For function f (x) = x 1/x , x ≥ 1, [JEE Adv 2009 P2]

for at least one x in interval [1, ∞), f(x + 2) – f(x)

for all x in the interval [1, ∞), f(x + 2) – f(x) > 2.

f′(x) is strictly decreasing in the interval [1, ∞).

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IIT JEE Advanced 2009 Paper 2

Section: Mathematics

Question : 44 of 57

Marks: +1, -0

For function f (x) =

x \cos \frac{1}{x}

Solution:

We have for

f (x) =

x \cos \frac{1}{x}

, x ≥ 1

f' (x) =

\cos \frac{1}{x} + \frac{1}{x} \sin \frac{1}{x}

→ 1 for x → ∞

Also, f' (x) =

\frac{1}{x} + \frac{1}{x} \sin \frac{1}{x} - \frac{1}{x^2} \sin \frac{1}{x} - \frac{1}{x^3} \cos \frac{1}{x}

-\frac{1}{x^3} \cos \frac{1}{x}

< 0 for x ≥ 1

⇒ f′(x) is decreasing for [1,∞)

⇒ f ′(x + 2) < f ′(x). Also,

\lim\limits_{x \to \infty}

f (x + 2) - f (x) =

\lim\limits_{x \to \infty} \left[ (x+2) \cos \frac{1}{x+2} - x \cos \frac{1}{x} \right]

= 2

Hence, f (x + 2) - f (x) > 2 ∀ x ≥ 1.

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