Standard +0.3 This is a straightforward application of fixed point iteration with standard functions. Part (a) requires sketching arccos with a horizontal shift (routine transformation), part (b) is simple sign-change verification by substitution, and part (c) involves calculator work to iterate twice. The question tests procedural competence rather than problem-solving insight, making it slightly easier than average for A-level.
11. (a) Given that \(0 \leqslant \mathrm { f } ( x ) \leqslant \pi\), sketch the graph of \(y = \mathrm { f } ( x )\) where
$$\mathrm { f } ( x ) = \arccos ( x - 1 ) , \quad 0 \leqslant x \leqslant 2$$
The equation \(\arccos ( x - 1 ) - \tan x = 0\) has a single root \(\alpha\).
(b) Show that \(0.9 < \alpha < 1.1\)
The iteration formula
$$x _ { n + 1 } = \arctan \left( \arccos \left( x _ { n } - 1 \right) \right)$$
can be used to find an approximation for \(\alpha\).
(c) Taking \(x _ { 0 } = 1.1\) find, to 3 decimal places, the values of \(x _ { 1 }\) and \(x _ { 2 }\)
Correct shape (curve becoming steeper at each end, decreasing)
M1
Correct shape anywhere. Ignore extra cycles. Curve should become steeper at each end.
Correct shape in correct position, no extra cycles
A1
Ignore axis labels
Part (b):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
\(f(0.9)=0.4108...,\quad f(1.1)=-0.4941...\)
M1
Substitutes \(x=0.9\) and \(x=1.1\), obtains at least one answer correct to 1 s.f.
Change of sign therefore \(0.9 < \alpha < 1.1\)
A1
Both values correct (1 s.f.), change of sign + conclusion. No contradictory statements.
Part (c):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
\(\arctan(\arccos(1.1-1))\)
M1
Attempt formula with \(x=1.1\). Implied by awrt 0.97 (radians) or awrt 89 (degrees)
\(x_1 = 0.974,\quad x_2 = 1.011\)
A1
Ignore subsequent iterations and labelling if answers are clearly second and third terms
## Question 11:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape (curve becoming steeper at each end, decreasing) | M1 | Correct shape anywhere. Ignore extra cycles. Curve should become steeper at each end. |
| Correct shape in correct position, no extra cycles | A1 | Ignore axis labels |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(0.9)=0.4108...,\quad f(1.1)=-0.4941...$ | M1 | Substitutes $x=0.9$ and $x=1.1$, obtains at least one answer correct to 1 s.f. |
| Change of sign therefore $0.9 < \alpha < 1.1$ | A1 | Both values correct (1 s.f.), change of sign + conclusion. No contradictory statements. |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\arctan(\arccos(1.1-1))$ | M1 | Attempt formula with $x=1.1$. Implied by awrt 0.97 (radians) or awrt 89 (degrees) |
| $x_1 = 0.974,\quad x_2 = 1.011$ | A1 | Ignore subsequent iterations and labelling if answers are clearly second and third terms |
11. (a) Given that $0 \leqslant \mathrm { f } ( x ) \leqslant \pi$, sketch the graph of $y = \mathrm { f } ( x )$ where
$$\mathrm { f } ( x ) = \arccos ( x - 1 ) , \quad 0 \leqslant x \leqslant 2$$
The equation $\arccos ( x - 1 ) - \tan x = 0$ has a single root $\alpha$.\\
(b) Show that $0.9 < \alpha < 1.1$
The iteration formula
$$x _ { n + 1 } = \arctan \left( \arccos \left( x _ { n } - 1 \right) \right)$$
can be used to find an approximation for $\alpha$.\\
(c) Taking $x _ { 0 } = 1.1$ find, to 3 decimal places, the values of $x _ { 1 }$ and $x _ { 2 }$
\hfill \mbox{\textit{Edexcel C34 2019 Q11 [6]}}