Let $f(x)$ be differentiable on the interval $(0, \infty)$ such that $f(1)=1$, and \[ \lim _{t \rightarrow x} \frac{t^{2} f(x)-x^{2} f(t)}{t-x}=1 \] for each $x>0$. Then $f(x)$ is

(A) $\frac{1}{3 x}+\frac{2 x^{2}}{3}$

(B) $\frac{-1}{3 x}+\frac{4 x^{2}}{3}$

(C) $\frac{-1}{x}+\frac{2}{x^{2}}$

JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Differential Equations

Let

f (x)

be differentiable on the interval

(0, \infty)

such that

f (1) = 1

, and

t \to x lim \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1

for each

x > 0

. Then

f (x)

Answer: (A)

Applying L'Hôpital's rule to the limit:

t \to x lim \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = t \to x lim \frac{2 t f ( x ) - x ^{2} f ^{'} ( t )}{1}

t = x

2 x f (x) - x^{2} f^{'} (x) = 1

This gives the differential equation:

x^{2} f^{'} (x) - 2 x f (x) + 1 = 0

Solving:

f (x) = c x^{2} + \frac{1}{3 x}

Using

f (1) = 1

c + \frac{1}{3} = 1

, so

c = \frac{2}{3}

Therefore,

f (x) = \frac{1}{3 x} + \frac{2 x ^{2}}{3}

JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Single correct option