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modulus

jacks

Well-known member
Apr 5, 2012
226
If [tex]y = \mid x \mid+\mid x-1 \mid+\mid x-3 \mid+\mid x-6 \mid+...........+\mid x-5151 \mid[/tex]


and [tex]m =[/tex] no. of terms in the expression [tex]y[/tex]


and [tex]n =[/tex] no. of integers for which [tex]y[/tex] has min. value


Then [tex]\displaystyle\frac{m+n-18}{10} =[/tex]
 

CaptainBlack

Well-known member
Jan 26, 2012
890
If [tex]y = \mid x \mid+\mid x-1 \mid+\mid x-3 \mid+\mid x-6 \mid+...........+\mid x-5151 \mid[/tex]


and [tex]m =[/tex] no. of terms in the expression [tex]y[/tex]


and [tex]n =[/tex] no. of integers for which [tex]y[/tex] has min. value


Then [tex]\displaystyle\frac{m+n-18}{10} =[/tex]
Maybe it's me, but I find that incomprehensible.

For a start why is \(m\) not \(5152\)?

Do you mean \(n\) to be the number of integers corresponding to a local minima of \(y\)? You can show that there is a global minimum and it achived this at two adjacent integer points.

(the slope is -5152 for -ve \(x\), and increases by 2 when we pass a integer argument moving to the right ...)

CB
 
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