2023 H2 Mathematics Paper 2 Question 1

Equations and Inequalities

Answers

(a)

(1,2.2).{{(1, 2.2).}}

(b)

(3x−5)(x−2)(x+1)(x−5).{{\displaystyle \frac{(3 x - 5)(x - 2)}{(x + 1)(x - 5)}.}}

x<−1,{{x < - 1,}} 53<x<2{{\frac{5}{3} < x < 2}} or x>5.{{x > 5.}}

Full solutions

(a)

From the graph, set of values of x{{x}} required is (1,2.2)  ■{{(1, 2.2) \; \blacksquare}}

(b)

x+25x2−4x−5+3=x+25+3x2−12x−15x2−4x−5=3x2−11x+10x2−4x−5=(3x−5)(x−2)(x+1)(x−5)  ■\begin{align*} & \frac{ x + 25 }{ x^2 - 4 x - 5 } + 3 \\ &= \frac{x + 25 + 3 x^2 - 12 x - 15}{x^2 - 4 x - 5} \\ &= \frac{ 3 x^2 - 11 x + 10 }{ x^2 - 4 x - 5 } \\ &= \frac{(3 x - 5)(x - 2)}{(x + 1)(x - 5)} \; \blacksquare \end{align*}
x+25x2−4x−5>−3x+25x2−4x−5+3>0(3x−5)(x−2)(x+1)(x−5)>0\begin{gather*} \frac{ x + 25 }{ x^2 - 4 x - 5 } > - 3 \\ \frac{ x + 25 }{ x^2 - 4 x - 5 } + 3 > 0 \\ \frac{(3 x - 5)(x - 2)}{(x + 1)(x - 5)} > 0 \end{gather*}

x<−1,53<x<2 or x>5  ■{x < - 1, \quad \frac{5}{3} < x < 2 \quad \text{ or } \quad x > 5 \; \blacksquare }

Question Commentary

Both parts should feel familiar to students who have diligently practiced their TYS: part (a) is reminiscent of the 2019 Paper 1 Question 4 where we used graphs to solve inequalities involving the modulus function, while part (b) is reminiscent of 2016 Paper 1 Question 1 where we solve inequalities involving rational functions algebraically.