| 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 | Standard +0.3 This is a multi-part question requiring quotient rule differentiation, odd function verification, and implicit differentiation. While it has several parts, each component is relatively standard C3 material with clear guidance ('show that' format). The algebraic manipulation is straightforward, and the implicit differentiation in part (iv) follows a structured approach. Slightly easier than average due to the scaffolded nature and routine techniques. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(-x) = \frac{-x}{\sqrt{2+(-x)^2}}\) | M1 | Substituting \(-x\) for \(x\) in \(f(x)\). Note: \(\frac{-x}{\sqrt{2+-x^2}}\), \(\frac{-x}{\sqrt{2+-(x^2)}}\) get M1A0. \(\frac{-x}{\sqrt{2-x^2}}\) gets M0A0 |
| \(= -\frac{x}{\sqrt{2+x^2}} = -f(x)\) | A1 | 1st line must be shown, must have \(f(-x) = -f(x)\) oe somewhere |
| Rotational symmetry of order 2 about O | B1 | Must have 'rotate' and 'O' and 'order 2 or 180 or \(\frac{1}{2}\) turn'. oe e.g. reflections in both \(x\)- and \(y\)-axes |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f'(x) = \frac{\sqrt{2+x^2}\cdot 1 - x\cdot\frac{1}{2}(2+x^2)^{-1/2}\cdot 2x}{(\sqrt{2+x^2})^2}\) | M1 | Quotient or product rule used. QR: condone \(udv \pm vdu\), but \(u\), \(v\) and denom must be correct |
| M1 | \(\frac{1}{2}u^{-1/2}\) or \(-\frac{1}{2}v^{-3/2}\) soi. \(x(-1/2)(2+x^2)^{-3/2}\cdot 2x + (2+x^2)^{-1/2}\) | |
| A1 | Correct expression. \(= (2+x^2)^{-3/2}(-x^2+2+x^2)\) | |
| \(= \frac{2+x^2-x^2}{(2+x^2)^{3/2}} = \frac{2}{(2+x^2)^{3/2}}\) | A1 | NB AG |
| When \(x=0\), \(f'(x) = 2/2^{3/2} = 1/\sqrt{2}\) | B1 | oe e.g. \(\sqrt{2}/2\), \(2^{-1/2}\), \(1/2^{1/2}\), but not \(2/2^{3/2}\). Allow isw on these seen |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = \int_0^1 \frac{x}{\sqrt{2+x^2}}\,dx\) | B1 | Correct integral and limits. Limits may be inferred from subsequent working, condone no \(dx\) |
| Let \(u = 2+x^2\), \(du = 2x\,dx\) | or \(v = \sqrt{2+x^2}\), \(dv = x(2+x^2)^{-1/2}dx\) | |
| \(= \int_2^3 \frac{1}{2}\cdot\frac{1}{\sqrt{u}}\,du\) | M1 | \(\int\frac{1}{2}\frac{1}{\sqrt{u}}[du]\) or \(= \int 1[dv]\) or \(k(2+x^2)^{1/2}\). Condone no \(du\) or \(dv\), but not \(\int\frac{1}{2}\frac{1}{\sqrt{u}}\,dx\) |
| \(= \left[u^{1/2}\right]_2^3\) | A1 | \([u^{1/2}]\) o.e. (but not \(1/u^{-1/2}\)) or \([v]\) or \(k=1\) |
| \(= \sqrt{3} - \sqrt{2}\) | A1cao | Must be exact. isw approximations |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y^2 = \frac{x^2}{2+x^2}\) | M1 | Squaring (correctly). Must show \(\left[\sqrt{(2+x^2)}\right]^2 = 2+x^2\) (o.e.) |
| \(\Rightarrow 1/y^2 = (2+x^2)/x^2 = 2/x^2 + 1\) | A1 | Or equivalent algebra NB AG. If argued backwards from given result without error, SCB1 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-2y^{-3}\,dy/dx = -4x^{-3}\) | B1B1 | LHS, RHS. Condone \(dy/dx\ {-2y^{-3}}\) unless pursued |
| \(\Rightarrow dy/dx = -4x^{-3}/-2y^{-3} = 2y^3/x^3\) | B1 | NB AG |
| Not possible to substitute \(x=0\) and \(y=0\) into this expression | B1 | soi (e.g. mention of \(0/0\)). Condone 'can't substitute \(x=0\)' o.e. (i.e. need not mention \(y=0\)). Condone also 'division by 0 is infinite' |
| [4] |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(-x) = \frac{-x}{\sqrt{2+(-x)^2}}$ | M1 | Substituting $-x$ for $x$ in $f(x)$. Note: $\frac{-x}{\sqrt{2+-x^2}}$, $\frac{-x}{\sqrt{2+-(x^2)}}$ get M1A0. $\frac{-x}{\sqrt{2-x^2}}$ gets M0A0 |
