A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched? | Interval | Function | |----------|----------| | (1) $(-\infty, \infty)$ | $x^{3}-3 x^{2}+3 x+3$ | | (2) $[2, \infty)$ | $2 x^{3}-3 x^{2}-12 x+6$ | | (3) $\left(-\infty, \frac{1}{3}\right]$ | $3 x^{2}-2 x+1$ | | (4) $(-\infty,-4]$ | $x^{3}+6 x^{2}+6$ |

(C) $\left(-\infty, \frac{1}{3}\right]$ and $3 x^{2}-2 x+1$

(A) $(-\infty, \infty)$ and $x^{3}-3 x^{2}+3 x+3$

(B) $[2, \infty)$ and $2 x^{3}-3 x^{2}-12 x+6$

(D) $(-\infty,-4]$ and $x^{3}+6 x^{2}+6$

JEE Main · 2005 · 08 May 2005 · Offline · Application of Derivatives

A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?

Answer: (C)

A function increases on an interval where its derivative is non-negative. For

3 x^{2} - 2 x + 1

the derivative is

6 x - 2

, which is non-negative only when

x \geq \frac{1}{3}

; this function increases on

[\frac{1}{3}, \infty)

, not on

(- \infty, \frac{1}{3}]

. The other three functions do satisfy the increasing condition on the intervals paired with them:

x^{3} - 3 x^{2} + 3 x + 3

has derivative

3 (x - 1)^{2} \geq 0

everywhere,

2 x^{3} - 3 x^{2} - 12 x + 6

has derivative

6 (x - 2) (x + 1) \geq 0

[2, \infty)

, and

x^{3} + 6 x^{2} + 6

has derivative

3 x (x + 4) \geq 0

(- \infty, - 4]

. The faulty pairing is therefore the one matching

3 x^{2} - 2 x + 1

with

(- \infty, \frac{1}{3}]

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option