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(A) Given the linear differential equation $f^{\prime}(x) + \frac{f(x)}{x} = 2$.The integrating factor is $e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.Mu

Vedclass · Accepted Answer

$\lim _{x \rightarrow 0+} f^{\prime}\left(\frac{1}{x}\right)=1$

Vedclass · Answer

(A) Given the linear differential equation $f^{\prime}(x) + \frac{f(x)}{x} = 2$.The integrating factor is $e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.Multiplying by the integrating factor,we get $\frac{d}{dx}(x f(x)) = 2x$.Integrating both sides,$x f(x) = x^2 + c$,which gives $f(x) = x + \frac{c}{x}$ for all $x \in (0, \infty)$.Given $f(1) 
eq 1$,we have $1 + c 
eq 1$,so $c 
eq 0$.Now,$f^{\prime}(x) = 1 - \frac{c}{x^2}$.Evaluating option $A$: $\lim _{x ightarrow 0+} f^{\prime}\left(\frac{1}{x}ight) = \lim _{x ightarrow 0+} (1 - c x^2) = 1$.Evaluating option $B$: $\lim _{x ightarrow 0+} x f\left(\frac{1}{x}ight) = \lim _{x ightarrow 0+} x \left(\frac{1}{x} + cxight) = \lim _{x ightarrow 0+} (1 + cx^2) = 1 
eq 2$.Evaluating option $C$: $\lim _{x ightarrow 0+} x^2 f^{\prime}(x) = \lim _{x ightarrow 0+} x^2 (1 - \frac{c}{x^2}) = \lim _{x ightarrow 0+} (x^2 - c) = -c 
eq 0$.Thus,option $A$ is correct.

Let $f:(0, \infty) \rightarrow \mathbb{R}$ 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

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