| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Standard +0.3 This is a standard C3 quotient rule question with routine follow-up parts. Part (i) requires straightforward application of the quotient rule to verify a given derivative. Parts (ii)-(v) involve standard techniques: evaluating endpoints for range, finding stationary points of the derivative, understanding inverse functions, and algebraic manipulation to find an inverse. All steps are procedural with no novel insight required, making this slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f'(x) = \frac{(x^2+1)4x-(2x^2-1)2x}{(x^2+1)^2}\) | M1, A1 | Quotient or product rule; correct expression |
| \(= \frac{4x^3+4x-4x^3+2x}{(x^2+1)^2} = \frac{6x}{(x^2+1)^2}\) * | E1 | www |
| When \(x>0\), \(6x>0\) and \((x^2+1)^2 > 0 \Rightarrow f'(x)>0\) | M1, E1 [5] | attempt to show/solve \(f'(x)>0\); numerator \(>0\) and denominator \(>0\), or solving \(6x>0 \Rightarrow x>0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(2) = \frac{8-1}{4+1} = 1\frac{2}{5}\) | B1 | |
| Range is \(-1 \leq y \leq 1\frac{2}{5}\) | B1 [2] | must be \(\leq\), \(y\) or \(f(x)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f'(x)\) max when \(f''(x)=0 \Rightarrow 6-18x^2=0\) | M1 | |
| \(x^2 = 1/3\), \(x = 1/\sqrt{3}\) | A1, M1 | \((\pm)1/\sqrt{3}\) oe (0.577 or better) |
| \(f'(x) = \frac{6/\sqrt{3}}{(1\frac{1}{3})^2} = \frac{6}{\sqrt{3}}\cdot\frac{9}{16} = \frac{9\sqrt{3}}{8} = 1.95\) | A1 [4] | substituting \(1/\sqrt{3}\) into \(f'(x)\); \(9\sqrt{3}/8\) o.e. or 1.95 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Domain is \(-1 < x < 1\frac{2}{5}\) | B1 | ft their 1.4 but not \(x \geq -1\) |
| Range is \(0 \leq y \leq 2\) | B1 | or \(0 \leq g(x) \leq 2\) (not f) |
| [Graph: reflection in \(y=x\)] | M1, A1cao [4] | Reasonable reflection in \(y=x\); from \((-1,0)\) to \((1.4, 2)\), through \((0, \sqrt{2}/2)\); allow omission of one of \(-1, 1.4, 2, \sqrt{2}/2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{2x^2-1}{x^2+1}\), swap \(x \leftrightarrow y\): \(x = \frac{2y^2-1}{y^2+1}\) | M1 | Attempt to invert (could start from g) |
| \(xy^2+x = 2y^2-1 \Rightarrow x+1 = 2y^2-xy^2 = y^2(2-x)\) | M1 | clearing fractions |
| \(y^2 = \frac{x+1}{2-x}\) | M1 | collecting terms in \(y^2\) and factorising |
| \(y = \sqrt{\frac{x+1}{2-x}}\) * | E1 [4] | www |
# Question 9:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x) = \frac{(x^2+1)4x-(2x^2-1)2x}{(x^2+1)^2}$ | M1, A1 | Quotient or product rule; correct expression |
| $= \frac{4x^3+4x-4x^3+2x}{(x^2+1)^2} = \frac{6x}{(x^2+1)^2}$ * | E1 | www |
| When $x>0$, $6x>0$ and $(x^2+1)^2 > 0 \Rightarrow f'(x)>0$ | M1, E1 [5] | attempt to show/solve $f'(x)>0$; numerator $>0$ and denominator $>0$, or solving $6x>0 \Rightarrow x>0$ |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(2) = \frac{8-1}{4+1} = 1\frac{2}{5}$ | B1 | |
| Range is $-1 \leq y \leq 1\frac{2}{5}$ | B1 [2] | must be $\leq$, $y$ or $f(x)$ |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x)$ max when $f''(x)=0 \Rightarrow 6-18x^2=0$ | M1 | |
| $x^2 = 1/3$, $x = 1/\sqrt{3}$ | A1, M1 | $(\pm)1/\sqrt{3}$ oe (0.577 or better) |
| $f'(x) = \frac{6/\sqrt{3}}{(1\frac{1}{3})^2} = \frac{6}{\sqrt{3}}\cdot\frac{9}{16} = \frac{9\sqrt{3}}{8} = 1.95$ | A1 [4] | substituting $1/\sqrt{3}$ into $f'(x)$; $9\sqrt{3}/8$ o.e. or 1.95 or better |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Domain is $-1 < x < 1\frac{2}{5}$ | B1 | ft their 1.4 but not $x \geq -1$ |
| Range is $0 \leq y \leq 2$ | B1 | or $0 \leq g(x) \leq 2$ (not f) |
| [Graph: reflection in $y=x$] | M1, A1cao [4] | Reasonable reflection in $y=x$; from $(-1,0)$ to $(1.4, 2)$, through $(0, \sqrt{2}/2)$; allow omission of one of $-1, 1.4, 2, \sqrt{2}/2$ |
## Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{2x^2-1}{x^2+1}$, swap $x \leftrightarrow y$: $x = \frac{2y^2-1}{y^2+1}$ | M1 | Attempt to invert (could start from g) |
| $xy^2+x = 2y^2-1 \Rightarrow x+1 = 2y^2-xy^2 = y^2(2-x)$ | M1 | clearing fractions |
| $y^2 = \frac{x+1}{2-x}$ | M1 | collecting terms in $y^2$ and factorising |
| $y = \sqrt{\frac{x+1}{2-x}}$ * | E1 [4] | www |9 Fig. 9 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 2 x ^ { 2 } - 1 } { x ^ { 2 } + 1 }$ for the domain $0 \leqslant x \leqslant 2$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b3e20ee-457c-46be-b2e5-12573bee2fbf-4_974_1211_358_466}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
(i) Show that $\mathrm { f } ^ { \prime } ( x ) = \frac { 6 x } { \left( x ^ { 2 } + 1 \right) ^ { 2 } }$, and hence that $\mathrm { f } ( x )$ is an increasing function for $x > 0$.\\
(ii) Find the range of $\mathrm { f } ( x )$.\\
(iii) Given that $\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 6 - 18 x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 3 } }$, find the maximum value of $\mathrm { f } ^ { \prime } ( x )$.
The function $\mathrm { g } ( x )$ is the inverse function of $\mathrm { f } ( x )$.\\
(iv) Write down the domain and range of $\mathrm { g } ( x )$. Add a sketch of the curve $y = \mathrm { g } ( x )$ to a copy of Fig. 9 .\\
(v) Show that $\mathrm { g } ( x ) = \sqrt { \frac { x + 1 } { 2 - x } }$.
\hfill \mbox{\textit{OCR MEI C3 2010 Q9 [19]}}