| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Region bounded by two curves |
| Difficulty | Moderate -0.3 This is a standard C2 integration question with routine differentiation, normal line, and area between curves. All parts follow textbook procedures with no novel problem-solving required, though the multi-part structure and fractional powers make it slightly more substantial than the most basic questions. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 3x^{\frac{1}{2}}\) | M1 | \(kx^{\frac{1}{2}}\) with or without \(+ c\) |
| \(= 6 \text{ (when } x = 4\text{)}\) | A1cao | Must be 6 and seen in (a)(i). \(6 + c\) is A0. 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(y\text{-coordinate of } A = 2 \times 4^{\frac{3}{2}} (= 16)\) | M1 | Substitute \(x = 4\) in \(y = 2x^{\frac{3}{2}}\) |
| \(6 \times m' = -1\) | M1 | \(m_1 \times m_2 = -1\) OE used with c's value of \(\frac{dy}{dx}\) when \(x = 4\). PI |
| \(y - 16 = m(x - 4)\) | m1 | dep on 1st M1 in (a)(ii). \(m\) must be numerical |
| \(y - 16 = -\frac{1}{6}(x - 4)\) | A1 | ACF. 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int 8x^{\frac{1}{2}} dx = \frac{8}{7/2}x^{\frac{1+1}{2}} \{+c\}\) | M1 | Index raised by 1 |
| \(= \frac{16}{3}x^{\frac{3}{2}} \{+c\}\) | A1 | Condone missing '\(+c\)'. Coefficient must be simplified. 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int 2x^{\frac{3}{2}} dx = \frac{2}{5/2} x^{\frac{5}{2}} \{+c\} \quad \{= \frac{4}{5}x^{\frac{5}{2}} [+c]\}\) | B1 | Can award for unsimplified form |
| \(\int_0^4 8x^{\frac{1}{2}} dx - \int_0^4 2x^{\frac{3}{2}} dx\) | M1 | Ignore limits here |
| \(= \frac{16}{3}(4)^{\frac{3}{2}} - 0 - \left[\frac{4}{5}(4)^{\frac{5}{2}} - 0\right]\) | M1 | F(4) − F(0) used in either; [F(0)=0 PI] Cand. must be using F(x) as a result of his/her integration in (b)(i) or in the (b)(ii) B1 line above |
| \(= \frac{256}{15}\) | A1 | Accept any value from 17.04 to 17.1 inclusive in place of 256/15. 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Translation | B1 | Accept 'translat...' as equivalent IT or Tr is NOT sufficient |
| \(\begin{bmatrix}-3\\0\end{bmatrix}\) | B1 | Accept equivalent in words provided linked to 'translation/move/shift' (B0B0 if >1 transformation). 2 marks total |
**4(a)(i)**
$\frac{dy}{dx} = 3x^{\frac{1}{2}}$ | M1 | $kx^{\frac{1}{2}}$ with or without $+ c$
$= 6 \text{ (when } x = 4\text{)}$ | A1cao | Must be 6 and seen in (a)(i). $6 + c$ is A0. 2 marks total
**4(a)(ii)**
$y\text{-coordinate of } A = 2 \times 4^{\frac{3}{2}} (= 16)$ | M1 | Substitute $x = 4$ in $y = 2x^{\frac{3}{2}}$
$6 \times m' = -1$ | M1 | $m_1 \times m_2 = -1$ OE used with c's value of $\frac{dy}{dx}$ when $x = 4$. PI
$y - 16 = m(x - 4)$ | m1 | dep on 1st M1 in (a)(ii). $m$ must be numerical
$y - 16 = -\frac{1}{6}(x - 4)$ | A1 | ACF. 4 marks total
**4(b)(i)**
$\int 8x^{\frac{1}{2}} dx = \frac{8}{7/2}x^{\frac{1+1}{2}} \{+c\}$ | M1 | Index raised by 1
$= \frac{16}{3}x^{\frac{3}{2}} \{+c\}$ | A1 | Condone missing '$+c$'. Coefficient must be simplified. 2 marks total
**4(b)(ii)**
$\int 2x^{\frac{3}{2}} dx = \frac{2}{5/2} x^{\frac{5}{2}} \{+c\} \quad \{= \frac{4}{5}x^{\frac{5}{2}} [+c]\}$ | B1 | Can award for unsimplified form
$\int_0^4 8x^{\frac{1}{2}} dx - \int_0^4 2x^{\frac{3}{2}} dx$ | M1 | Ignore limits here
$= \frac{16}{3}(4)^{\frac{3}{2}} - 0 - \left[\frac{4}{5}(4)^{\frac{5}{2}} - 0\right]$ | M1 | F(4) − F(0) used in either; [F(0)=0 PI] Cand. must be using F(x) as a result of his/her integration in (b)(i) or in the (b)(ii) B1 line above
$= \frac{256}{15}$ | A1 | Accept any value from 17.04 to 17.1 inclusive in place of 256/15. 4 marks total
**4(c)**
Translation | B1 | Accept 'translat...' as equivalent IT or Tr is NOT sufficient
$\begin{bmatrix}-3\\0\end{bmatrix}$ | B1 | Accept equivalent in words provided linked to 'translation/move/shift' (B0B0 if >1 transformation). 2 marks total
**Total for Q4: 14 marks**
---4 The diagram shows a sketch of the curves with equations $y = 2 x ^ { \frac { 3 } { 2 } }$ and $y = 8 x ^ { \frac { 1 } { 2 } }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{0e19665b-5ee5-49e4-8de2-6c8dd17f61eb-3_433_720_1452_644}
The curves intersect at the origin and at the point $A$, where $x = 4$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item For the curve $y = 2 x ^ { \frac { 3 } { 2 } }$, find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when $x = 4$.\\
(2 marks)
\item Find an equation of the normal to the curve $y = 2 x ^ { \frac { 3 } { 2 } }$ at the point $A$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find $\int 8 x ^ { \frac { 1 } { 2 } } \mathrm {~d} x$.
\item Find the area of the shaded region bounded by the two curves.
\end{enumerate}\item Describe a single geometrical transformation that maps the graph of $y = 2 x ^ { \frac { 3 } { 2 } }$ onto the graph of $y = 2 ( x + 3 ) ^ { \frac { 3 } { 2 } }$.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2009 Q4 [14]}}