Question

Let $f(x)$ be continuously 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. $-\frac{1}{x}+\frac{2}{x^{2}}$
• D. $\frac{1}{x}$

Answer 1

Answer: (A)

$1=\displaystyle\lim _{t \rightarrow x} \frac{l^{2} J(x)-x^{2} f(l)}{t-x}\left(\frac{0}{0}\right.$ form $)$
$=\displaystyle\lim _{t \rightarrow x} \frac{2 t f(x)-x^{2} f^{\prime}(t)}{1}$
$\Rightarrow 2 x f(x)-x^{2} f'(x)=1 $
$\Rightarrow x^{2} \frac{d y}{d x}-2 x y=-1, y=f(x)$
$\Rightarrow \frac{d y}{d x}-\frac{2}{x} y=-\frac{1}{x^{2}}$
which is linear equation whose
I.F. $=e^{-2 \log x}=\frac{1}{x^{2}}$.
Multiplying both sides of the last equation by I.F., we get
$\frac{d}{d x}\left(\frac{y}{x^{2}}\right)=-\frac{1}{x^{4}} $
$\Rightarrow \frac{y}{x^{2}}=\frac{1}{3 x^{3}}+K .$
Since $y=f(1)=1 .$
so $K=\frac{2}{3}$.
Thus $y=\frac{1}{3 x}+\frac{2}{3} x^{2}$