| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Challenging +1.2 This question requires finding a normal line equation (standard differentiation and perpendicular gradient), then setting up and evaluating a definite integral for area between curves. While it involves multiple steps across both parts, each component is a standard Core 2 technique with no novel insight required. The 'exact value' requirement adds minor complexity but is routine for this level. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = -2x+6\) and substitutes \(x=5\) to give gradient \(m = -4\) | M1 A1 | |
| Normal has gradient \(\frac{-1}{m} = \frac{1}{4}\) | M1 | |
| Equation of normal: \((y+3) = \frac{1}{4}(x-5)\) so \(x - 4y - 17 = 0\) | dM1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int -x^2+6x-8\,dx = -\frac{x^3}{3}+6\frac{x^2}{2}-8x\) | M1 | |
| The line meets the \(x\)-axis at \(17\) | B1 | |
| The curve meets the \(x\)-axis at \(4\) | B1 | |
| Uses correct limits for their integral: \(\left[-\frac{x^3}{3}+\frac{6x^2}{2}-8x\right]_4^5\) | M1 | |
| Finds area above line using area of triangle or integration \(= \frac{1}{2}\times 3\times("17"-5)\) | M1 | |
| Area of \(R = 18 + 1\frac{1}{3} = 19\frac{1}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Differentiates to give \(\frac{dy}{dx} = \pm 2x \pm 6\) and substitutes \(x = 5\) | M1 | |
| Obtains answer \(-4\) | A1 | |
| Uses negative reciprocal of their numerical \(\frac{dy}{dx}\) | M1 | Follow through; first M1 must have been awarded |
| Linear equation through point \((5, -3)\) with their changed gradient | dM1 | Dependent on first M; allow \((y+3) = 4(x-5)\) following answer of \(-4\) |
| Accept \(k(x - 4y - 17) = 0\) where \(k\) is positive or negative integer | A1 | Candidates working with gradient \(\pm 2\) score 0 marks in this part |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Integrates a quadratic expression correctly | M1 | If integrating (line \(-\) curve), follow through on new quadratic; coefficients must be correct |
| Obtains \(17\) for point where line meets \(x\)-axis | B1 | |
| Finds curve meets \(x\)-axis at \(4\) | B1 | Score for \(y=0 \Rightarrow x=2,4\); also allow as a limit in the integral; may score if \(4\) appears in correct place on diagram |
| Uses limits \(4\) and \(5\) in their integrated function | M1 | If candidate writes \(\int_4^5 \pm(-x^2+6x-8)\,dx = \pm\frac{4}{3}\) (from GC), allow this mark |
| Finds appropriate area above the line for their attempted integral | M1 | If integrating just curve, look for area of triangle \(= \frac{1}{2}\times 3 \times \text{"their }17-5\text{"}\); or \(\int_5^{"17"}\!\left(\frac{1}{4}x - \frac{17}{4}\right)dx = \left[\frac{1}{8}x^2 - \frac{17}{4}x\right]_5^{"17"}\); if integrating (line \(-\) curve) from \(4\) to \(5\), triangle \(= \frac{1}{2}\times\text{their }\frac{13}{4}\times\text{"their }17-4\text{"}\) |
| Correct work leading to \(19\frac{1}{3}\) | A1 | A candidate using GC for integration can potentially score M0 B1 B1 M1 M1 A0 |
## Question 14:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -2x+6$ and substitutes $x=5$ to give gradient $m = -4$ | M1 A1 | |
| Normal has gradient $\frac{-1}{m} = \frac{1}{4}$ | M1 | |
| Equation of normal: $(y+3) = \frac{1}{4}(x-5)$ so $x - 4y - 17 = 0$ | dM1 A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int -x^2+6x-8\,dx = -\frac{x^3}{3}+6\frac{x^2}{2}-8x$ | M1 | |
| The line meets the $x$-axis at $17$ | B1 | |
| The curve meets the $x$-axis at $4$ | B1 | |
| Uses correct limits for their integral: $\left[-\frac{x^3}{3}+\frac{6x^2}{2}-8x\right]_4^5$ | M1 | |
| Finds area above line using area of triangle or integration $= \frac{1}{2}\times 3\times("17"-5)$ | M1 | |
| Area of $R = 18 + 1\frac{1}{3} = 19\frac{1}{3}$ | A1 | |
# Question (a) [Tangent/Normal Problem]:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Differentiates to give $\frac{dy}{dx} = \pm 2x \pm 6$ and substitutes $x = 5$ | M1 | |
| Obtains answer $-4$ | A1 | |
| Uses negative reciprocal of their numerical $\frac{dy}{dx}$ | M1 | Follow through; first M1 must have been awarded |
| Linear equation through point $(5, -3)$ with their **changed** gradient | dM1 | Dependent on first M; allow $(y+3) = 4(x-5)$ following answer of $-4$ |
| Accept $k(x - 4y - 17) = 0$ where $k$ is positive or negative integer | A1 | Candidates working with gradient $\pm 2$ score 0 marks in this part |
---
# Question (b) [Area Problem]:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Integrates a quadratic expression correctly | M1 | If integrating (line $-$ curve), follow through on new quadratic; coefficients must be correct |
| Obtains $17$ for point where line meets $x$-axis | B1 | |
| Finds curve meets $x$-axis at $4$ | B1 | Score for $y=0 \Rightarrow x=2,4$; also allow as a limit in the integral; may score if $4$ appears in correct place on diagram |
| Uses limits $4$ and $5$ in their integrated function | M1 | If candidate writes $\int_4^5 \pm(-x^2+6x-8)\,dx = \pm\frac{4}{3}$ (from GC), allow this mark |
| Finds appropriate area above the line for their attempted integral | M1 | If integrating just curve, look for area of triangle $= \frac{1}{2}\times 3 \times \text{"their }17-5\text{"}$; or $\int_5^{"17"}\!\left(\frac{1}{4}x - \frac{17}{4}\right)dx = \left[\frac{1}{8}x^2 - \frac{17}{4}x\right]_5^{"17"}$; if integrating (line $-$ curve) from $4$ to $5$, triangle $= \frac{1}{2}\times\text{their }\frac{13}{4}\times\text{"their }17-4\text{"}$ |
| Correct work leading to $19\frac{1}{3}$ | A1 | A candidate using GC for integration can potentially score M0 B1 B1 M1 M1 A0 |
---14.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{53865e15-3838-4551-b507-fe49549b87db-40_456_689_269_623}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve $C$ with equation $y = - x ^ { 2 } + 6 x - 8$. The normal to $C$ at the point $P ( 5 , - 3 )$ is the line $l$, which is also shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $l$, giving your answer in the form $a x + b y + c = 0$, where $a$, b and $c$ are integers.
The finite region $R$, shown shaded in Figure 3, is bounded below by the line $l$ and the curve $C$, and is bounded above by the $x$-axis.
\item Find the exact value of the area of $R$.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q14 [11]}}