Differential Equations 2 Question 21
22. Let $f:(0, \infty) \rightarrow R$ be a differentiable function such that $f^{\prime}(x)=2-\frac{f(x)}{x}$ for all $x \in(0, \infty)$ and $f(1) \neq 1$. Then
(a) $\lim _{x \rightarrow 0+} f^{\prime} \frac{1}{x}=1$
(b) $\lim _{x \rightarrow 0+} x f \frac{1}{x}=2$
(c) $\lim _{x \rightarrow 0+} x^{2} f^{\prime}(x)=0$
(d) $|f(x)| \leq 2$ for all $x \in(0,2)$
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Answer:
Correct Answer: 22. (a)
Solution:
- Here, $f^{\prime}(x)=2-\frac{f(x)}{x}$
or $\frac{d y}{d x}+\frac{y}{x}=2$ [i.e. linear differential equation in $y$ ]
Integrating Factor, $IF=e^{\int \frac{1}{x} d x}=e^{\log x}=x$
$\therefore$ Required solution is $y \cdot(IF)=\int Q(IF) d x+C$
$$ \begin{array}{rlrl} \Rightarrow & & y(x) & =\int 2(x) d x+C \\ \Rightarrow & y x & =x^{2}+C \\ \therefore & y & =x+\frac{C}{x} \quad[\because C \neq 0, \text { as } f(1) \neq 1] \end{array} $$
(a) $\lim _{x \rightarrow 0^{+}} f^{\prime} \frac{1}{x}=\lim _{x \rightarrow 0^{+}}\left(1-C x^{2}\right)=1$
$\therefore$ Option (a) is correct.
(b) $\lim _{x \rightarrow 0^{+}} x f \frac{1}{x}=\lim _{x \rightarrow 0^{+}}\left(1+C x^{2}\right)=1$
$\therefore$ Option (b) is incorrect.
(c) $\lim _{x \rightarrow 0^{+}} x^{2} f^{\prime}(x)=\lim _{x \rightarrow 0^{+}}\left(x^{2}-C\right)=-C \neq 0$
$\therefore$ Option (c) is incorrect.
(d) $f(x)=x+\frac{C}{x}, C \neq 0$
For $C>0, \quad \lim _{x \rightarrow 0^{+}} f(x)=\infty$
$\therefore$ Function is not bounded in $(0,2)$.
$\therefore$ Option (d) is incorrect.
