QJEEAdv09Mac879: JEE Advanced 2009 Mathematics - Class 12 JEEAdv09Mac879 Solution: For the function

We have for

$f(x) = x\cos \left( {{1 \over x}} \right),x \ge 1$

$f'(x) = \cos \left( {{1 \over x}} \right) + {1 \over x}\sin \left( {{1 \over x}} \right) \to 1$ for $x \to \infty$

Also,

$f'(x) = \left( {{1 \over x}} \right) + {1 \over x}\sin \left( {{1 \over x}} \right) - {1 \over {{x^2}}}\sin \left( {{1 \over x}} \right) - {1 \over {{x^3}}}\cos \left( {{1 \over x}} \right)$

= - {1 \over {{x^3}}}\cos \left( {{1 \over x}} \right) < 0 for $x \ge 1$

$\Rightarrow f'(x)$ is decreasing for $[1,\infty )$

\Rightarrow f'(x + 2) < f'(x) . Also,

$\mathop {\lim }\limits_{x \to \infty } f(x + 2) - f(x) = \mathop {\lim }\limits_{x \to \infty } \left[ {(x + 2)\cos {1 \over {x + 2}} - x\cos {1 \over x}} \right]=2$

Hence, f(x + 2) - f(x) > 2\forall x \ge 1 .