Question 7 - A-Level Maths

OCR C3 — Question 7 11 marks

Exam Board	OCR
Module	C3 (Core Mathematics 3)
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Modulus function
Type	Sketch modulus functions involving quadratic or other non-linear
Difficulty	Standard +0.8 This question requires sketching a parabola and a V-shaped modulus function, identifying their intersections with axes in terms of a parameter, then solving a modulus equation by considering cases. It combines multiple skills (sketching, parametric analysis, case-work algebra) and requires careful handling of the absolute value, making it moderately challenging but within reach of a well-prepared C3 student.
Spec	1.02n Sketch curves: simple equations including polynomials 1.02s Modulus graphs: sketch graph of \|ax+b\|1.02t Solve modulus equations: graphically with modulus function

7. (i) 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.
(ii) Find the exact solutions of the equation $$4 - x ^ { 2 } = | 2 x - 1 |$$

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Question 7:

Part (i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Correct sketch of \(y = \	2x - a\	\) with $(0, 4a^2)$ labelled
Correct sketch of $y = 4a^2 - x^2$ with $(0,a)$, $(-2a,0)$, $(2a,0)$ labelled	B2

Part (ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$4 - x^2 = 2x - 1$	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
$x^2 - 2x - 3 = 0$
$(x+1)(x-3) = 0$	M1
$x < \frac{1}{2}\ \therefore x = -1$, giving $x = -1,\ -1+\sqrt{6}$	A1	(11)

# Question 7:

## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct sketch of $y = \|2x - a\|$ with $(0, 4a^2)$ labelled | B3 | |
| Correct sketch of $y = 4a^2 - x^2$ with $(0,a)$, $(-2a,0)$, $(2a,0)$ labelled | B2 | |

## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4 - x^2 = 2x - 1$ | 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 | |
| $x^2 - 2x - 3 = 0$ | | |
| $(x+1)(x-3) = 0$ | M1 | |
| $x < \frac{1}{2}\ \therefore x = -1$, giving $x = -1,\ -1+\sqrt{6}$ | A1 | **(11)** |

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7. (i) 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.\\
(ii) Find the exact solutions of the equation

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

\hfill \mbox{\textit{OCR C3  Q7 [11]}}

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