A) \[\frac{177}{11}\]
B) \[-\frac{177}{11}\]
C) \[-\frac{23}{11}\]
D) \[\frac{23}{11}\]
Correct Answer: C
Solution :
Given \[f(x)=7-20x+11{{x}^{2}}\] \[f'(x)=-20+22x\] Put \[f'(x)=0\] i.e., \[-20+22x=0\] Þ \[x=10/11\]and \[f''(x)=22>0\] Hence at \[x=10/11,\ \ \ f(x)\]will have minimum value, \[\therefore f\,\left( \frac{10}{11} \right)=7-\frac{200}{11}+\frac{100\times 11}{121}\]\[=7-\frac{200}{11}+\frac{100}{11}\]\[=-\frac{23}{11}\].
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