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 May 25th, 2019, 08:58 AM #1 Member   Joined: Jul 2017 From: KOLKATA Posts: 50 Thanks: 3 Number of Integral solutions if |x-2| -|x+1| =p has exactly one solution then the no. of integral values of p is May 25th, 2019, 09:26 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra $$x > 2 \implies |x-2|-|x+1|=(x-2)-(x+1)=-3=p$$ 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 $$p \in \{-2,-1,0,1,2\}$$ Tags integral, number, solutions Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Number Theory 5 December 31st, 2018 07:00 PM Nathalia Trigonometry 3 September 29th, 2016 11:22 PM ishaanmj007 Algebra 4 May 16th, 2015 05:38 PM rayman Applied Math 0 April 25th, 2013 01:12 AM earth Math Events 3 July 8th, 2009 10:14 PM

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