Edexcel C3 — Question 7 12 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSketch modulus functions involving quadratic or other non-linear
DifficultyStandard +0.8 This question requires sketching a parabola and a V-shaped modulus function, identifying their intercepts algebraically in terms of a parameter, then solving a modulus equation by considering cases. It combines multiple skills (sketching, parametric analysis, case-work for modulus) and requires careful algebraic manipulation, making it moderately challenging but within reach of a well-prepared C3 student.
Spec1.02l Modulus function: notation, relations, equations and inequalities1.02n Sketch curves: simple equations including polynomials1.02s Modulus graphs: sketch graph of |ax+b|
7. (a) Sketch on the same diagram the graphs of \(y = 4 a ^ { 2 } - x ^ { 2 }\) and \(y = | 2 x - a |\), where \(a\) is a positive constant. Show, in terms of \(a\), the coordinates of any points where each graph meets the coordinate axes.
(b) Find the exact solutions of the equation $$4 - x ^ { 2 } = | 2 x - 1 |$$
AnswerMarks Guidance
(a) Graph showing: \(y =2x - a \)
\(y = 4a^2 - x^2\)B3
(with points \((0, a)\), \((0, 4a^2)\), \((-2a, 0)\), \((\frac{a}{2}, 0)\), \((2a, 0)\) marked)
AnswerMarks Guidance
(b) \(4 - x^2 = 2x - 1\)M1
\(x^2 + 2x - 5 = 0\), \(x = \frac{-2 \pm \sqrt{4+20}}{2} = \frac{-2 \pm 2\sqrt{6}}{2}\)M1
\(x > \frac{1}{2}\) \(\therefore x = -1 + \sqrt{6}\)A1
\(4 - x^2 = -(2x - 1)\)M1
\(x^2 - 2x - 3 = 0\)M1
\((x+1)(x-3) = 0\)M1
\(x < \frac{1}{2}\) \(\therefore x = -1\), \(x = -1, -1 + \sqrt{6}\)A1 (12 marks) 
**(a)** Graph showing: $y = |2x - a|$ | B3
$y = 4a^2 - x^2$ | B3
(with points $(0, a)$, $(0, 4a^2)$, $(-2a, 0)$, $(\frac{a}{2}, 0)$, $(2a, 0)$ marked)

**(b)** $4 - x^2 = 2x - 1$ | M1
$x^2 + 2x - 5 = 0$, $x = \frac{-2 \pm \sqrt{4+20}}{2} = \frac{-2 \pm 2\sqrt{6}}{2}$ | M1
$x > \frac{1}{2}$ $\therefore x = -1 + \sqrt{6}$ | A1
$4 - x^2 = -(2x - 1)$ | M1
$x^2 - 2x - 3 = 0$ | M1
$(x+1)(x-3) = 0$ | M1
$x < \frac{1}{2}$ $\therefore x = -1$, $x = -1, -1 + \sqrt{6}$ | A1 | (12 marks)
7. (a) Sketch on the same diagram the graphs of $y = 4 a ^ { 2 } - x ^ { 2 }$ and $y = | 2 x - a |$, where $a$ is a positive constant. Show, in terms of $a$, the coordinates of any points where each graph meets the coordinate axes.\\
(b) Find the exact solutions of the equation

$$4 - x ^ { 2 } = | 2 x - 1 |$$

\hfill \mbox{\textit{Edexcel C3  Q7 [12]}}
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