Question:
Let \(f(x)\) be differentiable on the interval \((0,\infty)\) such that \(f(1)=1\) and
\[
\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1 \quad \text{for each } x>0.
\]
Then, \(f(x)\) is equal to
Let \(f(x)\) be differentiable on the interval \((0,\infty)\) such that \(f(1)=1\) and
\[
\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1 \quad \text{for each } x>0.
\]
Then, \(f(x)\) is equal to
Show Hint
Convert limit expressions into derivatives to form differential equations.
Updated On: Apr 16, 2026
- \(\frac{1}{3x} + \frac{2}{3}x^2\)
- \(-\frac{x}{3} + \frac{4x^2}{3}\)
- \(-\frac{1}{x}\)
- \(-\frac{1}{x} + \frac{2}{x^2}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
Use definition of derivative.
Step 1: Rewrite limit.
\[ \lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1 \] Rearrange: \[ = \lim_{t \to x} \frac{x^2 f(x) - x^2 f(t) + (t^2 - x^2)f(x)}{t-x} \] \[ = -x^2 f'(x) + 2x f(x) \]
Step 2: Form differential equation.
\[ - x^2 f'(x) + 2x f(x) = 1 \] \[ \Rightarrow f'(x) - \frac{2}{x}f(x) = -\frac{1}{x^2} \]
Step 3: Solve.
Integrating factor: \[ I.F = e^{\int -2/x dx} = x^{-2} \] \[ \frac{d}{dx}(f(x)x^{-2}) = -x^{-4} \] \[ f(x)x^{-2} = \frac{1}{3x^3} + C \] \[ f(x) = \frac{1}{3x} + Cx^2 \]
Step 4: Use \(f(1)=1\).
\[ 1 = \frac{1}{3} + C \Rightarrow C = \frac{2}{3} \] \[ f(x) = \frac{1}{3x} + \frac{2}{3}x^2 \]
Step 1: Rewrite limit.
\[ \lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1 \] Rearrange: \[ = \lim_{t \to x} \frac{x^2 f(x) - x^2 f(t) + (t^2 - x^2)f(x)}{t-x} \] \[ = -x^2 f'(x) + 2x f(x) \]
Step 2: Form differential equation.
\[ - x^2 f'(x) + 2x f(x) = 1 \] \[ \Rightarrow f'(x) - \frac{2}{x}f(x) = -\frac{1}{x^2} \]
Step 3: Solve.
Integrating factor: \[ I.F = e^{\int -2/x dx} = x^{-2} \] \[ \frac{d}{dx}(f(x)x^{-2}) = -x^{-4} \] \[ f(x)x^{-2} = \frac{1}{3x^3} + C \] \[ f(x) = \frac{1}{3x} + Cx^2 \]
Step 4: Use \(f(1)=1\).
\[ 1 = \frac{1}{3} + C \Rightarrow C = \frac{2}{3} \] \[ f(x) = \frac{1}{3x} + \frac{2}{3}x^2 \]
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