| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find stationary points - logarithmic functions |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard C3 techniques: verifying a point by substitution, finding stationary points by setting dy/dx=0, using the second derivative test, and integration by parts for ln x. All steps are routine textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x=1\), \(y = 3\ln 1 + 1 - 1^2 = 0\) | E1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{3}{x} + 1 - 2x\) | M1 | \(d/dx(\ln x) = 1/x\) |
| A1cao | ||
| At R, \(\frac{dy}{dx} = 0 \Rightarrow \frac{3}{x} + 1 - 2x = 0\) | M1 | re-arranging into a quadratic \(= 0\); SC1 for \(x=1.5\) unsupported, SC3 if verified |
| \(3 + x - 2x^2 = 0\); \((3-2x)(1+x) = 0\) | M1 | factorising or formula or completing square |
| \(x = 1.5\), (or \(-1\)) | A1 | |
| \(y = 3\ln 1.5 + 1.5 - 1.5^2 = 0.466\) (3 s.f.) | M1, A1cao | substituting their \(x\) |
| \(\frac{d^2y}{dx^2} = -\frac{3}{x^2} - 2\) | B1ft | ft their \(dy/dx\) on equivalent work |
| When \(x=1.5\), \(d^2y/dx^2 = -10/3 < 0 \Rightarrow\) max | E1 [9] | www – don't need to calculate \(10/3\); but condone rounding errors on 0.466 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(u = \ln x\), \(du/dx = 1/x\); \(dv/dx = 1\), \(v = x\) | M1 | parts |
| \(\int \ln x \, dx = x\ln x - \int x \cdot \frac{1}{x} \, dx\) | A1 | |
| \(= x\ln x - \int 1 \, dx = x\ln x - x + c\) | A1 | condone no \(c\); allow correct result to be quoted (SC3) |
| \(A = \int_1^{2.05}(3\ln x + x - x^2)\, dx\) | B1 | correct integral and limits (soi) |
| \(= \left[3x\ln x - 3x + \frac{1}{2}x^2 - \frac{1}{3}x^3\right]_1^{2.05}\) | B1ft | \(\left[3\times\text{their } x\ln x - x' + \frac{1}{2}x^2 - \frac{1}{3}x^3\right]\); substituting correct limits dep 1st B1 |
| \(= -2.5057 + 2.833\ldots = 0.33\) (2 s.f.) | M1dep, A1cao [7] |
## Question 8(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x=1$, $y = 3\ln 1 + 1 - 1^2 = 0$ | E1 [1] | |
---
## Question 8(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{3}{x} + 1 - 2x$ | M1 | $d/dx(\ln x) = 1/x$ |
| | A1cao | |
| At R, $\frac{dy}{dx} = 0 \Rightarrow \frac{3}{x} + 1 - 2x = 0$ | M1 | re-arranging into a quadratic $= 0$; SC1 for $x=1.5$ unsupported, SC3 if verified |
| $3 + x - 2x^2 = 0$; $(3-2x)(1+x) = 0$ | M1 | factorising or formula or completing square |
| $x = 1.5$, (or $-1$) | A1 | |
| $y = 3\ln 1.5 + 1.5 - 1.5^2 = 0.466$ (3 s.f.) | M1, A1cao | substituting their $x$ |
| $\frac{d^2y}{dx^2} = -\frac{3}{x^2} - 2$ | B1ft | ft their $dy/dx$ on equivalent work |
| When $x=1.5$, $d^2y/dx^2 = -10/3 < 0 \Rightarrow$ max | E1 [9] | www – don't need to calculate $10/3$; but condone rounding errors on 0.466 |
---
## Question 8(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $u = \ln x$, $du/dx = 1/x$; $dv/dx = 1$, $v = x$ | M1 | parts |
| $\int \ln x \, dx = x\ln x - \int x \cdot \frac{1}{x} \, dx$ | A1 | |
| $= x\ln x - \int 1 \, dx = x\ln x - x + c$ | A1 | condone no $c$; allow correct result to be quoted (SC3) |
| $A = \int_1^{2.05}(3\ln x + x - x^2)\, dx$ | B1 | correct integral and limits (soi) |
| $= \left[3x\ln x - 3x + \frac{1}{2}x^2 - \frac{1}{3}x^3\right]_1^{2.05}$ | B1ft | $\left[3\times\text{their } x\ln x - x' + \frac{1}{2}x^2 - \frac{1}{3}x^3\right]$; substituting correct limits dep 1st B1 |
| $= -2.5057 + 2.833\ldots = 0.33$ (2 s.f.) | M1dep, A1cao [7] | |
---8 Fig. 8 shows the curve $y = 3 \ln x + x - x ^ { 2 }$.\\
The curve crosses the $x$-axis at P and Q , and has a turning point at R . The $x$-coordinate of Q is approximately 2.05 .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{30d0d728-d6d6-4a54-baf9-a6df8646bf64-3_730_841_561_651}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Verify that the coordinates of P are $( 1,0 )$.\\
(ii) Find the coordinates of R , giving the $y$-coordinate correct to 3 significant figures.
Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$, and use this to verify that R is a maximum point.\\
(iii) Find $\int \ln x \mathrm {~d} x$.
Hence calculate the area of the region enclosed by the curve and the $x$-axis between P and Q , giving your answer to 2 significant figures.
\hfill \mbox{\textit{OCR MEI C3 2010 Q8 [17]}}