Discussion Forum

Let  $f:(0,\infty)\rightarrow R$ be a differentiable function such that   $f'(x)=2-\frac{f(x)}{x}$  for all  $x \epsilon (0,\infty)$  and $f(1)\neq 1$ Then

Answer:

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.