A) The cubic has minima at \[-\sqrt{\frac{p}{3}}\] and maxima at \[\sqrt{\frac{p}{3}}\]
B) The cubic has minima at \[\sqrt{\frac{p}{3}}\] and maxima at \[-\sqrt{\frac{p}{3}}\].
C) The cubic has maxima at both \[\sqrt{\frac{p}{3}}\,\,and\,\,-\sqrt{\frac{p}{3}}\]
D) The cubic has minima at both \[\sqrt{\frac{p}{3}}\,\,and\,\,-\sqrt{\frac{p}{3}}\]
Correct Answer: B
Solution :
Let \[\operatorname{y}={{x}^{3}}-px+\,\,q\,\,\Rightarrow \,\,\,\frac{dy}{dx}=3{{x}^{2}}-p\] For \[\frac{dy}{dx}=0\,\,\Rightarrow \,\,3{{x}^{2}}-p=0\Rightarrow x=\pm \,\sqrt{\frac{p}{3}}\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=6x\] \[\Rightarrow \,\,\frac{{{d}^{2}}y}{d{{x}^{2}}}{{\left| _{x=\sqrt{\frac{p}{3}}}\,=\,\,+ve\,\,and\,\,\frac{{{d}^{2}}y}{d{{x}^{2}}} \right|}_{x=-\sqrt{\frac{p}{3}}}}=-\,ve\] \[\therefore \, y\,has minima at x = \sqrt{\frac{p}{3}} and maxima at x =-\sqrt{\frac{p}{3}}\]
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