2 x^3 + 18 x^2 - 96x + 45 = 0 is an increasing function when - Vidyadip

MCQ

$2{x^3} + 18{x^2} - 96x + 45 = 0$ is an increasing function when

✓
$x \le - 8,\,x \ge 2$
B
$x < - 2,x \ge 8$
C
$x \le - 2,x \ge 8$
D
$0 \le x \le - 2$

✓

Answer

Correct option: A.

$x \le - 8,\,x \ge 2$

a
(a) $f'(x) = 6{x^2} + 36x - 96 > 0$, for increasing

==> $f'(x) = (x + 8)(x - 2) \ge 0$==>$x \ge 2,\;x \le - 8$.

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