A) 5
B) \[-5\]
C) \[\frac{1}{5}\]
D) None of these
Correct Answer: A
Solution :
We have the equation of curve \[y=-{{x}^{3}}+3{{x}^{2}}+2x-27\] \[\Rightarrow \] \[\frac{dy}{dx}=-3{{x}^{2}}+6x+2\] which is the slope of the curve Let \[Z=-3{{x}^{2}}+6x+2\] For maximum and minimum, put \[\frac{dZ}{dx}=0\] \[\Rightarrow \] \[-6x+6=0\] \[\Rightarrow \] \[x=1\] Now, \[\frac{{{d}^{2}}Z}{d{{x}^{2}}}=-6<0\] Thus, Z is maximum for\[x=1\]and maximum value of Z \[=-3{{(1)}^{2}}+6(1)+2\] \[=-3+8=5\] Hence, maximum slope of the given curve is 5.
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