Question

Let $f(x)$ be a differentiable function satisfying the equations $\lim_{t \to x} \dfrac{t^2 f(x)-x^2 f(t)}{t-x} = 3$ and $f(1)=2$. Find the value of $2f(2)$.

Hint:

Limits resembling derivative definitions often convert directly into differential equations.

Answer 1

Correct Answer: 23

Step 1: Rewrite the given limit in a suitable form

Given,

lim_t→x (t²f(x) − x²f(t)) / (t − x) = 3

Multiply numerator and denominator by −1:

lim_t→x (x²f(t) − t²f(x)) / (x − t) = 3

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Step 2: Use definition of derivative

As t → x,

x² · (f(t) − f(x)) / (x − t) − f(x) · (t² − x²) / (x − t) = 3

Taking limits:

x²f′(x) − f(x)(2x) = 3

x²f′(x) − 2x f(x) = 3

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Step 3: Form the differential equation

Divide throughout by x²:

f′(x) − (2/x)f(x) = 3/x²

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Step 4: Solve using integrating factor

Integrating factor (I.F.):

I.F. = e^{∫(−2/x)dx} = 1/x²

Multiplying throughout by I.F.:

d/dx [ f(x) / x² ] = 3 / x⁴

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Step 5: Integrate

f(x) / x² = ∫ 3x⁻⁴ dx

f(x) / x² = −1/x³ + C

f(x) = Cx² − 1/x

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Step 6: Use the given condition

Given f(1) = 2:

C − 1 = 2

C = 3

Thus,

f(x) = 3x² − 1/x

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Step 7: Find 2f(2)

f(2) = 3(4) − 1/2 = 12 − 1/2 = 23/2

2f(2) = 23

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Final Answer:

The value of 2f(2) is
23