| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Tangent or normal to curve |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining algebraic division/partial fractions with standard differentiation and finding a normal. Part (a) is routine algebraic manipulation given the form, and part (b) is standard calculus application. The question requires multiple techniques but no novel insight, 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Long division of \(x^4+x^3-3x^2+7x-6\) by \(x^2+x-6\), achieving quadratic quotient and linear/constant remainder | M1 | Look for quotient \(x^2(+..x)+A\) and remainder \((Cx)+D\) |
| Quotient \(= x^2+3\), Remainder \(= 4x+12\) | A1 | |
| \(\frac{x^4+x^3-3x^2+7x-6}{x^2+x-6} \equiv x^2+3+\frac{4(x+3)}{(x+3)(x-2)}\) | M1 | Factorise \(x^2+x-6\) and write in appropriate form |
| \(\equiv x^2+3+\frac{4}{(x-2)}\) | A1 | \(A=3\), \(B=4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^4+x^3-3x^2+7x-6 \equiv (x^2+A)(x^2+x-6)+B(x+3)\) | M1 | Multiply by \((x^2+x-6)\) and cancel |
| Compare terms/substitute two values AND solve simultaneously: \(x^2\Rightarrow A-6=-3\), \(x\Rightarrow A+B=7\), const \(\Rightarrow -6A+3B=-6\) | M1 | |
| \(A=3\), \(B=4\) | A1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = 2x - \frac{4}{(x-2)^2}\) | M1A1ft | Differentiate \(x^2+A+\frac{B}{x-2}\to 2x\pm\frac{B}{(x-2)^2}\); ft on their numerical \(A\), \(B\) |
| Substitute \(x=3\): \(f'(3) = 2\times3 - \frac{4}{(3-2)^2} = 2\) | M1 | Substitute \(x=3\) into \(f'(x)\) to find numerical gradient |
| Uses \(m = -\frac{1}{f'(3)} = -\frac{1}{2}\) with point \((3, f(3)) = (3, 16)\) to form equation of normal | M1 | Look for \(y - f(3) = -\frac{1}{f'(3)}(x-3)\) |
| \(y - 16 = -\frac{1}{2}(x-3)\) or equivalent | A1 | cso; e.g. \(2y+x-35=0\) also acceptable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = \frac{(x^2+x-6)(4x^3+3x^2-6x+7)-(x^4+x^3-3x^2+7x-6)(2x+1)}{(x^2+x-6)^2}\) | M1A1 | Use quotient rule with \(u=x^4+x^3-3x^2+7x-6\), \(v=x^2+x-6\) |
| Substitute \(x=3\) | M1 |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Long division of $x^4+x^3-3x^2+7x-6$ by $x^2+x-6$, achieving quadratic quotient and linear/constant remainder | M1 | Look for quotient $x^2(+..x)+A$ and remainder $(Cx)+D$ |
| Quotient $= x^2+3$, Remainder $= 4x+12$ | A1 | |
| $\frac{x^4+x^3-3x^2+7x-6}{x^2+x-6} \equiv x^2+3+\frac{4(x+3)}{(x+3)(x-2)}$ | M1 | Factorise $x^2+x-6$ and write in appropriate form |
| $\equiv x^2+3+\frac{4}{(x-2)}$ | A1 | $A=3$, $B=4$ |
### Alternative (equating terms):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^4+x^3-3x^2+7x-6 \equiv (x^2+A)(x^2+x-6)+B(x+3)$ | M1 | Multiply by $(x^2+x-6)$ and cancel |
| Compare terms/substitute two values AND solve simultaneously: $x^2\Rightarrow A-6=-3$, $x\Rightarrow A+B=7$, const $\Rightarrow -6A+3B=-6$ | M1 | |
| $A=3$, $B=4$ | A1, A1 | |
---
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 2x - \frac{4}{(x-2)^2}$ | M1A1ft | Differentiate $x^2+A+\frac{B}{x-2}\to 2x\pm\frac{B}{(x-2)^2}$; ft on their numerical $A$, $B$ |
| Substitute $x=3$: $f'(3) = 2\times3 - \frac{4}{(3-2)^2} = 2$ | M1 | Substitute $x=3$ into $f'(x)$ to find numerical gradient |
| Uses $m = -\frac{1}{f'(3)} = -\frac{1}{2}$ with point $(3, f(3)) = (3, 16)$ to form equation of normal | M1 | Look for $y - f(3) = -\frac{1}{f'(3)}(x-3)$ |
| $y - 16 = -\frac{1}{2}(x-3)$ or equivalent | A1 | cso; e.g. $2y+x-35=0$ also acceptable |
### Alternative (b) using quotient rule:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = \frac{(x^2+x-6)(4x^3+3x^2-6x+7)-(x^4+x^3-3x^2+7x-6)(2x+1)}{(x^2+x-6)^2}$ | M1A1 | Use quotient rule with $u=x^4+x^3-3x^2+7x-6$, $v=x^2+x-6$ |
| Substitute $x=3$ | M1 | |6.
$$f ( x ) = \frac { x ^ { 4 } + x ^ { 3 } - 3 x ^ { 2 } + 7 x - 6 } { x ^ { 2 } + x - 6 } , \quad x > 2 , x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Given that
$$\frac { x ^ { 4 } + x ^ { 3 } - 3 x ^ { 2 } + 7 x - 6 } { x ^ { 2 } + x - 6 } \equiv x ^ { 2 } + A + \frac { B } { x - 2 }$$
find the values of the constants $A$ and $B$.
\item Hence or otherwise, using calculus, find an equation of the normal to the curve with equation $y = \mathrm { f } ( x )$ at the point where $x = 3$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2016 Q6 [9]}}