| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with logarithmic form |
| Difficulty | Standard +0.3 This is a straightforward multi-part C3 question testing standard techniques: finding asymptotes by solving a simple equation, applying quotient rule differentiation (shown result), solving for stationary points, and using reverse chain rule for a logarithmic integral. All parts follow routine procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Asymptote when \(1 + 2x^3 = 0\) | M1 | |
| \(2x^3 = -1\) | ||
| \(x = -\frac{1}{\sqrt[3]{2}}\) | A1 | oe, condone \(\pm\frac{1}{\sqrt[3]{2}}\) if positive root is rejected |
| \(= -0.794\) | A1cao [3] | Must be to 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{(1+2x^3).2x - x^2.6x^2}{(1+2x^3)^2}\) | M1 | Quotient or product rule (\(udv-vdu\) M0); \(2x(1+2x^3)^{-1} + x^2(-1)(1+2x^3)^{-2}.6x^2\) allow one slip on derivatives |
| \(= \frac{2x+4x^4-6x^4}{(1+2x^3)^2}\) | A1 | Correct expression – condone missing bracket if intention implied by following line |
| \(= \frac{2x-2x^4}{(1+2x^3)^2}\) | E1 | |
| \(dy/dx = 0\) when \(2x(1-x^3)=0\) | M1 | Derivative \(= 0\) |
| \(x=0,\ y=0\) | B1 B1 | \(x=0\) or \(1\) – allow unsupported answers |
| or \(x=1,\ y=\frac{1}{3}\) | B1 B1 [8] | \(y=0\) and \(\frac{1}{3}\); SC\(-1\) for setting denom \(=0\) or extra solutions (e.g. \(x=-1\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = \int_0^1 \frac{x^2}{1+2x^3}\,dx\) | M1 | Correct integral and limits – allow \(\int_1^0\) |
| Either: \(= \left[\frac{1}{6}\ln(1+2x^3)\right]_0^1\) | M1 | \(k\ln(1+2x^3)\) |
| A1 | \(k = \frac{1}{6}\) | |
| M1 | Substituting limits dep previous M1 | |
| \(= \frac{1}{6}\ln 3\) | E1 [5] | www |
| Or: let \(u = 1+2x^3 \Rightarrow du = 6x^2\,dx\) | ||
| \(A = \int_1^3 \frac{1}{6}\cdot\frac{1}{u}\,du\) | M1 | \(\frac{1}{6u}\) |
| \(= \left[\frac{1}{6}\ln u\right]_1^3\) | A1 | \(\frac{1}{6}\ln u\) |
| M1 | Substituting correct limits (but must have used substitution) | |
| \(= \frac{1}{6}\ln 3\) | E1 [5] | www |
## Question 2:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Asymptote when $1 + 2x^3 = 0$ | M1 | |
| $2x^3 = -1$ | | |
| $x = -\frac{1}{\sqrt[3]{2}}$ | A1 | oe, condone $\pm\frac{1}{\sqrt[3]{2}}$ if positive root is rejected |
| $= -0.794$ | A1cao [3] | Must be to 3 s.f. |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{(1+2x^3).2x - x^2.6x^2}{(1+2x^3)^2}$ | M1 | Quotient or product rule ($udv-vdu$ M0); $2x(1+2x^3)^{-1} + x^2(-1)(1+2x^3)^{-2}.6x^2$ allow one slip on derivatives |
| $= \frac{2x+4x^4-6x^4}{(1+2x^3)^2}$ | A1 | Correct expression – condone missing bracket if intention implied by following line |
| $= \frac{2x-2x^4}{(1+2x^3)^2}$ | E1 | |
| $dy/dx = 0$ when $2x(1-x^3)=0$ | M1 | Derivative $= 0$ |
| $x=0,\ y=0$ | B1 B1 | $x=0$ or $1$ – allow unsupported answers |
| or $x=1,\ y=\frac{1}{3}$ | B1 B1 [8] | $y=0$ and $\frac{1}{3}$; SC$-1$ for setting denom $=0$ or extra solutions (e.g. $x=-1$) |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = \int_0^1 \frac{x^2}{1+2x^3}\,dx$ | M1 | Correct integral and limits – allow $\int_1^0$ |
| **Either:** $= \left[\frac{1}{6}\ln(1+2x^3)\right]_0^1$ | M1 | $k\ln(1+2x^3)$ |
| | A1 | $k = \frac{1}{6}$ |
| | M1 | Substituting limits dep previous M1 |
| $= \frac{1}{6}\ln 3$ | E1 [5] | www |
| **Or:** let $u = 1+2x^3 \Rightarrow du = 6x^2\,dx$ | | |
| $A = \int_1^3 \frac{1}{6}\cdot\frac{1}{u}\,du$ | M1 | $\frac{1}{6u}$ |
| $= \left[\frac{1}{6}\ln u\right]_1^3$ | A1 | $\frac{1}{6}\ln u$ |
| | M1 | Substituting correct limits (but must have used substitution) |
| $= \frac{1}{6}\ln 3$ | E1 [5] | www |
---2 Fig. 7 shows the curve $y = \frac { x ^ { 2 } } { 1 + 2 x ^ { 3 } }$. It is undefined at $x = a$; the line $x = a$ is a vertical asymptote.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{00c12cc4-f7ee-4219-8d34-a1854284f65d-1_647_1027_832_534}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Calculate the value of $a$, giving your answer correct to 3 significant figures.\\
(ii) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x - 2 x ^ { 4 } } { \left( 1 + 2 x ^ { 3 } \right) ^ { 2 } }$. Hence determine the coordinates of the turning points of the curve.\\
(iii) Show that the area of the region between the curve and the $x$-axis from $x = 0$ to $x = 1$ is $\frac { 1 } { 6 } \ln 3$.
\hfill \mbox{\textit{OCR MEI C3 Q2 [16]}}