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Differential Equations question

2019 · 10 Jan · Shift 2 · Q24
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  5. /2019 · 10 Jan · Shift 2 · Q24

Differential Equations question

2019 · 10 Jan · Shift 2 · Q24

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let f be a differentiable function such that f '(x) = 7 - 34f(x)x,{3 \over 4}{{f\left( x \right)} \over x},43​xf(x)​,(x > 0) and f(1) eee 4. Then lim⁡x→0′ \mathop {\lim }\limits_{x \to 0'} \,x→0′lim​ xf (1x)\left( {{1 \over x}} \right)(x1​) :
  1. A
    does not exist
  2. B
    exists and equals 47{4 \over 7}74​
  3. C
    exists and equals 4
  4. D
    exists and equals 0
View written solutionFree

Correct answer: C

  1. Given differential equation

We are given f′(x)=7−34f(x)x,x>0,f'(x)=7-\frac{3}{4}\frac{f(x)}{x}, \qquad x>0,f′(x)=7−43​xf(x)​,x>0, and f(1)=4.f(1)=4.f(1)=4.

We need to find lim⁡x→0+xf ⁣(1x).\lim_{x\to 0^+} x f\!\left(\frac1x\right).limx→0+​xf(x1​).

  1. Rewrite the differential equation in standard linear form

Bring the term involving f(x)f(x)f(x) to the left: f′(x)+34xf(x)=7.f'(x)+\frac{3}{4x}f(x)=7.f′(x)+4x3​f(x)=7.

This is a first-order linear differential equation.

  1. Find the integrating factor

The integrating factor is I.F.=e∫34x dx=e34ln⁡x=x3/4,\text{I.F.}=e^{\int \frac{3}{4x}\,dx}=e^{\frac34\ln x}=x^{3/4},I.F.=e∫4x3​dx=e43​lnx=x3/4, using x>0x>0x>0.

Multiply the equation by x3/4x^{3/4}x3/4: x3/4f′(x)+34x−1/4f(x)=7x3/4.x^{3/4}f'(x)+\frac{3}{4}x^{-1/4}f(x)=7x^{3/4}.x3/4f′(x)+43​x−1/4f(x)=7x3/4.

The left side is ddx(x3/4f(x))=7x3/4.\frac{d}{dx}\left(x^{3/4}f(x)\right)=7x^{3/4}.dxd​(x3/4f(x))=7x3/4.

  1. Integrate

Integrating, x3/4f(x)=7∫x3/4 dx+C.x^{3/4}f(x)=7\int x^{3/4}\,dx + C.x3/4f(x)=7∫x3/4dx+C.

Now, ∫x3/4 dx=x7/47/4=47x7/4.\int x^{3/4}\,dx=\frac{x^{7/4}}{7/4}=\frac{4}{7}x^{7/4}.∫x3/4dx=7/4x7/4​=74​x7/4.

So, x3/4f(x)=7⋅47x7/4+C=4x7/4+C.x^{3/4}f(x)=7\cdot \frac{4}{7}x^{7/4}+C=4x^{7/4}+C.x3/4f(x)=7⋅74​x7/4+C=4x7/4+C.

Hence, f(x)=4x+Cx−3/4.f(x)=4x+Cx^{-3/4}.f(x)=4x+Cx−3/4.

  1. Use the initial condition

Given f(1)=4f(1)=4f(1)=4, 4=4(1)+C(1)−3/4=4+C.4=4(1)+C(1)^{-3/4}=4+C.4=4(1)+C(1)−3/4=4+C. Therefore, C=0.C=0.C=0.

So the function is f(x)=4x.f(x)=4x.f(x)=4x.

  1. Evaluate the required limit

We need lim⁡x→0+xf(1x).\lim_{x\to 0^+} x f\left(\frac1x\right).limx→0+​xf(x1​).

Since f(t)=4tf(t)=4tf(t)=4t for all t>0t>0t>0, f(1x)=4x.f\left(\frac1x\right)=\frac{4}{x}.f(x1​)=x4​.

Thus, xf(1x)=x⋅4x=4.x f\left(\frac1x\right)=x\cdot \frac{4}{x}=4.xf(x1​)=x⋅x4​=4.

Therefore, lim⁡x→0+xf(1x)=4.\lim_{x\to 0^+} x f\left(\frac1x\right)=4.limx→0+​xf(x1​)=4.

  1. Check options
  • A: does not exist — false
  • B: exists and equals 47\frac4774​ — false
  • C: exists and equals 444 — true
  • D: exists and equals 000 — false

So the correct option is C.

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