| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| 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 Further Maths implicit differentiation question requiring standard techniques: differentiate cos y = x implicitly to get dy/dx, then differentiate again using the chain rule and quotient rule to obtain the given second derivative formula. Part (ii) is direct substitution. While it's Further Maths content, the execution is mechanical with no novel insight required, making it slightly easier than average overall. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-\sin y \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = -(\sin y)^{-1}\) | M1 A1 | Differentiates implicitly once. |
| \(\frac{d^2y}{dx^2} = (\sin y)^{-2}\cos y\left(\frac{dy}{dx}\right) = -\cot y\left(\frac{dy}{dx}\right)^2\) | M1 A1 | Differentiates again, AG. |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = -\left(\sin\frac{\pi}{3}\right)^{-1} = -\frac{2}{\sqrt{3}}\) | B1 | |
| \(\frac{d^2y}{dx^2} = -\frac{1}{\sqrt{3}}\left(-\frac{2}{\sqrt{3}}\right)^2 = -\frac{4}{3\sqrt{3}} = -\frac{4}{9}\sqrt{3}\) | B1 | AEF, must be exact. |
| Total: 2 |
## Question 1:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-\sin y \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = -(\sin y)^{-1}$ | M1 A1 | Differentiates implicitly once. |
| $\frac{d^2y}{dx^2} = (\sin y)^{-2}\cos y\left(\frac{dy}{dx}\right) = -\cot y\left(\frac{dy}{dx}\right)^2$ | M1 A1 | Differentiates again, AG. |
| **Total: 4** | | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = -\left(\sin\frac{\pi}{3}\right)^{-1} = -\frac{2}{\sqrt{3}}$ | B1 | |
| $\frac{d^2y}{dx^2} = -\frac{1}{\sqrt{3}}\left(-\frac{2}{\sqrt{3}}\right)^2 = -\frac{4}{3\sqrt{3}} = -\frac{4}{9}\sqrt{3}$ | B1 | AEF, must be exact. |
| **Total: 2** | | |
---1 A curve $C$ has equation $\cos y = x$, for $- \pi < x < \pi$.\\
(i) Use implicit differentiation to show that
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - \cot y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 }$$
(ii) Hence find the exact value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at the point $\left( \frac { 1 } { 2 } , \frac { 1 } { 3 } \pi \right)$ on $C$.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q1 [6]}}