| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Standard +0.8 This is a two-part implicit differentiation question requiring (a) first derivative verification using product rule and chain rule, and (b) finding the second derivative by differentiating the first derivative implicitly again. While the techniques are standard for Further Maths, the algebraic manipulation for part (b) is moderately demanding, and implicit second derivatives require careful bookkeeping. This is above average difficulty but not exceptional for Further Pure. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3xy^2y' + y^3\) | B1 | Differentiates \(xy^3\). |
| \(-4(x^3y'+3x^2y)=0\) | B1 | Differentiates \(x^3y\). |
| \(3(-1)(1)y'+1^3-4\bigl((-1)^3y'+3(-1)^2(1)\bigr)=0\) leading to \(y'=11\) | B1 | Substitutes \((-1,1)\). AG. \(y'(3xy^2-4x^3)=12x^2y-y^3\) |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3xy^2 - 4x^3)y'' + y'(6xyy' + 3y^2 - 12x^2)\) | B1 B1 | Differentiates \((3xy^2 - 4x^3)y'\) |
| \(+3y^2y' - 12(x^2y' + 2xy) = 0\) | B1 | Differentiates \(y^3 - 12x^2y\) |
| \(3xy^2y'' - 4x^3y'' + 6xy(y')^2 + 6y^2y' - 24x^2y' - 24xy = 0\) | ||
| \(y'' = \frac{(3xy^2 - 4x^3)(12x^2y' + 24xy - 3y^2y') - (12x^2y - y^3)(6xyy' + 3y^2 - 12x^2)}{(3xy^2 - 4x^3)^2}\) | ||
| \(y'' + 11(-75) + 3(11) - 12(9) = 0\) | M1 | Substitutes \((-1, 1)\) |
| \(y'' = 900\) | A1 | |
| 5 |
**Question 3(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3xy^2y' + y^3$ | B1 | Differentiates $xy^3$. |
| $-4(x^3y'+3x^2y)=0$ | B1 | Differentiates $x^3y$. |
| $3(-1)(1)y'+1^3-4\bigl((-1)^3y'+3(-1)^2(1)\bigr)=0$ leading to $y'=11$ | B1 | Substitutes $(-1,1)$. AG. $y'(3xy^2-4x^3)=12x^2y-y^3$ |
| **Total** | **3** | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3xy^2 - 4x^3)y'' + y'(6xyy' + 3y^2 - 12x^2)$ | B1 B1 | Differentiates $(3xy^2 - 4x^3)y'$ |
| $+3y^2y' - 12(x^2y' + 2xy) = 0$ | B1 | Differentiates $y^3 - 12x^2y$ |
| $3xy^2y'' - 4x^3y'' + 6xy(y')^2 + 6y^2y' - 24x^2y' - 24xy = 0$ | | |
| $y'' = \frac{(3xy^2 - 4x^3)(12x^2y' + 24xy - 3y^2y') - (12x^2y - y^3)(6xyy' + 3y^2 - 12x^2)}{(3xy^2 - 4x^3)^2}$ | | |
| $y'' + 11(-75) + 3(11) - 12(9) = 0$ | M1 | Substitutes $(-1, 1)$ |
| $y'' = 900$ | A1 | |
| | **5** | |
---3 The curve $C$ has equation
$$x y ^ { 3 } - 4 x ^ { 3 } y = 3$$
\begin{enumerate}[label=(\alph*)]
\item Show that, at the point $( - 1,1 )$ on $C , \frac { \mathrm { dy } } { \mathrm { dx } } = 11$.
\item Find the value of $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ at the point $( - 1,1 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{37db1c60-0f94-413f-b29b-5872975eee9e-06_535_1584_276_276}
The diagram shows the curve with equation $\mathrm { y } = \frac { \ln \mathrm { x } } { \mathrm { x } ^ { 2 } }$ for $x \geqslant 2$, together with a set of $( N - 2 )$ rectangles\\
of unit width.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q3 [8]}}