Answer:
Option A
Explanation:
Here, $f'(x)=2-\frac{f(x)}{x}$
or $\frac{\text{d}y}{\text{d}x}+\frac{y}{x}=2$
[i.e , linear differential equation in y]
Integrationg Factor. IF
$=\int_{e}^{1/x} dx=e^{\log x}=x$
$\therefore$ Required solution is
$y.(IF)=\int_{}^{} Q(IF)dx+C$
$\Rightarrow$ $y(x)=\int_{}^{}2(x)dx+C $
$\Rightarrow$ $yx= x^{2}+C$
$\therefore$ $y=x+\frac{C}{x}$ $[\therefore C\neq 0,asf(1)\neq1]$
(a) $\lim_{x \rightarrow 0+}f'(\frac{1}{x})=\lim_{x \rightarrow 0+}(1-Cx^{2})=1$
$\therefore$ Option (a) is correct.
(b) $\lim_{x \rightarrow 0+}xf(\frac{1}{x})=\lim_{x \rightarrow 0+}(1+Cx^{2})=1$
$\therefore$ Option (b) is incorrect.
(c) $\lim_{x \rightarrow 0+}x^{2}f'(x)=\lim_{x \rightarrow 0+}(x^{2}-C)=-C\neq 0$
$\therefore$ Option (c) is incorrect.
(d) $f(x)=x+\frac{C}{x},C\neq0$
For C>0, $\lim_{x \rightarrow 0+}f(x)=\infty$
$\therefore$ Function is not bounded in (0,2)
$\therefore$ Option (d) is incorrect.
