| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Differentiation of Simplified Fractions |
| Difficulty | Standard +0.3 This C1 question requires expanding and simplifying the fraction before differentiating term-by-term, then finding a tangent equation. While it involves multiple steps (algebraic manipulation, differentiation, point-gradient form), each technique is standard and the question provides clear scaffolding. Slightly above average difficulty due to the initial algebraic manipulation required, but well within typical C1 expectations. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x^2+4)(x-3) = x^3 - 3x^2 + 4x - 12\) | M1 | Attempt to multiply out numerator to get a cubic with 4 terms and at least 2 correct |
| \(\frac{dy}{dx} = \frac{2x(3x^2-6x+4)-2(x^3-3x^2+4x-12)}{(2x)^2}\) | M1A1 | M1: Correct application of quotient rule; A1: Correct derivative |
| \(= \frac{4x^3 - 6x^2 + 24}{4x^2} = x - \frac{3}{2} + \frac{6}{x^2}\) | ddM1A1 | M1: Collects terms and divides by denominator. Dependent on both previous method marks. A1: \(x - \frac{3}{2} + \frac{6}{x^2}\) oe and isw. Accept \(1x\) or even \(1x^1\) but not \(\frac{2x}{2}\) and not \(x^0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \left(\frac{x}{2}+\frac{2}{x}\right)(x-3)\) or \((x^2+4)\left(\frac{1}{2}-\frac{3}{2x}\right)\) | M1 | Divides one bracket by \(2x\) |
| \(\frac{dy}{dx} = (x-3)\left(\frac{1}{2}-\frac{2}{x^2}\right)+\left(\frac{x}{2}+\frac{2}{x}\right)\) or \(\frac{dy}{dx}=(x^2+4)\frac{3}{2x^2}+2x\left(\frac{1}{2}-\frac{3}{2x}\right)\) | M1A1 | M1: Correct application of product rule; A1: Correct derivative |
| \(= \frac{3}{2} + \frac{6}{x^2} + x - 3 = x - \frac{3}{2} + \frac{6}{x^2}\) | ddM1A1 | M1: Expands and collects terms. Dependent on both previous method marks. A1: \(x-\frac{3}{2}+\frac{6}{x^2}\) oe and isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x^2+4)(x-3) = x^3-3x^2+4x-12\) | M1 | Attempt to multiply out numerator to get cubic with 4 terms and at least 2 correct |
| \(\frac{dy}{dx} = (x^3-3x^2+4x-12)\times(-\frac{1}{2}x^{-2})+\frac{1}{2}x^{-1}(3x^2-6x+4)\) | M1A1 | M1: Correct application of product rule; A1: Correct derivative |
| \(= x - \frac{3}{2} + \frac{6}{x^2}\) | ddM1A1 | ddM1: Expands and collects terms. Dependent on both previous method marks. A1: \(x-\frac{3}{2}+\frac{6}{x^2}\) oe and isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \left(\frac{x}{2}+\frac{2}{x}\right)(x-3)\) or \((x^2+4)\left(\frac{1}{2}-\frac{3}{2x}\right)\) | M1 | Divides one bracket by \(2x\) |
| \(= \frac{x^2}{2} - \frac{3}{2}x + 2 - 6x^{-1}\) | M1A1 | M1: Expands; A1: Correct expression |
| \(\frac{dy}{dx} = x - \frac{3}{2} + \frac{6}{x^2}\) | ddM1A1 | ddM1: \(x^n \rightarrow x^{n-1}\) or \(2 \rightarrow 0\). Dependent on both previous method marks. A1: \(x-\frac{3}{2}+\frac{6}{x^2}\) oe and isw. Accept \(1x\) or even \(1x^1\) but not \(\frac{2x}{2}\). If they lose previous A1 because of incorrect constant only then allow recovery here for correct derivative. |
## Question 6(a) Appendix — Alternative Methods:
### Way 2 (Quotient Rule):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x^2+4)(x-3) = x^3 - 3x^2 + 4x - 12$ | M1 | Attempt to multiply out numerator to get a cubic with 4 terms and at least 2 correct |
| $\frac{dy}{dx} = \frac{2x(3x^2-6x+4)-2(x^3-3x^2+4x-12)}{(2x)^2}$ | M1A1 | M1: Correct application of quotient rule; A1: Correct derivative |
| $= \frac{4x^3 - 6x^2 + 24}{4x^2} = x - \frac{3}{2} + \frac{6}{x^2}$ | ddM1A1 | M1: Collects terms and divides by denominator. Dependent on both previous method marks. A1: $x - \frac{3}{2} + \frac{6}{x^2}$ oe and **isw**. Accept $1x$ or even $1x^1$ but not $\frac{2x}{2}$ and not $x^0$ |
### Way 3 (Product Rule):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \left(\frac{x}{2}+\frac{2}{x}\right)(x-3)$ or $(x^2+4)\left(\frac{1}{2}-\frac{3}{2x}\right)$ | M1 | Divides one bracket by $2x$ |
| $\frac{dy}{dx} = (x-3)\left(\frac{1}{2}-\frac{2}{x^2}\right)+\left(\frac{x}{2}+\frac{2}{x}\right)$ or $\frac{dy}{dx}=(x^2+4)\frac{3}{2x^2}+2x\left(\frac{1}{2}-\frac{3}{2x}\right)$ | M1A1 | M1: Correct application of product rule; A1: Correct derivative |
| $= \frac{3}{2} + \frac{6}{x^2} + x - 3 = x - \frac{3}{2} + \frac{6}{x^2}$ | ddM1A1 | M1: Expands and collects terms. Dependent on both previous method marks. A1: $x-\frac{3}{2}+\frac{6}{x^2}$ oe and **isw** |
### Way 4 (Product Rule variant):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x^2+4)(x-3) = x^3-3x^2+4x-12$ | M1 | Attempt to multiply out numerator to get cubic with 4 terms and at least 2 correct |
| $\frac{dy}{dx} = (x^3-3x^2+4x-12)\times(-\frac{1}{2}x^{-2})+\frac{1}{2}x^{-1}(3x^2-6x+4)$ | M1A1 | M1: Correct application of product rule; A1: Correct derivative |
| $= x - \frac{3}{2} + \frac{6}{x^2}$ | ddM1A1 | ddM1: Expands and collects terms. Dependent on both previous method marks. A1: $x-\frac{3}{2}+\frac{6}{x^2}$ oe and **isw** |
### Way 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \left(\frac{x}{2}+\frac{2}{x}\right)(x-3)$ or $(x^2+4)\left(\frac{1}{2}-\frac{3}{2x}\right)$ | M1 | Divides one bracket by $2x$ |
| $= \frac{x^2}{2} - \frac{3}{2}x + 2 - 6x^{-1}$ | M1A1 | M1: Expands; A1: Correct expression |
| $\frac{dy}{dx} = x - \frac{3}{2} + \frac{6}{x^2}$ | ddM1A1 | ddM1: $x^n \rightarrow x^{n-1}$ or $2 \rightarrow 0$. Dependent on both previous method marks. A1: $x-\frac{3}{2}+\frac{6}{x^2}$ oe and **isw**. Accept $1x$ or even $1x^1$ but not $\frac{2x}{2}$. If they lose previous A1 because of incorrect constant only then allow recovery here for correct derivative. |\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { \left( x ^ { 2 } + 4 \right) ( x - 3 ) } { 2 x } , \quad x \neq 0$$
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in its simplest form.\\
(b) Find an equation of the tangent to $C$ at the point where $x = - 1$
Give your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{Edexcel C1 2015 Q6 [10]}}