Standard +0.3 This question requires solving a simple modulus equation to find x = 3 or x = -5, then substituting into a modulus expression. While it involves multiple modulus operations, the steps are straightforward with no conceptual difficulty beyond basic modulus understanding. Slightly above average due to the two-stage process and need to handle two cases carefully.
Solve 3-term quadratic equation or a pair of linear equations
M1
For M1, must square both sides when attempting a quadratic equation
Obtain \(x = -5\) and \(x = 3\)
A1
Substitute (at least) one value of \(x\) (less than 4) into \(\
x+4\
- \
Obtain \(-8\) and \(6\) and no others
A1
Total
4
## Question 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| Solve 3-term quadratic equation or a pair of linear equations | M1 | For M1, must square both sides when attempting a quadratic equation |
| Obtain $x = -5$ and $x = 3$ | A1 | |
| Substitute (at least) one value of $x$ (less than 4) into $\|x+4\| - \|x-4\|$, showing correct evaluation of modulus and producing only one answer in each case | M1 | |
| Obtain $-8$ and $6$ and no others | A1 | |
| **Total** | **4** | |
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2 It is given that $x$ satisfies the equation $| x + 1 | = 4$. Find the possible values of
$$| x + 4 | - | x - 4 | .$$
\hfill \mbox{\textit{CAIE P2 2017 Q2 [4]}}