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Indefinite Integrals question

2020 · 5 Sep · Shift 1 · Q24
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  5. /2020 · 5 Sep · Shift 1 · Q24

Indefinite Integrals question

2020 · 5 Sep · Shift 1 · Q24

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If ∫(e2x+2ex−e−x−1)e(ex+e−x)dx\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}∫(e2x+2ex−e−x−1)e(ex+e−x)dx=g(x)e(ex+e−x)+cg\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + cg(x)e(ex+e−x)+c where c is a constant of integration, then g(0) is equal to :
  1. A
    1
  2. B
    2
  3. C
    e
  4. D
    e2
View written solutionFree

Correct answer: B

  1. We need to evaluate
∫(e2x+2ex−e−x−1)e(ex+e−x) dx\int \left(e^{2x}+2e^x-e^{-x}-1\right)e^{\left(e^x+e^{-x}\right)}\,dx∫(e2x+2ex−e−x−1)e(ex+e−x)dx

in the form

g(x)e(ex+e−x)+c.g(x)e^{\left(e^x+e^{-x}\right)}+c.g(x)e(ex+e−x)+c.

So we want a function g(x)g(x)g(x) such that

ddx[g(x)e(ex+e−x)]=(e2x+2ex−e−x−1)e(ex+e−x).\frac{d}{dx}\left[g(x)e^{\left(e^x+e^{-x}\right)}\right] = \left(e^{2x}+2e^x-e^{-x}-1\right)e^{\left(e^x+e^{-x}\right)}.dxd​[g(x)e(ex+e−x)]=(e2x+2ex−e−x−1)e(ex+e−x).
  1. Differentiate using product rule:
ddx[g(x)e(ex+e−x)]=e(ex+e−x)[g′(x)+g(x)ddx(ex+e−x)].\frac{d}{dx}\left[g(x)e^{\left(e^x+e^{-x}\right)}\right] = e^{\left(e^x+e^{-x}\right)}\left[g'(x)+g(x)\frac{d}{dx}(e^x+e^{-x})\right].dxd​[g(x)e(ex+e−x)]=e(ex+e−x)[g′(x)+g(x)dxd​(ex+e−x)].

Now,

ddx(ex+e−x)=ex−e−x.\frac{d}{dx}(e^x+e^{-x})=e^x-e^{-x}.dxd​(ex+e−x)=ex−e−x.

Hence,

ddx[g(x)e(ex+e−x)]=e(ex+e−x)[g′(x)+g(x)(ex−e−x)].\frac{d}{dx}\left[g(x)e^{\left(e^x+e^{-x}\right)}\right] = e^{\left(e^x+e^{-x}\right)}\left[g'(x)+g(x)(e^x-e^{-x})\right].dxd​[g(x)e(ex+e−x)]=e(ex+e−x)[g′(x)+g(x)(ex−e−x)].
  1. Compare with the integrand:
e(ex+e−x)[g′(x)+g(x)(ex−e−x)]=(e2x+2ex−e−x−1)e(ex+e−x).e^{\left(e^x+e^{-x}\right)}\left[g'(x)+g(x)(e^x-e^{-x})\right] = \left(e^{2x}+2e^x-e^{-x}-1\right)e^{\left(e^x+e^{-x}\right)}.e(ex+e−x)[g′(x)+g(x)(ex−e−x)]=(e2x+2ex−e−x−1)e(ex+e−x).

So,

g′(x)+g(x)(ex−e−x)=e2x+2ex−e−x−1.g'(x)+g(x)(e^x-e^{-x})=e^{2x}+2e^x-e^{-x}-1.g′(x)+g(x)(ex−e−x)=e2x+2ex−e−x−1.
  1. We now try a simple form for g(x)g(x)g(x). Since the right side contains terms like e2x,ex,e−x,1e^{2x}, e^x, e^{-x}, 1e2x,ex,e−x,1, try
g(x)=ex+kg(x)=e^x+kg(x)=ex+k

for some constant kkk.

Then

g′(x)=ex.g'(x)=e^x.g′(x)=ex.

Also,

g(x)(ex−e−x)=(ex+k)(ex−e−x)=e2x−1+kex−ke−x.g(x)(e^x-e^{-x})=(e^x+k)(e^x-e^{-x})=e^{2x}-1+ke^x-ke^{-x}.g(x)(ex−e−x)=(ex+k)(ex−e−x)=e2x−1+kex−ke−x.

Therefore,

g′(x)+g(x)(ex−e−x)=ex+e2x−1+kex−ke−x=e2x+(k+1)ex−ke−x−1.g'(x)+g(x)(e^x-e^{-x}) = e^x+e^{2x}-1+ke^x-ke^{-x} = e^{2x}+(k+1)e^x-ke^{-x}-1.g′(x)+g(x)(ex−e−x)=ex+e2x−1+kex−ke−x=e2x+(k+1)ex−ke−x−1.
  1. Match coefficients with
e2x+2ex−e−x−1.e^{2x}+2e^x-e^{-x}-1.e2x+2ex−e−x−1.

Thus,

k+1=2⇒k=1,k+1=2 \Rightarrow k=1,k+1=2⇒k=1,

and also coefficient of e−xe^{-x}e−x becomes −k=−1-k=-1−k=−1, which is satisfied.

So,

g(x)=ex+1.g(x)=e^x+1.g(x)=ex+1.
  1. Now compute g(0)g(0)g(0):
g(0)=e0+1=1+1=2.g(0)=e^0+1=1+1=2.g(0)=e0+1=1+1=2.

Therefore the correct answer is

2.\boxed{2}.2​.
  1. Option check:
  • A: 111 — incorrect
  • B: 222 — correct
  • C: eee — incorrect
  • D: e2e^2e2 — incorrect
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