Differentiation and its Application: For the function f(x)=xcosfrac1x,quad xge1, | JEE Advanced 2009 Paper 2

JEE Advanced 2009 Paper 2Question 1

For the function $$f(x)=x\cos\frac1x,\quad x\ge1,$$

Differentiation and its Application

More than one option may be correct.

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

Solution

Concept. Show $f'$ is strictly decreasing with limit $1$; then $f'>1$, and the mean value theorem settles (A), (C). Formulas used. (1) MVT: $f(x+2)-f(x)=2f'(c)$ for some $c\in(x,x+2)$Step 1 - $f'$. $$f'(x)=\cos\frac1x+\frac1x\sin\frac1x\ \to\ 1+0=1\quad(x\to\infty).$$ (B) true. Step 2 - $f''$. $$f''(x)=\frac{\sin\frac1x}{x^2}-\frac{\sin\frac1x}{x^2}-\frac{\cos\frac1x}{x^3}=-\frac{\cos\frac1x}{x^3}<0\quad(x\ge1).$$ So $f'$ is strictly decreasing: (D) true. Decreasing to the limit $1$ means $f'(x)>1$. Step 3 - (A), (C). By (1), $f(x+2)-f(x)=2f'(c)>2$ for every $x$. (C) true, (A) false. Answer: (B), (C), (D)

Differentiation and its Application - JEE Mathematics question (JEE Advanced 2009 Paper 2)

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