| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Tangent or normal to curve |
| Difficulty | Standard +0.3 Part (a) requires algebraic manipulation with partial fractions and common denominators—routine but multi-step. Part (b) is standard differentiation (quotient rule) and finding a tangent line equation. This is slightly easier than average as it's methodical work with no novel insight required, though the algebra in part (a) needs care. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02y Partial fractions: decompose rational functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(x) = 3 - \frac{x-1}{x-3} + \frac{x+11}{(2x)(x-3)}\) | B1 | |
| \(= \frac{3(2x^2 - 5x - 3) - (x-1)(2x+1) + (x+11)}{(2x)(x-3)}\) | M1 A1 | |
| \(= \frac{4x^2 - 13x + 3}{(2x)(x-3)} = \frac{(4x-1)(x-3)}{(2x)(x-3)} = \frac{4x-1}{2x+1}\) | M1 A1 | |
| (b) \(f'(x) = \frac{4x(2x+1) - (4x-1) \times 2}{(2x+1)^2} = \frac{6}{(2x+1)^2}\) | M1 A1 | |
| \(x = -2 \Rightarrow y = 3\), grad \(= \frac{2}{3}\) | A1 | |
| \(\therefore y - 3 = \frac{2}{3}(x + 2)\) | M1 | |
| \(3y - 9 = 2x + 4\) | A1 | |
| \(2x - 3y + 13 = 0\) | A1 | (10 marks) |
**(a)** $f(x) = 3 - \frac{x-1}{x-3} + \frac{x+11}{(2x)(x-3)}$ | B1
$= \frac{3(2x^2 - 5x - 3) - (x-1)(2x+1) + (x+11)}{(2x)(x-3)}$ | M1 A1
$= \frac{4x^2 - 13x + 3}{(2x)(x-3)} = \frac{(4x-1)(x-3)}{(2x)(x-3)} = \frac{4x-1}{2x+1}$ | M1 A1
**(b)** $f'(x) = \frac{4x(2x+1) - (4x-1) \times 2}{(2x+1)^2} = \frac{6}{(2x+1)^2}$ | M1 A1
$x = -2 \Rightarrow y = 3$, grad $= \frac{2}{3}$ | A1
$\therefore y - 3 = \frac{2}{3}(x + 2)$ | M1
$3y - 9 = 2x + 4$ | A1
$2x - 3y + 13 = 0$ | A1 | (10 marks)5.
$$f ( x ) = 3 - \frac { x - 1 } { x - 3 } + \frac { x + 11 } { 2 x ^ { 2 } - 5 x - 3 } , \quad x \in \mathbb { R } , \quad x < - 1$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$f ( x ) = \frac { 4 x - 1 } { 2 x + 1 }$$
\item Find an equation for the tangent to the curve $y = \mathrm { f } ( x )$ at the point where $x = - 2$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q5 [10]}}