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Limits Continuity and Differentiability question

2024 · 5 Apr · Shift 1 · Q55
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  5. /2024 · 5 Apr · Shift 1 · Q55

Limits Continuity and Differentiability question

2024 · 5 Apr · Shift 1 · Q55

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1
Let fff be a differentiable function in the interval (0,∞)(0, \infty)(0,∞) such that f(1)=1f(1)=1f(1)=1 and lim⁡t→xt2f(x)−x2f(t)t−x=1\lim_{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1t→xlim​t−xt2f(x)−x2f(t)​=1 for each x>0x\gt 0x>0. Then 2f(2)+3f(3)2 f(2)+3 f(3)2f(2)+3f(3) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 24

  1. We are given, for each fixed x>0x>0x>0,
lim⁡t→xt2f(x)−x2f(t)t−x=1.\lim_{t\to x}\frac{t^2f(x)-x^2f(t)}{t-x}=1.t→xlim​t−xt2f(x)−x2f(t)​=1.

Also, fff is differentiable on (0,∞)(0,\infty)(0,∞) and f(1)=1f(1)=1f(1)=1.

We must find 2f(2)+3f(3)2f(2)+3f(3)2f(2)+3f(3).

  1. Since fff is differentiable, as t→xt\to xt→x we can use
f(t)=f(x)+f′(x)(t−x)+o(t−x).f(t)=f(x)+f'(x)(t-x)+o(t-x).f(t)=f(x)+f′(x)(t−x)+o(t−x).

Also,

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

Now expand the numerator:

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

Substitute the expansions:

t2f(x)=(x2+2x(t−x)+o(t−x))f(x)=x2f(x)+2xf(x)(t−x)+o(t−x),t^2f(x)=\big(x^2+2x(t-x)+o(t-x)\big)f(x) = x^2f(x)+2xf(x)(t-x)+o(t-x),t2f(x)=(x2+2x(t−x)+o(t−x))f(x)=x2f(x)+2xf(x)(t−x)+o(t−x),

and

x2f(t)=x2(f(x)+f′(x)(t−x)+o(t−x))=x2f(x)+x2f′(x)(t−x)+o(t−x).x^2f(t)=x^2\big(f(x)+f'(x)(t-x)+o(t-x)\big) = x^2f(x)+x^2f'(x)(t-x)+o(t-x).x2f(t)=x2(f(x)+f′(x)(t−x)+o(t−x))=x2f(x)+x2f′(x)(t−x)+o(t−x).

Therefore,

t2f(x)−x2f(t)=(2xf(x)−x2f′(x))(t−x)+o(t−x).t^2f(x)-x^2f(t)=\big(2xf(x)-x^2f'(x)\big)(t-x)+o(t-x).t2f(x)−x2f(t)=(2xf(x)−x2f′(x))(t−x)+o(t−x).

Dividing by t−xt-xt−x and taking limit gives

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

So fff satisfies the differential equation

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

Equivalently,

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

The integrating factor is

I.F.=e∫−2x dx=e−2ln⁡x=x−2.I.F.=e^{\int -\frac{2}{x}\,dx}=e^{-2\ln x}=x^{-2}.I.F.=e∫−x2​dx=e−2lnx=x−2.

Multiply the equation 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-hand side is

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

Hence,

ddx(f(x)x−2)=−x−4.\frac{d}{dx}\big(f(x)x^{-2}\big)=-x^{-4}.dxd​(f(x)x−2)=−x−4.

Integrating,

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

Thus,

f(x)=x2(13x−3+C)=13x+Cx2.f(x)=x^2\left(\frac{1}{3}x^{-3}+C\right)=\frac{1}{3x}+Cx^2.f(x)=x2(31​x−3+C)=3x1​+Cx2.
  1. Use the condition f(1)=1f(1)=1f(1)=1:
1=f(1)=13+C.1=f(1)=\frac{1}{3}+C.1=f(1)=31​+C.

So,

C=23.C=\frac{2}{3}.C=32​.

Therefore,

f(x)=13x+23x2.f(x)=\frac{1}{3x}+\frac{2}{3}x^2.f(x)=3x1​+32​x2.
  1. Compute f(2)f(2)f(2) and f(3)f(3)f(3):
f(2)=16+23⋅4=16+83=176,f(2)=\frac{1}{6}+\frac{2}{3}\cdot 4=\frac{1}{6}+\frac{8}{3}=\frac{17}{6},f(2)=61​+32​⋅4=61​+38​=617​, f(3)=19+23⋅9=19+6=559.f(3)=\frac{1}{9}+\frac{2}{3}\cdot 9=\frac{1}{9}+6=\frac{55}{9}.f(3)=91​+32​⋅9=91​+6=955​.

Now,

2f(2)+3f(3)=2⋅176+3⋅559=173+553=723=24.2f(2)+3f(3)=2\cdot \frac{17}{6}+3\cdot \frac{55}{9} =\frac{17}{3}+\frac{55}{3} =\frac{72}{3}=24.2f(2)+3f(3)=2⋅617​+3⋅955​=317​+355​=372​=24.
  1. Final answer:
24\boxed{24}24​

This matches the stored correct answer.

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