| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Solve equation with reciprocal functions |
| Difficulty | Standard +0.3 Part (i) requires converting tan to sin/cos, rearranging to a quadratic in sin x, and solving—a standard multi-step technique. Part (ii) involves routine substitution into a sinusoidal model and finding max/min values. While multi-part with several marks, all techniques are textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals1.05q Trig in context: vectors, kinematics, forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses \(\tan x = \frac{\sin x}{\cos x} \Rightarrow 5\sin x \times \frac{\sin x}{\cos x}+13=\cos x\) | B1 | |
| \(5\sin^2 x+13\cos x = \cos^2 x \Rightarrow 5(1-\cos^2 x)+13\cos x=\cos^2 x\) | M1 | Uses \(\tan x=\frac{\sin x}{\cos x}\), \(\sin^2 x+\cos^2 x=1\) and multiplies by \(\cos x\); allow slips in coefficients but trig terms must be correct |
| \(\Rightarrow 6\cos^2 x-13\cos x-5=0\) | A1 | The "\(=0\)" may be implied |
| \((3\cos x+1)(2\cos x-5)=0 \Rightarrow \cos x = -\frac{1}{3}\) | M1 | Solves 3TQ in \(\cos x\) leading to at least one value |
| \(\Rightarrow x=1.91\) | A1 | awrt \(1.91\) following \(\cos x=-\frac{1}{3}\); no other values in range; note: from \(6\cos^2 x-13\cos x-5=0\) directly to 1.91 scores M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(20=10+12\sin(6k+18)^\circ \Rightarrow \sin(6k+18)^\circ = \frac{5}{6}\) | M1 | Proceeds to \(\sin(6k+18)^\circ=c\); may be implied by \(6k+18=\arcsin\frac{B}{A}\) |
| \((6k+18)=56.4, 123.6\) | dM1 | Takes arcsin leading to value for \(6k+18\); accept radian values (0.985, 2.16); allow awrt 2 s.f. |
| \(k=6.41, 17.59\) | A1, A1 | Must be in degrees; SC allow 6.4 and 17.6 both given if no more accurate answers stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(22^\circ\)C | B1 | Condone just 22 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{"6.41"}t+18=90 \Rightarrow t=11.23\); Time of day \(=11{:}14\) | M1, A1 | Sets their \(\text{"6.41"}t+18=90\); note \(\text{"6.41"}t+18=\frac{\pi}{2}\) is M0; cao time of day \(=11{:}14\) o.e. (e.g. 11h 14m acceptable) |
# Question 8:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $\tan x = \frac{\sin x}{\cos x} \Rightarrow 5\sin x \times \frac{\sin x}{\cos x}+13=\cos x$ | B1 | |
| $5\sin^2 x+13\cos x = \cos^2 x \Rightarrow 5(1-\cos^2 x)+13\cos x=\cos^2 x$ | M1 | Uses $\tan x=\frac{\sin x}{\cos x}$, $\sin^2 x+\cos^2 x=1$ and multiplies by $\cos x$; allow slips in coefficients but trig terms must be correct |
| $\Rightarrow 6\cos^2 x-13\cos x-5=0$ | A1 | The "$=0$" may be implied |
| $(3\cos x+1)(2\cos x-5)=0 \Rightarrow \cos x = -\frac{1}{3}$ | M1 | Solves 3TQ in $\cos x$ leading to at least one value |
| $\Rightarrow x=1.91$ | A1 | awrt $1.91$ following $\cos x=-\frac{1}{3}$; no other values in range; note: from $6\cos^2 x-13\cos x-5=0$ directly to 1.91 scores M0A0 |
## Part (ii)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $20=10+12\sin(6k+18)^\circ \Rightarrow \sin(6k+18)^\circ = \frac{5}{6}$ | M1 | Proceeds to $\sin(6k+18)^\circ=c$; may be implied by $6k+18=\arcsin\frac{B}{A}$ |
| $(6k+18)=56.4, 123.6$ | dM1 | Takes arcsin leading to value for $6k+18$; accept radian values (0.985, 2.16); allow awrt 2 s.f. |
| $k=6.41, 17.59$ | A1, A1 | Must be in degrees; SC allow 6.4 and 17.6 both given if no more accurate answers stated |
## Part (ii)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $22^\circ$C | B1 | Condone just 22 |
## Part (ii)(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{"6.41"}t+18=90 \Rightarrow t=11.23$; Time of day $=11{:}14$ | M1, A1 | Sets their $\text{"6.41"}t+18=90$; note $\text{"6.41"}t+18=\frac{\pi}{2}$ is M0; cao time of day $=11{:}14$ o.e. (e.g. 11h 14m acceptable) |\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.\\
(i) Solve, for $0 < x \leqslant \pi$, the equation
\end{enumerate}
$$5 \sin x \tan x + 13 = \cos x$$
giving your answer in radians to 3 significant figures.\\
(ii) The temperature inside a greenhouse is monitored on one particular day.
The temperature, $H ^ { \circ } \mathrm { C }$, inside the greenhouse, $t$ hours after midnight, is modelled by the equation
$$H = 10 + 12 \sin ( k t + 18 ) ^ { \circ } \quad 0 \leqslant t < 24$$
where $k$ is a constant.\\
Use the equation of the model to answer parts (a) to (c).\\
Given that
\begin{itemize}
\item the temperature inside the greenhouse was $20 ^ { \circ } \mathrm { C }$ at 6 am
\item $0 < k < 20$\\
(a) find all possible values for $k$, giving each answer to 2 decimal places.
\end{itemize}
Given further that $0 < k < 10$\\
(b) find the maximum temperature inside the greenhouse,\\
(c) find the time of day at which this maximum temperature occurs.
Give your answer to the nearest minute.
\hfill \mbox{\textit{Edexcel P2 2024 Q8 [12]}}