step 5: Compute the required values.

f(0)=e^{0}\sin 0=0

. Differentiating,

f'(x)=e^{x}(\sin 2x+2\cos 2x)

, so

f'(0)=1\cdot(0+2)=2

\therefore f(0)+f'(0)=0+2=2

Correct answer: (3)

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