| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Difficulty | Standard +0.3 This is a standard C3 question testing quotient rule differentiation, finding x-intercepts, and volume of revolution. Part (i) is routine verification, part (ii) requires finding where y=0 (straightforward), and part (iii) is a standard volume of revolution integral with substitution u=4ln(x)+3. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07q Product and quotient rules: differentiation4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Attempt use of quotient rule | M1 | allow for numerator 'wrong way round'; or equiv |
| Obtain \(\frac{(4\ln x + 3) \frac{1}{4} - (4\ln x - 3) \frac{1}{4}}{(4\ln x + 3)^2}\) | A1 | or equiv |
| Confirm \(\frac{24}{x(4\ln x + 3)^2}\) | A1 | 3 AG; necessary detail required |
| (ii) Identify \(\ln x = \frac{3}{4}\) | B1 | or equiv |
| State or imply \(x = e^{\frac{3}{4}}\) | B1 | |
| Substitute \(e^{\frac{3}{4}}\) completely in expression for derivative | M1 | and deal with \(\ln e^{\frac{3}{4}}\) term |
| Obtain \(\frac{3}{2}e^{-\frac{3}{4}}\) | A1 | 4 or exact (single term) equiv |
| (iii) State or imply \(\int \frac{4\pi}{x(4\ln x + 3)^2} dx\) | B1 | |
| Obtain integral of form \(k \frac{4\ln x - 3}{4\ln x + 3}\) | ||
| or \(k(4\ln x + 3)^{-1}\) | *M1 | any constant \(k\) |
| Substitute both limits and subtract right way round | M1 | dep *M |
| Obtain \(\frac{4}{21}\pi\) | A1 | 4 or exact equiv |
(i) Attempt use of quotient rule | M1 | allow for numerator 'wrong way round'; or equiv
Obtain $\frac{(4\ln x + 3) \frac{1}{4} - (4\ln x - 3) \frac{1}{4}}{(4\ln x + 3)^2}$ | A1 | or equiv
Confirm $\frac{24}{x(4\ln x + 3)^2}$ | A1 | 3 AG; necessary detail required
(ii) Identify $\ln x = \frac{3}{4}$ | B1 | or equiv
State or imply $x = e^{\frac{3}{4}}$ | B1 |
Substitute $e^{\frac{3}{4}}$ completely in expression for derivative | M1 | and deal with $\ln e^{\frac{3}{4}}$ term
Obtain $\frac{3}{2}e^{-\frac{3}{4}}$ | A1 | 4 or exact (single term) equiv
(iii) State or imply $\int \frac{4\pi}{x(4\ln x + 3)^2} dx$ | B1 |
Obtain integral of form $k \frac{4\ln x - 3}{4\ln x + 3}$ | |
or $k(4\ln x + 3)^{-1}$ | *M1 | any constant $k$
Substitute both limits and subtract right way round | M1 | dep *M
Obtain $\frac{4}{21}\pi$ | A1 | 4 or exact equiv
---8 (i) Given that $\mathrm { y } = \frac { 4 \ln \mathrm { x } - 3 } { 4 \ln \mathrm { x } + 3 }$, show that $\frac { \mathrm { dy } } { \mathrm { dx } } = \frac { 24 } { \mathrm { x } ( 4 \ln \mathrm { x } + 3 ) ^ { 2 } }$.\\
(ii) Find the exact value of the gradient of the curve $y = \frac { 4 \ln x - 3 } { 4 \ln x + 3 }$ at the point where it crosses the $x$-axis.\\
(iii)\\
\includegraphics[max width=\textwidth, alt={}, center]{133c38fb-307f-4f20-86cb-1bd57cc4f870-3_524_830_941_699}
The diagram shows part of the curve with equation
$$\mathrm { y } = \frac { 2 } { \mathrm { x } ^ { \frac { 1 } { 2 } } ( 4 \ln \mathrm { x } + 3 ) }$$
The region shaded in the diagram is bounded by the curve and the lines $x = 1 , x = e$ and $y = 0$. Find the exact volume of the solid produced when this shaded region is rotated completely about the x -axis.
\hfill \mbox{\textit{OCR C3 2007 Q8 [11]}}