1if x<-2
and ure taking k as 1 or 0
you get
x=0 and x=-1
arent these >-2 and contrary to given condition???
find the values of k for which
l x-1l + l x -2 l + lx + 1 l + l x +2 l = 4k has integral solutions
ans in my book is 2 , 3 ,4 ,5 ..............why not one? ( matrix match type)
srry i hv edited it now ..........plz proceed now
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UP 0 DOWN 0 0 5
5 Answers
1x<-1
1-x+2-x-x-1-x+4=4k
6-4x=4k
x=(4k+6)/4
No solution
-1<x<1
1-x-x+2+x+1-x+4=4k
8-2x=4k
4-x=2k
x=4-2k
k=0,1,2,3......
1<x<2
x-1-x+2+x+1-x+4=4k
6=4k
No solution
2<x<4
x-1+x-2+x+1-x+4=4k
2x+2=4k
x=2k-1
k=0,1,2...
x>4
x-1+x-2+x+1+x-4=4k
4x-6=4k
x=4k+6/4
No solution
Final ans k=0,1,2....
1i think that u've got the q wrong.
In the last modulus it should be |x-2|
that gives answers as 2,3,4,5
1lets see the case
-2 > x / -2 = x
then
-x+1 - x + 2 -x -1- x-2 =4k
=> -4x = 4k
=> x = -k
then k =0,1 ,2 ,3 ,4 ,5 ?
plz tell me where i m mistaken
1
1just divide the problem into domains madhu,
catagorize them as (-∞,-2) (-2,-1)(-1,+1)(+1,+2)(+2,+∞)
and solve for each case, that shud simplify the working..
cheers!!