Question
Let $$f\left( x \right)$$ be differentiable on the interval $$\left( {0,\,\infty } \right)$$ such that $$f\left( 1 \right) = 1,$$ and $$\mathop {\lim }\limits_{t \to x} \frac{{{t^2}f\left( x \right) - {x^2}f\left( t \right)}}{{t - x}} = 1$$ for each $$x > 0.$$ Then $$f\left( x \right)$$ is-
A.
$$\frac{1}{{3x}} + \frac{{2{x^2}}}{3}$$
B.
$$\frac{{ - 1}}{{3x}} + \frac{{4{x^2}}}{3}$$
C.
$$\frac{{ - 1}}{x} + \frac{2}{{{x^2}}}$$
D.
$$\frac{1}{x}$$
Answer :
$$\frac{1}{{3x}} + \frac{{2{x^2}}}{3}$$
Solution :
Given that $$f\left( x \right)$$ is differentiable on $$\left( {0,\,\infty } \right)$$ with
$$f\left( 1 \right) = 1\,{\text{and}}\mathop {\lim }\limits_{t \to x} \frac{{{t^2}f\left( x \right) - {x^2}f\left( t \right)}}{{t - x}} = 1$$ for each $$x>0$$
$$\eqalign{
& \Rightarrow \mathop {\lim }\limits_{t \to x} \frac{{2t\,f\left( x \right) - {x^2}f'\left( t \right)}}{1} = 1\,\,\left[ {{\text{using L'Hospital rule}}} \right] \cr
& \Rightarrow 2x\,f\left( x \right) - {x^2}\,f'\left( x \right) = 1 \cr
& \Rightarrow f'\left( x \right) - \frac{2}{x}f\left( x \right) = - \frac{1}{{{x^2}}}\left[ {{\text{Linear differential equation}}} \right] \cr} $$
Integrating factor
$$\eqalign{
& {e^{\int { - \,\frac{2}{x}dx} }} = {e^{ - 2\,\log \,x}} = {e^{\log \frac{1}{{{x^2}}}}} = \frac{1}{{{x^2}}} \cr
& \therefore \,\,{\text{Solution is}}\,f\left( x \right) \times \frac{1}{{{x^2}}} = \int {\left( { - \frac{1}{{{x^2}}}} \right) \times \frac{1}{{{x^2}}}dx} \cr
& \Rightarrow \frac{{f\left( x \right)}}{{{x^2}}} = \frac{1}{{3{x^3}}} + C \cr
& \Rightarrow f\left( x \right) = C{x^2} + \frac{1}{{3x}} \cr
& {\text{Also }}f\left( 1 \right) = 1 \cr
& \Rightarrow 1 = C + \frac{1}{3} \cr
& \Rightarrow C = \frac{2}{3} \cr
& \therefore f\left( x \right) = \frac{2}{3}{x^2} + \frac{1}{{3x}} \cr} $$