| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Deduce related integral from numerical approximation |
| Difficulty | Standard +0.3 This is a multi-part question requiring trapezium rule application, geometric reasoning to relate two regions, and rearranging to integrate with respect to y. While it involves several steps, each component is straightforward: part (i) is routine numerical approximation, part (ii) requires recognizing that A + B = rectangle (a standard insight), and part (iii) is a guided integration after rearrangement. The question is slightly easier than average due to its structured guidance and use of standard C2 techniques. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.5 \times 2.5 \times (1 + 2(-3 + 2\sqrt{6.5}) + 3)\) | M1* | Attempt \(y\)-values at \(x = 0, 2.5, 5\) only. M0 if additional \(y\)-values found, unless not used. \(y_1\) can be exact or decimal (2.1 or better). Allow M1 for using incorrect function as long as still clearly \(y\)-values intended to be original function eg \(-3 + 2\sqrt{x+4}\) (from \(\sqrt{(x+4)} = \sqrt{x} + \sqrt{4}\)) |
| \(= 10.2\) | M1d* | Attempt correct trapezium rule, inc \(h = 2.5\). Fully correct structure reqd, including placing of \(y\)-values. The 'big brackets' must be seen, or implied by later working. Could be implied by stating general rule in terms of \(y_0\) etc, as long as these have been attempted elsewhere and clearly labelled. Using \(x\)-values is M0. Can give M1, even if error in evaluating \(y\)-values as long correct intention is clear |
| A1 | Obtain 10.2, or better. Allow answers in range [10.24, 10.25] if \(>3\)sf. A0 if exact surd value given as final answer. Answer only is 0/3. Using 2 separate trapezia can get full marks. Using anything other than 2 strips of width 2.5 is M0. Using trapezium rule on result of an integration attempt is 0/3 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((5 \times 3) - 10.2 = 4.8\) | M1 | Attempt area of rectangle \(-\) their (i). As long as \(0 <\) their (i) \(< 15\) |
| A1FT | Obtain 4.8, or better. Allow for exact surd value as well. Allow answers in range [4.75, 4.80] if \(> 2\)sf | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = \frac{1}{4}(y^2 + 6y - 7)\) | M1 | Attempt to write as \(x = f(y)\). Must be correct order of operations, but allow slip with inverse operations eg \(+/-\), and omitting to square the \(\frac{1}{2}\). Allow \(y^2 + 9\) from an attempt to square \(y+3\), even if \((y+3)^2\) is not seen explicitly first. Allow maximum of 1 error |
| A1 | Obtain \(x = \frac{1}{4}(y^2 + 6y - 7)\) aef. Allow A1 as soon as any correct equation seen in format \(x = f(y)\), eg \(x = \frac{1}{4}(y+3)^2 - 4\) or \(x = \frac{1}{4}(y^2+6y+9) - 4\), and isw subsequent error | |
| \(\text{area} = \left[\frac{1}{12}y^3 + \frac{3}{4}y^2 - \frac{7}{4}y\right]_1^3\) | M1* | Attempt integration of \(f(y)\). Expand bracket and increase in power by 1 for at least two terms (allow if constant term disappears). Independent of rearrangement attempt so M0M1 is possible. Can gain M1 if their \(f(y)\) has only two terms, as long as both increase in power by 1. Allow M1 for \(k(y+3)^3\), any numerical \(k\), as the integral of \((y+3)^2\) or M1 for \(k(\frac{1}{2}(y+3))^3\) from \((\frac{1}{2}(y+3))^2\) oe if their power is other than 2 |
| A1 | Obtain \(\frac{1}{12}y^3 + \frac{3}{4}y^2 - \frac{7}{4}y\) aef. Or \(\frac{1}{12}(y+3)^3 - 4y\). A0 if constant term becomes \(-\frac{7}{4}x\) not \(-\frac{7}{4}y\) | |
| B1 | State or imply limits are \(y = 1, 3\). Stated, or just used as limits in definite integral. Allow B1 even if limits used incorrectly (eg wrong order, or addition). Allow B1 even if constant term is \(-\frac{7}{4}x\) (or their \(cx\)) | |
