| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Moderate -0.8 This is a straightforward C2 differentiation and integration question with standard techniques throughout. All parts involve routine application of power rule for differentiation/integration, finding normal lines (negative reciprocal of gradient), and basic area calculations. The multi-part structure guides students through each step with no novel problem-solving required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dy}{dx} = 4\times\frac{1}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = 2x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}}\) | M1, A1A1 | A power decreased by 1. A1 for each correct term — Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| At \(P(4,0)\), \(\frac{dy}{dx} = \frac{2}{\sqrt{4}} - \frac{3}{2}\times 2\) | M1 | Attempts \(\frac{dy}{dx}\) when \(x=4\) |
| \(= 1-3 = -2\) | A1 | AG — Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Gradient of normal \(= \frac{1}{2}\) | M1 | Use of or stating \(m\times m' = -1\) |
| Equation of normal is \(y - 0 = m[x-4]\) | M1 | \(m\) numerical; can be awarded even if \(m=-2\) |
| \(y - 0 = \frac{1}{2}(x-4) \Rightarrow 2y = x-4\) | A1 | ACF of the equation — Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| At \(Q\), \(x=0\), \(2y = 0-4\), \(y_Q = -2\) | M1, A1F | PI. Ft on a linear equation for normal provided \(y_Q\) is negative and prev A1 is lost |
| Area of triangle \(OPQ = 0.5\times4\times\ | y_Q\ | \) |
| \(= 4\) | B1F | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = 0 \Rightarrow 2x^{-\frac{1}{2}} = \frac{3}{2}x^{\frac{1}{2}}\) | M1 | Puts c's \(\frac{dy}{dx}=0\) and a 1st step in attempt to solve |
| \(2 = \frac{3}{2}x\); \(\Rightarrow x = \frac{4}{3}\) | m1, A1 | Valid method to \(ax=b\). Condone 1.3 or better — Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int\left(4x^{\frac{1}{2}} - x^{\frac{3}{2}}\right)dx = 4\frac{x^{\frac{3}{2}}}{1.5} - \frac{x^{\frac{5}{2}}}{2.5}\{+c\}\) | M1, A1,A1 | One power correct. Condone absence of "\(+c\)" |
| \(= \frac{8}{3}x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}}\{+c\}\) | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Area under curve \(= 4\frac{4^{\frac{3}{2}}}{1.5} - \frac{4^{\frac{5}{2}}}{2.5} - \{0\}\) | M1 | \(F(4) - \{F(0)\}\) |
| Total area \(= F(4) +\) area triangle \(OPQ\) | m1 | |
| Total area \(= \frac{128}{15} + 4 = \frac{188}{15} = 12.5\ (3\ldots)\) | A1 | Accept 3 sf if clear — Total: 3 |
## Question 5:
### Part (a)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx} = 4\times\frac{1}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = 2x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}}$ | M1, A1A1 | A power decreased by 1. A1 for each correct term — **Total: 3** |
### Part (a)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| At $P(4,0)$, $\frac{dy}{dx} = \frac{2}{\sqrt{4}} - \frac{3}{2}\times 2$ | M1 | Attempts $\frac{dy}{dx}$ when $x=4$ |
| $= 1-3 = -2$ | A1 | AG — **Total: 2** |
### Part (a)(iii)
| Working | Mark | Guidance |
|---------|------|----------|
| Gradient of normal $= \frac{1}{2}$ | M1 | Use of or stating $m\times m' = -1$ |
| Equation of normal is $y - 0 = m[x-4]$ | M1 | $m$ numerical; can be awarded even if $m=-2$ |
| $y - 0 = \frac{1}{2}(x-4) \Rightarrow 2y = x-4$ | A1 | ACF of the equation — **Total: 3** |
### Part (a)(iv)
| Working | Mark | Guidance |
|---------|------|----------|
| At $Q$, $x=0$, $2y = 0-4$, $y_Q = -2$ | M1, A1F | PI. Ft on a linear equation for normal provided $y_Q$ is negative and prev A1 is lost |
| Area of triangle $OPQ = 0.5\times4\times\|y_Q\|$ | | Ft on c's negative $y_Q$ |
| $= 4$ | B1F | **Total: 3** |
### Part (a)(v)
| Working | Mark | Guidance |
|---------|------|----------|
| $2x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = 0 \Rightarrow 2x^{-\frac{1}{2}} = \frac{3}{2}x^{\frac{1}{2}}$ | M1 | Puts c's $\frac{dy}{dx}=0$ and a 1st step in attempt to solve |
| $2 = \frac{3}{2}x$; $\Rightarrow x = \frac{4}{3}$ | m1, A1 | Valid method to $ax=b$. Condone 1.3 or better — **Total: 3** |
### Part (b)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $\int\left(4x^{\frac{1}{2}} - x^{\frac{3}{2}}\right)dx = 4\frac{x^{\frac{3}{2}}}{1.5} - \frac{x^{\frac{5}{2}}}{2.5}\{+c\}$ | M1, A1,A1 | One power correct. Condone absence of "$+c$" |
| $= \frac{8}{3}x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}}\{+c\}$ | | **Total: 3** |
### Part (b)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| Area under curve $= 4\frac{4^{\frac{3}{2}}}{1.5} - \frac{4^{\frac{5}{2}}}{2.5} - \{0\}$ | M1 | $F(4) - \{F(0)\}$ |
| Total area $= F(4) +$ area triangle $OPQ$ | m1 | |
| Total area $= \frac{128}{15} + 4 = \frac{188}{15} = 12.5\ (3\ldots)$ | A1 | Accept 3 sf if clear — **Total: 3** |
---5 A curve, drawn from the origin $O$, crosses the $x$-axis at the point $P ( 4,0 )$.\\
The normal to the curve at $P$ meets the $y$-axis at the point $Q$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{14c2acbb-5f3e-40e2-8b88-162341ab9f71-3_526_629_916_813}
The curve, defined for $x \geqslant 0$, has equation
$$y = 4 x ^ { \frac { 1 } { 2 } } - x ^ { \frac { 3 } { 2 } }$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(3 marks)
\item Show that the gradient of the curve at $P ( 4,0 )$ is - 2 .
\item Find an equation of the normal to the curve at $P ( 4,0 )$.
\item Find the $y$-coordinate of $Q$ and hence find the area of triangle $O P Q$.
\item The curve has a maximum point $M$. Find the $x$-coordinate of $M$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find $\int \left( 4 x ^ { \frac { 1 } { 2 } } - x ^ { \frac { 3 } { 2 } } \right) \mathrm { d } x$.
\item Find the total area of the region bounded by the curve and the lines $P Q$ and $Q O$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2008 Q5 [20]}}