A) \[(-\infty ,\,-2)\cup (4,\infty )\]
B) \[(-2,\infty )\]
C) (?2, 4)
D) \[(-\infty ,\,4)\]
Correct Answer: A
Solution :
\[f(x)={{x}^{3}}-3{{x}^{2}}-24x+5\] For increasing, \[f'(x)>0\], \[3{{x}^{2}}-6x-24>0\] Þ \[{{x}^{2}}-2x-8>0\] \[{{x}^{2}}-4x+2x-8>0\] Þ \[(x+2)(x-4)>0\] \[x\in (-\infty ,\,-2)\cup (4,\infty )\].
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