| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find normal equation at point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question with standard techniques. Part (a) requires applying the product rule and chain rule systematically to differentiate implicitly, then rearranging—routine for A-level. Part (b) involves substituting a point to find the gradient, taking the negative reciprocal, and writing the normal equation using y-y₁=m(x-x₁). Both parts are textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3x^2 - 6xy - 3x^2\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\) | M1* | Attempt implicit differentiation. Either of the two \(\frac{dy}{dx}\) terms correct, allowing sign errors |
| Correct derivative www | A1 | Condone no \(= 0\) on RHS. Condone \(\frac{dy}{dx} = \ldots\) as long as not used |
| \(3x^2 - 6xy + (2y - 3x^2)\frac{dy}{dx} = 0\) OR \(-3x^2\frac{dy}{dx} + 2y\frac{dy}{dx} = 6xy - 3x^2\) | M1d* | Attempt to make \(\frac{dy}{dx}\) the subject. Either collect like terms on each side or take out a common factor of \(\frac{dy}{dx}\). Must have two terms involving \(\frac{dy}{dx}\) and two terms without \(\frac{dy}{dx}\) |
| \((2y - 3x^2)\frac{dy}{dx} = 6xy - 3x^2\), \(\frac{dy}{dx} = \frac{6xy - 3x^2}{2y - 3x^2}\) A.G. | A1 | Obtain correct \(\frac{dy}{dx}\) having collected like terms on either side and taken out a common factor (possibly with both steps done in one go) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 9\) | B1 | Obtain gradient of 9. Could be implied by \(-\frac{1}{9}\) |
| \(m' = -\frac{1}{9}\) | B1FT | Correct gradient of normal. FT their \(m\) |
| \(y - 2 = -\frac{1}{9}(x-1)\) | M1 | Attempt equation of normal using \((1,2)\) and their normal gradient (M0 if using gradient of tangent). Gradient must be numerical |
| \(x + 9y = 19\) | A1 | Obtain correct three term equation. Any correct equiv |
| [4] |
## Question 5:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $3x^2 - 6xy - 3x^2\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$ | M1* | Attempt implicit differentiation. Either of the two $\frac{dy}{dx}$ terms correct, allowing sign errors |
| Correct derivative www | A1 | Condone no $= 0$ on RHS. Condone $\frac{dy}{dx} = \ldots$ as long as not used |
| $3x^2 - 6xy + (2y - 3x^2)\frac{dy}{dx} = 0$ OR $-3x^2\frac{dy}{dx} + 2y\frac{dy}{dx} = 6xy - 3x^2$ | M1d* | Attempt to make $\frac{dy}{dx}$ the subject. Either collect like terms on each side or take out a common factor of $\frac{dy}{dx}$. Must have two terms involving $\frac{dy}{dx}$ and two terms without $\frac{dy}{dx}$ |
| $(2y - 3x^2)\frac{dy}{dx} = 6xy - 3x^2$, $\frac{dy}{dx} = \frac{6xy - 3x^2}{2y - 3x^2}$ **A.G.** | A1 | Obtain correct $\frac{dy}{dx}$ having collected like terms on either side and taken out a common factor (possibly with both steps done in one go) |
| **[4]** | | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 9$ | B1 | Obtain gradient of 9. Could be implied by $-\frac{1}{9}$ |
| $m' = -\frac{1}{9}$ | B1FT | Correct gradient of normal. FT their $m$ |
| $y - 2 = -\frac{1}{9}(x-1)$ | M1 | Attempt equation of normal using $(1,2)$ and their normal gradient (M0 if using gradient of tangent). Gradient must be numerical |
| $x + 9y = 19$ | A1 | Obtain correct three term equation. Any correct equiv |
| **[4]** | | |
---5 A curve has equation $x ^ { 3 } - 3 x ^ { 2 } y + y ^ { 2 } + 1 = 0$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 x y - 3 x ^ { 2 } } { 2 y - 3 x ^ { 2 } }$.
\item Find the equation of the normal to the curve at the point ( 1,2 ).
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2019 Q5 [8]}}