Difference between revisions of "MIE 2016/Day 1/Problem 8"
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The minimum value of this function is when <math>x</math> is in the middle between 1 and 2017. <math>\frac{2017+1}{2}=1009</math> so inputting <math>1009</math> as <math>x</math>: | The minimum value of this function is when <math>x</math> is in the middle between 1 and 2017. <math>\frac{2017+1}{2}=1009</math> so inputting <math>1009</math> as <math>x</math>: | ||
<cmath>f(1009)=\sqrt{1008+1007+\dots+1+0+1+\dots+1007+1008}=\sqrt{1008\times 1009}\approx 1008</cmath> | <cmath>f(1009)=\sqrt{1008+1007+\dots+1+0+1+\dots+1007+1008}=\sqrt{1008\times 1009}\approx 1008</cmath> | ||
+ | Thus the answer is <math>\boxed{\text{B}}</math>. | ||
==See Also== | ==See Also== |
Latest revision as of 13:01, 9 May 2024
Problem 8
Let . The minimum value of is in the interval:
(a)
(b)
(c)
(d)
(e)
Solution
The minimum value of this function is when is in the middle between 1 and 2017. so inputting as : Thus the answer is .