| \(= \frac{15}{4} - (-\frac{11}{12})\) | M1d* | Attempt correct use of limits. Correct order and subtraction. Allow M1 (BOD) if \(y\) limits used in \(-\frac{7}{4}x\) (or their \(cx\)), but M0 if \(x = 0, 5\) used. Minimum of two terms in \(y\). Only term allowed in \(x\) is their \(c\) becoming \(cx\). Allow processing errors eg \((\frac{1}{12} \times 3)^3\) for \(\frac{1}{12} \times 3^3\). Answer is given so M0 if \(\frac{14}{3}\) appears with no evidence of use of limits. Minimum working required is \(\frac{15}{4} + \frac{11}{12}\). Allow M1 if using decimals (0.92 or better for \(\frac{11}{12}\)). M0 if using lower limit as \(y=0\), even if \(y=3\) is also used. Limits must be from attempt at \(y\)-values, so M0 if using 0 and 5 |
| \(= \frac{14}{3}\) AG | A1 | Obtain \(\frac{14}{3}\). Must come from exact working ie fractions or recurring decimals - correct notation required so A0 for 0.9166... A0 if \(-\frac{7}{4}x\) seen in solution |
| [7] | SR for candidates who find the exact area by first integrating onto the \(x\)-axis: B4 obtain area between curve and \(x\)-axis as \(10\frac{1}{3}\); B1 subtract from 15 to obtain \(\frac{14}{3}\); And, if seen in the solution, M1A1 for \(x = f(y)\) as above |
## Question 9(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.5 \times 2.5 \times (1 + 2(-3 + 2\sqrt{6.5}) + 3)$ | M1* | Attempt $y$-values at $x = 0, 2.5, 5$ only. M0 if additional $y$-values found, unless not used. $y_1$ can be exact or decimal (2.1 or better). Allow M1 for using incorrect function as long as still clearly $y$-values intended to be original function eg $-3 + 2\sqrt{x+4}$ (from $\sqrt{(x+4)} = \sqrt{x} + \sqrt{4}$) |
| $= 10.2$ | M1d* | Attempt correct trapezium rule, inc $h = 2.5$. Fully correct structure reqd, including placing of $y$-values. The 'big brackets' must be seen, or implied by later working. Could be implied by stating general rule in terms of $y_0$ etc, as long as these have been attempted elsewhere and clearly labelled. Using $x$-values is M0. Can give M1, even if error in evaluating $y$-values as long correct intention is clear |
| | A1 | Obtain 10.2, or better. Allow answers in range [10.24, 10.25] if $>3$sf. A0 if exact surd value given as final answer. Answer only is 0/3. Using 2 separate trapezia can get full marks. Using anything other than 2 strips of width 2.5 is M0. Using trapezium rule on result of an integration attempt is 0/3 |
| | [3] | |
---
## Question 9(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(5 \times 3) - 10.2 = 4.8$ | M1 | Attempt area of rectangle $-$ their (i). As long as $0 <$ their (i) $< 15$ |
| | A1FT | Obtain 4.8, or better. Allow for exact surd value as well. Allow answers in range [4.75, 4.80] if $> 2$sf |
| | [2] | |
---
## Question 9(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = \frac{1}{4}(y^2 + 6y - 7)$ | M1 | Attempt to write as $x = f(y)$. Must be correct order of operations, but allow slip with inverse operations eg $+/-$, and omitting to square the $\frac{1}{2}$. Allow $y^2 + 9$ from an attempt to square $y+3$, even if $(y+3)^2$ is not seen explicitly first. Allow maximum of 1 error |
| | A1 | Obtain $x = \frac{1}{4}(y^2 + 6y - 7)$ aef. Allow A1 as soon as any correct equation seen in format $x = f(y)$, eg $x = \frac{1}{4}(y+3)^2 - 4$ or $x = \frac{1}{4}(y^2+6y+9) - 4$, and isw subsequent error |
