If x is an integer such that ∣2x−5∣+∣x+1∣≤11 then how many such integer values of x are possible?
Solution
We will consider 4 cases for this inequality:
Case 1:
2x-5 is +ve but x+1 is -ve.
2x-5 >= 0 ==> x >= 2.5 and,
x+1 < 0 ==> x < -1. There are no possible values of x which lie in both the ranges.
Case 2:
2x-5 is +ve and x+1 is +ve.
x >= 2.5 and x >= -1.
This gives the inequality => 2x-5 + x+1 <= 11
==> x<= 5. Possible values for x are 3, 4 and 5.
Case 3:
2x-5 is -ve and x+1 is +ve
x < 2.5 and x >= -1.
This gives the inequality => -(2x-5) + x+1 <= 11
x >= -5. Possible values of x are -1, 0, 1 and 2.
Case 4:
2x-5 is -ve and x+1 is -ve.
x < 2.5 and x < -1.
This gives the inequality => -(2x-5) - (x+1) <= 11
==> x >= -2.33
Possible values for x = -2.
Total values of x possible are 8.
