Question

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

• A. $\frac{1}{3x}+\frac{2x^2}{3}$
• B. $-\frac{1}{3x}+\frac{4x^2}{3}$
• C. $-\frac{1}{x}+\frac{2}{x^2}$
• D. $\frac{1}{x}$

Answer 1

Answer: (A)

$1=\underset{t\to x}{\lim }\frac{l^{2}J(x)-x^{2}f(l)}{t-x}(\frac{0}{0}$ form $)$
$=\underset{t\to x}{\lim }\frac{2tf(x)-x^2{f}^{\prime }(t)}{1}$
$\Rightarrow 2xf(x)-x^2{f}^{\prime }(x)=1$
$\Rightarrow x^2\frac{dy}{dx}-2xy=-1,y=f(x)$
$\Rightarrow \frac{dy}{dx}-\frac{2}{x}y=-\frac{1}{x^2}$
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
$\frac{d}{dx}\left(\frac{y}{x^2}\right)=-\frac{1}{x^4}$
$\Rightarrow \frac{y}{x^2}=\frac{1}{3x^3}+K.$
Since $y=f(1)=1.$
so $K=\frac{2}{3}$.
Thus $y=\frac{1}{3x}+\frac{2}{3}{x}^2$