| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Combined region areas |
| Difficulty | Standard +0.8 This is a multi-part question requiring: (a) understanding logarithm domains, (b) solving when y=0, (c) solving simultaneous logarithmic equations, and (d) comparing curved area to trapezoid approximation. Part (d) requires integration of logarithmic functions and careful geometric reasoning, making this harder than standard area-under-curve questions but not requiring novel mathematical insight. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.06f Laws of logarithms: addition, subtraction, power rules1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 1\) | B1 | cao or \((1, 0)\). Need not see \(x =\) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\log_2\!\left(x - \tfrac{3}{2}\right) = -3\) | M1 | Setting \(\log_2\!\left(x - \tfrac{3}{2}\right) + 3\) equal to zero and isolating \(\log_2\!\left(x - \tfrac{3}{2}\right)\) term |
| \(x - \tfrac{3}{2} = 2^{-3}\) | M1 | Correctly removing logs. After sensible work |
| \(x = 2^{-3} + \tfrac{3}{2} \Rightarrow x = 1.625\) | A1 | cao (o.e. exact answer e.g. \(\tfrac{13}{8}\)). Need not see \(x =\). Condone 1.63 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\log_2 x = \log_2(x^2)\) seen | B1 | Using the power law for logarithms |
| \(\log_2\!\left(\dfrac{x^2}{x - \frac{3}{2}}\right) = 3\) | M1 | Using subtraction or addition law for logarithms. oe e.g. using \(3 = \log_2 8\) etc |
| \(x^2 = 8\!\left(x - \tfrac{3}{2}\right) \Rightarrow x^2 - 8x + 12 (= 0)\) | M1 | Removing logs correctly and rearranging to a three-term quadratic in \(x\). After sensible work |
| \((x-2)(x-6) = 0\); therefore the \(x\) coordinate of \(C\) is 2 | A1 | AG so sufficient working must be shown. If solving BC then need to see \(x = 2\) and \(x = 6\) with \(x = 2\) chosen as \(x\) coordinate |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y\)-coordinate of \(C\) is \(2\) | B1 | soi; NB \(x\)-coordinate of \(C\) is \(2\) |
| \(\text{Area} = \frac{1}{2}(1.625-1)(2)\) | M1* | Correct expression for area of triangle \(ABC\) with their \(x\)-coordinate of \(B\) from (b) and their \(y\)-coordinate; \(0.625\), Need \(x=1\) |
| \(0.656 - 0.625\) | M1dep* | Difference between their value and \(0.656\) is calculated |
| Under-estimate by \(0.031\) (units²) | A1 | cao (both numerical value and 'under-estimate' required); Allow \(0.026\), Allow \(4.73\%\) or \(3.96\%\) |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 1$ | B1 | cao or $(1, 0)$. Need not see $x =$ |
| **[1]** | | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\log_2\!\left(x - \tfrac{3}{2}\right) = -3$ | M1 | Setting $\log_2\!\left(x - \tfrac{3}{2}\right) + 3$ equal to zero and isolating $\log_2\!\left(x - \tfrac{3}{2}\right)$ term |
| $x - \tfrac{3}{2} = 2^{-3}$ | M1 | Correctly removing logs. After sensible work |
| $x = 2^{-3} + \tfrac{3}{2} \Rightarrow x = 1.625$ | A1 | cao (o.e. exact answer e.g. $\tfrac{13}{8}$). Need not see $x =$. Condone 1.63 |
| **[3]** | | |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\log_2 x = \log_2(x^2)$ seen | B1 | Using the power law for logarithms |
| $\log_2\!\left(\dfrac{x^2}{x - \frac{3}{2}}\right) = 3$ | M1 | Using subtraction or addition law for logarithms. oe e.g. using $3 = \log_2 8$ etc |
| $x^2 = 8\!\left(x - \tfrac{3}{2}\right) \Rightarrow x^2 - 8x + 12 (= 0)$ | M1 | Removing logs correctly and rearranging to a three-term quadratic in $x$. After sensible work |
| $(x-2)(x-6) = 0$; therefore the $x$ coordinate of $C$ is 2 | A1 | AG so sufficient working must be shown. If solving BC then need to see $x = 2$ and $x = 6$ with $x = 2$ chosen as $x$ coordinate |
| **[4]** | | |
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y$-coordinate of $C$ is $2$ | B1 | soi; **NB** $x$-coordinate of $C$ is $2$ |
| $\text{Area} = \frac{1}{2}(1.625-1)(2)$ | M1* | Correct expression for area of triangle $ABC$ with their $x$-coordinate of $B$ from **(b)** and their $y$-coordinate; $0.625$, Need $x=1$ |
| $0.656 - 0.625$ | M1dep* | Difference between their value and $0.656$ is calculated |
| Under-estimate by $0.031$ (units²) | A1 | cao (both numerical value and 'under-estimate' required); Allow $0.026$, Allow $4.73\%$ or $3.96\%$ |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-05_538_531_264_246}
The shape $A B C$ shown in the diagram is a student's design for the sail of a small boat.\\
The curve $A C$ has equation $y = 2 \log _ { 2 } x$ and the curve $B C$ has equation $y = \log _ { 2 } \left( x - \frac { 3 } { 2 } \right) + 3$.
\begin{enumerate}[label=(\alph*)]
\item State the $x$-coordinate of point $A$.
\item Determine the $x$-coordinate of point $B$.
\item By solving an equation involving logarithms, show that the $x$-coordinate of point $C$ is 2 .
It is given that, correct to 3 significant figures, the area of the sail is 0.656 units $^ { 2 }$.
\item Calculate by how much the area is over-estimated or under-estimated when the curved edges of the sail are modelled as straight lines.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q6 [12]}}