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Application of Integration question

2009 · Shift 1 · Q29
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  5. /2009 · Shift 1 · Q29

Application of Integration question

2009 · Shift 1 · Q29

JEE AdvancedMathematicsApplication of IntegrationMCQ+3 / −1
Let fff be a non-negative function defined on the interval [0,1][0,1][0,1]. If ∫0x1−(f′(t))2dt=∫0xf(t)dt,0≤x≤1\int\limits_0^x {\sqrt {1 - {{(f'(t))}^2}dt} = \int\limits_0^x {f(t)dt,0 \le x \le 1} }0∫x​1−(f′(t))2dt​=0∫x​f(t)dt,0≤x≤1, and f(0)=0f(0) = 0f(0)=0, then
  1. A
    f(12)<12f\left( {{1 \over 2}} \right) \lt {1 \over 2}f(21​)<21​ and f(13)>13f\left( {{1 \over 3}} \right) \gt {1 \over 3}f(31​)>31​
  2. B
    f(12)>12f\left( {{1 \over 2}} \right) \gt {1 \over 2}f(21​)>21​ and f(13)>13f\left( {{1 \over 3}} \right) \gt {1 \over 3}f(31​)>31​
  3. C
    f(12)<12f\left( {{1 \over 2}} \right) \lt {1 \over 2}f(21​)<21​ and f(13)<13f\left( {{1 \over 3}} \right) \lt {1 \over 3}f(31​)<31​
  4. D
    f(12)>12f\left( {{1 \over 2}} \right) \gt {1 \over 2}f(21​)>21​ and f(13)<13f\left( {{1 \over 3}} \right) \lt {1 \over 3}f(31​)<31​
View written solutionFree

Correct answer: C

  1. Interpret the given condition carefully

    The statement is intended as

    ∫0x1−(f′(t))2 dt=∫0xf(t) dt,0≤x≤1,\int_0^x \sqrt{1-(f'(t))^2}\,dt = \int_0^x f(t)\,dt, \qquad 0\le x\le 1,∫0x​1−(f′(t))2​dt=∫0x​f(t)dt,0≤x≤1,

    with f(0)=0f(0)=0f(0)=0 and f≥0f\ge 0f≥0 on [0,1][0,1][0,1].

  2. Differentiate both sides with respect to xxx

    Since the equality holds for every x∈[0,1]x\in[0,1]x∈[0,1], by the Fundamental Theorem of Calculus,

    1−(f′(x))2=f(x).\sqrt{1-(f'(x))^2}=f(x).1−(f′(x))2​=f(x).

    Because f(x)≥0f(x)\ge 0f(x)≥0, squaring gives

    1−(f′(x))2=f(x)21-(f'(x))^2=f(x)^21−(f′(x))2=f(x)2

    or

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

  3. Differentiate this relation

    Differentiate:

    2f′(x)f′′(x)+2f(x)f′(x)=0.2f'(x)f''(x)+2f(x)f'(x)=0.2f′(x)f′′(x)+2f(x)f′(x)=0.

    Hence

    2f′(x)(f′′(x)+f(x))=0.2f'(x)(f''(x)+f(x))=0.2f′(x)(f′′(x)+f(x))=0.

    A better way is to use the original unsquared equation:

    f(x)=1−(f′(x))2.f(x)=\sqrt{1-(f'(x))^2}.f(x)=1−(f′(x))2​.

    Differentiate both sides:

    f′(x)=−2f′(x)f′′(x)21−(f′(x))2=−f′(x)f′′(x)f(x).f'(x)=\frac{-2f'(x)f''(x)}{2\sqrt{1-(f'(x))^2}} =-\frac{f'(x)f''(x)}{f(x)}.f′(x)=21−(f′(x))2​−2f′(x)f′′(x)​=−f(x)f′(x)f′′(x)​.

    So either f′(x)=0f'(x)=0f′(x)=0, or

    f(x)=−f′′(x).f(x)=-f''(x).f(x)=−f′′(x).

    Thus the function satisfies the differential equation

    f′′+f=0.f''+f=0.f′′+f=0.

  4. Solve the differential equation

    General solution:

    f(x)=Asin⁡x+Bcos⁡x.f(x)=A\sin x+B\cos x.f(x)=Asinx+Bcosx.

    Using f(0)=0f(0)=0f(0)=0 gives

    B=0.B=0.B=0.

    So

    f(x)=Asin⁡x.f(x)=A\sin x.f(x)=Asinx.

    Then

    f′(x)=Acos⁡x.f'(x)=A\cos x.f′(x)=Acosx.

    Substitute into

    (f′)2+f2=1:(f')^2+f^2=1:(f′)2+f2=1: A2cos⁡2x+A2sin⁡2x=1A^2\cos^2 x + A^2\sin^2 x =1A2cos2x+A2sin2x=1 A2=1.A^2=1.A2=1.

    Since f(x)≥0f(x)\ge 0f(x)≥0 on [0,1][0,1][0,1] and sin⁡x>0\sin x>0sinx>0 for x∈(0,1]x\in(0,1]x∈(0,1], we must take

    A=1.A=1.A=1.

    Therefore,

    f(x)=sin⁡x.f(x)=\sin x.f(x)=sinx.

  5. Compare values with xxx

    We now check f(12)=sin⁡(12)f\left(\frac12\right)=\sin\left(\frac12\right)f(21​)=sin(21​) and f(13)=sin⁡(13)f\left(\frac13\right)=\sin\left(\frac13\right)f(31​)=sin(31​).

    Use the standard fact:

    • For 0<x<π20<x<\frac\pi20<x<2π​, we have sin⁡x<x\sin x < xsinx<x.

    Since

    0<13<12<π2,0<\frac13<\frac12<\frac\pi2,0<31​<21​<2π​,

    it follows that

    sin⁡(12)<12,sin⁡(13)<13.\sin\left(\frac12\right)<\frac12, \qquad \sin\left(\frac13\right)<\frac13.sin(21​)<21​,sin(31​)<31​.

    Hence,

    f(12)<12,f(13)<13.f\left(\frac12\right)<\frac12, \qquad f\left(\frac13\right)<\frac13.f(21​)<21​,f(31​)<31​.

  6. Evaluate the options

    • A: first true, second false
    • B: both false
    • C: both true
    • D: both false

    Therefore the correct option is

    C.\boxed{\text{C}}.C​.

  7. Comparison with stored answer

    Stored correct answer: C

    My derived answer: C

    So they agree.

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