1997 Paper 1 Q4

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Year: 1997
Paper: 1
Question Number: 4

Course: LFM Pure
Section: Modulus function

Difficulty: 1500.0 Banger: 1500.0

modulus function piecewise linear case analysis equation solving absolute value equation

Problem

Find all the solutions of the equation \[|x+1|-|x|+3|x-1|-2|x-2|=x+2.\]

Solution

Case 1: \(x \leq -1\) \begin{align*} && -1-x+x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && -x-2 &= x + 2 \\ \Leftrightarrow && x = -2 \end{align*} Case \(-1 < x \leq 0\): \begin{align*} && x+1+x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && x&= x + 2 \\ \end{align*} No solutions Case \(0 < x \leq 1\): \begin{align*} && x+1-x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && -x&= x + 2 \\ \end{align*} No solutions Case \(1 < x \leq 2\): \begin{align*} && x+1-x+3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && 5x-6&= x + 2 \\ \Leftrightarrow && x = 2 \end{align*} Case \(2 < x\): \begin{align*} && x+1-x+3(x-1)-2(x-2) &= x + 2 \\ \Leftrightarrow && x+2&= x + 2 \\ \end{align*} Therefore the solutions are \(x \in \{-2\} \cup [2, \infty)\)

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Difficulty Rating: 1500.0

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Banger Rating: 1500.0

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Problem source

Find all the solutions of the equation
\[|x+1|-|x|+3|x-1|-2|x-2|=x+2.\]

Solution source

Case 1: $x \leq -1$

\begin{align*}
&& -1-x+x-3(x-1)+2(x-2) &= x + 2 \\
\Leftrightarrow && -x-2 &= x + 2 \\
\Leftrightarrow && x = -2
\end{align*}

Case $-1 < x \leq 0$:
\begin{align*}
&& x+1+x-3(x-1)+2(x-2) &= x + 2 \\
\Leftrightarrow && x&= x + 2 \\
\end{align*}
No solutions

Case $0 < x \leq 1$:
\begin{align*}
&& x+1-x-3(x-1)+2(x-2) &= x + 2 \\
\Leftrightarrow && -x&= x + 2 \\
\end{align*}
No solutions

Case $1 < x \leq 2$:
\begin{align*}
&& x+1-x+3(x-1)+2(x-2) &= x + 2 \\
\Leftrightarrow && 5x-6&= x + 2 \\
\Leftrightarrow && x = 2
\end{align*}

Case $2 < x$:
\begin{align*}
&& x+1-x+3(x-1)-2(x-2) &= x + 2 \\
\Leftrightarrow && x+2&= x + 2 \\
\end{align*}

Therefore the solutions are $x \in \{-2\} \cup [2, \infty)$

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