| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Moderate -0.3 This is a standard multi-part C3 question testing quotient rule differentiation, stationary points, odd/even functions, and integration by substitution. All parts follow routine procedures with no novel problem-solving required. Part (i) is a 'show that' derivative calculation, parts (ii-iii) are textbook applications, and parts (iv-v) use standard u-substitution. Slightly easier than average due to the guided 'show that' structure and straightforward algebraic manipulations. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(x) = \frac{x}{x^2+1} \Rightarrow f'(x) = \frac{(x^2+1)\cdot 1 - x\cdot 2x}{(x^2+1)^2}\) | M1, A1 | Formula, middle section |
| \(= \frac{1-x^2}{(x^2+1)^2}\) | E1 | answer |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f'(x) = \frac{1-x^2}{(x^2+1)^2} = 0 \Rightarrow 1-x^2 = 0 \Rightarrow x = \pm 1\) | ||
| When \(x=1\), \(f(x) = \frac{1}{1+1} = \frac{1}{2}\), i.e. \(\left(1, \frac{1}{2}\right)\) | E1 | Substitute, find \(f(x)\) |
| Other stationary point: \(x=-1\), \(f(x) = \frac{-1}{1+1} = -\frac{1}{2}\), i.e. \(\left(-1, -\frac{1}{2}\right)\) | B1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The graph is odd. | B1 | |
| \(f(-x) = \frac{-x}{(-x)^2+1} = -\frac{x}{x^2+1} = -f(x)\) | B1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(u = x^2 \Rightarrow du = 2x\,dx\) | M1 | |
| When \(x=1\), \(u=1\); when \(x=4\), \(u=16\) | B1, B1 | |
| \(\Rightarrow \int_1^4 \frac{x}{x^2+1}\,dx = \int_1^{16}\frac{1}{u+1}\cdot\frac{1}{2}\,du\) | A1, A1 | |
| \(\Rightarrow a=1,\ b=16,\ k=\frac{1}{2}\) | ||
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Because the function is odd, the area in \([-1,0]\) is equal in magnitude but opposite in sign to the area in \([0,1]\) | B1 | |
| Shaded area \(= \int_1^4\frac{x}{x^2+1}\,dx + 2\int_0^1\frac{x}{x^2+1}\,dx\) | M1 | |
| \(= \int_1^{16}\frac{1}{u+1}\cdot\frac{1}{2}\,du + 2\int_0^1\frac{1}{u+1}\cdot\frac{1}{2}\,du\) | A1 | |
| \(= \frac{1}{2}\left[\ln(u+1)\right]_1^{16} + 2\cdot\frac{1}{2}\left[\ln(u+1)\right]_0^1\) | M1, A1 | \(\ln(u+1)\) |
| \(= \frac{1}{2}\ln\frac{17}{2} + 2\cdot\frac{1}{2}\ln 2 = \frac{1}{2}\left(\ln\frac{17}{2}+\ln 4\right) = \frac{1}{2}\ln 34 \approx 1.763\) | A1 | |
| Total: 6 |
## Question 8:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x) = \frac{x}{x^2+1} \Rightarrow f'(x) = \frac{(x^2+1)\cdot 1 - x\cdot 2x}{(x^2+1)^2}$ | M1, A1 | Formula, middle section |
| $= \frac{1-x^2}{(x^2+1)^2}$ | E1 | answer |
| **Total: 3** | | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f'(x) = \frac{1-x^2}{(x^2+1)^2} = 0 \Rightarrow 1-x^2 = 0 \Rightarrow x = \pm 1$ | | |
| When $x=1$, $f(x) = \frac{1}{1+1} = \frac{1}{2}$, i.e. $\left(1, \frac{1}{2}\right)$ | E1 | Substitute, find $f(x)$ |
| Other stationary point: $x=-1$, $f(x) = \frac{-1}{1+1} = -\frac{1}{2}$, i.e. $\left(-1, -\frac{1}{2}\right)$ | B1 | |
| **Total: 2** | | |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| The graph is odd. | B1 | |
| $f(-x) = \frac{-x}{(-x)^2+1} = -\frac{x}{x^2+1} = -f(x)$ | B1 | |
| **Total: 2** | | |
### Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $u = x^2 \Rightarrow du = 2x\,dx$ | M1 | |
| When $x=1$, $u=1$; when $x=4$, $u=16$ | B1, B1 | |
| $\Rightarrow \int_1^4 \frac{x}{x^2+1}\,dx = \int_1^{16}\frac{1}{u+1}\cdot\frac{1}{2}\,du$ | A1, A1 | |
| $\Rightarrow a=1,\ b=16,\ k=\frac{1}{2}$ | | |
| **Total: 5** | | |
### Part (v):
| Answer | Mark | Guidance |
|--------|------|----------|
| Because the function is odd, the area in $[-1,0]$ is equal in magnitude but opposite in sign to the area in $[0,1]$ | B1 | |
| Shaded area $= \int_1^4\frac{x}{x^2+1}\,dx + 2\int_0^1\frac{x}{x^2+1}\,dx$ | M1 | |
| $= \int_1^{16}\frac{1}{u+1}\cdot\frac{1}{2}\,du + 2\int_0^1\frac{1}{u+1}\cdot\frac{1}{2}\,du$ | A1 | |
| $= \frac{1}{2}\left[\ln(u+1)\right]_1^{16} + 2\cdot\frac{1}{2}\left[\ln(u+1)\right]_0^1$ | M1, A1 | $\ln(u+1)$ |
| $= \frac{1}{2}\ln\frac{17}{2} + 2\cdot\frac{1}{2}\ln 2 = \frac{1}{2}\left(\ln\frac{17}{2}+\ln 4\right) = \frac{1}{2}\ln 34 \approx 1.763$ | A1 | |
| **Total: 6** | | |
---8 You are given that $\mathrm { f } ( x ) = \frac { x } { x ^ { 2 } + 1 }$ for all real values of $x$.\\
(i) Show that $\mathrm { f } ^ { \prime } ( x ) = \frac { 1 - x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 2 } }$.\\
(ii) Hence show that there is a stationary value at $\left( 1 , \frac { 1 } { 2 } \right)$ and find the coordinates of the other stationary point.\\
(iii) The graph of the curve is shown in Fig. 8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-3_518_892_1612_705}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
State whether the curve is odd or even and prove the result algebraically.\\
(iv) Show that $\int _ { 1 } ^ { 4 } \frac { x } { x ^ { 2 } + 1 } \mathrm {~d} x = \int _ { a } ^ { b } k \frac { 1 } { u + 1 } \mathrm {~d} u$, where the values of $a , b$ and $k$ are to be determined.\\
(v) Hence find the area of the shaded region in Fig. 8.
\hfill \mbox{\textit{OCR MEI C3 Q8 [18]}}