Question:
If \(f(x)\) is a differentiable function in the interval \((0, \infty)\) such that \(f(1)=1\) and \(\lim_{t \to x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1\), for each \(x\gt 0\), then \(f\left(\frac{3}{2}\right)\) is equal to :
If \(f(x)\) is a differentiable function in the interval \((0, \infty)\) such that \(f(1)=1\) and \(\lim_{t \to x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1\), for each \(x\gt 0\), then \(f\left(\frac{3}{2}\right)\) is equal to :
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Differentiate the numerator and denominator in \(t\) (L'Hospital) to get a linear differential equation in \(f\).
Updated On: Nov 11, 2026
- \(\frac{13}{6}\)
- \(\frac{23}{18}\)
- \(\frac{25}{9}\)
- \(\frac{31}{18}\)
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The Correct Option is D
Solution and Explanation
The limit is of the form \(\frac00\); apply L'Hospital's rule (differentiate numerator and denominator w.r.t. \(t\)).
\(\lim_{t\to x}\frac{2t f(x)-x^2 f'(t)}{1}=2x f(x)-x^2 f'(x)=1\).
So \(f'(x)-\frac{2}{x}f(x)=-\frac{1}{x^2}\), a linear differential equation with integrating factor \(e^{-\int 2/x\,dx}=x^{-2}\).
\(\frac{d}{dx}\left(\frac{f(x)}{x^2}\right)=-\frac{1}{x^4}\Rightarrow \frac{f(x)}{x^2}=\frac{1}{3x^3}+C\), so \(f(x)=\frac{1}{3x}+Cx^2\).
\(f(1)=1\Rightarrow \frac13+C=1\Rightarrow C=\frac23\).
\(f\left(\frac32\right)=\frac{1}{3}\cdot\frac{2}{3}+\frac23\cdot\frac94=\frac29+\frac32=\frac{4+27}{18}=\frac{31}{18}\).
Answer: (4) \(\frac{31}{18}\)
\(\lim_{t\to x}\frac{2t f(x)-x^2 f'(t)}{1}=2x f(x)-x^2 f'(x)=1\).
So \(f'(x)-\frac{2}{x}f(x)=-\frac{1}{x^2}\), a linear differential equation with integrating factor \(e^{-\int 2/x\,dx}=x^{-2}\).
\(\frac{d}{dx}\left(\frac{f(x)}{x^2}\right)=-\frac{1}{x^4}\Rightarrow \frac{f(x)}{x^2}=\frac{1}{3x^3}+C\), so \(f(x)=\frac{1}{3x}+Cx^2\).
\(f(1)=1\Rightarrow \frac13+C=1\Rightarrow C=\frac23\).
\(f\left(\frac32\right)=\frac{1}{3}\cdot\frac{2}{3}+\frac23\cdot\frac94=\frac29+\frac32=\frac{4+27}{18}=\frac{31}{18}\).
Answer: (4) \(\frac{31}{18}\)
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