| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Show gradient condition leads to equation |
| Difficulty | Standard +0.3 This is a straightforward differentiation question requiring chain rule for ln and standard trig identities. Part (i) involves differentiating a composite function and solving dy/dx=0, while part (ii) uses the double angle formula sin(2x)=2sin(x)cos(x). All techniques are standard P3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.07a Derivative as gradient: of tangent to curve1.07b Gradient as rate of change: dy/dx notation1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{6x}{x^2-5} - 8x\) | M1 A1 | M1: differentiating \(3\ln(x^2-5) \to \frac{Ax}{x^2-5}\); A1: correct expression |
| Stationary point: \(\frac{6x}{x^2-5} - 8x = 0 \Rightarrow x^2 = \frac{23}{4} \Rightarrow x = \frac{\sqrt{23}}{2}\) only | dM1 A1 | dM1: proceeds to quadratic/cubic; A1: \(x=\frac{\sqrt{23}}{2}\) only; \(x=0\) and \(x=-\frac{\sqrt{23}}{2}\) must be rejected |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 4 - 24\sin x\cos x\) | M1 | Differentiating using chain rule with \(\sin^2 x \to \ldots\sin x\cos x\) |
| \(\frac{dy}{dx} = 4 - 12\sin 2x\) | dM1 A1 | Uses \(\sin 2x = 2\sin x\cos x\); condone slip writing \(A+B\sin x\) if coefficient of \(\sin x\cos x\) has been halved |
| Maximum gradient \(= 16\) | A1 |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{6x}{x^2-5} - 8x$ | M1 A1 | M1: differentiating $3\ln(x^2-5) \to \frac{Ax}{x^2-5}$; A1: correct expression |
| Stationary point: $\frac{6x}{x^2-5} - 8x = 0 \Rightarrow x^2 = \frac{23}{4} \Rightarrow x = \frac{\sqrt{23}}{2}$ only | dM1 A1 | dM1: proceeds to quadratic/cubic; A1: $x=\frac{\sqrt{23}}{2}$ only; $x=0$ and $x=-\frac{\sqrt{23}}{2}$ must be rejected |
### Part (ii)(a) and (b) - marked together:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 4 - 24\sin x\cos x$ | M1 | Differentiating using chain rule with $\sin^2 x \to \ldots\sin x\cos x$ |
| $\frac{dy}{dx} = 4 - 12\sin 2x$ | dM1 A1 | Uses $\sin 2x = 2\sin x\cos x$; condone slip writing $A+B\sin x$ if coefficient of $\sin x\cos x$ has been halved |
| Maximum gradient $= 16$ | A1 | |6. (i) The curve $C _ { 1 }$ has equation
$$y = 3 \ln \left( x ^ { 2 } - 5 \right) - 4 x ^ { 2 } + 15 \quad x > \sqrt { 5 }$$
Show that $C _ { 1 }$ has a stationary point at $x = \frac { \sqrt { p } } { 2 }$ where $p$ is a constant to be found.\\
(ii) A different curve $C _ { 2 }$ has equation
$$y = 4 x - 12 \sin ^ { 2 } x$$
\begin{enumerate}[label=(\alph*)]
\item Show that, for this curve,
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = A + B \sin 2 x$$
where $A$ and $B$ are constants to be found.
\item Hence, state the maximum gradient of this curve.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q6 [8]}}