| $\text{area} = \left[\frac{1}{12}y^3 + \frac{3}{4}y^2 - \frac{7}{4}y\right]_1^3$ | M1* | Attempt integration of $f(y)$. Expand bracket and increase in power by 1 for at least two terms (allow if constant term disappears). Independent of rearrangement attempt so M0M1 is possible. Can gain M1 if their $f(y)$ has only two terms, as long as both increase in power by 1. Allow M1 for $k(y+3)^3$, any numerical $k$, as the integral of $(y+3)^2$ or M1 for $k(\frac{1}{2}(y+3))^3$ from $(\frac{1}{2}(y+3))^2$ oe if their power is other than 2 |
| | A1 | Obtain $\frac{1}{12}y^3 + \frac{3}{4}y^2 - \frac{7}{4}y$ aef. Or $\frac{1}{12}(y+3)^3 - 4y$. A0 if constant term becomes $-\frac{7}{4}x$ not $-\frac{7}{4}y$ |
| | B1 | State or imply limits are $y = 1, 3$. Stated, or just used as limits in definite integral. Allow B1 even if limits used incorrectly (eg wrong order, or addition). Allow B1 even if constant term is $-\frac{7}{4}x$ (or their $cx$) |
| $= \frac{15}{4} - (-\frac{11}{12})$ | M1d* | Attempt correct use of limits. Correct order and subtraction. Allow M1 (BOD) if $y$ limits used in $-\frac{7}{4}x$ (or their $cx$), but M0 if $x = 0, 5$ used. Minimum of two terms in $y$. Only term allowed in $x$ is their $c$ becoming $cx$. Allow processing errors eg $(\frac{1}{12} \times 3)^3$ for $\frac{1}{12} \times 3^3$. Answer is given so M0 if $\frac{14}{3}$ appears with no evidence of use of limits. Minimum working required is $\frac{15}{4} + \frac{11}{12}$. Allow M1 if using decimals (0.92 or better for $\frac{11}{12}$). M0 if using lower limit as $y=0$, even if $y=3$ is also used. Limits must be from attempt at $y$-values, so M0 if using 0 and 5 |
| $= \frac{14}{3}$ **AG** | A1 | Obtain $\frac{14}{3}$. Must come from exact working ie fractions or recurring decimals - correct notation required so A0 for 0.9166... A0 if $-\frac{7}{4}x$ seen in solution |
| | [7] | **SR** for candidates who find the exact area by first integrating onto the $x$-axis: **B4** obtain area between curve and $x$-axis as $10\frac{1}{3}$; **B1** subtract from 15 to obtain $\frac{14}{3}$; And, if seen in the solution, **M1A1** for $x = f(y)$ as above |
The pages you've shared only contain **Appendix 1** (Guidance for Marking C2) and the OCR contact/back page — these are **not question-specific mark scheme pages**. They contain only general marking guidance (accuracy rules, extra solutions policy, solving equations guidance, etc.) and no individual question mark allocations.
To extract question-by-question mark scheme content in the format you've requested, I would need the **main mark scheme pages** for the 4722 June 2014 paper, which would typically show:
- Question numbers
- Working/answers
- Mark codes (M1, A1, B1, etc.)
- Examiner guidance notes
Could you share those pages? They would typically appear **before** the Appendix page shown here.
9\\
\includegraphics[max width=\textwidth, alt={}, center]{9e95415c-00f5-4b52-a443-0b946602b3b4-4_387_624_287_717}
The diagram shows part of the curve $y = - 3 + 2 \sqrt { x + 4 }$. The point $P ( 5,3 )$ lies on the curve. Region $A$ is bounded by the curve, the $x$-axis, the $y$-axis and the line $x = 5$. Region $B$ is bounded by the curve, the $y$-axis and the line $y = 3$.\\
(i) Use the trapezium rule, with 2 strips each of width 2.5 , to find an approximate value for the area of region $A$, giving your answer correct to 3 significant figures.\\
(ii) Use your answer to part (i) to deduce an approximate value for the area of region $B$.\\
(iii) By first writing the equation of the curve in the form $x = \mathrm { f } ( y )$, use integration to show that the exact area of region $B$ is $\frac { 14 } { 3 }$.
\section*{END OF QUESTION PAPER}
\section*{OCR $^ { \text {N } }$}
\hfill \mbox{\textit{OCR C2 2014 Q9 [12]}}