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

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

(B) $\lim _{x \rightarrow 0+} x f\left(\frac{1}{x}\right)=2$

(C) $\lim _{x \rightarrow 0+} x^{2} f^{\prime}(x)=0$

(D) $|f(x)| \leq 2$ for all $x \in(0,2)$

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Differential Equations

Let

f : (0, \infty) \to R

be a differentiable function such that

f^{'} (x) = 2 - \frac{f ( x )}{x}

for all

x \in (0, \infty)

and

f (1) \neq = 1

. Then

Answer: (A)

The differential equation is

f^{'} (x) + \frac{1}{x} f (x) = 2

. This is a first-order linear ODE.

Integrating factor:

e^{\int \frac{1}{x} d x} = x

.

Multiplying:

x f^{'} (x) + f (x) = 2 x

, so

(x f (x))^{'} = 2 x

.

Integrating:

x f (x) = x^{2} + c

, thus

f (x) = x + \frac{c}{x}

.

Since

f (1) \neq = 1

, we have

c \neq = 0

.

(A)

f^{'} (x) = 1 - \frac{c}{x ^{2}}

, so

lim_{x \to 0 +} f^{'} (\frac{1}{x}) = lim_{x \to 0 +} (1 - c x^{2}) = 1

. True.

(B)

lim_{x \to 0 +} x f (\frac{1}{x}) = lim_{x \to 0 +} x (\frac{1}{x} + c x) = lim_{x \to 0 +} (1 + c x^{2}) = 1 \neq = 2

. False.

(C)

lim_{x \to 0 +} x^{2} f^{'} (x) = lim_{x \to 0 +} x^{2} (1 - \frac{c}{x ^{2}}) = lim_{x \to 0 +} (x^{2} - c) = - c \neq = 0

. False.

(D)

f (x) = x + \frac{c}{x}

is unbounded as

x \to 0 +

for

c \neq = 0

. False.

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · One or more correct options