A) \[(-1,3)\]
B) \[(-3,3)\]
C) \[(-4,4)\]
D) \[(-2,2)\]
Correct Answer: A
Solution :
Given, \[f(x)={{x}^{3}}-3{{x}^{2}}-9x+5\] \[f'(x)=3{{x}^{2}}-6x-9\] For maxima or minima, \[f'(x)=0\] \[\Rightarrow \] \[3{{x}^{2}}-6x-9=0\] \[\Rightarrow \] \[3({{x}^{2}}-2x-3)=0\] \[\Rightarrow \] \[{{x}^{2}}-2x-3=0\] \[\Rightarrow \] \[{{x}^{2}}-3x+x-3=0\] \[\Rightarrow \] \[x(x-3)+1(x-3)=0\] \[\Rightarrow \] \[(x+1)(x-3)=0\] \[\Rightarrow \] \[x=-1,3\] \[\therefore \] Intervals are \[(-\infty ,-1),\,(-1,3)\] and \[(3,\,\infty )\].Intervals | Sign of \[f'(x)\] | Nature of \[f'(x)\] |
\[(-\infty ,-1)\] | \[+ve\] | Strictly increasing |
\[(-1,3)\] | \[-ve\] | Strictly decreasing |
\[(3,\infty )\] | \[+ve\] | Strictly increasing |
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