| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Integration using chain rule reversal |
| Difficulty | Standard +0.3 This is a straightforward application of the chain rule for differentiation followed by recognition of the reverse chain rule for integration. Part (i) requires chain rule with logarithms (standard P3 content), while part (ii) is a scaffolded 'hence' question where the derivative is given and students simply need to recognize the pattern to integrate. The definite integral evaluation is routine. This is slightly easier than average as it's highly structured with clear signposting. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{d}{dx}\ln(\sin^2 3x) = \dfrac{1}{\sin^2 3x}\times 2\sin 3x \times 3\cos 3x = 6\cot 3x\) | M1 A1 | M1: Attempts to differentiate ln function; award for \(\dfrac{1}{\sin^2 3x}\times\ldots\) A1: \(6\cot 3x\) oe. Accept \(\dfrac{6\cos 3x}{\sin 3x}\) or \(\dfrac{6}{\tan 3x}\) etc. Not \(6\tan^{-1}3x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{d}{dx}(3x^2-4)^6 = 36x(3x^2-4)^5\) | M1 A1 | M1: Achieves \(Ax(3x^2-4)^5\) where \(A\) is a constant. A1: \(36x(3x^2-4)^5\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\displaystyle\int x(3x^2-4)^5\,dx = \dfrac{1}{36}(3x^2-4)^6\) | B1ft | ft from (a) provided of correct form \(\dfrac{1}{A}(3x^2-4)^6\) |
| \(\displaystyle\int_0^{\sqrt{2}} x(3x^2-4)^5\,dx = \left[\dfrac{1}{36}(3x^2-4)^6\right]_0^{\sqrt{2}} = \dfrac{1}{36}(2)^6 - \dfrac{1}{36}(-4)^6 = -112\) | M1 A1cso | M1: Substitutes both limits and subtracts into expression of form \(D(3x^2-4)^6\). A1: \(-112\) (isw if made positive after correct answer seen) |
## Question 3:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{d}{dx}\ln(\sin^2 3x) = \dfrac{1}{\sin^2 3x}\times 2\sin 3x \times 3\cos 3x = 6\cot 3x$ | M1 A1 | M1: Attempts to differentiate ln function; award for $\dfrac{1}{\sin^2 3x}\times\ldots$ A1: $6\cot 3x$ oe. Accept $\dfrac{6\cos 3x}{\sin 3x}$ or $\dfrac{6}{\tan 3x}$ etc. Not $6\tan^{-1}3x$ |
### Part (ii)(a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{d}{dx}(3x^2-4)^6 = 36x(3x^2-4)^5$ | M1 A1 | M1: Achieves $Ax(3x^2-4)^5$ where $A$ is a constant. A1: $36x(3x^2-4)^5$ oe |
### Part (ii)(b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\displaystyle\int x(3x^2-4)^5\,dx = \dfrac{1}{36}(3x^2-4)^6$ | B1ft | ft from (a) provided of correct form $\dfrac{1}{A}(3x^2-4)^6$ |
| $\displaystyle\int_0^{\sqrt{2}} x(3x^2-4)^5\,dx = \left[\dfrac{1}{36}(3x^2-4)^6\right]_0^{\sqrt{2}} = \dfrac{1}{36}(2)^6 - \dfrac{1}{36}(-4)^6 = -112$ | M1 A1cso | M1: Substitutes both limits and subtracts into expression of form $D(3x^2-4)^6$. A1: $-112$ (isw if made positive after correct answer seen) |
---\begin{enumerate}
\item (i) Find $\frac { \mathrm { d } } { \mathrm { d } x } \ln \left( \sin ^ { 2 } 3 x \right)$ writing your answer in simplest form.\\
(ii) (a) Find $\frac { \mathrm { d } } { \mathrm { d } x } \left( 3 x ^ { 2 } - 4 \right) ^ { 6 }$\\
(b) Hence show that
\end{enumerate}
$$\int _ { 0 } ^ { \sqrt { 2 } } x \left( 3 x ^ { 2 } - 4 \right) ^ { 5 } \mathrm {~d} x = R$$
where $R$ is an integer to be found.\\
(Solutions relying on calculator technology are not acceptable.)
\hfill \mbox{\textit{Edexcel P3 2023 Q3 [7]}}