| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Moderate -0.3 This is a standard multi-part question on tangent lines and area between curves. Part (a) requires equating derivative to given gradient (routine), part (b) is algebraic verification, and part (c) is straightforward integration of a cubic polynomial. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dy}{dx} = 3x^2 - 14 = -2\) or \(\frac{d}{dx}(\pm(\text{curve}-\text{line})) = \pm(3x^2-12)=0\) | M1 | Differentiates cubic to achieve \(px^2+q\) and sets equal to \(-2\). Setting linear and cubic equal and solving is M0, but differentiating \(\pm(\text{curve}-\text{line})\) and setting to 0 is M1. |
| \(3x^2 - 14 = -2 \Rightarrow 3x^2 = 12 \Rightarrow x = \ldots\) | dM1 | Proceeds from equation to find real value of \(x\) with at least one intermediate step. |
| \(x = 2\) only * | A1* | 2 only (solution of \(-2\) need not be found). \(\pm2\) without selecting 2/rejecting \(-2\) is A0*. |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Substitutes \(x=-4\) into \(y=x^3-14x+23\) and \(y=-2x+7\) | M1 | Mark general method; do not be concerned by slips. Either: substitutes \(x=-4\) into both equations (sight of \(-4\) embedded sufficient); or substitutes \(x=-4\) into one to find \(y\) then uses to find \(x\) in other; or equates/rearranges and substitutes/factors. |
| Correct solution + conclusion * | A1* | Correct solution with conclusion. Either finds \(y=15\) when \(x=-4\) for both; or finds \(y=15\) using one and verifies in other; or verifies \(x=-4\) is solution of cubic\(-\)linear\(=0\); or solves and finds linear factor \((x+4)\) leading to \(x=-4\). Must have conclusion: tick, QED, underline, "proven" etc. Note: stating coordinates \((-4,15)\) is insufficient. |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int(x^3-14x+23)\,dx = \frac{x^4}{4} - 7x^2 + 23x\ (+c)\) | M1 | Integrates curve (or curve\(-\)line in Alt method), achieving at least two terms with correct index from \(\ldots x^4 \pm \ldots x^2 \pm \ldots x\) |
| \(\frac{x^4}{4} - 7x^2 + 23x\) | A1 | Correct integration (ignore constant). Also accept \(\frac{x^4}{4}-6x^2+16x\) for Alt method. |
| \(= \left(\frac{2^4}{4}-7\times2^2+23\times2\right) - \left(\frac{(-4)^4}{4}-7(-4)^2+23(-4)\right)\) | dM1 | Correct limits of 2 and \(-4\) used for integral. Evidence of substituting 2 and \(-4\) must be seen (e.g. \(22-(-140)\)). Dependent on first M. |
| Area \(= \int_{-4}^{2}(x^3-14x+23)\,dx - \frac{1}{2}\times(15+3)\times(2--4)\) | dM1 | Correct strategy for shaded area. Dependent on first M only but limits must be correct. Method for any areas (trapezium/rectangle/triangle) must be correct; adding/subtracting must be correct. |
| Area \(= 108\) * | A1* | 108 from rigorous argument. Integral sign and \(dx\) must appear at least once each side of expression, and should not be present after integration completed. |
## Question 8:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx} = 3x^2 - 14 = -2$ or $\frac{d}{dx}(\pm(\text{curve}-\text{line})) = \pm(3x^2-12)=0$ | M1 | Differentiates cubic to achieve $px^2+q$ and sets equal to $-2$. Setting linear and cubic equal and solving is M0, but differentiating $\pm(\text{curve}-\text{line})$ and setting to 0 is M1. |
| $3x^2 - 14 = -2 \Rightarrow 3x^2 = 12 \Rightarrow x = \ldots$ | dM1 | Proceeds from equation to find real value of $x$ with at least one intermediate step. |
| $x = 2$ only * | A1* | 2 only (solution of $-2$ need not be found). $\pm2$ without selecting 2/rejecting $-2$ is A0*. |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| Substitutes $x=-4$ into $y=x^3-14x+23$ and $y=-2x+7$ | M1 | Mark general method; do not be concerned by slips. Either: substitutes $x=-4$ into both equations (sight of $-4$ embedded sufficient); or substitutes $x=-4$ into one to find $y$ then uses to find $x$ in other; or equates/rearranges and substitutes/factors. |
| Correct solution + conclusion * | A1* | **Correct solution with conclusion**. Either finds $y=15$ when $x=-4$ for both; or finds $y=15$ using one and verifies in other; or verifies $x=-4$ is solution of cubic$-$linear$=0$; or solves and finds linear factor $(x+4)$ leading to $x=-4$. **Must have conclusion**: tick, QED, underline, "proven" etc. Note: stating coordinates $(-4,15)$ is insufficient. |
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $\int(x^3-14x+23)\,dx = \frac{x^4}{4} - 7x^2 + 23x\ (+c)$ | M1 | Integrates curve (or curve$-$line in Alt method), achieving at least two terms with correct index from $\ldots x^4 \pm \ldots x^2 \pm \ldots x$ |
| $\frac{x^4}{4} - 7x^2 + 23x$ | A1 | Correct integration (ignore constant). Also accept $\frac{x^4}{4}-6x^2+16x$ for Alt method. |
| $= \left(\frac{2^4}{4}-7\times2^2+23\times2\right) - \left(\frac{(-4)^4}{4}-7(-4)^2+23(-4)\right)$ | dM1 | Correct limits of 2 and $-4$ used for integral. Evidence of substituting 2 and $-4$ must be seen (e.g. $22-(-140)$). Dependent on first M. |
| Area $= \int_{-4}^{2}(x^3-14x+23)\,dx - \frac{1}{2}\times(15+3)\times(2--4)$ | dM1 | Correct strategy for shaded area. **Dependent on first M only but limits must be correct**. Method for any areas (trapezium/rectangle/triangle) must be correct; adding/subtracting must be correct. |
| Area $= 108$ * | A1* | 108 from rigorous argument. **Integral sign and $dx$ must appear at least once** each side of expression, and should not be present after integration completed. |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{23689deb-7eed-4022-848f-1278231a4056-20_915_924_303_580}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Figure 3 shows a sketch of the curve $C$ with equation
$$y = x ^ { 3 } - 14 x + 23$$
The line $l$ is the tangent to $C$ at the point $A$, also shown in Figure 3.\\
Given that $l$ has equation $y = - 2 x + 7$
\begin{enumerate}[label=(\alph*)]
\item show, using calculus, that the $x$ coordinate of $A$ is 2
The line $l$ cuts $C$ again at the point $B$.
\item Verify that the $x$ coordinate of $B$ is - 4
The finite region, $R$, shown shaded in Figure 3, is bounded by $C$ and $l$.\\
Using algebraic integration,
\item show that the area of $R$ is 108
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2024 Q8 [10]}}