| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Moderate -0.8 This is a straightforward multi-part C2 question testing standard differentiation and integration of powers. Part (a) is direct application of power rule, parts (b)(i)-(ii) are routine tangent equations, part (c) is reverse power rule, and part (d) requires calculating area using integration minus a triangle—all standard textbook techniques with no problem-solving insight required. |
| Spec | 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 integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dy}{dx}=12-5x^{\frac{2}{3}}\) | M1, A1 | \(kx^{\frac{2}{3}}\) term; ACF |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| When \(x=0\), \(\frac{dy}{dx}=12\) | B1F | Ft on \(y'\) evaluated correctly at \(x=0\) |
| Equation of tangent at \(O\) is \(y=12x\) | B1F | OE Ft on \(y'(0)\) provided \(y'(0)>0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| When \(x=8\), \(\frac{dy}{dx}=12-5\times(8)^{\frac{2}{3}}\) | M1 | Attempt to find \(\frac{dy}{dx}\) when \(x=8\) |
| \(y-0=y'(8)[x-8]\) | m1 | \(y=y'(8)[x-8]\) OE |
| \(y=-8(x-8) \Rightarrow y+8x=64\) | A1 | CSO AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int\left(12x-3x^{\frac{5}{3}}\right)dx=\frac{12x^2}{2}-\frac{3x^{\frac{8}{3}}}{\frac{8}{3}}\) (+c) | M1 | \(kx^{\frac{5}{3}+1}\) term after integrating |
| \(=6x^2-\frac{9}{8}x^{\frac{8}{3}}\) (+c) | B1, A1 | B1 for \(6x^2\); A1 for \(-\frac{9}{8}x^{\frac{8}{3}}\) OE |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int_0^8\left(12x-3x^{\frac{5}{3}}\right)dx=6\times8^2-\frac{9}{8}\times(8)^{\frac{8}{3}}\) | M1 | \(\pm F(8)\{-F(0)\}\) PI following integration |
| \(=384-288=96\) | A1 | PI by correct final answer |
| At \(P\): \(12x+8x=64\), \((x_P=3.2)\), \(y_P=38.4\) | M1, A1 | Solving \(y+8x=64\) and \(y=kx\), \(k>0\); \(y_P=38.4\) OE |
| Area of triangle \(OPA=\frac{1}{2}\times8\times y_P\) | M1 | Need perpendicular height linked to \(y_P>0\) |
| Area of shaded region \(=\) Area \(\triangle OPA - \int_0^8\left(12x-3x^{\frac{5}{3}}\right)dx\) | M1 | M0 if evaluated to value \(<0\) |
| \(=153.6-96=57.6\) | A1 | OE e.g. \(\frac{288}{5}\) |
# Question 9:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx}=12-5x^{\frac{2}{3}}$ | M1, A1 | $kx^{\frac{2}{3}}$ term; ACF |
## Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| When $x=0$, $\frac{dy}{dx}=12$ | B1F | Ft on $y'$ evaluated correctly at $x=0$ |
| Equation of tangent at $O$ is $y=12x$ | B1F | OE Ft on $y'(0)$ provided $y'(0)>0$ |
## Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| When $x=8$, $\frac{dy}{dx}=12-5\times(8)^{\frac{2}{3}}$ | M1 | Attempt to find $\frac{dy}{dx}$ when $x=8$ |
| $y-0=y'(8)[x-8]$ | m1 | $y=y'(8)[x-8]$ OE |
| $y=-8(x-8) \Rightarrow y+8x=64$ | A1 | CSO AG |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $\int\left(12x-3x^{\frac{5}{3}}\right)dx=\frac{12x^2}{2}-\frac{3x^{\frac{8}{3}}}{\frac{8}{3}}$ (+c) | M1 | $kx^{\frac{5}{3}+1}$ term after integrating |
| $=6x^2-\frac{9}{8}x^{\frac{8}{3}}$ (+c) | B1, A1 | B1 for $6x^2$; A1 for $-\frac{9}{8}x^{\frac{8}{3}}$ OE |
## Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $\int_0^8\left(12x-3x^{\frac{5}{3}}\right)dx=6\times8^2-\frac{9}{8}\times(8)^{\frac{8}{3}}$ | M1 | $\pm F(8)\{-F(0)\}$ PI following integration |
| $=384-288=96$ | A1 | PI by correct final answer |
| At $P$: $12x+8x=64$, $(x_P=3.2)$, $y_P=38.4$ | M1, A1 | Solving $y+8x=64$ and $y=kx$, $k>0$; $y_P=38.4$ OE |
| Area of triangle $OPA=\frac{1}{2}\times8\times y_P$ | M1 | Need perpendicular height linked to $y_P>0$ |
| Area of shaded region $=$ Area $\triangle OPA - \int_0^8\left(12x-3x^{\frac{5}{3}}\right)dx$ | M1 | M0 if evaluated to value $<0$ |
| $=153.6-96=57.6$ | A1 | OE e.g. $\frac{288}{5}$ |9 The diagram shows part of a curve crossing the $x$-axis at the origin $O$ and at the point $A ( 8,0 )$. Tangents to the curve at $O$ and $A$ meet at the point $P$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{02e5dfac-18d7-480d-ac23-dfd2ca348cba-5_547_536_497_760}
The curve has equation
$$y = 12 x - 3 x ^ { \frac { 5 } { 3 } }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at the point $O$ and hence write down an equation of the tangent at $O$.
\item Show that the equation of the tangent at $A ( 8,0 )$ is $y + 8 x = 64$.
\end{enumerate}\item Find $\int \left( 12 x - 3 x ^ { \frac { 5 } { 3 } } \right) \mathrm { d } x$.
\item Calculate the area of the shaded region bounded by the curve from $O$ to $A$ and the tangents $O P$ and $A P$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2012 Q9 [17]}}