QJEEAdv16Macbc7: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Macbc7 Solution: Let be a differentiable function such that for all and . Then

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

$\Rightarrow f'(x) = {1 \over x}f(x) = 2$ 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.