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

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

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[- 2, \infty), 2 x^{3} - 3 x^{2} - 12 x + 6

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[\frac{1}{3}, \infty), 3 x^{2} - 2 x + 1

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(- \infty, - 4], x^{3} + 6 x^{2} + 6

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Solution

The correct option is D $[\frac{1}{3}, \infty), 3 x^{2} - 2 x + 1$
Consider A
$f (x) = x^{3} - 3 x^{2} + 3 x + 3$
$f^{'} (x) = 3 x^{2} - 6 x + 3$
$= 3 (x^{2} - 2 x + 1)$
$= 3 (x - 1)^{2}$
Hence, $f^{'} (x) > 0$ for all $x ϵ R$ .
Hence, $f (x)$ is increasing for $x ϵ R$ .
Consider B
$f (x) = 2 x^{3} - 3 x^{2} - 12 x + 6$
$f^{'} (x) = 6 x^{2} - 6 x - 12$
$= 6 (x^{2} - x - 2)$
$= 6 (x - 2) (x + 1)$
Now for increasing function
$f^{'} (x) > 0$
Or
$(x - 2) (x + 1) > 0$
$x > 2$ and $x < - 1$
Hence for $f (x)$ to be increasing $x ϵ (- \infty, - 1) \cup (2, \infty)$
Hence, incorrect option is B.
Consider C
$f (x) = 3 x^{2} - 2 x + 1$
$f^{'} (x) = 6 x - 2$
Hence for increasing function
$f^{'} (x) > 0$
Or
$x > \frac{1}{3}$
Consider D
$f (x) = x^{3} + 6 x^{2} + 6$
$f^{'} (x) = 3 x^{2} + 12 x$
$f^{'} (x) > 0$ implies
$3 x (x + 4) > 0$
$x > 0$ or $x < - 4$
Hence for $f (x)$ to be increasing $x ϵ (- \infty, - 4) \cup (0 \infty)$ .

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(Q)	$(\frac{- 10}{3} x y^{3}) \times (\frac{6}{5} x^{3} y)$	(2)	$- x^{3} + x^{2} + 3 x - 6$
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Q. Find the intervals in which the following functions are increasing or decreasing.
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Question 24
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(iv) $3 x^{3} - x^{2} - 3 x + 1$

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