| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Determine increasing/decreasing intervals |
| Difficulty | Standard +0.3 Part (i) requires standard product rule application and factoring to find g(x), then solving a straightforward inequality. Part (ii) involves implicit differentiation with trigonometric functions and simplification using the given constraint. While multi-step, these are routine C3 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.07q Product and quotient rules: differentiation1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 2x(x^2-1)^5 \Rightarrow \frac{dy}{dx} = (x^2-1)^5 \times 2 + 2x \times 10x(x^2-1)^4\) | M1, A1 | M1: product rule to form \(A(x^2-1)^5 + Bx^n(x^2-1)^4\), \(n=1\) or \(2\), \(A,B>0\) |
| \(\frac{dy}{dx} = (x^2-1)^4(2x^2-2+20x^2) = (x^2-1)^4(22x^2-2)\) | M1, A1 | Factorises out \((x^2-1)^4\) and simplifies |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx}=0 \Rightarrow (22x^2-2)=0 \Rightarrow x = \pm\frac{1}{\sqrt{11}}\) | M1, A1 | M1: sets \(\frac{dy}{dx}=0\); A1: both critical values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \ln(\sec 2y) \Rightarrow \frac{dx}{dy} = \frac{1}{\sec 2y}\times 2\sec 2y\tan 2y\) | B1 | |
| \(\Rightarrow \frac{dy}{dx} = \frac{1}{2\tan 2y} = \frac{1}{2\sqrt{\sec^2 2y -1}} = \frac{1}{2\sqrt{e^{2x}-1}}\) | M1, M1, A1 | M1: reciprocal; M1: use of \(\sec^2-1=\tan^2\) and \(\sec 2y = e^x\); A1: correct final form |
| Alt 2: \(y = \frac{1}{2}\arccos(e^{-x}) \Rightarrow \frac{dy}{dx} = -\frac{1}{2}\times\frac{1}{\sqrt{1-(e^{-x})^2}}\times -e^{-x} \Rightarrow \frac{1}{2\sqrt{e^{2x}-1}}\) | B1M1M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 2(x^2-1)^5 + 20x^2(x^2-1)^4\) | A1 | Any unsimplified but correct form |
| Factor of \((x^2-1)^4\) taken out | M1 | Look for \(A(x^2-1)^5 \pm Bx^n(x^2-1)^4 = (x^2-1)^4\{A(x^2-1) \pm Bx^n\}\) |
| \(\frac{dy}{dx} = (x^2-1)^4(22x^2-2)\) | A1 | Expect \(g(x)\) simplified; accept \(\frac{dy}{dx} = (x^2-1)^4 \cdot 2(11x^2-1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Sets \(\frac{dy}{dx} = 0\) and finds critical values | M1 | \(\frac{dy}{dx} = 0\) rearranged \(\div (x^2-1)^4\); g(x) should be at least 2TQ with real roots |
| \(x \ldots \frac{1}{\sqrt{11}}\), \(x \ldots -\frac{1}{\sqrt{11}}\) | A1 | Exact equivalent only; condone \(x \ldots \frac{\sqrt{11}}{11}\), \(x \ldots -\frac{\sqrt{11}}{11}\); accept \(\left(-\infty, -\frac{\sqrt{11}}{11}\right] \cup \left[\frac{\sqrt{11}}{11}, \infty\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct line involving \(\frac{dy}{dx}\) or \(\frac{dx}{dy}\) | B1 | Accept \(\frac{dx}{dy} = \frac{1}{\sec 2y} \times 2\sec 2y \tan 2y\); or \(\frac{dx}{dy} = -\frac{1}{\cos 2y} \times -2\sin 2y\); or \(2\sec 2y \tan 2y \frac{dy}{dx} = e^x\) |
| Inverts \(\frac{dx}{dy}\) to get \(\frac{dy}{dx}\) | M1 | Variables on rhs must be consistent |
| Uses \(\tan^2 2y = \pm 1 \pm \sec^2 2y\) and \(\sec 2y = e^x\) | M1 | To achieve \(\frac{dy}{dx}\) or \(\frac{dx}{dy}\) in terms of \(x\); may condone \(\sec^2 2y = (e^x)^2\) appearing as \(e^{x^2}\) |
| \(\frac{dy}{dx} = \frac{1}{2\sqrt{e^{2x}-1}}\) | A1 | Do not allow if simplified to \(\frac{1}{2e^x - 1}\); condone \(\pm\frac{1}{2\sqrt{e^{2x}-1}}\) but not \(-\frac{1}{2\sqrt{e^{2x}-1}}\) |
