Edexcel P3 2018 Specimen — Question 5 6 marks

Exam BoardEdexcel
ModuleP3 (Pure Mathematics 3)
Year2018
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeShow derivative equals given algebraic form
DifficultyModerate -0.3 This is a straightforward quotient rule application with polynomial functions. While it requires careful algebraic manipulation and simplification over 6 marks, it's a standard textbook exercise with no conceptual difficulty—students simply apply the quotient rule formula and simplify to match the given form. Slightly easier than average due to its routine nature.
Spec1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation

5. Given that $$y = \frac { 5 x ^ { 2 } - 10 x + 9 } { ( x - 1 ) ^ { 2 } } \quad x \neq 1$$ show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { ( x - 1 ) ^ { 3 } }\), where \(k\) is a constant to be found.
(6)

Question 5:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Differentiates numerator to \(10x - 10\), denominator to \(2(x-1)\)B1
\(\frac{dy}{dx} = \frac{(x-1)^2(10x-10) - (5x^2-10x+9)\cdot 2(x-1)}{(x-1)^4}\)M1 A1 Uses quotient rule; fully correct
Takes out common factor \((x-1)\) from numeratorM1 Reach form \(\frac{g(x)}{(x-1)^3}\)
Simplifies numerator: \(\{10x^2 - 20x + 10 - 10x^2 + 20x - 18\}\)M1 Multiplying out and collecting terms
\(\frac{dy}{dx} = \frac{-8}{(x-1)^3}\)A1
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiates numerator to $10x - 10$, denominator to $2(x-1)$ | B1 | |
| $\frac{dy}{dx} = \frac{(x-1)^2(10x-10) - (5x^2-10x+9)\cdot 2(x-1)}{(x-1)^4}$ | M1 A1 | Uses quotient rule; fully correct |
| Takes out common factor $(x-1)$ from numerator | M1 | Reach form $\frac{g(x)}{(x-1)^3}$ |
| Simplifies numerator: $\{10x^2 - 20x + 10 - 10x^2 + 20x - 18\}$ | M1 | Multiplying out and collecting terms |
| $\frac{dy}{dx} = \frac{-8}{(x-1)^3}$ | A1 | |

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5. Given that

$$y = \frac { 5 x ^ { 2 } - 10 x + 9 } { ( x - 1 ) ^ { 2 } } \quad x \neq 1$$

show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { ( x - 1 ) ^ { 3 } }$, where $k$ is a constant to be found.\\
(6)\\

\hfill \mbox{\textit{Edexcel P3 2018 Q5 [6]}}