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Differential Equations question

2016 · 9 Apr · Shift 1 · Q41
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  5. /2016 · 9 Apr · Shift 1 · Q41

Differential Equations question

2016 · 9 Apr · Shift 1 · Q41

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If f(x) is a differentiable function in the interval (0, ∞\infty∞) such that f (1) = 1 and lim⁡t→x\mathop {\lim }\limits_{t \to x}t→xlim​ t2f(x)−x2f(t)t−x=1,{{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,t−xt2f(x)−x2f(t)​=1, for each x > 0, then f(32)f\left( {\frac{3}{2}} \right)f(23​) equal to :
  1. A
    136{{13} \over 6}613​
  2. B
    2318{{23} \over 18}1823​
  3. C
    259{{25} \over 9}925​
  4. D
    3118{{31} \over 18}1831​
View written solutionFree

Correct answer: D

  1. Interpret the given limit

We are given, for each x>0x>0x>0,

lim⁡t→xt2f(x)−x2f(t)t−x=1.\lim_{t\to x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1.t→xlim​t−xt2f(x)−x2f(t)​=1.

Since fff is differentiable, we can evaluate this limit using expansion near t=xt=xt=x.

  1. Expand the numerator near t=xt=xt=x

Let

N(t)=t2f(x)−x2f(t).N(t)=t^2 f(x)-x^2 f(t).N(t)=t2f(x)−x2f(t).

As t→xt\to xt→x,

t2−x2=(t−x)(t+x).t^2-x^2=(t-x)(t+x).t2−x2=(t−x)(t+x).

So,

N(t)=f(x)(t2−x2)+x2(f(x)−f(t)).N(t)=f(x)(t^2-x^2)+x^2\bigl(f(x)-f(t)\bigr).N(t)=f(x)(t2−x2)+x2(f(x)−f(t)).

Hence,

N(t)t−x=f(x)(t+x)+x2⋅f(x)−f(t)t−x.\frac{N(t)}{t-x}=f(x)(t+x)+x^2\cdot \frac{f(x)-f(t)}{t-x}.t−xN(t)​=f(x)(t+x)+x2⋅t−xf(x)−f(t)​.

Now let t→xt\to xt→x:

  • t+x→2xt+x \to 2xt+x→2x
  • f(x)−f(t)t−x→−f′(x)\dfrac{f(x)-f(t)}{t-x} \to -f'(x)t−xf(x)−f(t)​→−f′(x)

Therefore,

lim⁡t→xt2f(x)−x2f(t)t−x=2xf(x)−x2f′(x).\lim_{t\to x}\frac{t^2 f(x)-x^2 f(t)}{t-x}=2x f(x)-x^2 f'(x).t→xlim​t−xt2f(x)−x2f(t)​=2xf(x)−x2f′(x).

Given that this limit equals 111, we get

2xf(x)−x2f′(x)=1.2x f(x)-x^2 f'(x)=1.2xf(x)−x2f′(x)=1.

Rearranging,

x2f′(x)−2xf(x)=−1.x^2 f'(x)-2x f(x)=-1.x2f′(x)−2xf(x)=−1.
  1. Solve the differential equation

Write it as

f′(x)−2xf(x)=−1x2.f'(x)-\frac{2}{x}f(x)=-\frac{1}{x^2}.f′(x)−x2​f(x)=−x21​.

This is a linear differential equation.

Its integrating factor is

I.F.=e∫−2/x dx=e−2ln⁡x=x−2.I.F.=e^{\int -2/x\,dx}=e^{-2\ln x}=x^{-2}.I.F.=e∫−2/xdx=e−2lnx=x−2.

Multiplying throughout by x−2x^{-2}x−2,

x−2f′(x)−2x−3f(x)=−x−4.x^{-2}f'(x)-2x^{-3}f(x)=-x^{-4}.x−2f′(x)−2x−3f(x)=−x−4.

The left side is

ddx(f(x)x−2).\frac{d}{dx}\left(f(x)x^{-2}\right).dxd​(f(x)x−2).

So,

ddx(f(x)x2)=−1x4.\frac{d}{dx}\left(\frac{f(x)}{x^2}\right)=-\frac{1}{x^4}.dxd​(x2f(x)​)=−x41​.

Integrating,

f(x)x2=∫−x−4 dx=13x−3+C.\frac{f(x)}{x^2}=\int -x^{-4}\,dx=\frac{1}{3}x^{-3}+C.x2f(x)​=∫−x−4dx=31​x−3+C.

Thus,

f(x)=x2(13x3+C)=13x+Cx2.f(x)=x^2\left(\frac{1}{3x^3}+C\right)=\frac{1}{3x}+Cx^2.f(x)=x2(3x31​+C)=3x1​+Cx2.
  1. Use the condition f(1)=1f(1)=1f(1)=1

Given f(1)=1f(1)=1f(1)=1,

1=13+C  ⟹  C=23.1=\frac13 + C \implies C=\frac23.1=31​+C⟹C=32​.

So,

f(x)=13x+23x2.f(x)=\frac{1}{3x}+\frac23 x^2.f(x)=3x1​+32​x2.
  1. Find f(32)f\left(\frac32\right)f(23​)
f(32)=13⋅32+23(32)2.f\left(\frac32\right)=\frac{1}{3\cdot \frac32}+\frac23\left(\frac32\right)^2.f(23​)=3⋅23​1​+32​(23​)2.

Now,

13⋅32=192=29,\frac{1}{3\cdot \frac32}=\frac{1}{\frac92}=\frac{2}{9},3⋅23​1​=29​1​=92​,

and

23(32)2=23⋅94=32.\frac23\left(\frac32\right)^2=\frac23\cdot \frac94=\frac{3}{2}.32​(23​)2=32​⋅49​=23​.

Therefore,

f(32)=29+32=418+2718=3118.f\left(\frac32\right)=\frac29+\frac32=\frac{4}{18}+\frac{27}{18}=\frac{31}{18}.f(23​)=92​+23​=184​+1827​=1831​.
  1. Compare with options

Thus,

f(32)=3118.f\left(\frac32\right)=\frac{31}{18}.f(23​)=1831​.

So the correct option is D.

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