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JEE Advance - Mathematics (2009 - Paper 2 Offline - No. 10)

For the function $$$f\left( x \right) = x\cos \,{1 \over x},x \ge 1,$$$
for at least one $$x$$ in the interval $$\left[ {1,\infty } \right)$$, $$f\left( {x + 2} \right) - f\left( x \right) < 2$$
$$\mathop {\lim }\limits_{x \to \infty } f'\left( x \right) = 1$$
for all $$x$$ in the interval $$\left[ {1,\infty } \right)f\left( {x + 2} \right) - f\left( x \right) > 2$$
$$f'(x)$$ is strictly decreasing in the interval $$\left[ {1,\infty } \right)$$

Explanation

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

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