| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Standard +0.3 This is a straightforward product rule application with exponential and trigonometric functions, followed by routine algebraic manipulation using standard identities (cos²x = 1/sec²x, sin2x). Part (b) requires solving a simple trigonometric equation. Slightly above average due to the algebraic manipulation required, but all techniques are standard for this level. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Step 1 | M1 | Use correct product rule or quotient rule, and attempt chain rule: \(ke^{-4x}\tan x + e^{-4x}\sec^2 x\) or \(\frac{e^{4x}\sec^2 x - \tan x(ke^{4x})}{(e^{4x})^2}\). Need \(\frac{d(\tan x)}{dx} = \sec^2 x\) and attempt at \(ke^{-4x}\) where \(k\neq 1\). |
| Step 2 | A1 | Obtain correct derivative in any form: \(-4e^{-4x}\tan x + e^{-4x}\sec^2 x\) or \(\frac{e^{4x}\sec^2 x - \tan x(4e^{4x})}{(e^{4x})^2}\) |
| Step 3 | M1 | Use trig formulae to express derivative in the form \(ke^{-4x}\sin x\cos x\sec^2 x + ae^{-4x}\sec^2 x\) or \(ke^{-4x}\frac{\sin x\cos x}{\cos x\cos x} + ae^{-4x}\sec^2 x\) or \(\sec^2 x(ke^{-4x}\sin x\cos x + ae^{-4x})\). Need \(\frac{\tan x}{\sec^2 x} = \sin x\cos x\) or \(\tan x = \frac{\sin x}{\cos x}\cdot\frac{\cos x}{\cos x}\). Allow \(\frac{1}{\cos^2 x}\) instead of \(\sec^2 x\). M1 independent of previous M1 but expression must be of appropriate form. |
| Step 4 | A1 | Obtain correct answer with \(a=1\) and \(b=-2\): \(\sec^2 x(1-2\sin 2x)e^{-4x}\). At least one line of trig working required from \(-4e^{-4x}\tan x + e^{-4x}\sec^2 x\) to given answer. If only error: \(4\sin x\cos x = 4\sin 2x\) → M1 A1 M1 A0. |
| Answer | Marks | Guidance |
|---|---|---|
| Step 1 | M1 | Equate derivative to zero and use correct method to solve for \(x\): \(\sin 2x = \frac{1}{2}\), hence \(x = \frac{1}{2}\sin^{-1}\frac{1}{2}\) or \(x = \tan^{-1}(2\pm\sqrt{3})\). Allow M1 for correct method for non-exact value. |
| Step 2 | A1 | Obtain answer e.g. \(x = \frac{1}{12}\pi\). \([0.262 \text{ M1 A0}]\) |
| Step 3 | A1 FT | Obtain second answer e.g. \(x = \frac{5}{12}\pi\) and no other in given interval. FT \(\frac{\pi - \text{their } 2x}{2}\) if exact values; \(x\) must be \(<\frac{\pi}{2}\). Ignore answers outside given interval. Treat answers in degrees as misread: \(15°\), \(75°\). SC: No values found for \(a\) and \(b\) in 4(a) but chooses values in 4(b): max M1 for \(x\). |
## Question 4(a):
**Step 1** | M1 | Use correct product rule or quotient rule, and attempt chain rule: $ke^{-4x}\tan x + e^{-4x}\sec^2 x$ or $\frac{e^{4x}\sec^2 x - \tan x(ke^{4x})}{(e^{4x})^2}$. Need $\frac{d(\tan x)}{dx} = \sec^2 x$ and attempt at $ke^{-4x}$ where $k\neq 1$.
**Step 2** | A1 | Obtain correct derivative in any form: $-4e^{-4x}\tan x + e^{-4x}\sec^2 x$ or $\frac{e^{4x}\sec^2 x - \tan x(4e^{4x})}{(e^{4x})^2}$
**Step 3** | M1 | Use trig formulae to express derivative in the form $ke^{-4x}\sin x\cos x\sec^2 x + ae^{-4x}\sec^2 x$ or $ke^{-4x}\frac{\sin x\cos x}{\cos x\cos x} + ae^{-4x}\sec^2 x$ or $\sec^2 x(ke^{-4x}\sin x\cos x + ae^{-4x})$. Need $\frac{\tan x}{\sec^2 x} = \sin x\cos x$ or $\tan x = \frac{\sin x}{\cos x}\cdot\frac{\cos x}{\cos x}$. Allow $\frac{1}{\cos^2 x}$ instead of $\sec^2 x$. M1 independent of previous M1 but expression must be of appropriate form.
**Step 4** | A1 | Obtain correct answer with $a=1$ and $b=-2$: $\sec^2 x(1-2\sin 2x)e^{-4x}$. At least one line of trig working required from $-4e^{-4x}\tan x + e^{-4x}\sec^2 x$ to given answer. If only error: $4\sin x\cos x = 4\sin 2x$ → M1 A1 M1 A0.
**Total: 4 marks**
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## Question 4(b):
**Step 1** | M1 | Equate derivative to zero and use correct method to solve for $x$: $\sin 2x = \frac{1}{2}$, hence $x = \frac{1}{2}\sin^{-1}\frac{1}{2}$ or $x = \tan^{-1}(2\pm\sqrt{3})$. Allow M1 for correct method for non-exact value.
**Step 2** | A1 | Obtain answer e.g. $x = \frac{1}{12}\pi$. $[0.262 \text{ M1 A0}]$
**Step 3** | A1 FT | Obtain second answer e.g. $x = \frac{5}{12}\pi$ and no other in given interval. FT $\frac{\pi - \text{their } 2x}{2}$ if exact values; $x$ must be $<\frac{\pi}{2}$. Ignore answers outside given interval. Treat answers in degrees as misread: $15°$, $75°$. SC: No values found for $a$ and $b$ in **4(a)** but chooses values in **4(b)**: max **M1** for $x$.
**Total: 3 marks**4 The curve $y = \mathrm { e } ^ { - 4 x } \tan x$ has two stationary points in the interval $0 \leqslant x < \frac { 1 } { 2 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item Obtain an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and show it can be written in the form $\sec ^ { 2 } x ( a + b \sin 2 x ) \mathrm { e } ^ { - 4 x }$, where $a$ and $b$ are constants.
\item Hence find the exact $x$-coordinates of the two stationary points.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q4 [7]}}