| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Inverse trigonometric function equations |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on inverse trigonometric functions. Part (a) is a standard sketch, (b) requires simple rearrangement and exact value recall (tan(π/3)=√3), (c) is routine substitution to verify an inequality, and (d) is calculator-based iteration. All parts are textbook exercises requiring recall and basic manipulation rather than problem-solving or insight, making it slightly easier than average. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| [Curve sketch] | M1A1 | M1: Curve not a straight line through \((0,0)\) in quadrants 1 and 3 only. A1: Grad \(\to 0\) as \(x \to \pm\infty\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(3\arctan(x+1) - \pi = 0 \Rightarrow \arctan(x+1) = \frac{\pi}{3}\) | M1 | Substitutes \(g(x+1) = \arctan(x+1)\) in \(3g(x+1)-\pi=0\) and makes \(\arctan(x+1)\) the subject. Do not condone missing brackets unless later work implies their presence. |
| \(\Rightarrow x = \tan\left(\frac{\pi}{3}\right) - 1 = \sqrt{3}-1\) | dM1A1 | dM1: Takes tan and makes \(x\) the subject, allow \(x = \sqrt{3}\pm1\). Note \(\tan\left(\frac{\pi}{3}\right)\) need not be evaluated for this mark. A1: \(\sqrt{3}-1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Sub \(x=5\) and \(x=6\) into \(\pm\left(\arctan x - 4 + \frac{1}{2}x\right) \Rightarrow -0.126\ldots, +0.405\ldots\) | M1 | Obtains at least one answer correct to 1sf |
| Both values correct (to 1sf), change of sign + conclusion | A1 | Allow equivalent statements. Mark may be withheld if contradictory statements present. |
| Answer | Marks | Guidance |
|---|---|---|
| \(x_1 = 8 - 2\arctan 5\) | M1 | Score for \(x_1 = 8 - 2\arctan 5 = \ldots\) May be implied by awrt 5.3 (radians) or awrt \(-149\) (degrees) for \(x_1\) |
| \(x_1 = 5.253, \quad x_2 = 5.235\) | A1 | \(x_1\) awrt 5.253, \(x_2\) awrt 5.235. Ignore any subsequent iterations and ignore labelling if answers are clearly the second and third terms. |
## Question 10(a):
[Curve sketch] | M1A1 | M1: Curve not a straight line through $(0,0)$ in quadrants 1 and 3 only. A1: Grad $\to 0$ as $x \to \pm\infty$
---
## Question 10(b):
$3\arctan(x+1) - \pi = 0 \Rightarrow \arctan(x+1) = \frac{\pi}{3}$ | M1 | Substitutes $g(x+1) = \arctan(x+1)$ in $3g(x+1)-\pi=0$ and makes $\arctan(x+1)$ the subject. Do not condone missing brackets unless later work implies their presence.
$\Rightarrow x = \tan\left(\frac{\pi}{3}\right) - 1 = \sqrt{3}-1$ | dM1A1 | dM1: Takes tan and makes $x$ the subject, allow $x = \sqrt{3}\pm1$. Note $\tan\left(\frac{\pi}{3}\right)$ need not be evaluated for this mark. A1: $\sqrt{3}-1$
---
## Question 10(c):
Sub $x=5$ **and** $x=6$ into $\pm\left(\arctan x - 4 + \frac{1}{2}x\right) \Rightarrow -0.126\ldots, +0.405\ldots$ | M1 | Obtains at least one answer correct to 1sf
Both values correct (to 1sf), change of sign + conclusion | A1 | Allow equivalent statements. Mark may be withheld if contradictory statements present.
---
## Question 10(d):
$x_1 = 8 - 2\arctan 5$ | M1 | Score for $x_1 = 8 - 2\arctan 5 = \ldots$ May be implied by awrt 5.3 (radians) or awrt $-149$ (degrees) for $x_1$
$x_1 = 5.253, \quad x_2 = 5.235$ | A1 | $x_1$ awrt 5.253, $x_2$ awrt 5.235. Ignore any subsequent iterations and ignore labelling if answers are clearly the second and third terms.
---10. (a) Given that $- \frac { \pi } { 2 } < \mathrm { g } ( x ) < \frac { \pi } { 2 }$, sketch the graph of $y = \mathrm { g } ( x )$ where
$$\mathrm { g } ( x ) = \arctan x , \quad x \in \mathbb { R }$$
(b) Find the exact value of $x$ for which
$$3 g ( x + 1 ) - \pi = 0$$
The equation $\arctan x - 4 + \frac { 1 } { 2 } x = 0$ has a positive root at $x = \alpha$ radians.\\
(c) Show that $5 < \alpha < 6$
The iteration formula
$$x _ { n + 1 } = 8 - 2 \arctan x _ { n }$$
can be used to find an approximation for $\alpha$\\
(d) Taking $x _ { 0 } = 5$, use this formula to find $x _ { 1 }$ and $x _ { 2 }$, giving each answer to 3 decimal places.
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel C34 2016 Q10 [9]}}