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Definite Integration question

2022 · 28 Jul · Shift 1 · Q37
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  5. /2022 · 28 Jul · Shift 1 · Q37

Definite Integration question

2022 · 28 Jul · Shift 1 · Q37

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The minimum value of the twice differentiable function f(x)=∫0xex−tf′(t)dt−(x2−x+1)exf(x)=\int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}f(x)=0∫x​ex−tf′(t)dt−(x2−x+1)ex, x∈Rx \in \mathbf{R}x∈R, is :
  1. A
    −2e-\frac{2}{\sqrt{\mathrm{e}}}−e​2​
  2. B
    −2e-2 \sqrt{\mathrm{e}}−2e​
  3. C
    −e-\sqrt{\mathrm{e}}−e​
  4. D
    2e\frac{2}{\sqrt{\mathrm{e}}}e​2​
View written solutionFree

Correct answer: A

  1. Given equation

We have

f(x)=∫0xex−tf′(t) dt−(x2−x+1)ex.f(x)=\int_0^x e^{x-t}f'(t)\,dt-(x^2-x+1)e^x.f(x)=∫0x​ex−tf′(t)dt−(x2−x+1)ex.

We need the minimum value of f(x)f(x)f(x).

  1. Simplify the integral using integration by parts / Leibniz form

Let

I(x)=∫0xex−tf′(t) dt.I(x)=\int_0^x e^{x-t}f'(t)\,dt.I(x)=∫0x​ex−tf′(t)dt.

Observe that

ddt(ex−tf(t))=ex−tf′(t)−ex−tf(t).\frac{d}{dt}\big(e^{x-t}f(t)\big)=e^{x-t}f'(t)-e^{x-t}f(t).dtd​(ex−tf(t))=ex−tf′(t)−ex−tf(t).

So,

ex−tf′(t)=ddt(ex−tf(t))+ex−tf(t).e^{x-t}f'(t)=\frac{d}{dt}\big(e^{x-t}f(t)\big)+e^{x-t}f(t).ex−tf′(t)=dtd​(ex−tf(t))+ex−tf(t).

Integrating from 000 to xxx,

I(x)=[ex−tf(t)]0x+∫0xex−tf(t) dt=f(x)−exf(0)+∫0xex−tf(t) dt.I(x)=\Big[e^{x-t}f(t)\Big]_0^x+\int_0^x e^{x-t}f(t)\,dt = f(x)-e^x f(0)+\int_0^x e^{x-t}f(t)\,dt.I(x)=[ex−tf(t)]0x​+∫0x​ex−tf(t)dt=f(x)−exf(0)+∫0x​ex−tf(t)dt.

Substitute into the given equation:

f(x)=f(x)−exf(0)+∫0xex−tf(t) dt−(x2−x+1)ex.f(x)=f(x)-e^x f(0)+\int_0^x e^{x-t}f(t)\,dt-(x^2-x+1)e^x.f(x)=f(x)−exf(0)+∫0x​ex−tf(t)dt−(x2−x+1)ex.

Hence,

0=−exf(0)+∫0xex−tf(t) dt−(x2−x+1)ex.0=-e^x f(0)+\int_0^x e^{x-t}f(t)\,dt-(x^2-x+1)e^x.0=−exf(0)+∫0x​ex−tf(t)dt−(x2−x+1)ex.

So,

∫0xex−tf(t) dt=ex(f(0)+x2−x+1).\int_0^x e^{x-t}f(t)\,dt=e^x\big(f(0)+x^2-x+1\big).∫0x​ex−tf(t)dt=ex(f(0)+x2−x+1).

Divide by exe^xex:

∫0xe−tf(t) dt=f(0)+x2−x+1.\int_0^x e^{-t}f(t)\,dt=f(0)+x^2-x+1.∫0x​e−tf(t)dt=f(0)+x2−x+1.

Now put x=0x=0x=0:

0=f(0)+1  ⟹  f(0)=−1.0=f(0)+1 \implies f(0)=-1.0=f(0)+1⟹f(0)=−1.

Therefore,

∫0xe−tf(t) dt=x2−x.\int_0^x e^{-t}f(t)\,dt=x^2-x.∫0x​e−tf(t)dt=x2−x.
  1. Differentiate to find f(x)f(x)f(x)

Differentiate both sides with respect to xxx:

e−xf(x)=2x−1.e^{-x}f(x)=2x-1.e−xf(x)=2x−1.

Thus,

f(x)=ex(2x−1).f(x)=e^x(2x-1).f(x)=ex(2x−1).
  1. Find the minimum of f(x)=ex(2x−1)f(x)=e^x(2x-1)f(x)=ex(2x−1)

Differentiate:

f′(x)=ex(2x−1)+2ex=ex(2x+1).f'(x)=e^x(2x-1)+2e^x=e^x(2x+1).f′(x)=ex(2x−1)+2ex=ex(2x+1).

Critical point occurs when

2x+1=0  ⟹  x=−12.2x+1=0 \implies x=-\frac12.2x+1=0⟹x=−21​.

Second derivative:

f′′(x)=ex(2x+1)+2ex=ex(2x+3).f''(x)=e^x(2x+1)+2e^x=e^x(2x+3).f′′(x)=ex(2x+1)+2ex=ex(2x+3).

At x=−12x=-\frac12x=−21​,

f′′(−12)=e−1/2(2)>0,f''\left(-\frac12\right)=e^{-1/2}(2)>0,f′′(−21​)=e−1/2(2)>0,

so this is a minimum.

Now evaluate:

f(−12)=e−1/2(2⋅−12−1)=e−1/2(−2)=−2e.f\left(-\frac12\right)=e^{-1/2}\left(2\cdot -\frac12-1\right)=e^{-1/2}(-2)=-\frac{2}{\sqrt e}.f(−21​)=e−1/2(2⋅−21​−1)=e−1/2(−2)=−e​2​.
  1. Option check

Minimum value is

−2e.-\frac{2}{\sqrt e}.−e​2​.

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.

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