| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard techniques. Part (a) involves one application of the product rule and chain rule to verify a given derivative value. Part (b) requires differentiating again and substituting known values. While it's a Further Maths question, the mechanics are routine with no conceptual challenges beyond applying the implicit differentiation algorithm twice, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{d}{dx}\big((x+1)y\big) = (x+1)y' + y\) | B1 | Differentiates \((x+1)y\) correctly. |
| \(\dfrac{d}{dx}\big(y^2\big) = 2yy'\) | B1 | Differentiates \(y^2\) correctly. |
| \((1)y' - 2 + 2(-2)y' = 0\) leading to \(y' = -\dfrac{2}{3}\) | B1 | Substitutes \((0, -2)\), AG. |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((x+1)y''+2y'\) | B1 | Differentiates \((x+1)y'+y\) |
| \(+2yy''+2(y')^2=0\) | B1 | Differentiates \(2yy'\) |
| \(y''+2(-\frac{2}{3})+2(-2)y''+2(-\frac{2}{3})^2=0\) | M1 | Substitutes \((0,-2)\) into their expression with \(y''\) |
| \(y''=-\frac{4}{27}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y'=\frac{-y}{x+1+2y}\), \(\quad y''=\frac{(x+1+2y)(-y')+y(1+2y')}{(x+1+2y)^2}\) | B1 B1 | B1 for \((x+1+2y)(-y')\) in numerator; B1 for \(y(1+2y')\) in numerator |
| \(y''=\frac{(0+1+2(-2))(\frac{2}{3})-2(1+2(-\frac{2}{3}))}{(0+1+2(-2))^2}\) | M1 | Substitutes \((0,-2)\) and \(y'=-\frac{2}{3}\) into their expression |
| \(y''=-\frac{4}{27}\) | A1 | |
| 4 |
## Question 2:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{d}{dx}\big((x+1)y\big) = (x+1)y' + y$ | B1 | Differentiates $(x+1)y$ correctly. |
| $\dfrac{d}{dx}\big(y^2\big) = 2yy'$ | B1 | Differentiates $y^2$ correctly. |
| $(1)y' - 2 + 2(-2)y' = 0$ leading to $y' = -\dfrac{2}{3}$ | B1 | Substitutes $(0, -2)$, AG. |
| **Total: 3** | | |
## Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(x+1)y''+2y'$ | **B1** | Differentiates $(x+1)y'+y$ |
| $+2yy''+2(y')^2=0$ | **B1** | Differentiates $2yy'$ |
| $y''+2(-\frac{2}{3})+2(-2)y''+2(-\frac{2}{3})^2=0$ | **M1** | Substitutes $(0,-2)$ into their expression with $y''$ |
| $y''=-\frac{4}{27}$ | **A1** | |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $y'=\frac{-y}{x+1+2y}$, $\quad y''=\frac{(x+1+2y)(-y')+y(1+2y')}{(x+1+2y)^2}$ | **B1 B1** | B1 for $(x+1+2y)(-y')$ in numerator; B1 for $y(1+2y')$ in numerator |
| $y''=\frac{(0+1+2(-2))(\frac{2}{3})-2(1+2(-\frac{2}{3}))}{(0+1+2(-2))^2}$ | **M1** | Substitutes $(0,-2)$ and $y'=-\frac{2}{3}$ into their expression |
| $y''=-\frac{4}{27}$ | **A1** | |
| | **4** | |
---2 A curve has equation
$$( x + 1 ) y + y ^ { 2 } = 2$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { dy } } { \mathrm { dx } } = - \frac { 2 } { 3 }$ at the point $( 0 , - 2 )$.
\item Find the value of $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ at the point $( 0 , - 2 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q2 [7]}}