2018 H2 Maths Paper 1, Question 4

Since $2x^{2} + 3x - 2 = (2x - 1)(x + 2)$, this is zero at $x = - 2$ and $x = \frac{1}{2}$, so $2x^{2} + 3x - 2 \geq 0$ for $x \leq - 2$ or $x \geq \frac{1}{2}$, and $2x^{2} + 3x - 2 < 0$ for $- 2 < x < \frac{1}{2}$.

Case 1: $x \leq - 2$ or $x \geq \frac{1}{2}$. Here $\left| 2x^{2} + 3x - 2 \right| = 2x^{2} + 3x - 2$, so

Since $- 1 - \sqrt{3} \approx - 2.73 \leq - 2$ and $- 1 + \sqrt{3} \approx 0.73 \geq \frac{1}{2}$, both roots lie in this case.

Case 2: $- 2 < x < \frac{1}{2}$. Here $\left| 2x^{2} + 3x - 2 \right| = - \left( 2x^{2} + 3x - 2 \right)$, so

giving $x = 0\ \left( \text{in range} \right)$ or $x = - 1\ \left( \text{in range} \right)$; both lie in $- 2 < x < \frac{1}{2}$.

(All four values also satisfy $2 - x \geq 0$, needed since the left side is non-negative.)

The original parabola $2x^{2} + 3x - 2$ has vertex at $x = - \frac{3}{4}$, with value $2\left( - \frac{3}{4} \right)^{2} + 3\left( - \frac{3}{4} \right) - 2 = - \frac{25}{8}$; after reflection, the middle hump of $y = \left| 2x^{2} + 3x - 2 \right|$ has a maximum at $\left( - \frac{3}{4},\ \frac{25}{8} \right)$. The graph touches the $x$-axis (corners, not smooth turning points) at $( - 2,0)$ and $\left( \frac{1}{2},0 \right)$, and its $y$-intercept is $(0,2)$.

The line $y = 2 - x$ has intercepts $(0,2)$ and $(2,0)$.

Using the roots from (i), the graphs meet where $y = 2 - x$, at

Hence: the inequality $\left| 2x^{2} + 3x - 2 \right| < 2 - x$ holds exactly where the curve lies strictly below the line. Reading this off the graph, the curve dips below the line on two intervals between consecutive intersections: from $- 1 - \sqrt{3}$ to $- 1$, and from $0$ to $- 1 + \sqrt{3}$ (checking, e.g., $x = - 2$ gives curve $\text{=}0 < 4\text{=}$ line, and $x = 0.3$ gives curve $\text{≈}0.92 < 1.7\text{≈}$ line, confirming both intervals; on $- 1 < x < 0$ the curve sits above the line, e.g. at $x = - 0.5$, curve $\text{=}3 > 2.5\text{=}$ line).