Pre-U Pre-U 9794/2 2010 June — Question 2 5 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
Year2010
SessionJune
Marks5
TopicLaws of Logarithms
TypeSolve log equation with domain restrictions
DifficultyStandard +0.8 This requires manipulating logarithm laws to combine logs, converting the inequality to exponential form, then solving a rational inequality with careful domain restrictions (ensuring arguments are positive). The multiple algebraic steps, factorization, and sign analysis of the rational expression make this moderately challenging, though the techniques are all standard A-level material.
Spec1.02g Inequalities: linear and quadratic in single variable1.06f Laws of logarithms: addition, subtraction, power rules

Solve the inequality $$\log_3(2x^2 - x) - \log_3(2x^2 - 3x + 1) > 1.$$ [5]

AnswerMarks Guidance
Correctly combine the logs \(\log_3\left(\frac{2x^2 - x}{2x^2 - 3x + 1}\right) > 1\)M1
Simplify \(\log_3\frac{x}{x-1} > 1\)A1
Rewrite as \(\frac{x}{x-1} > 3\)M1
Solve: if \(x > 1\) then \(x < \frac{3}{2}\); if \(x < 1\), then \(x > \frac{2}{3}\)M1
The solution set is \(1 < x < \frac{3}{2}\)A1 [5]
*Note: Equivalent marks for candidates who do not divide out the algebraic fraction and attempt to solve a quadratic inequality.*
Correctly combine the logs $\log_3\left(\frac{2x^2 - x}{2x^2 - 3x + 1}\right) > 1$ | M1 |
Simplify $\log_3\frac{x}{x-1} > 1$ | A1 |
Rewrite as $\frac{x}{x-1} > 3$ | M1 |
Solve: if $x > 1$ then $x < \frac{3}{2}$; if $x < 1$, then $x > \frac{2}{3}$ | M1 |
The solution set is $1 < x < \frac{3}{2}$ | A1 | [5]
*Note: Equivalent marks for candidates who do not divide out the algebraic fraction and attempt to solve a quadratic inequality.*
Solve the inequality
$$\log_3(2x^2 - x) - \log_3(2x^2 - 3x + 1) > 1.$$ [5]

\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2010 Q2 [5]}}