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What are the no. of positive integral solutions of ||x - 1| - 2| + x = 3 ?

My effort

||x - 1| - 2| = 3 - x

|x - 1| - 2 = 3 - x OR |x - 1| - 2 = x - 3

|x - 1| = 5 - x OR |x - 1| = x - 1

x - 1 = 5 - x OR x - 1 = x - 5 OR x - 1 $\geq$ 0

2x = 6 OR x $\geq$ 1

Therefore, x $\geq$ 1

But the any value of x greater than 1 except 3 does not satisfy the equation. Where have I gone wrong?

Community
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kartikeykant18
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  • "But the any value of x greater than 1 except 3 does not satisfy the equation. Where have I gone wrong? "

    I see 1 and 2 also satisfy your equation.

     – nature1729 May 02 '14 at 14:50
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    Further when you wrote, $||x - 1| - 2| = 3 - x$ it means $x \leq 3$. – nature1729 May 02 '14 at 14:55

2 Answers2

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You should start with expanding the innermost absolute value: $$ \vert \vert x-1 \vert -2 \vert = \begin{cases} \vert x-3 \vert & x \geqslant 1 \\ \vert -1-x \vert & x < 1 \end{cases} = \begin{cases} x-3 & x \geqslant 1, x > 3 \\ 3-x & x \geqslant 1, x \leqslant 3 \\ 1+x & x < 1, x \geqslant -1 \\ -1-x & x < 1, x < -1 \end{cases} = \begin{cases} x-3 & x > 3 \\ 3-x & 1 \leqslant x \leqslant 3\\ 1+x & -1 \leqslant x < 1 \\ -1-x & x < -1 \end{cases} $$ You now seek solutions for each of these branches being equal to $3-x$ within the appropriate domain, giving you three integral solutions $x=1, x=2, x=3$.

Sasha
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  • But why the answer is not coming by my way?? after all this is what we have been taught.... How do you decide whether to start solving from inside or outside??? – kartikeykant18 May 02 '14 at 14:48
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    In concluding $x\geqslant 1$ you fail to take into account the assumptions you made when expanding the absolute value function. Once you add those you will get the correct answer. In my opinion, it is easier to keep track of these assumptions when expanding from the inside. – Sasha May 02 '14 at 14:50
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\sum_{x\ =\ 0}^{\infty} \delta_{\verts{\vphantom{\large A}\verts{x\ -\ 1}\ -\ 2}\ +\ x,3}} =\sum_{x\ =\ 0}^{\infty}\oint_{\verts{z}\ =\ 1} {1 \over z^{-\verts{\verts{\vphantom{\Large A}x\ -\ 1}\ -\ 2}\ -\ x\ +\ 4}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1}{1 \over z^{4}} \sum_{x\ =\ 0}^{\infty}z^{\verts{\verts{\vphantom{\Large A}x\ -\ 1}\ -\ 2}\ +\ x} \,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ 1}{1 \over z^{4}}\pars{% z + z^{3} + \sum_{x\ =\ 2}^{\infty}z^{\verts{x\ -\ 3}\ +\ x}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1}{1 \over z^{4}}\pars{% z + z^{3} + z^{3} + z^{3} + \sum_{x\ =\ 4}^{\infty}z^{2x - 3}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1}\pars{% {1 \over z^{3}} + {3 \over z} + \sum_{x\ =\ 4}^{\infty}{1 \over z^{7 - 2x}}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\overbrace{\oint_{\verts{z}\ =\ 1}{1 \over z^{3}}\,{\dd z \over 2\pi\ic}} ^{\ds{=\ \dsc{0}}}\ +\ \overbrace{\oint_{\verts{z}\ =\ 1}{3 \over z}\,{\dd z \over 2\pi\ic}} ^{\ds{=\ \dsc{3}}}\ +\ \sum_{x\ =\ 4}^{\infty}\overbrace{% \oint_{\verts{z}\ =\ 1}{1 \over z^{7 - 2x}}\,{\dd z \over 2\pi\ic}} ^{\ds{=\ \dsc{\delta_{x,3}}}}\ =\ \color{#66f}{\LARGE 3} \end{align}

Felix Marin
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