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 14:01, 9 May 2024
Problem 8
Let 
. The minimum value of 
is in the interval:
(a) ![$(-\infty,1008]$](http://latex.artofproblemsolving.com/9/c/3/9c3f0cd9ce7d4e89be5190f6dc60180d3363e96c.png)
(b) ![$(1008,1009]$](http://latex.artofproblemsolving.com/7/5/b/75bb7a0261dc6254465d88e70e4e960d0168dc6c.png)
(c) ![$(1009,1010]$](http://latex.artofproblemsolving.com/e/c/f/ecf6f844c56ebe845852158ed002a6a68904b550.png)
(d) ![$(1010,1011]$](http://latex.artofproblemsolving.com/0/c/8/0c8b0bdbefe87669231f4d789d7b60c2861cbd5d.png)
(e) 
Solution
The minimum value of this function is when 
is in the middle between 1 and 2017. 
so inputting 
as :
![\[f(1009)=\sqrt{1008+1007+\dots+1+0+1+\dots+1007+1008}=\sqrt{1008\times 1009}\approx 1008\]](http://latex.artofproblemsolving.com/6/9/a/69af9b97d515fdad9dc2c43cbb785cc552697ee7.png)
Thus the answer is 
.
