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Definite Integration question

2021 · 27 Jul · Shift 1 · Q44
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Definite Integration question

2021 · 27 Jul · Shift 1 · Q44

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
Let F:[3,5]→RF:[3,5] \to RF:[3,5]→R be a twice differentiable function on (3, 5) such that F(x)=e−x∫3x(3t2+2t+4F′(t))dtF(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt}F(x)=e−x3∫x​(3t2+2t+4F′(t))dt. If F′(4)=αeβ−224(eβ−4)2F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}F′(4)=(eβ−4)2αeβ−224​, then α\alphaα+β\betaβ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 16

  1. Given integral equation

We have

F(x)=e−x∫3x(3t2+2t+4F′(t))dt.F(x)=e^{-x}\int_3^x \left(3t^2+2t+4F'(t)\right)dt.F(x)=e−x∫3x​(3t2+2t+4F′(t))dt.

Let

I(x)=∫3x(3t2+2t+4F′(t))dt.I(x)=\int_3^x \left(3t^2+2t+4F'(t)\right)dt.I(x)=∫3x​(3t2+2t+4F′(t))dt.

Then

F(x)=e−xI(x).F(x)=e^{-x}I(x).F(x)=e−xI(x).
  1. Differentiate the equation

Using product rule and Fundamental Theorem of Calculus,

F′(x)=−e−xI(x)+e−x(3x2+2x+4F′(x)).F'(x)=-e^{-x}I(x)+e^{-x}\left(3x^2+2x+4F'(x)\right).F′(x)=−e−xI(x)+e−x(3x2+2x+4F′(x)).

But since F(x)=e−xI(x)F(x)=e^{-x}I(x)F(x)=e−xI(x), we get

F′(x)=−F(x)+e−x(3x2+2x+4F′(x)).F'(x)=-F(x)+e^{-x}(3x^2+2x+4F'(x)).F′(x)=−F(x)+e−x(3x2+2x+4F′(x)).

Multiply by exe^xex if needed, but a cleaner way is to first rewrite the original equation as

exF(x)=∫3x(3t2+2t+4F′(t))dt.e^xF(x)=\int_3^x (3t^2+2t+4F'(t))dt.exF(x)=∫3x​(3t2+2t+4F′(t))dt.

Differentiating,

ex(F+F′)=3x2+2x+4F′(x).e^x(F+F')=3x^2+2x+4F'(x).ex(F+F′)=3x2+2x+4F′(x).

So,

exF(x)+(ex−4)F′(x)=3x2+2x.e^xF(x)+(e^x-4)F'(x)=3x^2+2x.exF(x)+(ex−4)F′(x)=3x2+2x.

Hence,

F′(x)=3x2+2x−exF(x)ex−4.(1)F'(x)=\frac{3x^2+2x-e^xF(x)}{e^x-4}. \qquad (1)F′(x)=ex−43x2+2x−exF(x)​.(1)
  1. Find a differential equation for FFF

From

ex(F+F′)=3x2+2x+4F′,e^x(F+F')=3x^2+2x+4F',ex(F+F′)=3x2+2x+4F′,

we get

exF+(ex−4)F′=3x2+2x.e^xF+(e^x-4)F'=3x^2+2x.exF+(ex−4)F′=3x2+2x.

Now check x=3x=3x=3 in the original equation:

F(3)=e−3∫33(⋯ )dt=0.F(3)=e^{-3}\int_3^3(\cdots)dt=0.F(3)=e−3∫33​(⋯)dt=0.

So initial condition is

F(3)=0.F(3)=0.F(3)=0.

Thus FFF satisfies the linear ODE

(ex−4)F′(x)+exF(x)=3x2+2x.(e^x-4)F'(x)+e^xF(x)=3x^2+2x.(ex−4)F′(x)+exF(x)=3x2+2x.

Notice the left side is exactly

ddx[(ex−4)F(x)],\frac{d}{dx}\left[(e^x-4)F(x)\right],dxd​[(ex−4)F(x)],

since

ddx[(ex−4)F]=(ex−4)F′+exF.\frac{d}{dx}[(e^x-4)F]=(e^x-4)F'+e^xF.dxd​[(ex−4)F]=(ex−4)F′+exF.

Therefore,

ddx[(ex−4)F(x)]=3x2+2x.\frac{d}{dx}\left[(e^x-4)F(x)\right]=3x^2+2x.dxd​[(ex−4)F(x)]=3x2+2x.

Integrating,

(ex−4)F(x)=x3+x2+C.(e^x-4)F(x)=x^3+x^2+C.(ex−4)F(x)=x3+x2+C.

Use F(3)=0F(3)=0F(3)=0:

(e3−4)⋅0=27+9+C  ⟹  C=−36.(e^3-4)\cdot 0=27+9+C \implies C=-36.(e3−4)⋅0=27+9+C⟹C=−36.

So,

(ex−4)F(x)=x3+x2−36.(e^x-4)F(x)=x^3+x^2-36.(ex−4)F(x)=x3+x2−36.

Hence,

F(x)=x3+x2−36ex−4.F(x)=\frac{x^3+x^2-36}{e^x-4}.F(x)=ex−4x3+x2−36​.
  1. Compute F′(4)F'(4)F′(4)

Let

N(x)=x3+x2−36,D(x)=ex−4.N(x)=x^3+x^2-36, \qquad D(x)=e^x-4.N(x)=x3+x2−36,D(x)=ex−4.

Then

F(x)=N(x)D(x).F(x)=\frac{N(x)}{D(x)}.F(x)=D(x)N(x)​.

Using quotient rule,

F′(x)=N′(x)D(x)−N(x)D′(x)D(x)2.F'(x)=\frac{N'(x)D(x)-N(x)D'(x)}{D(x)^2}.F′(x)=D(x)2N′(x)D(x)−N(x)D′(x)​.

Now,

N′(x)=3x2+2x,N'(x)=3x^2+2x,N′(x)=3x2+2x,

so at x=4x=4x=4,

N(4)=64+16−36=44,N(4)=64+16-36=44,N(4)=64+16−36=44, N′(4)=3(16)+8=56,N'(4)=3(16)+8=56,N′(4)=3(16)+8=56, D(4)=e4−4,D(4)=e^4-4,D(4)=e4−4, D′(4)=e4.D'(4)=e^4.D′(4)=e4.

Therefore,

F′(4)=56(e4−4)−44e4(e4−4)2F'(4)=\frac{56(e^4-4)-44e^4}{(e^4-4)^2}F′(4)=(e4−4)256(e4−4)−44e4​ =56e4−224−44e4(e4−4)2=\frac{56e^4-224-44e^4}{(e^4-4)^2}=(e4−4)256e4−224−44e4​ =12e4−224(e4−4)2.=\frac{12e^4-224}{(e^4-4)^2}.=(e4−4)212e4−224​.
  1. Compare with given form

Given

F′(4)=αeβ−224(eβ−4)2.F'(4)=\frac{\alpha e^{\beta}-224}{(e^{\beta}-4)^2}.F′(4)=(eβ−4)2αeβ−224​.

Comparing,

α=12,β=4.\alpha=12, \qquad \beta=4.α=12,β=4.

Thus,

α+β=12+4=16.\alpha+\beta=12+4=16.α+β=12+4=16.
  1. Final answer
16\boxed{16}16​

The derived answer matches the stored correct answer.

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