| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Normal or tangent line problems |
| Difficulty | Standard +0.3 This is a straightforward two-part question combining standard differentiation of a surd function to find a normal line equation, followed by integration using a simple substitution u = 4x - 7. Both techniques are routine P3/C3 content with no novel problem-solving required. The algebraic integration requirement and multi-step nature elevate it slightly above trivial, but it remains easier than average A-level questions. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = (4x-7)^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = 2(4x-7)^{-\frac{1}{2}}\) | M1 A1 | M1: achieves \(a(4x\pm7)^{-\frac{1}{2}}\); A1: correct derivative (index need not be processed for M1) |
| At \((8,5)\): gradient of tangent \(= 2(4\times8-7)^{-\frac{1}{2}} = \frac{2}{5}\) | dM1 | Substitutes \(x=8\) into their derivative; dependent on first M1 |
| Equation of \(l\): \(y - 5 = -\frac{5}{2}(x-8)\) | ddM1 | Full method for normal: negative reciprocal gradient, uses point \((8,5)\); dependent on both previous M marks |
| \(5x + 2y - 50 = 0\) | A1* | Correct equation in required form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int(4x-7)^{\frac{1}{2}}\,dx = \left[\frac{(4x-7)^{\frac{3}{2}}}{6}\right]\) | M1, A1 | Correct integration |
| Complete area \(= \int_{\frac{7}{4}}^{8}(4x-7)^{\frac{1}{2}}\,dx = \left[\frac{(4x-7)^{\frac{3}{2}}}{6}\right]_{\frac{7}{4}}^{8} + 5\) | dM1 | Applies limits and adds rectangle/triangle area of 5 |
| \(= \frac{155}{6}\) | A1 | Correct final answer |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = (4x-7)^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = 2(4x-7)^{-\frac{1}{2}}$ | M1 A1 | M1: achieves $a(4x\pm7)^{-\frac{1}{2}}$; A1: correct derivative (index need not be processed for M1) |
| At $(8,5)$: gradient of tangent $= 2(4\times8-7)^{-\frac{1}{2}} = \frac{2}{5}$ | dM1 | Substitutes $x=8$ into their derivative; dependent on first M1 |
| Equation of $l$: $y - 5 = -\frac{5}{2}(x-8)$ | ddM1 | Full method for normal: negative reciprocal gradient, uses point $(8,5)$; dependent on both previous M marks |
| $5x + 2y - 50 = 0$ | A1* | Correct equation in required form |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int(4x-7)^{\frac{1}{2}}\,dx = \left[\frac{(4x-7)^{\frac{3}{2}}}{6}\right]$ | M1, A1 | Correct integration |
| Complete area $= \int_{\frac{7}{4}}^{8}(4x-7)^{\frac{1}{2}}\,dx = \left[\frac{(4x-7)^{\frac{3}{2}}}{6}\right]_{\frac{7}{4}}^{8} + 5$ | dM1 | Applies limits and adds rectangle/triangle area of 5 |
| $= \frac{155}{6}$ | A1 | Correct final answer |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5a695b86-1660-4c06-ac96-4cdb07af9a2e-18_856_990_246_539}
\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 part of the curve with equation
$$y = \sqrt { 4 x - 7 }$$
The line $l$, shown in Figure 3, is the normal to the curve at the point $P ( 8,5 )$
\begin{enumerate}[label=(\alph*)]
\item Use calculus to show that an equation of $l$ is
$$5 x + 2 y - 50 = 0$$
The region $R$, shown shaded in Figure 3, is bounded by the curve, the $x$-axis and $l$.
\item Use algebraic integration to find the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2024 Q6 [9]}}