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May 25th, 2019, 07:58 AM   #1
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Number of Integral solutions

if |x-2| -|x+1| =p has exactly one solution then the no. of integral values of p is
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May 25th, 2019, 08:26 PM   #2
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$$x > 2 \implies |x-2|-|x+1|=(x-2)-(x+1)=-3=p$$
Thus, $p=-3$ gives us an infinite number of solutions. Since we require only one solution, we must have $p\ne-3$.

$$x < -1 \implies |x-2|-|x+1|=(2-x)+(x+1)=03=p$$
Thus, $p=3$ gives us an infinite number of solutions. Since we require only one solution, we must have $p\ne 3$.

\begin{align}-1 \le x \le 2 \implies |x-2|-|x+1|&=p \\ (2-x)-(x+1)&=p \\1-2x&=p \\ x &= \frac12(1-p)\end{align}
This holds as long as \begin{alignat}{2}-1 &\le \frac12(1-p) &&\le 2 \\ -2 &\le 1-p &&\le 4 \\ -3 &\le -p &&\le 3 \end{alignat}
But we already eliminated the limiting values, so there are five possible integer values of $p$. Specifically $$p \in \{-2,-1,0,1,2\}$$
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