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Application of Integration question

2018 · Shift 1 · Q24
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  5. /2018 · Shift 1 · Q24

Application of Integration question

2018 · Shift 1 · Q24

JEE AdvancedMathematicsApplication of IntegrationMultiple correct+4 / −1
Let f : [0, ∞\infty∞) →\to→ R be a continuous function such that f(x)=1−2x+∫0xex−tf(t)dtf(x) = 1 - 2x + \int_0^x {{e^{x - t}}f(t)dt}f(x)=1−2x+∫0x​ex−tf(t)dt for all x ∈\in∈[0, ∞\infty∞). Then, which of the following statement(s) is (are) TRUE?
  1. A
    The curve y = f(x) passes through the point (1, 2)
  2. B
    The curve y = f(x) passes through the point (2, −-− 1)
  3. C
    The area of the region {(x,y)∈[0,1]×R:f(x)≤y≤1−x2}\{ (x,y) \in [0,1] \times R:f(x) \le y \le \sqrt {1 - {x^2}} \}{(x,y)∈[0,1]×R:f(x)≤y≤1−x2​} is π−24{{\pi - 2} \over 4}4π−2​
  4. D
    The area of the region {(x,y)∈[0,1]×R:f(x)≤y≤1−x2}\{ (x,y) \in [0,1] \times R:f(x) \le y \le \sqrt {1 - {x^2}} \}{(x,y)∈[0,1]×R:f(x)≤y≤1−x2​} is π−14{{\pi - 1} \over 4}4π−1​
View written solutionFree

Correct answer: B, C

  1. Given integral equation

    We have f(x)=1−2x+∫0xex−tf(t) dt,x≥0.f(x)=1-2x+\int_0^x e^{x-t}f(t)\,dt, \qquad x\ge 0.f(x)=1−2x+∫0x​ex−tf(t)dt,x≥0.

    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. Then f(x)=1−2x+I(x).f(x)=1-2x+I(x).f(x)=1−2x+I(x).

  2. Differentiate to get a differential equation

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

    Since f(x)=1−2x+I(x),f(x)=1-2x+I(x),f(x)=1−2x+I(x), differentiating gives f′(x)=−2+I′(x).f'(x)=-2+I'(x).f′(x)=−2+I′(x).

    Substitute I′(x)=f(x)+I(x)I'(x)=f(x)+I(x)I′(x)=f(x)+I(x) and I(x)=f(x)−1+2xI(x)=f(x)-1+2xI(x)=f(x)−1+2x: f′(x)=−2+f(x)+I(x)f'(x)=-2+f(x)+I(x)f′(x)=−2+f(x)+I(x) =−2+f(x)+f(x)−1+2x=-2+f(x)+f(x)-1+2x=−2+f(x)+f(x)−1+2x =2f(x)+2x−3.=2f(x)+2x-3.=2f(x)+2x−3.

    So fff satisfies f′(x)−2f(x)=2x−3.f'(x)-2f(x)=2x-3.f′(x)−2f(x)=2x−3.

  3. Initial condition

    Put x=0x=0x=0 in the original equation: f(0)=1−0+∫00e0−tf(t) dt=1.f(0)=1-0+\int_0^0 e^{0-t}f(t)\,dt=1.f(0)=1−0+∫00​e0−tf(t)dt=1.

    Hence, f(0)=1.f(0)=1.f(0)=1.

  4. Solve the differential equation

    We solve f′(x)−2f(x)=2x−3.f'(x)-2f(x)=2x-3.f′(x)−2f(x)=2x−3.

    Integrating factor is e−2x.e^{-2x}.e−2x.

    Then ddx(f(x)e−2x)=(2x−3)e−2x.\frac{d}{dx}\big(f(x)e^{-2x}\big)=(2x-3)e^{-2x}.dxd​(f(x)e−2x)=(2x−3)e−2x.

    But it is quicker to try a particular solution of the form fp(x)=ax+b.f_p(x)=ax+b.fp​(x)=ax+b.

    Then fp′(x)−2fp(x)=a−2(ax+b)=(−2a)x+(a−2b).f_p'(x)-2f_p(x)=a-2(ax+b)=(-2a)x+(a-2b).fp′​(x)−2fp​(x)=a−2(ax+b)=(−2a)x+(a−2b).

    Compare with 2x−32x-32x−3: −2a=2  ⟹  a=−1,-2a=2\implies a=-1,−2a=2⟹a=−1, a−2b=−3  ⟹  −1−2b=−3  ⟹  b=1.a-2b=-3\implies -1-2b=-3\implies b=1.a−2b=−3⟹−1−2b=−3⟹b=1.

    So a particular solution is fp(x)=1−x.f_p(x)=1-x.fp​(x)=1−x.

    Homogeneous solution: fh(x)=Ce2x.f_h(x)=Ce^{2x}.fh​(x)=Ce2x.

    Therefore, f(x)=Ce2x+1−x.f(x)=Ce^{2x}+1-x.f(x)=Ce2x+1−x.

    Using f(0)=1f(0)=1f(0)=1: C+1=1  ⟹  C=0.C+1=1\implies C=0.C+1=1⟹C=0.

    Hence, f(x)=1−x.\boxed{f(x)=1-x}.f(x)=1−x​.

  5. Check options A and B

    • At x=1x=1x=1: f(1)=1−1=0.f(1)=1-1=0.f(1)=1−1=0. So the curve passes through (1,0)(1,0)(1,0), not (1,2)(1,2)(1,2). Therefore, A is false.

    • At x=2x=2x=2: f(2)=1−2=−1.f(2)=1-2=-1.f(2)=1−2=−1. So the curve passes through (2,−1)(2,-1)(2,−1). Therefore, B is true.

  6. Area in options C and D

    The region is {(x,y)∈[0,1]×R:f(x)≤y≤1−x2}.\{(x,y)\in [0,1]\times \mathbb R: f(x)\le y\le \sqrt{1-x^2}\}.{(x,y)∈[0,1]×R:f(x)≤y≤1−x2​}.

    Since f(x)=1−xf(x)=1-xf(x)=1−x, area is A=∫01(1−x2−(1−x))dx.A=\int_0^1\left(\sqrt{1-x^2}-(1-x)\right)dx.A=∫01​(1−x2​−(1−x))dx.

    Split it: A=∫011−x2 dx−∫01(1−x) dx.A=\int_0^1\sqrt{1-x^2}\,dx-\int_0^1(1-x)\,dx.A=∫01​1−x2​dx−∫01​(1−x)dx.

    Now, ∫011−x2 dx=π4\int_0^1\sqrt{1-x^2}\,dx=\frac{\pi}{4}∫01​1−x2​dx=4π​ because it is the area of a quarter unit circle.

    Also, ∫01(1−x) dx=[x−x22]01=1−12=12.\int_0^1(1-x)\,dx=\left[x-\frac{x^2}{2}\right]_0^1=1-\frac12=\frac12.∫01​(1−x)dx=[x−2x2​]01​=1−21​=21​.

    Therefore, A=π4−12=π−24.A=\frac{\pi}{4}-\frac12=\frac{\pi-2}{4}.A=4π​−21​=4π−2​.

    So C is true and D is false.

  7. Final conclusion

    The true statements are B, C.\boxed{\text{B, C}}.B, C​.

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