For the function \[ f(x)=x \cos \frac{1}{x}, \quad x \geq 1, \]

(B) $\lim _{x \rightarrow \infty} f^{\prime}(x)=1$; (C) for all $x$ in the interval $[1, \infty), f(x+2)-f(x)>2$; (D) $f^{\prime}(x)$ is strictly decreasing in the interval $[1, \infty)$

(A) for at least one $x$ in the interval $[1, \infty), f(x+2)-f(x)<2$

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Application of Derivatives

WhatsApp Email

For the function

f (x) = x cos \frac{1}{x}, x \geq 1,

(A)for at least one $x$ in the interval $[1, \infty), f (x + 2) - f (x) < 2$
(B) $lim_{x \to \infty} f^{'} (x) = 1$ ✓ Correct
(C)for all $x$ in the interval $[1, \infty), f (x + 2) - f (x) > 2$ ✓ Correct
(D) $f^{'} (x)$ is strictly decreasing in the interval $[1, \infty)$ ✓ Correct

Answer: (B), (C), (D)

Solution

Given

f (x) = x cos (1/ x)

for

x \geq 1

f^{'} (x) = cos (1/ x) + \frac{1}{x} sin (1/ x)

.

As

x \to \infty

cos (1/ x) \to 1

and

\frac{1}{x} sin (1/ x) \to 0

, so

lim_{x \to \infty} f^{'} (x) = 1

. Option (B) is correct.

f^{''} (x) = \frac{1}{x ^{2}} sin (1/ x) - \frac{1}{x ^{2}} sin (1/ x) - \frac{1}{x ^{3}} cos (1/ x) = - \frac{1}{x ^{3}} cos (1/ x) < 0

for

x \geq 1

.

Thus

f^{'} (x)

is strictly decreasing. Option (D) is correct.

Since

f^{'} (x)

is decreasing and approaches 1, we have

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

for all

x \geq 1

.

By the mean value theorem,

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

for some

c \in (x, x + 2)

.

Since

f^{'} (c) > 1

for all

c \geq 1

, we have

f (x + 2) - f (x) > 2

. Option (C) is correct.

Option (A) is false since (C) holds for all

x

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · One or more correct options

Practise more: all JEE Advanced PYQs on Application of Derivatives →

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Application of Derivatives

Solution

More JEE Advanced questions on Application of Derivatives