Question

If the equation |x – 2| – |x + 1| = p has exactly one solution, then find the number of integral values of p.

Answer 1

Let f(x) = |x – 2| – |x + 1|

Clearly, Range of the function f(x) is [–3, 3] 

Thus, –3 ≤ p ≤ 3

Therefore, the number of integral values of p is 7,

where p = –3, –2, –1, 0, 1, 2, 3.