Given that |2x−5|+|x+1|+|x|<10 and we need to find how many integer values can x take

When we have multiple Absolute values in a problem then we need to reduce the number of cases. Let's see how to do that.

Lets find out the points on the number line where the values inside each absolute value will change sign.
We can do that by putting each term inside the absolute value individually equal to zero.

For |2x-5| -> 2x-5 = 0 => x = \(\frac{5}{2}\) = 2.5
For |x+1| -> x+1 = 0 => x = -1
For |x| -> x = 0



Case 1: x < -1
Let's take any value of x < -1, lets say -2 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*-2)-5 = -9 => -ve =>|2x-5| = -(2x-5)
For |x+1| -> -2+1 = -1 => -ve => |x+1| = -(x+1)
For |x| -> -2 => -ve => |x| = -x

|2x−5|+|x+1|+|x|<10 => -(2x-5) - (x+1) -x < 10
=> -2x + 5 -x - 1 -x < 10
=> -4x < 10 -4
=> -4x < 6 => x > \(\frac{-6}{4}\) => x > -1.5
And conditions was x < -1
So, intersection will be -1.5 < x < -1 => NO Integer solution in this range

Case 2: -1 ≤ x < 0
Let's take any value of x which satisfies -1 ≤ x < 0, lets say -0.5 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*-0.5)-5 => -ve =>|2x-5| = -(2x-5)
For |x+1| -> -0.5+1 => +ve => |x+1| = x+1
For |x| -> -0.5 => -ve => |x| = -x

|2x−5|+|x+1|+|x|<10 => -(2x-5) + (x+1) -x < 10
=> -2x + 5 + x + 1 -x < 10
=> -2x < 10 -6
=> -2x < 4 => x > \(\frac{-4}{2}\) => x > -2
And conditions was -1 ≤ x < 0
So, intersection will be -1 ≤ x < 0 => x=-1 is an integer solution in this range

Case 3: 0 ≤ x < 2.5
Let's take any value of x which satisfies 0 ≤ x < 2.5, lets say 1 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*1)-5 => -ve =>|2x-5| = -(2x-5)
For |x+1| -> 1+1 => +ve => |x+1| = x+1
For |x| -> 1 => +ve => |x| = x

|2x−5|+|x+1|+|x|<10 => -(2x-5) + (x+1) + x < 10
=> -2x + 5 + x + 1 + x < 10
=> 6 < 10
Which is always true
So, entire range 0 ≤ x < 2.5 is possible
x=0, 1, 2 are integer solution in this range

Case 4: x ≥ 2.5
Let's take any value of x which satisfies x ≥ 2.5, lets say 10 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*10)-5 => +ve =>|2x-5| = 2x-5
For |x+1| -> 10+1 => +ve => |x+1| = x+1
For |x| -> 10 => +ve => |x| = x

|2x−5|+|x+1|+|x|<10 => 2x-5 + (x+1) + x < 10
=> 2x - 5 + x + 1 + x < 10
=> 4x - 4 < 10
=> 4x < 14
=> x < \(\frac{14}{4}\)
=> x < 3.5
And conditions was x ≥ 2.5
So, intersection will be 2.5 ≤ x < 3.5=> x=3 is an integer solution in this range

So, integer values which are possible are -1 , 0, 1, 2, 3
=> 5 values

So, Answer will be D
Hope it helps!

Watch the following video to learn how to Solve Inequality + Absolute value Problems