| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch modulus functions involving quadratic or other non-linear |
| Difficulty | Standard +0.3 This is a standard Further Maths modulus question requiring systematic case analysis of |f(x)| = g(x), sketching both graphs, and solving an inequality. While it involves a quadratic inside the modulus and multiple steps, the techniques are routine for FP2 students: split into cases where the expression is positive/negative, solve resulting quadratics, sketch by reflecting negative portions, and read off the inequality solution from the graph. The algebra is straightforward with no conceptual surprises, making this slightly easier than average overall. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b|1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x^2 + 6x - 5 = 5 - 2x\) | M1 | |
| \(2x^2 + 8x - 10 = 0 \Rightarrow x^2 + 4x - 5 = 0\) | ||
| \((x+5)(x-1)=0\) or by formula | M1 | Obtaining 3-term quadratic and solving |
| \(x=-5,\ x=1\) | A1 | |
| \(-2x^2 - 6x + 5 = 5 - 2x\) | M1 | Considering reflected part of quadratic |
| \(2x^2 + 4x = 0\) | A1 | Correct 2-term quadratic |
| \(x=0,\ x=-2\) | A1 | |
| Line drawn, negative gradient, crossing positive \(y\)-axis | B1 | |
| Quadratic curve with reflected part, correct shape, crossing \(y\)-axis at same point as line, pointed where it meets \(x\)-axis | B1 | |
| \(x\)-coordinates of intersections shown: \(-5,\ -2,\ 0,\ 1\) | B1ft | Follow through on \(x\)-coords from (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x < -5\) | B1 | |
| \(-2 < x < 0\) | B1 | |
| \(x > 1\) | B1 | |
| Special case: deduct last B mark earned if \(\leqslant\) or \(\geqslant\) used |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x^2 + 6x - 5 = 5 - 2x$ | M1 | |
| $2x^2 + 8x - 10 = 0 \Rightarrow x^2 + 4x - 5 = 0$ | | |
| $(x+5)(x-1)=0$ or by formula | M1 | Obtaining 3-term quadratic and solving |
| $x=-5,\ x=1$ | A1 | |
| $-2x^2 - 6x + 5 = 5 - 2x$ | M1 | Considering reflected part of quadratic |
| $2x^2 + 4x = 0$ | A1 | Correct 2-term quadratic |
| $x=0,\ x=-2$ | A1 | |
| Line drawn, negative gradient, crossing positive $y$-axis | B1 | |
| Quadratic curve with reflected part, correct shape, crossing $y$-axis at same point as line, pointed where it meets $x$-axis | B1 | |
| $x$-coordinates of intersections shown: $-5,\ -2,\ 0,\ 1$ | B1ft | Follow through on $x$-coords from (a) |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x < -5$ | B1 | |
| $-2 < x < 0$ | B1 | |
| $x > 1$ | B1 | |
| Special case: deduct last B mark earned if $\leqslant$ or $\geqslant$ used | | |\begin{enumerate}
\item (a) Use algebra to find the exact solutions of the equation
\end{enumerate}
$$\left| 2 x ^ { 2 } + 6 x - 5 \right| = 5 - 2 x$$
(b) On the same diagram, sketch the curve with equation $y = \left| 2 x ^ { 2 } + 6 x - 5 \right|$ and the line with equation $y = 5 - 2 x$, showing the $x$-coordinates of the points where the line crosses the curve.\\
(c) Find the set of values of $x$ for which
$$\left| 2 x ^ { 2 } + 6 x - 5 \right| > 5 - 2 x$$
\hfill \mbox{\textit{Edexcel FP2 2013 Q6 [12]}}