2021 H2 Maths Paper 1 Q2 — Worked Solution

Part (a)

2 marks

The curve $y = \left| x^{2} - 5 \right|$ meets the $x$ -axis at $x = \pm \sqrt{5}$ and has a local maximum of $5$ at $x = 0$ ; the line $y = 2x - 1$ has $x$ -intercept $\frac{1}{2}$ and $y$ -intercept $- 1$ .

Part (b)

4 marks

Approach

Since $\left| x^{2} - 5 \right| \geq 0$ always, a solution can only occur where $2x - 1 \geq 0$ . We remove the modulus by splitting into the two cases $x^{2} \geq 5$ and $x^{2} < 5$ , solve the resulting quadratic in each case, then check every root against its case condition.

\left| x^{2} - 5 \right| = 2x - 1

Since the left side is non-negative, we require $2x - 1 \geq 0$ , i.e. $x \geq \frac{1}{2}$ , for a solution to exist.

Case 1: $x^{2} \geq 5$ . Here $\left| x^{2} - 5 \right| = x^{2} - 5$ , so

x^{2} - 5 = 2x - 1

\Rightarrow x^{2} - 2x - 4 = 0

\Rightarrow x = \frac{2 \pm \sqrt{4 + 16}}{2} = 1 \pm \sqrt{5}

Testing against $x^{2} \geq 5$ : $x = 1 - \sqrt{5} \approx - 1.24$ fails, since $\left( 1 - \sqrt{5} \right)^{2} \approx 1.53 < 5$ (rej); $x = 1 + \sqrt{5} \approx 3.24$ satisfies it.

Case 2: $x^{2} < 5$ . Here $\left| x^{2} - 5 \right| = 5 - x^{2}$ , so

5 - x^{2} = 2x - 1

\Rightarrow x^{2} + 2x - 6 = 0

\Rightarrow x = \frac{- 2 \pm \sqrt{4 + 24}}{2} = - 1 \pm \sqrt{7}

Testing against $x^{2} < 5$ : $x = - 1 - \sqrt{7} \approx - 3.65$ fails, since $\left( - 1 - \sqrt{7} \right)^{2} \approx 13.3 \geq 5$ (rej); $x = \sqrt{7} - 1 \approx 1.65$ satisfies it.

Both surviving roots also satisfy $x \geq \frac{1}{2}$ .

To check your answer: at

x = 1 + \sqrt{5}

\left| x^{2} - 5 \right| = 1 + 2\sqrt{5} \approx 5.47

and

2x - 1 \approx 5.47

✓; at

x = \sqrt{7} - 1

\left| x^{2} - 5 \right| = 2\sqrt{7} - 3 \approx 2.29

and

2x - 1 \approx 2.29

✓.

$\therefore$ $x = 1 + \sqrt{5}$ or $x = \sqrt{7} - 1$ .

2021 H2 Maths Paper 1, Question 2

Part (a)

Part (b)

This is how we mark, and how we teach.