IIT-JEE 2009 · Paper 2 · Mathematics Q26

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MathematicsMultiple correct+4 / -1Question 26 of 57

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

for at least one

x

in the interval

[1,\infty)

f(x+2)-f(x)<2

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

for all

x

in the interval

[1,\infty)

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

f'(x)

is strictly decreasing in the interval

[1,\infty)

Answer and worked solution

Official answer: B, C, D

Derivatives. $f'(x)=\cos\frac1x+\frac1x\sin\frac1x$ , and
$f''(x)=-\frac1{x^3}\cos\frac1x<0\quad\text{for }x\ge1\ \left(0<\tfrac1x\le1<\tfrac\pi2\right).$

(D) $f'$ is strictly decreasing. ✓
(B) As $x\to\infty$ , $f'(x)\to1$ . ✓
(C) $f'$ decreases to 1, so $f'(x)>1$ ; by the mean value theorem $f(x+2)-f(x)=2f'(\xi)>2$ for every $x$ . ✓
(A) is then false.

(B), (C), (D)

Topics:Rolle's and mean value theorems Monotonicity Limits

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IIT-JEE 2009 · Paper 2, Mathematics question 26