A) \[3,\,-\frac{1}{2}\]
B) \[3,\frac{1}{3}\]
C) \[-3,\,-\frac{1}{3}\]
D) None of these
Correct Answer: B
Solution :
Let \[y=\frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1}\] Þ \[\frac{dy}{dx}=\frac{({{x}^{2}}+x+1)\,(2x-1)-({{x}^{2}}-x+1)\,(2x+1)}{{{({{x}^{2}}+x+1)}^{2}}}\] Þ\[\frac{dy}{dx}=\frac{2{{x}^{2}}-2}{{{({{x}^{2}}+x+1)}^{2}}}=0\]Þ\[2{{x}^{2}}-2=0\]Þ\[x=-1,\,+1\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{4\,(-{{x}^{3}}+3x+1)}{{{x}^{2}}+x+1}\] At \[x=-1\], \[\frac{{{d}^{2}}y}{d{{x}^{2}}}<0,\] the function will occupy maximum value, \ \[f(-1)=3\] and at \[x=1\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}>0,\] the function will occupy minimum value \ \[f(1)=\frac{1}{3}\].
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