$$f'(x) = 2 - {{f(x)} \over x}$$

$$ \Rightarrow f'(x) = {1 \over x}f(x) = 2$$ is linear differential equation.

Hence, $$I.F. = {e^{\int {{1 \over x}dx} }} = {e^{\ln x}} = x$$

Thus the solution is given by

$$x\,.\,f(x) = \int {2x\,dx + \lambda } $$

i.e., $$xf(x) = {x^2} + \lambda $$

As f(1) $$\ne$$ 1, we have $$\lambda$$ $$\ne$$ 0

$$\therefore$$ $$f(x) = x + {\lambda \over x}$$, $$\lambda$$ $$\ne$$ 0

Thus, $$f'(x) = 1 - {\lambda \over {{x^2}}}$$, $$\lambda$$ $$\ne$$ 0

Now, $$\mathop {\lim }\limits_{x \to {0^ + }} f'\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 - \lambda {x^2}) = 1$$

$$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} x\left( {{1 \over x} + \lambda x} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 + \lambda {x^2}) = 1$$

$$\mathop {\lim }\limits_{x \to {0^ + }} {x^2}f'(x) = \mathop {\lim }\limits_{x \to {0^ + }} {x^2}\left( {1 - {\lambda \over {{x^2}}}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} ({x^2} - \lambda ) = - \lambda $$

Again, $$\mathop {\lim }\limits_{x \to {0^ + }} f(x) \to \infty $$

Hence the function is not bounded.

Note that $$\lambda$$ can be $$-$$ve or +ve.