| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Show dy/dx equals given expression |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C3 techniques (odd functions, quotient rule differentiation, integration, implicit differentiation). Each part is routine: (i) is algebraic verification, (ii) is quotient rule application, (iii) is standard integration, (iv) is algebraic manipulation then implicit differentiation. While multi-step, all techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| 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 integrals1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(-x) = \frac{-x}{\sqrt{2+(-x)^2}}\) | M1 | Substituting \(-x\) for \(x\) in \(f(x)\); \(\frac{-x}{\sqrt{2+-x^2}}, \frac{-x}{\sqrt{2+-(x^2)}}, \frac{-x}{\sqrt{2+(-x)^2}}\) M1A0 |
| \(= -\frac{x}{\sqrt{2+x^2}} = -f(x)\) | A1 | 1st line must be shown, must have \(f(-x) = -f(x)\) oe somewhere; \(\frac{-x}{\sqrt{2-x^2}}\) M0A0 |
| Rotational symmetry of order 2 about O | B1 [3] | 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 |
| 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 |
| M1 | \(\frac{1}{2}u^{-1/2}\) or \(-\frac{1}{2}v^{-3/2}\) soi | |
| Correct expression | A1 | \(x(-1/2)(2+x^2)^{-3/2}\cdot 2x + (2+x^2)^{-1/2}\) \(= (2+x^2)^{-3/2}(-x^2+2+x^2)\) |
| \(= \frac{2+x^2-x^2}{(2+x^2)^{3/2}} = \frac{2}{(2+x^2)^{3/2}}\) * | A1 NB AG | |
| When \(x=0\), \(f'(x) = \frac{2}{2^{3/2}} = \frac{1}{\sqrt{2}}\) | B1 [5] | oe e.g. \(\frac{\sqrt{2}}{2}\), \(2^{-1/2}\), \(\frac{1}{2^{1/2}}\), but not \(\frac{2}{2^{3/2}}\); allow isw on these seen |
| 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 \(u = 2+x^2\), \(du = 2x\,dx\) | ||
| \(= \int_2^3 \frac{1}{2}\cdot\frac{1}{\sqrt{u}}\,du\) | M1 | \(\int\frac{1}{2}\cdot\frac{1}{\sqrt{u}}\,du\) or \(= \int 1[dv]\) or \(k(2+x^2)^{1/2}\); condone no \(du\) or \(dv\), but not \(\int\frac{1}{2}\cdot\frac{1}{\sqrt{u}}\,dx\) |
| \(= \left[u^{1/2}\right]_2^3\) | A1 | \([u^{1/2}]\) oe (but not \(1/u^{-1/2}\)) or \([v]\) or \(k=1\) |
| \(= \sqrt{3} - \sqrt{2}\) | A1cao [4] | Must be exact; isw approximations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y^2 = \frac{x^2}{2+x^2}\) | M1 | Squaring (correctly); must show \(\left[\sqrt{(2+x^2)}\right]^2 + 2 + x^2\) (oe) |
| \(\Rightarrow 1/y^2 = (2+x^2)/x^2 = 2/x^2 + 1\) * | A1 [2] | Or equivalent algebra NB AG; if argued backwards from given result without error, SCB1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-2y^{-3}\frac{dy}{dx} = -4x^{-3}\) | B1B1 | LHS, RHS; condone \(\frac{dy}{dx} - 2y^{-3}\) unless pursued |
| \(\Rightarrow \frac{dy}{dx} = -4x^{-3}/-2y^{-3} = 2y^3/x^3\) * | B1 NB AG | |
| Not possible to substitute \(x=0\) and \(y=0\) into this expression | B1 [4] | soi (e.g. mention of \(0/0\)); condone 'can't substitute \(x=0\)' oe (need not mention \(y=0\)); condone also 'division by 0 is infinite' |
## Question 3:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(-x) = \frac{-x}{\sqrt{2+(-x)^2}}$ | M1 | Substituting $-x$ for $x$ in $f(x)$; $\frac{-x}{\sqrt{2+-x^2}}, \frac{-x}{\sqrt{2+-(x^2)}}, \frac{-x}{\sqrt{2+(-x)^2}}$ M1A0 |
| $= -\frac{x}{\sqrt{2+x^2}} = -f(x)$ | A1 | 1st line must be shown, must have $f(-x) = -f(x)$ oe somewhere; $\frac{-x}{\sqrt{2-x^2}}$ M0A0 |
| Rotational symmetry of order 2 about O | B1 [3] | 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 |
### 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 |
| | M1 | $\frac{1}{2}u^{-1/2}$ or $-\frac{1}{2}v^{-3/2}$ soi |
| Correct expression | A1 | $x(-1/2)(2+x^2)^{-3/2}\cdot 2x + (2+x^2)^{-1/2}$ $= (2+x^2)^{-3/2}(-x^2+2+x^2)$ |
| $= \frac{2+x^2-x^2}{(2+x^2)^{3/2}} = \frac{2}{(2+x^2)^{3/2}}$ * | A1 **NB AG** | |
| When $x=0$, $f'(x) = \frac{2}{2^{3/2}} = \frac{1}{\sqrt{2}}$ | B1 [5] | oe e.g. $\frac{\sqrt{2}}{2}$, $2^{-1/2}$, $\frac{1}{2^{1/2}}$, but not $\frac{2}{2^{3/2}}$; allow isw on these seen |
### 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 $u = 2+x^2$, $du = 2x\,dx$ | | |
| $= \int_2^3 \frac{1}{2}\cdot\frac{1}{\sqrt{u}}\,du$ | M1 | $\int\frac{1}{2}\cdot\frac{1}{\sqrt{u}}\,du$ or $= \int 1[dv]$ or $k(2+x^2)^{1/2}$; condone no $du$ or $dv$, but not $\int\frac{1}{2}\cdot\frac{1}{\sqrt{u}}\,dx$ |
| $= \left[u^{1/2}\right]_2^3$ | A1 | $[u^{1/2}]$ oe (but not $1/u^{-1/2}$) or $[v]$ or $k=1$ |
| $= \sqrt{3} - \sqrt{2}$ | A1cao [4] | Must be exact; isw approximations |
### Part (iv)(A):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y^2 = \frac{x^2}{2+x^2}$ | M1 | Squaring (correctly); must show $\left[\sqrt{(2+x^2)}\right]^2 + 2 + x^2$ (oe) |
| $\Rightarrow 1/y^2 = (2+x^2)/x^2 = 2/x^2 + 1$ * | A1 [2] | Or equivalent algebra **NB AG**; if argued backwards from given result without error, SCB1 |
### Part (iv)(B):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-2y^{-3}\frac{dy}{dx} = -4x^{-3}$ | B1B1 | LHS, RHS; condone $\frac{dy}{dx} - 2y^{-3}$ unless pursued |
| $\Rightarrow \frac{dy}{dx} = -4x^{-3}/-2y^{-3} = 2y^3/x^3$ * | B1 **NB AG** | |
| Not possible to substitute $x=0$ and $y=0$ into this expression | B1 [4] | soi (e.g. mention of $0/0$); condone 'can't substitute $x=0$' oe (need not mention $y=0$); condone also 'division by 0 is infinite' |3 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]{e0636807-d5bf-43c2-a484-68245e639cee-3_476_674_498_708}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Show algebraically that $\mathrm { f } ( x )$ is an odd function. Interpret this result geometrically.
\item 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.
\item Find the exact area of the region bounded by the curve, the $x$-axis and the line $x = 1$.
\item (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.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 Q3 [18]}}