minimum value of |x|+|x-1|+|x-2|
if |x|=x ,x>=0
|x|=-x,x

• draw the graph,it can be seen that
  minimum value of |x|+|x-1|+|x-2| will be at x=1
  and value is =1+0+1=2
  ans- 2
• 10 years agoHelpfull: Yes(51) No(3)
• if we take x=0 then we get |x|=x so |x|+|x-1|+|x-2|=x+x-1+x-2=3x-3 X=0,,so minmum value is -3.
• 10 years agoHelpfull: Yes(3) No(9)
• best way to solve is eliminate yhe option.
• 10 years agoHelpfull: Yes(1) No(1)
• min value is 2. for x=1 , bcz |-2|=2 nd |-3|=3 for x=0 ;
• 10 years agoHelpfull: Yes(1) No(0)
• 6 use differnciation for both case
• 10 years agoHelpfull: Yes(0) No(0)
• |X|= -X, IF X
• 10 years agoHelpfull: Yes(0) No(0)
• can u plz send me the graph of dis prob @rakesh email-id= senaelitmus@gmail.com
• 10 years agoHelpfull: Yes(0) No(0)
• = -3x + 3 if x
• 10 years agoHelpfull: Yes(0) No(1)
• -3x+3 if x
• 10 years agoHelpfull: Yes(0) No(1)
• Here the each term of series are in ap..such that..
  1) 6,11,16...,996 2) 7,12,17....,997. 3) 8,13,18,....,998 4) 9,14,19,.....,999.
  Here if we add each term of each series in position wise then we get a number which is divisible by 10.
  So take any series we get similar ans.
  a/c to a.p,
  996=6+(n-1)5
  198=(n-1)
  So, n =199. We can't take 1,2,3,4 as it contain 1.
  so required ans is 199.
• 8 years agoHelpfull: Yes(0) No(0)