A) \[\left( -\infty ,\,-2 \right)\cup (4,\infty )\]
B) \[\left( -2,\infty \right)\]
C) \[\left( -2,4 \right)\]
D) \[\left( -\infty ,4 \right)\]
Correct Answer: A
Solution :
\[f(x)={{x}^{3}}-3{{x}^{2}}-24x+5\] |
For increasing, \[f'(x)>0,\] \[3{{x}^{2}}-6x-24>0\] |
\[\Rightarrow \,{{x}^{2}}-2x-8>0\Rightarrow {{x}^{2}}-4x+2x-8>0\] |
\[\Rightarrow \,\left( x+2 \right)\left( x-4 \right)>0.\] |
Now, by the sign scheme for \[3{{x}^{2}}-6x-24,\] |
\[\Rightarrow \,\,x\in \left( -\infty ,-2 \right)\cup \left( 4,\infty \right)\] |
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