2013 H2 Mathematics Paper 1 Question 2

Equations and Inequalities

Answers

(−∞,3−23]∪[3+23,∞){\left( -\infty, 3 - 2 \sqrt{3} \right]}\cup\allowbreak { \left[ 3 + 2 \sqrt{3}, \infty \right)}

Full solutions

y=x2+x+1x−1(x−1)y=x2+x+1x2+x(1−y)+(1+y)=0\begin{gather*} y = \frac{x^2 + x + 1}{x-1} \\ (x-1)y = x^2 + x + 1 \\ x^2 + x(1-y) + (1+y) = 0 \\ \end{gather*}
For set of values y{y} can take,
discriminant≥0(1−y)2−4(1)(1+y)≥0−3−6y+y2≥0\begin{gather*} \textrm{discriminant} \geq 0 \\ (1-y)^2 - 4(1)(1+y) \geq 0 \\ - 3 - 6 y + y^2 \geq 0 \\ \end{gather*}
y≤6−62−4(−3)2or   y≥6+62−4(−3)2y≤3−23or   y≥3+23\begin{align*} &y \leq \frac{6-\sqrt{6^2-4(-3)}}{2} &\textrm{or }& \; y \geq \frac{6+\sqrt{6^2-4(-3)}}{2} \\ &y \leq 3 - 2 \sqrt{3} &\textrm{or }& \; y \geq 3 + 2 \sqrt{3} \end{align*}
Hence set of values that y{y} can take:
(−∞,3−23]∪[3+23,∞)  ■\left( -\infty, 3 - 2 \sqrt{3} \right] \cup \left[ 3 + 2 \sqrt{3}, \infty \right) \; \blacksquare