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Solutions of |f(x)|=g(x) from an eᶜᵒˢ Integral | JEE

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

Question

\displaystyle\int e^{\,1-\cos x}(1+x\sin x)\,dx=f(x)\,e^{\,1-g(x)}+c

, then the number of solutions of

|f(x)|=g(x)

1

2

correct

3

Dzero

Solution

Step 1: Split the integral into two parts:

I=\int e^{\,1-\cos x}\,dx+\int x\,\sin x\,e^{\,1-\cos x}\,dx

Step 2: Integrate the second part by parts. Since

\dfrac{d}{dx}e^{\,1-\cos x}=\sin x\,e^{\,1-\cos x}

, take

u=x,\ dv=\sin x\,e^{\,1-\cos x}\,dx\Rightarrow v=e^{\,1-\cos x}

\int x\,\sin x\,e^{\,1-\cos x}\,dx=x\,e^{\,1-\cos x}-\int e^{\,1-\cos x}\,dx

Step 3: Add the first part; the two

\displaystyle\int e^{\,1-\cos x}\,dx

terms cancel:

I=\int e^{\,1-\cos x}\,dx+\left[x\,e^{\,1-\cos x}-\int e^{\,1-\cos x}\,dx\right]=x\,e^{\,1-\cos x}+c

Step 4: Match the stated form

f(x)\,e^{\,1-g(x)}+c

f(x)=x,\qquad g(x)=\cos x

Step 5: Solve

|f(x)|=g(x)

, i.e.

|x|=\cos x

. The V-shaped graph

y=|x|

meets the curve

y=\cos x

at one point with

x>0

and one with

x<0

, giving exactly two solutions. Correct answer: (2)

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