Question 7 - A-Level Maths

Edexcel C3 — Question 7 14 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Fixed Point Iteration
Type	Solve trigonometric equation via iteration
Difficulty	Standard +0.3 This is a straightforward multi-part question testing standard C3 techniques: solving a basic inverse trig equation, sketching curves, using a graph to locate a root, and applying a given iterative formula. All steps are routine with no novel problem-solving required, making it slightly easier than average.
Spec	1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

7. (a) Solve the equation $$\pi - 3 \arccos \theta = 0$$ (b) Sketch on the same diagram the curves $y = \arccos ( x - 1 ) , 0 \leq x \leq 2$ and $y = \sqrt { x + 2 } , x \geq - 2$. Given that $\alpha$ is the root of the equation $$\arccos ( x - 1 ) = \sqrt { x + 2 }$$ (c) show that $0 < \alpha < 1$,
(d) use the iterative formula $$x _ { n + 1 } = 1 + \cos \sqrt { x _ { n } + 2 }$$ with $x _ { 0 } = 1$ to find $\alpha$ correct to 3 decimal places. END

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Answer	Marks	Guidance
(a)	$\arccos\theta = \frac{\pi}{3}, \quad \theta = \cos\frac{\pi}{3} = \frac{1}{2}$	M1 A1
(b)	[Graph showing $y = \sqrt{x+2}$ and $y = \arccos(x-1)$ intersecting]	B2
B3
(c)	Let $f(x) = \arccos(x-1) - \sqrt{x+2}$
$f(0) = 1.7, \quad f(1) = -0.16$	M1 A1
Sign change, $f(x)$ continuous $\therefore$ root	A1
(d)	$x_1 = 0.83944, \quad x_2 = 0.88598, \quad x_3 = 0.87233$
$x_4 = 0.87632, \quad x_5 = 0.87515, \quad x_6 = 0.87549$	M1 A2
$\alpha = 0.875$ (3dp)	A1
(14)

(a) | $\arccos\theta = \frac{\pi}{3}, \quad \theta = \cos\frac{\pi}{3} = \frac{1}{2}$ | M1 A1 |

(b) | [Graph showing $y = \sqrt{x+2}$ and $y = \arccos(x-1)$ intersecting] | B2 |
| | B3 |

(c) | Let $f(x) = \arccos(x-1) - \sqrt{x+2}$ | |
| $f(0) = 1.7, \quad f(1) = -0.16$ | M1 A1 |
| Sign change, $f(x)$ continuous $\therefore$ root | A1 |

(d) | $x_1 = 0.83944, \quad x_2 = 0.88598, \quad x_3 = 0.87233$ | |
| $x_4 = 0.87632, \quad x_5 = 0.87515, \quad x_6 = 0.87549$ | M1 A2 |
| $\alpha = 0.875$ (3dp) | A1 |
| | (14) |

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7. (a) Solve the equation

$$\pi - 3 \arccos \theta = 0$$

(b) Sketch on the same diagram the curves $y = \arccos ( x - 1 ) , 0 \leq x \leq 2$ and $y = \sqrt { x + 2 } , x \geq - 2$.

Given that $\alpha$ is the root of the equation

$$\arccos ( x - 1 ) = \sqrt { x + 2 }$$

(c) show that $0 < \alpha < 1$,\\
(d) use the iterative formula

$$x _ { n + 1 } = 1 + \cos \sqrt { x _ { n } + 2 }$$

with $x _ { 0 } = 1$ to find $\alpha$ correct to 3 decimal places.

END

\hfill \mbox{\textit{Edexcel C3  Q7 [14]}}

This paper (7 questions)

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Q1 7 Q2 9 Q3 9 Q4 10 Q5 13 Q6 13 Q7 14

Answer	Marks	Guidance
(a)	\(\arccos\theta = \frac{\pi}{3}, \quad \theta = \cos\frac{\pi}{3} = \frac{1}{2}\)	M1 A1
(b)	[Graph showing \(y = \sqrt{x+2}\) and \(y = \arccos(x-1)\) intersecting]	B2
B3
(c)	Let \(f(x) = \arccos(x-1) - \sqrt{x+2}\)
\(f(0) = 1.7, \quad f(1) = -0.16\)	M1 A1
Sign change, \(f(x)\) continuous \(\therefore\) root	A1
(d)	\(x_1 = 0.83944, \quad x_2 = 0.88598, \quad x_3 = 0.87233\)
\(x_4 = 0.87632, \quad x_5 = 0.87515, \quad x_6 = 0.87549\)	M1 A2
\(\alpha = 0.875\) (3dp)	A1
(14)