| 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 Further Maths implicit differentiation question requiring both first and second derivatives. Part (a) is routine verification using the product rule, but part (b) requires careful application of implicit differentiation twice, substituting known values, and managing algebraic complexity. The second derivative calculation is non-trivial and error-prone, placing it moderately above average difficulty. |
| Spec | 1.07s Parametric and implicit differentiation1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3xy^2y' + y^3\) | B1 | Differentiates \(xy^3\) |
| \(-4(x^3y' + 3x^2y) = 0\) | B1 | Differentiates \(x^3y\) |
| \(3(-1)(1)y' + 1^3 - 4\!\left((-1)^3 y' + 3(-1)^2(1)\right) = 0\) 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 $x^3y$ |
| $3(-1)(1)y' + 1^3 - 4\!\left((-1)^3 y' + 3(-1)^2(1)\right) = 0$ 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]{59982339-c496-4bd7-8dcd-9b257f3afc02-06_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]}}