2017 H2 Maths Paper 1 Q5 — Worked Solution

Part (i)

4 marks

Let $f(x) = x^{3} + ax^{2} + bx + c$ . By the remainder theorem, dividing by $(x - 1)$ , $(x - 2)$ and $(x - 3)$ gives remainders $f(1)$ , $f(2)$ and $f(3)$ respectively. Using the given values:

f(1) = 1 + a + b + c = 8

f(2) = 8 + 4a + 2b + c = 12

f(3) = 27 + 9a + 3b + c = 25

Simplifying each:

a + b + c = 7

4a + 2b + c = 4

9a + 3b + c = - 2

By GC (solving the system), $a = - \frac{3}{2}$ , $b = \frac{3}{2}$ , $c = 7$ .

To check your answer:

f(1) = 1 - \frac{3}{2} + \frac{3}{2} + 7 = 8

, which matches. ✓

$\therefore$ $a = - \frac{3}{2}$ , $b = \frac{3}{2}$ , $c = 7$ .

Part (ii)

3 marks

Since $f(x) = x^{3} - \frac{3}{2}x^{2} + \frac{3}{2}x + 7$ ,

\frac{dy}{dx}

= 3x^{2} - 3x + \frac{3}{2}

Completing the square:

3x^{2} - 3x + \frac{3}{2}

= 3\left( x^{2} - x \right) + \frac{3}{2}

= 3\left( x - \frac{1}{2} \right)^{2} - \frac{3}{4} + \frac{3}{2}

= 3\left( x - \frac{1}{2} \right)^{2} + \frac{3}{4}

Since $\left( x - \frac{1}{2} \right)^{2} \geq 0$ for all real $x$ ,

\frac{dy}{dx} = 3\left( x - \frac{1}{2} \right)^{2} + \frac{3}{4} \geq \frac{3}{4} > 0\ \ \text{for all }x.

$\therefore$ the gradient of the curve is always positive. (shown)

Explanation: Since the gradient is always positive, the curve is strictly increasing for all $x$ , so it can cross the $x$ -axis at most once. A cubic with real coefficients has at least one real root, so $f(x) = 0$ has exactly one real root.

Approach

The root is not a nice rational number (there is no factor of $f(x)$ among $\text{±}1,\ \pm 7$ etc., since $f(1) = 8 \neq 0$ and $f( - 1) = - 1.5 - 1.5 - 1.5 + 7 \neq 0$ ), so we solve $f(x) = 0$ numerically.

By GC, solving $x^{3} - \frac{3}{2}x^{2} + \frac{3}{2}x + 7 = 0$ gives $x = - 1.33$ (3 s.f.).

$\therefore$ the equation $f(x) = 0$ has only one real root, $x = - 1.33$ (3 s.f.).

Part (iii)

3 marks

Approach

The tangent is parallel to $y = 2x - 3$ when the gradient of the curve equals the gradient of the line, $2$ .

\frac{dy}{dx} = 2

3\left( x - \frac{1}{2} \right)^{2} + \frac{3}{4} = 2

\left( x - \frac{1}{2} \right)^{2} = \frac{5}{12}

x - \frac{1}{2} = \pm \sqrt{\frac{5}{12}}

x = \frac{1}{2} \pm \frac{\sqrt{15}}{6}

$\therefore$ the tangent is parallel to $y = 2x - 3$ at $x = \frac{1}{2} + \frac{\sqrt{15}}{6}$ and $x = \frac{1}{2} - \frac{\sqrt{15}}{6}$ .

2017 H2 Maths Paper 1, Question 5

Part (i)

Part (ii)

Part (iii)

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