# Question 7:
## Part (i)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 2x(x^2-1)^5 \Rightarrow \frac{dy}{dx} = (x^2-1)^5 \times 2 + 2x \times 10x(x^2-1)^4$ | M1, A1 | M1: product rule to form $A(x^2-1)^5 + Bx^n(x^2-1)^4$, $n=1$ or $2$, $A,B>0$ |
| $\frac{dy}{dx} = (x^2-1)^4(2x^2-2+20x^2) = (x^2-1)^4(22x^2-2)$ | M1, A1 | Factorises out $(x^2-1)^4$ and simplifies |
## Part (i)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx}=0 \Rightarrow (22x^2-2)=0 \Rightarrow x = \pm\frac{1}{\sqrt{11}}$ | M1, A1 | M1: sets $\frac{dy}{dx}=0$; A1: both critical values |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \ln(\sec 2y) \Rightarrow \frac{dx}{dy} = \frac{1}{\sec 2y}\times 2\sec 2y\tan 2y$ | B1 | |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{2\tan 2y} = \frac{1}{2\sqrt{\sec^2 2y -1}} = \frac{1}{2\sqrt{e^{2x}-1}}$ | M1, M1, A1 | M1: reciprocal; M1: use of $\sec^2-1=\tan^2$ and $\sec 2y = e^x$; A1: correct final form |
| **Alt 2:** $y = \frac{1}{2}\arccos(e^{-x}) \Rightarrow \frac{dy}{dx} = -\frac{1}{2}\times\frac{1}{\sqrt{1-(e^{-x})^2}}\times -e^{-x} \Rightarrow \frac{1}{2\sqrt{e^{2x}-1}}$ | B1M1M1A1 | |
# Question A1 (Differentiation):
## Part (i)(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 2(x^2-1)^5 + 20x^2(x^2-1)^4$ | A1 | Any unsimplified but correct form |
| Factor of $(x^2-1)^4$ taken out | M1 | Look for $A(x^2-1)^5 \pm Bx^n(x^2-1)^4 = (x^2-1)^4\{A(x^2-1) \pm Bx^n\}$ |
| $\frac{dy}{dx} = (x^2-1)^4(22x^2-2)$ | A1 | Expect $g(x)$ simplified; accept $\frac{dy}{dx} = (x^2-1)^4 \cdot 2(11x^2-1)$ |
## Part (i)(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sets $\frac{dy}{dx} = 0$ and finds critical values | M1 | $\frac{dy}{dx} = 0$ rearranged $\div (x^2-1)^4$; g(x) should be at least 2TQ with real roots |
| $x \ldots \frac{1}{\sqrt{11}}$, $x \ldots -\frac{1}{\sqrt{11}}$ | A1 | Exact equivalent only; condone $x \ldots \frac{\sqrt{11}}{11}$, $x \ldots -\frac{\sqrt{11}}{11}$; accept $\left(-\infty, -\frac{\sqrt{11}}{11}\right] \cup \left[\frac{\sqrt{11}}{11}, \infty\right)$ |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct line involving $\frac{dy}{dx}$ or $\frac{dx}{dy}$ | B1 | Accept $\frac{dx}{dy} = \frac{1}{\sec 2y} \times 2\sec 2y \tan 2y$; or $\frac{dx}{dy} = -\frac{1}{\cos 2y} \times -2\sin 2y$; or $2\sec 2y \tan 2y \frac{dy}{dx} = e^x$ |
| Inverts $\frac{dx}{dy}$ to get $\frac{dy}{dx}$ | M1 | Variables on rhs must be consistent |
| Uses $\tan^2 2y = \pm 1 \pm \sec^2 2y$ **and** $\sec 2y = e^x$ | M1 | To achieve $\frac{dy}{dx}$ or $\frac{dx}{dy}$ in terms of $x$; may condone $\sec^2 2y = (e^x)^2$ appearing as $e^{x^2}$ |
| $\frac{dy}{dx} = \frac{1}{2\sqrt{e^{2x}-1}}$ | A1 | Do not allow if simplified to $\frac{1}{2e^x - 1}$; condone $\pm\frac{1}{2\sqrt{e^{2x}-1}}$ but not $-\frac{1}{2\sqrt{e^{2x}-1}}$ |
---\begin{enumerate}
\item (i) Given $y = 2 x \left( x ^ { 2 } - 1 \right) ^ { 5 }$, show that\\
(a) $\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { g } ( x ) \left( x ^ { 2 } - 1 \right) ^ { 4 }$ where $\mathrm { g } ( x )$ is a function to be determined.\\
(b) Hence find the set of values of $x$ for which $\frac { \mathrm { d } y } { \mathrm {~d} x } \geqslant 0$\\
(ii) Given
\end{enumerate}
$$x = \ln ( \sec 2 y ) , \quad 0 < y < \frac { \pi } { 4 }$$
find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ as a function of $x$ in its simplest form.\\
\hfill \mbox{\textit{Edexcel C3 2017 Q7 [10]}}