| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Sign Change with Function Evaluation |
| Difficulty | Standard +0.3 This is a standard C3 question testing routine techniques: sign change verification (straightforward substitution), rearranging equations (basic algebra with logs), iteration (mechanical application), and integration to find area. Part (ii) of the second question requires understanding that one-one functions don't intersect y=x unless they're self-inverse, leading to an inequality—this adds slight conceptual depth but remains within typical C3 scope. Overall slightly easier than average due to the guided, multi-part structure with standard methods. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Attempt relevant calculations with 5.2 and 5.3 | M1 | |
| Obtain correct values | A1 | \(x\) |
| 5.2 | 2.83 | 2.87 |
| 5.3 | 2.89 | 2.88 |
| Conclude appropriately | A1 | Total: 3 marks [AG; comparing \(y\) values or noting sign change in difference in \(y\) values or equiv] |
| (ii) Equate expressions and attempt rearrangement to \(x =\) | M1 | |
| Obtain \(x = \frac{2}{3}\ln(3x + 8)\) | A1 | Total: 2 marks [AG; necessary detail required] |
| (iii) Obtain correct first iterate | B1 | |
| Carry out correct process to find at least two iterates in all | M1 | |
| Obtain 5.29 | A1 | Total: 3 marks [must be exactly 2 decimal places; 5.2→5.2687→5.2832→5.2863→5.2869; 5.25→5.2793→5.2855→5.2868→5.2870; 5.3→5.2898→5.2877→5.2872→5.2871] |
| (iv) Obtain integral of form \(k(3x + 8)^{\frac{1}{4}}\) | M1 | |
| Obtain integral of form \(ke^{\frac{1}{3}x}\) | M1 | |
| Obtain \(\frac{1}{4}(3x + 8)^{\frac{4}{3}} - 5e^{\frac{1}{3}x}\) | A1 | [or equiv] |
| Apply limits 0 and their answer to (iii) | M1 | [applied to difference of two integrals] |
| Obtain 3.78 | A1 | Total: 5 marks [or greater accuracy] |
**(i)** Attempt relevant calculations with 5.2 and 5.3 | M1 |
Obtain correct values | A1 | $x$ | $y_1$ | $y_2$ | $y_1 - y_2$
| 5.2 | 2.83 | 2.87 | –0.04 |
| 5.3 | 2.89 | 2.88 | 0.006 |
Conclude appropriately | A1 | Total: 3 marks [AG; comparing $y$ values or noting sign change in difference in $y$ values or equiv]
**(ii)** Equate expressions and attempt rearrangement to $x =$ | M1 |
Obtain $x = \frac{2}{3}\ln(3x + 8)$ | A1 | Total: 2 marks [AG; necessary detail required]
**(iii)** Obtain correct first iterate | B1 |
Carry out correct process to find at least two iterates in all | M1 |
Obtain 5.29 | A1 | Total: 3 marks [must be exactly 2 decimal places; 5.2→5.2687→5.2832→5.2863→5.2869; 5.25→5.2793→5.2855→5.2868→5.2870; 5.3→5.2898→5.2877→5.2872→5.2871]
**(iv)** Obtain integral of form $k(3x + 8)^{\frac{1}{4}}$ | M1 |
Obtain integral of form $ke^{\frac{1}{3}x}$ | M1 |
Obtain $\frac{1}{4}(3x + 8)^{\frac{4}{3}} - 5e^{\frac{1}{3}x}$ | A1 | [or equiv]
Apply limits 0 and their answer to (iii) | M1 | [applied to difference of two integrals]
Obtain 3.78 | A1 | Total: 5 marks [or greater accuracy]8\\
\includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-3_588_915_954_614}
The diagram shows part of each of the curves $y = e ^ { \frac { 1 } { 5 } x }$ and $y = \sqrt [ 3 ] { } ( 3 x + 8 )$. The curves meet, as shown in the diagram, at the point $P$. The region $R$, shaded in the diagram, is bounded by the two curves and by the $y$-axis.\\
(i) Show by calculation that the $x$-coordinate of $P$ lies between 5.2 and 5.3.\\
(ii) Show that the $x$-coordinate of $P$ satisfies the equation $x = \frac { 5 } { 3 } \ln ( 3 x + 8 )$.\\
(iii) Use an iterative formula, based on the equation in part (ii), to find the $x$-coordinate of $P$ correct to 2 decimal places.\\
(iv) Use integration, and your answer to part (iii), to find an approximate value of the area of the region $R$.\\
\includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-4_625_647_264_749}
The function f is defined by $\mathrm { f } ( x ) = \sqrt { } ( m x + 7 ) - 4$, where $x \geqslant - \frac { 7 } { m }$ and $m$ is a positive constant. The diagram shows the curve $y = \mathrm { f } ( x )$.\\
(i) A sequence of transformations maps the curve $y = \sqrt { } x$ to the curve $y = \mathrm { f } ( x )$. Give details of these transformations.\\
(ii) Explain how you can tell that f is a one-one function and find an expression for $\mathrm { f } ^ { - 1 } ( x )$.\\
(iii) It is given that the curves $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$ do not meet. Explain how it can be deduced that neither curve meets the line $y = x$, and hence determine the set of possible values of $m$. [5]
\hfill \mbox{\textit{OCR C3 2005 Q8 [13]}}