| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a standard C1 tangent line question requiring differentiation to find gradient, point-slope form for the equation, then basic integration and area calculation. All techniques are routine textbook exercises with clear scaffolding across multiple parts, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\dfrac{dy}{dx} = -2x - 9x^2\) | M1 A1 | One term correct; all correct (no \(+c\) etc) |
| When \(x=-2\), \(\dfrac{dy}{dx} = (4-36=)-32\) | A1 | |
| \(y = \text{"their"}-32x+c\) and attempt to find \(c\) using \(x=-2\) and \(y=24\) | m1 | Or \(y-24=\text{"their"}-32(x--2)\) |
| \(y = -32x - 40\) | A1 | Must write in this form; no ISW here |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y=0 \Rightarrow x = -\dfrac{5}{4}\) OE | B1F | Strict FT from their answer to (a)(i) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(4x - \dfrac{x^3}{3} - \dfrac{3x^4}{4}\ (+c)\) | M1 A1 | Two terms correct; all correct |
| \(\left[4\times1 - \dfrac{1^3}{3} - \dfrac{3\times1^4}{4}\right] - \left[4\times(-2) - \dfrac{(-2)^3}{3} - \dfrac{3(-2)^4}{4}\right]\) | m1 | "their" \(F(1) - F(-2)\) |
| \(\left[4 - \dfrac{1}{3} - \dfrac{3}{4}\right] - \left[-8 + \dfrac{8}{3} - \dfrac{48}{4}\right]\) | A1 | Correct with powers of 1 and \((-2)\) and minus signs handled correctly |
| \(= 20\tfrac{1}{4}\) | A1 | \(20.25\), \(\dfrac{81}{4}\), \(\dfrac{243}{12}\) OE |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Area of missing triangle \(= \left(\tfrac{1}{2}\times24\times\tfrac{3}{4}\right) = 9\) | B1 | Or correct single equivalent fraction |
| Area of region \(=\) "their" (b)(i) \(-\) "their" \(\triangle\) | M1 | "their" \((20\tfrac{1}{4}-9)\) |
| \(= 11\tfrac{1}{4}\) | A1 | \(11.25\), \(\dfrac{45}{4}\), \(\dfrac{135}{12}\) OE |
# Question 7:
## Part (a)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $\dfrac{dy}{dx} = -2x - 9x^2$ | M1 A1 | One term correct; all correct (no $+c$ etc) |
| When $x=-2$, $\dfrac{dy}{dx} = (4-36=)-32$ | A1 | |
| $y = \text{"their"}-32x+c$ and attempt to find $c$ using $x=-2$ and $y=24$ | m1 | Or $y-24=\text{"their"}-32(x--2)$ |
| $y = -32x - 40$ | A1 | Must write in this form; no ISW here |
## Part (a)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $y=0 \Rightarrow x = -\dfrac{5}{4}$ OE | B1F | Strict FT from their answer to (a)(i) |
## Part (b)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $4x - \dfrac{x^3}{3} - \dfrac{3x^4}{4}\ (+c)$ | M1 A1 | Two terms correct; all correct |
| $\left[4\times1 - \dfrac{1^3}{3} - \dfrac{3\times1^4}{4}\right] - \left[4\times(-2) - \dfrac{(-2)^3}{3} - \dfrac{3(-2)^4}{4}\right]$ | m1 | "their" $F(1) - F(-2)$ |
| $\left[4 - \dfrac{1}{3} - \dfrac{3}{4}\right] - \left[-8 + \dfrac{8}{3} - \dfrac{48}{4}\right]$ | A1 | Correct with powers of 1 and $(-2)$ and minus signs handled correctly |
| $= 20\tfrac{1}{4}$ | A1 | $20.25$, $\dfrac{81}{4}$, $\dfrac{243}{12}$ OE |
## Part (b)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| Area of missing triangle $= \left(\tfrac{1}{2}\times24\times\tfrac{3}{4}\right) = 9$ | B1 | Or correct single equivalent fraction |
| Area of region $=$ "their" (b)(i) $-$ "their" $\triangle$ | M1 | "their" $(20\tfrac{1}{4}-9)$ |
| $= 11\tfrac{1}{4}$ | A1 | $11.25$, $\dfrac{45}{4}$, $\dfrac{135}{12}$ OE |
---7 The diagram shows the sketch of a curve and the tangent to the curve at $P$.\\
\includegraphics[max width=\textwidth, alt={}, center]{0d5b9235-af2b-4fd5-8fcf-b2b45e3c0a3c-14_519_817_356_614}
The curve has equation $y = 4 - x ^ { 2 } - 3 x ^ { 3 }$ and the point $P ( - 2,24 )$ lies on the curve. The tangent at $P$ crosses the $x$-axis at $Q$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the equation of the tangent to the curve at the point $P$, giving your answer in the form $y = m x + c$.
\item Hence find the $x$-coordinate of $Q$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find $\int _ { - 2 } ^ { 1 } \left( 4 - x ^ { 2 } - 3 x ^ { 3 } \right) \mathrm { d } x$.
\item The point $R ( 1,0 )$ lies on the curve. Calculate the area of the shaded region bounded by the curve and the lines $P Q$ and $Q R$.\\[0pt]
[3 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2016 Q7 [14]}}