| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Finding maximum/minimum on curve |
| Difficulty | Standard +0.3 This is a standard C3 calculus question requiring routine techniques: finding asymptotes by solving a simple equation, applying quotient rule differentiation, solving dy/dx=0 for turning points, and integration by substitution (recognizing the numerator relates to the derivative of the denominator). All steps are textbook exercises 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: differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Asymptote when \(1 + 2x^3 = 0\) | M1 | |
| \(2x^3 = -1\) | ||
| \(x = -\dfrac{1}{\sqrt[3]{2}}\) | A1 | oe, condone \(+\dfrac{1}{\sqrt[3]{2}}\) if positive root is rejected |
| \(= -0.794\) | A1cao [3] | must be to 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \dfrac{(1+2x^3)\cdot 2x - x^2 \cdot 6x^2}{(1+2x^3)^2}\) | M1 | Quotient or product rule (\(udv - vdu\) M0); \(2x(1+2x^3)^{-1} + x^2(-1)(1+2x^3)^{-2} \cdot 6x^2\) allow one slip on derivatives |
| \(= \dfrac{2x + 4x^4 - 6x^4}{(1+2x^3)^2}\) | A1 | correct expression – condone missing bracket if intention implied by following line |
| \(= \dfrac{2x - 2x^4}{(1+2x^3)^2}\) * | 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 |
| \(A = \int_{0}^{1} \dfrac{x^2}{1+2x^3}\, dx\) | M1 | Correct integral and limits – allow \(\int_{1}^{0}\) |
| Either: \(= \left[\frac{1}{6}\ln(1+2x^3)\right]_{0}^{1}\) | M1, A1 | \(k\ln(1+2x^3)\); \(k = 1/6\) |
| substituting limits | M1 | dep previous M1 |
| \(= \dfrac{1}{6}\ln 3\) * | E1 | 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 | \(\frac{1}{6}\ln u\) |
| substituting correct limits | M1 | (but must have used substitution) |
| \(= \dfrac{1}{6}\ln 3\) * | E1 [5] | www |
## Question 3:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Asymptote when $1 + 2x^3 = 0$ | M1 | |
| $2x^3 = -1$ | | |
| $x = -\dfrac{1}{\sqrt[3]{2}}$ | A1 | oe, condone $+\dfrac{1}{\sqrt[3]{2}}$ if positive root is rejected |
| $= -0.794$ | A1cao [3] | must be to 3 s.f. |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \dfrac{(1+2x^3)\cdot 2x - x^2 \cdot 6x^2}{(1+2x^3)^2}$ | M1 | Quotient or product rule ($udv - vdu$ M0); $2x(1+2x^3)^{-1} + x^2(-1)(1+2x^3)^{-2} \cdot 6x^2$ allow one slip on derivatives |
| $= \dfrac{2x + 4x^4 - 6x^4}{(1+2x^3)^2}$ | A1 | correct expression – condone missing bracket if intention implied by following line |
| $= \dfrac{2x - 2x^4}{(1+2x^3)^2}$ * | 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 |
|---|---|---|
| $A = \int_{0}^{1} \dfrac{x^2}{1+2x^3}\, dx$ | M1 | Correct integral and limits – allow $\int_{1}^{0}$ |
| **Either:** $= \left[\frac{1}{6}\ln(1+2x^3)\right]_{0}^{1}$ | M1, A1 | $k\ln(1+2x^3)$; $k = 1/6$ |
| substituting limits | M1 | dep previous M1 |
| $= \dfrac{1}{6}\ln 3$ * | E1 | 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 | $\frac{1}{6}\ln u$ |
| substituting correct limits | M1 | (but must have used substitution) |
| $= \dfrac{1}{6}\ln 3$ * | E1 [5] | www |3 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]{9e68f5e0-3394-4962-acd9-25bb31f09f2b-3_654_1034_463_531}
\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 Q3 [16]}}