2024 H2 Mathematics Paper 1 Question 1

Equations and Inequalities

Answers

$a = 2,$ $b = - 1,$ $c = - 1.$

Since the graph has a vertical asymptote $x = - \frac{1}{2},$

\begin{array}{lcr} a ({(- \frac{1}{2})}^{2}) + b (- \frac{1}{2}) + c = 0 \\ \frac{1}{4} a - \frac{1}{2} b + c = 0 \end{array}

Since the graph passes through $(- 2, - \frac{1}{9}),$

\begin{array}{lcr} \frac{- 2 + 1}{a ({(- 2)}^{2}) + b (- 2) + c} = - \frac{1}{9} \\ - \frac{1}{4 a - 2 b + c} = - \frac{1}{9} \\ 4 a - 2 b + c = 9 \end{array}

\begin{aligned} y & = \frac{x + 1}{a x^{2} + b x + c} \\ \frac{d y}{d x} & = \frac{a x^{2} + b x + c - (x + 1) (2 a x + b)}{{(a x^{2} + b x + c)}^{2}} \end{aligned}

Since $(- 2, - \frac{1}{9})$ is a turning point,

\begin{array}{lcr} \frac{d y}{d x} = 0 \\ a x^{2} + b x + c - (x + 1) (2 a x + b) = 0 \\ a ({(- 2)}^{2}) + b (- 2) + c - (- 2 + 1) (2 a (- 2) + b) = 0 \\ - b + c = 0 \end{array}

Solving $(2), (5)$ and $(9)$ simultaneously,

a = 2, b = - 1, c = - 1 ∎