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Differential Equations question

2024 · Shift 1 · Q18
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  5. /2024 · Shift 1 · Q18

Differential Equations question

2024 · Shift 1 · Q18

JEE AdvancedMathematicsDifferential EquationsMCQ+3 / −1
Let f(x)f(x)f(x) be a continuously differentiable function on the interval (0,∞)(0, \infty)(0,∞) such that f(1)=2f(1)=2f(1)=2 and lim⁡t→xt10f(x)−x10f(t)t9−x9=1\lim\limits_{t \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1t→xlim​t9−x9t10f(x)−x10f(t)​=1 for each x>0x\gt 0x>0. Then, for all x>0,f(x)x\gt 0, f(x)x>0,f(x) is equal to :
  1. A
    3111x−911x10\frac{31}{11 x}-\frac{9}{11} x^{10}11x31​−119​x10
  2. B
    911x+1311x10\frac{9}{11 x}+\frac{13}{11} x^{10}11x9​+1113​x10
  3. C
    −911x+3111x10\frac{-9}{11 x}+\frac{31}{11} x^{10}11x−9​+1131​x10
  4. D
    1311x+911x10\frac{13}{11 x}+\frac{9}{11} x^{10}11x13​+119​x10
View written solutionFree

Correct answer: B

  1. Given limit condition

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

lim⁡t→xt10f(x)−x10f(t)t9−x9=1.\lim_{t\to x}\frac{t^{10}f(x)-x^{10}f(t)}{t^9-x^9}=1.t→xlim​t9−x9t10f(x)−x10f(t)​=1.

Also, f(1)=2f(1)=2f(1)=2.

We need to determine f(x)f(x)f(x).

  1. Rewrite the limit in a form suitable for differentiation

Fix x>0x>0x>0. As t→xt\to xt→x, both numerator and denominator go to 000, so we expand using differentiability.

Let us write:

t10f(x)−x10f(t)t9−x9=1.\frac{t^{10}f(x)-x^{10}f(t)}{t^9-x^9}=1.t9−x9t10f(x)−x10f(t)​=1.

Taking limit as t→xt\to xt→x.

Now,

t10f(x)−x10f(t)=f(x)(t10−x10)−x10(f(t)−f(x)).t^{10}f(x)-x^{10}f(t)=f(x)(t^{10}-x^{10})-x^{10}(f(t)-f(x)).t10f(x)−x10f(t)=f(x)(t10−x10)−x10(f(t)−f(x)).

Thus,

lim⁡t→xf(x)(t10−x10)−x10(f(t)−f(x))t9−x9=1.\lim_{t\to x}\frac{f(x)(t^{10}-x^{10})-x^{10}(f(t)-f(x))}{t^9-x^9}=1.t→xlim​t9−x9f(x)(t10−x10)−x10(f(t)−f(x))​=1.

Using standard limits,

lim⁡t→xt10−x10t9−x9=10x99x8=109x,\lim_{t\to x}\frac{t^{10}-x^{10}}{t^9-x^9}=\frac{10x^9}{9x^8}=\frac{10}{9}x,t→xlim​t9−x9t10−x10​=9x810x9​=910​x,

and

lim⁡t→xf(t)−f(x)t9−x9=lim⁡t→xf(t)−f(x)t−x⋅t−xt9−x9=f′(x)9x8.\lim_{t\to x}\frac{f(t)-f(x)}{t^9-x^9} = \lim_{t\to x}\frac{f(t)-f(x)}{t-x}\cdot \frac{t-x}{t^9-x^9} = \frac{f'(x)}{9x^8}.t→xlim​t9−x9f(t)−f(x)​=t→xlim​t−xf(t)−f(x)​⋅t9−x9t−x​=9x8f′(x)​.

Therefore,

f(x)⋅109x−x10⋅f′(x)9x8=1.f(x)\cdot \frac{10}{9}x - x^{10}\cdot \frac{f'(x)}{9x^8}=1.f(x)⋅910​x−x10⋅9x8f′(x)​=1.

Simplify:

109xf(x)−x29f′(x)=1.\frac{10}{9}x f(x)-\frac{x^2}{9}f'(x)=1.910​xf(x)−9x2​f′(x)=1.

Multiply by 999:

10xf(x)−x2f′(x)=9.10x f(x)-x^2 f'(x)=9.10xf(x)−x2f′(x)=9.

So the differential equation is

x2f′(x)−10xf(x)=−9.x^2 f'(x)-10x f(x)=-9.x2f′(x)−10xf(x)=−9.

Or,

f′(x)−10xf(x)=−9x2.f'(x)-\frac{10}{x}f(x)=-\frac{9}{x^2}.f′(x)−x10​f(x)=−x29​.
  1. Solve the linear differential equation

We solve

f′(x)−10xf(x)=−9x2.f'(x)-\frac{10}{x}f(x)=-\frac{9}{x^2}.f′(x)−x10​f(x)=−x29​.

This is a first-order linear ODE with

P(x)=−10x,Q(x)=−9x2.P(x)=-\frac{10}{x}, \qquad Q(x)=-\frac{9}{x^2}.P(x)=−x10​,Q(x)=−x29​.

The integrating factor is

I.F.=e∫−10/x dx=e−10ln⁡x=x−10\text{I.F.}=e^{\int -10/x\,dx}=e^{-10\ln x}=x^{-10}I.F.=e∫−10/xdx=e−10lnx=x−10

for x>0x>0x>0.

Multiplying the equation by x−10x^{-10}x−10:

x−10f′(x)−10xx−10f(x)=−9x−12.x^{-10}f'(x)-\frac{10}{x}x^{-10}f(x)=-9x^{-12}.x−10f′(x)−x10​x−10f(x)=−9x−12.

The left-hand side becomes

ddx(f(x)x−10)=−9x−12.\frac{d}{dx}\left(f(x)x^{-10}\right)=-9x^{-12}.dxd​(f(x)x−10)=−9x−12.

Integrate:

f(x)x−10=∫−9x−12 dx=−9⋅x−11−11+C=911x−11+C.f(x)x^{-10}=\int -9x^{-12}\,dx = -9\cdot \frac{x^{-11}}{-11}+C=\frac{9}{11}x^{-11}+C.f(x)x−10=∫−9x−12dx=−9⋅−11x−11​+C=119​x−11+C.

Hence,

f(x)=x10(911x−11+C)=911x+Cx10.f(x)=x^{10}\left(\frac{9}{11}x^{-11}+C\right)=\frac{9}{11x}+Cx^{10}.f(x)=x10(119​x−11+C)=11x9​+Cx10.
  1. Use the condition f(1)=2f(1)=2f(1)=2

Substitute x=1x=1x=1:

f(1)=911+C=2.f(1)=\frac{9}{11}+C=2.f(1)=119​+C=2.

So,

C=2−911=22−911=1311.C=2-\frac{9}{11}=\frac{22-9}{11}=\frac{13}{11}.C=2−119​=1122−9​=1113​.

Therefore,

f(x)=911x+1311x10.f(x)=\frac{9}{11x}+\frac{13}{11}x^{10}.f(x)=11x9​+1113​x10.
  1. Match with the options

This is exactly:

911x+1311x10\boxed{\frac{9}{11x}+\frac{13}{11}x^{10}}11x9​+1113​x10​

which corresponds to Option B.

  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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