Applications Of Derivatives Question 133
Question: The minimum value of the expression $ 7-20x+11x^{2} $ is
Options:
A) $ \frac{177}{11} $
B) $ -\frac{177}{11} $
C) $ -\frac{23}{11} $
D) $ \frac{23}{11} $
Show Answer
Answer:
Correct Answer: C
Solution:
Given $ f(x)=7-20x+11x^{2} $
$ f’(x)=-20+22x $ Put $ f’(x)=0 $ i.e., $ -20+22x=0 $
therefore $ x=10/11 $ and $ f’’(x)=22>0 $
Hence at $ x=10/11,\ \ \ f(x) $ will have minimum value,
$ \therefore f,( \frac{10}{11} )=7-\frac{200}{11}+\frac{100\times 11}{121} $
$ =7-\frac{200}{11}+\frac{100}{11} $
$ =-\frac{23}{11} $ .
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