| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| 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 to combine fractions with different denominators and simplify—a standard C3 skill. Part (b) is routine differentiation (likely quotient rule) followed by finding the normal gradient and equation. This is a straightforward multi-part question testing standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(x^2 - 9 = (x+3)(x-3)\) | B1 | |
| \(\frac{4x-5}{(2x+1)(x-3)} - \frac{2x}{(x+3)(x-3)}\) | ||
| \(= \frac{(4x-5)(x+3)}{(2x+1)(x-3)(x+3)} - \frac{2x(2x+1)}{(2x+1)(x+3)(x-3)}\) | M1 | |
| \(= \frac{5x-15}{(2x+1)(x-3)(x+3)}\) | M1A1 | |
| \(= \frac{5(x-3)}{(2x+1)(x-3)(x+3)} = \frac{5}{(2x+1)(x+3)}\) | A1* | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(f(x) = \frac{5}{2x^2+7x+3}\) | ||
| \(f'(x) = \frac{-5(4x+7)}{(2x^2+7x+3)^2}\) | M1M1A1 | |
| \(f'(-1) = -\frac{15}{4}\) | M1A1 | |
| Uses \(m_1m_2=-1\) to give gradient of normal \(= \frac{4}{15}\) | M1 | |
| \(\frac{y-(-\frac{5}{2})}{x-(-1)} = \frac{4}{15}\) | M1 | |
| \(y + \frac{5}{2} = \frac{4}{15}(x+1)\) or equivalent | A1 | (8 marks) |
## Question 7:
### Part (a):
| Working | Mark | Notes |
|---------|------|-------|
| $x^2 - 9 = (x+3)(x-3)$ | B1 | |
| $\frac{4x-5}{(2x+1)(x-3)} - \frac{2x}{(x+3)(x-3)}$ | | |
| $= \frac{(4x-5)(x+3)}{(2x+1)(x-3)(x+3)} - \frac{2x(2x+1)}{(2x+1)(x+3)(x-3)}$ | M1 | |
| $= \frac{5x-15}{(2x+1)(x-3)(x+3)}$ | M1A1 | |
| $= \frac{5(x-3)}{(2x+1)(x-3)(x+3)} = \frac{5}{(2x+1)(x+3)}$ | A1* | (5 marks) |
### Part (b):
| Working | Mark | Notes |
|---------|------|-------|
| $f(x) = \frac{5}{2x^2+7x+3}$ | | |
| $f'(x) = \frac{-5(4x+7)}{(2x^2+7x+3)^2}$ | M1M1A1 | |
| $f'(-1) = -\frac{15}{4}$ | M1A1 | |
| Uses $m_1m_2=-1$ to give gradient of normal $= \frac{4}{15}$ | M1 | |
| $\frac{y-(-\frac{5}{2})}{x-(-1)} = \frac{4}{15}$ | M1 | |
| $y + \frac{5}{2} = \frac{4}{15}(x+1)$ or equivalent | A1 | (8 marks) |
---7.
$$f ( x ) = \frac { 4 x - 5 } { ( 2 x + 1 ) ( x - 3 ) } - \frac { 2 x } { x ^ { 2 } - 9 } , \quad x \neq \pm 3 , x \neq - \frac { 1 } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$f ( x ) = \frac { 5 } { ( 2 x + 1 ) ( x + 3 ) }$$
The curve $C$ has equation $y = \mathrm { f } ( x )$. The point $P \left( - 1 , - \frac { 5 } { 2 } \right)$ lies on $C$.
\item Find an equation of the normal to $C$ at $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2011 Q7 [13]}}