| $= -\frac{x}{\sqrt{2+x^2}} = -f(x)$ | A1 | 1st line must be shown, must have $f(-x) = -f(x)$ oe somewhere |
| Rotational symmetry of order 2 about O | B1 | Must have 'rotate' and 'O' and 'order 2 or 180 or $\frac{1}{2}$ turn'. oe e.g. reflections in both $x$- and $y$-axes |
| **[3]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x) = \frac{\sqrt{2+x^2}\cdot 1 - x\cdot\frac{1}{2}(2+x^2)^{-1/2}\cdot 2x}{(\sqrt{2+x^2})^2}$ | M1 | Quotient or product rule used. QR: condone $udv \pm vdu$, but $u$, $v$ and denom must be correct |
| | M1 | $\frac{1}{2}u^{-1/2}$ or $-\frac{1}{2}v^{-3/2}$ soi. $x(-1/2)(2+x^2)^{-3/2}\cdot 2x + (2+x^2)^{-1/2}$ |
| | A1 | Correct expression. $= (2+x^2)^{-3/2}(-x^2+2+x^2)$ |
| $= \frac{2+x^2-x^2}{(2+x^2)^{3/2}} = \frac{2}{(2+x^2)^{3/2}}$ | A1 | **NB AG** |
| When $x=0$, $f'(x) = 2/2^{3/2} = 1/\sqrt{2}$ | B1 | oe e.g. $\sqrt{2}/2$, $2^{-1/2}$, $1/2^{1/2}$, but not $2/2^{3/2}$. Allow isw on these seen |
| **[5]** | | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = \int_0^1 \frac{x}{\sqrt{2+x^2}}\,dx$ | B1 | Correct integral and limits. Limits may be inferred from subsequent working, condone no $dx$ |
| Let $u = 2+x^2$, $du = 2x\,dx$ | | or $v = \sqrt{2+x^2}$, $dv = x(2+x^2)^{-1/2}dx$ |
| $= \int_2^3 \frac{1}{2}\cdot\frac{1}{\sqrt{u}}\,du$ | M1 | $\int\frac{1}{2}\frac{1}{\sqrt{u}}[du]$ or $= \int 1[dv]$ or $k(2+x^2)^{1/2}$. Condone no $du$ or $dv$, but not $\int\frac{1}{2}\frac{1}{\sqrt{u}}\,dx$ |
| $= \left[u^{1/2}\right]_2^3$ | A1 | $[u^{1/2}]$ o.e. (but not $1/u^{-1/2}$) or $[v]$ or $k=1$ |
| $= \sqrt{3} - \sqrt{2}$ | A1cao | Must be exact. isw approximations |
| **[4]** | | |
### Part (iv)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y^2 = \frac{x^2}{2+x^2}$ | M1 | Squaring (correctly). Must show $\left[\sqrt{(2+x^2)}\right]^2 = 2+x^2$ (o.e.) |
| $\Rightarrow 1/y^2 = (2+x^2)/x^2 = 2/x^2 + 1$ | A1 | Or equivalent algebra **NB AG**. If argued backwards from given result without error, SCB1 |
| **[2]** | | |
### Part (iv)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-2y^{-3}\,dy/dx = -4x^{-3}$ | B1B1 | LHS, RHS. Condone $dy/dx\ {-2y^{-3}}$ unless pursued |
| $\Rightarrow dy/dx = -4x^{-3}/-2y^{-3} = 2y^3/x^3$ | B1 | **NB AG** |
| Not possible to substitute $x=0$ and $y=0$ into this expression | B1 | soi (e.g. mention of $0/0$). Condone 'can't substitute $x=0$' o.e. (i.e. need not mention $y=0$). Condone also 'division by 0 is infinite' |
| **[4]** | | |
---2 Fig. 8 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 + x ^ { 2 } } }$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{93ee09be-f014-4dd7-a8da-8646837b17a5-1_471_674_761_719}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Show algebraically that $\mathrm { f } ( x )$ is an odd function. Interpret this result geometrically.\\
(ii) Show that $\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \left( 2 + x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$. Hence find the exact gradient of the curve at the origin.\\
(iii) Find the exact area of the region bounded by the curve, the $x$-axis and the line $x = 1$.\\
(iv) $( A )$ Show that if $y = \frac { x } { \sqrt { 2 + x ^ { 2 } } }$, then $\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1$.\\
(B) Differentiate $\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1$ implicitly to show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y ^ { 3 } } { x ^ { 3 } }$. Explain why this expression cannot be used to find the gradient of the curve at the origin.
\hfill \mbox{\textit{OCR MEI C3 Q2 [18]}}