| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find equation of normal |
| Difficulty | Moderate -0.3 This is a straightforward application of the quotient rule to find dy/dx, evaluate at x=3, then use the normal gradient formula. It requires multiple standard steps (differentiation, substitution, finding normal gradient, equation of line) but each step is routine with no conceptual challenges or novel insights required. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 7\) at point \(P\) | B1 | Awarded for seeing \(y=7\) when \(x=3\); may be embedded in equation |
| \(\frac{dy}{dx} = \frac{(x-2)^2 \times 3 - (3x-2) \times 2(x-2)}{(x-2)^4}\) | M1 | Application of Quotient Rule; if quoted must be correct. Accept implied by \(u=3x-2\), \(v=(x-2)^2\) with \(\frac{vu'-uv'}{v^2}\) |
| Correct unsimplified derivative | A1 | Accept product rule form \(3(x-2)^{-2} - 2(3x-2)(x-2)^{-3}\) or partial fractions \(-3(x-2)^{-2}-8(x-2)^{-3}\) |
| Sub \(x=3\) into \(\frac{dy}{dx} = (-11)\) | M1 | Substitute \(x=3\) into their derivative to find numerical value |
| \(\frac{1}{11} = \frac{y-7}{x-3} \Rightarrow x - 11y + 74 = 0\) | M1A1 cso | Uses \(x=3\), \(y=7\), and \(-\frac{dx}{dy}\) to form normal equation; if \(y=mx+c\) used must reach value of \(c\) |
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 7$ at point $P$ | B1 | Awarded for seeing $y=7$ when $x=3$; may be embedded in equation |
| $\frac{dy}{dx} = \frac{(x-2)^2 \times 3 - (3x-2) \times 2(x-2)}{(x-2)^4}$ | M1 | Application of Quotient Rule; if quoted must be correct. Accept implied by $u=3x-2$, $v=(x-2)^2$ with $\frac{vu'-uv'}{v^2}$ |
| Correct unsimplified derivative | A1 | Accept product rule form $3(x-2)^{-2} - 2(3x-2)(x-2)^{-3}$ or partial fractions $-3(x-2)^{-2}-8(x-2)^{-3}$ |
| Sub $x=3$ into $\frac{dy}{dx} = (-11)$ | M1 | Substitute $x=3$ into their derivative to find numerical value |
| $\frac{1}{11} = \frac{y-7}{x-3} \Rightarrow x - 11y + 74 = 0$ | M1A1 cso | Uses $x=3$, $y=7$, and $-\frac{dx}{dy}$ to form normal equation; if $y=mx+c$ used must reach value of $c$ |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { 3 x - 2 } { ( x - 2 ) ^ { 2 } } , \quad x \neq 2$$
The point $P$ on $C$ has $x$ coordinate 3\\
Find an equation of the normal to $C$ at the point $P$ in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\
\hfill \mbox{\textit{Edexcel C34 2015 Q1 [6]}}