| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative satisfies condition |
| Difficulty | Standard +0.3 This is a standard multi-part C3/C4 question requiring product rule application, solving a trigonometric equation, algebraic manipulation to show a given result, and numerical methods (iteration with interval verification). While it involves multiple techniques, each step follows routine procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sqrt{\frac{3}{2}}\) or \(\frac{\sqrt{3}}{\sqrt{2}}\) or \(\sqrt{1.5}\) or \(\frac{\sqrt{6}}{2}\) | B1 | Exact equivalent; no others inside range; ignore solutions outside range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y=(2x^2-3)\tan\left(\frac{1}{2}x\right) \Rightarrow \frac{dy}{dx} = 4x\tan\left(\frac{1}{2}x\right)+(2x^2-3)\times\frac{1}{2}\sec^2\left(\frac{1}{2}x\right)\) | M1A1A1 | M1: product rule attempt; A1: one term correct; A1: fully correct derivative |
| \(8\alpha\frac{\sin\left(\frac{1}{2}\alpha\right)}{\cos\left(\frac{1}{2}\alpha\right)}+(2\alpha^2-3)\times\frac{1}{\cos^2\left(\frac{1}{2}\alpha\right)}=0\) | M1 | Using \(\tan\left(\frac{1}{2}x\right)=\frac{\sin(\frac{1}{2}x)}{\cos(\frac{1}{2}x)}\) and \(\sec^2\left(\frac{1}{2}x\right)=\frac{1}{\cos^2(\frac{1}{2}x)}\) and setting \(\frac{dy}{dx}=0\) |
| \(8\alpha\sin\left(\frac{1}{2}\alpha\right)\cos\left(\frac{1}{2}\alpha\right)+(2\alpha^2-3)=0\) | dM1 | Dependent on both M's; multiplying by \(\cos^2\left(\frac{1}{2}x\right)\) and using \(2\sin\left(\frac{1}{2}x\right)\cos\left(\frac{1}{2}x\right)=\sin x\) |
| \(2\alpha^2 - 3 + 4\alpha\sin\alpha = 0\) | A1* | Printed answer; no errors including bracketing errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_2 = \frac{3}{(2\times0.7+4\sin 0.7)}\) | M1 | Substituting \(x_1=0.7\) into iterative formula |
| \(x_2 = 0.7544,\ x_3 = 0.7062\) | A1 | awrt 4 dp |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Chooses interval \([0.72825, 0.72835]\) | M1 | Or smaller interval containing the root |
| \(2\times0.72825^2-3+4\times0.72825\sin0.72825 = -0.0005 < 0\) | A1 | Both values substituted into suitable function e.g. \(\pm(2\alpha^2-3+4\sin\alpha)\), calculations correct to 1sf, reason (change of sign) and conclusion |
| \(2\times0.72835^2-3+4\times0.72835\sin0.72835 = 0.00026 > 0\) + Reason + Conclusion |
# Question 11:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{\frac{3}{2}}$ or $\frac{\sqrt{3}}{\sqrt{2}}$ or $\sqrt{1.5}$ or $\frac{\sqrt{6}}{2}$ | B1 | Exact equivalent; no others inside range; ignore solutions outside range |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y=(2x^2-3)\tan\left(\frac{1}{2}x\right) \Rightarrow \frac{dy}{dx} = 4x\tan\left(\frac{1}{2}x\right)+(2x^2-3)\times\frac{1}{2}\sec^2\left(\frac{1}{2}x\right)$ | M1A1A1 | M1: product rule attempt; A1: one term correct; A1: fully correct derivative |
| $8\alpha\frac{\sin\left(\frac{1}{2}\alpha\right)}{\cos\left(\frac{1}{2}\alpha\right)}+(2\alpha^2-3)\times\frac{1}{\cos^2\left(\frac{1}{2}\alpha\right)}=0$ | M1 | Using $\tan\left(\frac{1}{2}x\right)=\frac{\sin(\frac{1}{2}x)}{\cos(\frac{1}{2}x)}$ and $\sec^2\left(\frac{1}{2}x\right)=\frac{1}{\cos^2(\frac{1}{2}x)}$ and setting $\frac{dy}{dx}=0$ |
| $8\alpha\sin\left(\frac{1}{2}\alpha\right)\cos\left(\frac{1}{2}\alpha\right)+(2\alpha^2-3)=0$ | dM1 | Dependent on both M's; multiplying by $\cos^2\left(\frac{1}{2}x\right)$ and using $2\sin\left(\frac{1}{2}x\right)\cos\left(\frac{1}{2}x\right)=\sin x$ |
| $2\alpha^2 - 3 + 4\alpha\sin\alpha = 0$ | A1* | Printed answer; no errors including bracketing errors |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_2 = \frac{3}{(2\times0.7+4\sin 0.7)}$ | M1 | Substituting $x_1=0.7$ into iterative formula |
| $x_2 = 0.7544,\ x_3 = 0.7062$ | A1 | awrt 4 dp |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Chooses interval $[0.72825, 0.72835]$ | M1 | Or smaller interval containing the root |
| $2\times0.72825^2-3+4\times0.72825\sin0.72825 = -0.0005 < 0$ | A1 | Both values substituted into suitable function e.g. $\pm(2\alpha^2-3+4\sin\alpha)$, calculations correct to 1sf, reason (change of sign) and conclusion |
| $2\times0.72835^2-3+4\times0.72835\sin0.72835 = 0.00026 > 0$ + Reason + Conclusion | | |11.
$$y = \left( 2 x ^ { 2 } - 3 \right) \tan \left( \frac { 1 } { 2 } x \right) , \quad 0 < x < \pi$$
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $x$ when $y = 0$
Given that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ when $x = \alpha$,
\item show that
$$2 \alpha ^ { 2 } - 3 + 4 \alpha \sin \alpha = 0$$
The iterative formula
$$x _ { n + 1 } = \frac { 3 } { \left( 2 x _ { n } + 4 \sin x _ { n } \right) }$$
can be used to find an approximation for $\alpha$.
\item Taking $x _ { 1 } = 0.7$, find the values of $x _ { 2 }$ and $x _ { 3 }$, giving each answer to 4 decimal places.
\item By choosing a suitable interval, show that $\alpha = 0.7283$ to 4 decimal places.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{29b56d51-120a-4275-a761-8b8aed7bca54-38_2253_50_314_1977}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q11 [11]}}