Edexcel C3 2010 June — Question 8 7 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2010
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicPolynomial Division & Manipulation
TypeSimple Algebraic Fraction Simplification
DifficultyModerate -0.8 Part (a) is routine factorization of quadratics and cancellation, a standard C3 skill. Part (b) applies basic logarithm laws (difference of logs, exponential form) in a straightforward way. This is easier than average A-level questions as it requires only direct application of well-practiced techniques with no problem-solving insight needed.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.06f Laws of logarithms: addition, subtraction, power rules
8. (a) Simplify fully $$\frac { 2 x ^ { 2 } + 9 x - 5 } { x ^ { 2 } + 2 x - 15 }$$ Given that $$\ln \left( 2 x ^ { 2 } + 9 x - 5 \right) = 1 + \ln \left( x ^ { 2 } + 2 x - 15 \right) , \quad x \neq - 5$$ (b) find \(x\) in terms of e.
Question 8:
Part (a)
AnswerMarks Guidance
\(\dfrac{(x+5)(2x-1)}{(x+5)(x-3)} = \dfrac{(2x-1)}{(x-3)}\)M1 B1 A1 aef M1: attempt to factorise numerator; B1: correct factorisation of denominator \((x+5)(x-3)\); A1: correct simplified form
Part (b)
AnswerMarks Guidance
\(\ln\!\left(\dfrac{2x^2+9x-5}{x^2+2x-15}\right) = 1\)M1 Uses correct log law to combine at least two terms
\(\dfrac{2x^2+9x-5}{x^2+2x-15} = e\)dM1 Remove ln correctly; anti-ln of 1 is e; dependent on previous M
\(\dfrac{2x-1}{x-3} = e \Rightarrow 3e - 1 = x(e-2)\)M1 Collect \(x\) terms and factorise
\(x = \dfrac{3e-1}{e-2}\)A1 aef cso Also accept \(\dfrac{3e^1-1}{e^1-2}\) or \(\dfrac{1-3e}{2-e}\); must be in terms of e; decimal answer \(9.9610559\ldots\) not sufficient
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## Question 8:

### Part (a)
| $\dfrac{(x+5)(2x-1)}{(x+5)(x-3)} = \dfrac{(2x-1)}{(x-3)}$ | M1 B1 A1 aef | M1: attempt to factorise numerator; B1: correct factorisation of denominator $(x+5)(x-3)$; A1: correct simplified form |

### Part (b)
| $\ln\!\left(\dfrac{2x^2+9x-5}{x^2+2x-15}\right) = 1$ | M1 | Uses correct log law to combine at least two terms |
| $\dfrac{2x^2+9x-5}{x^2+2x-15} = e$ | dM1 | Remove ln correctly; anti-ln of 1 is e; dependent on previous M |
| $\dfrac{2x-1}{x-3} = e \Rightarrow 3e - 1 = x(e-2)$ | M1 | Collect $x$ terms and factorise |
| $x = \dfrac{3e-1}{e-2}$ | A1 aef cso | Also accept $\dfrac{3e^1-1}{e^1-2}$ or $\dfrac{1-3e}{2-e}$; must be in terms of e; decimal answer $9.9610559\ldots$ not sufficient |

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To extract mark scheme content, please share the actual mark scheme pages containing the questions, answers, mark allocations, and guidance notes.
8. (a) Simplify fully

$$\frac { 2 x ^ { 2 } + 9 x - 5 } { x ^ { 2 } + 2 x - 15 }$$

Given that

$$\ln \left( 2 x ^ { 2 } + 9 x - 5 \right) = 1 + \ln \left( x ^ { 2 } + 2 x - 15 \right) , \quad x \neq - 5$$

(b) find $x$ in terms of e.\\

\hfill \mbox{\textit{Edexcel C3 2010 Q8 [7]}}
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