2019 H2 Maths Paper 1 Q3 — Worked Solution

Part (i)

2 marks

Approach

To complete the cube on $x^{3} - 3x^{2} + 3x$ , we compare with the expansion of $(x - 1)^{3} = x^{3} - 3x^{2} + 3x - 1$ , then account for the constant term left over.

Factoring out $2$ from every term,

f(x) = 2x^{3} - 6x^{2} + 6x - 12

= 2\left( x^{3} - 3x^{2} + 3x - 6 \right)

Since $(x - 1)^{3} = x^{3} - 3x^{2} + 3x - 1$ ,

x^{3} - 3x^{2} + 3x - 6 = (x - 1)^{3} - 1 - 6 + 1

Note: the constant $- 1$ from $(x - 1)^{3}$ has to be removed before the $- 6$ is added, since $(x - 1)^{3}$ already carries a $- 1$ .

x^{3} - 3x^{2} + 3x - 6 = (x - 1)^{3} - 5

Therefore

f(x) = 2\{(x - 1)^{3} - 5\}

$\therefore$ $p = 2$ , $q = - 1$ , $r = - 5$ .

To check your answer:

f(1) = 2(1) - 6(1) + 6(1) - 12 = - 10

, and

2\{(1 - 1)^{3} - 5\} = 2( - 5) = - 10

. ✓

Part (ii)

3 marks

Since $f(x) = 2\{(x - 1)^{3} - 5\}$ , we build up $y = f(x)$ from $y = x^{3}$ in the order the brackets suggest: first the translation inside the cube, then the translation added to it, then the stretch multiplying the whole bracket.

Starting from $y = x^{3}$ :

Translate $1$ unit in the positive $x\text{-}$ direction, giving $y = (x - 1)^{3}$ .
Translate $5$ units in the negative $y\text{-}$ direction, giving $y = (x - 1)^{3} - 5$ .
Stretch with scale factor $2$ parallel to the $y\text{-}$ axis (relative to the $x$ -axis), giving $y = 2\{(x - 1)^{3} - 5\} = f(x)$ .

*Working with the expanded form $y = 2(x - 1)^{3} - 10$ gives an equivalent sequence: translate $1$ unit in the positive $x\text{-}$ direction, stretch with scale factor $2$ parallel to the $y\text{-}$ axis, then translate $10$ units in the negative $y\text{-}$ direction.

$\therefore$ the sequence is: translate by $1$ unit in the positive $x$ -direction, translate by $5$ units in the negative $y$ -direction, then stretch with scale factor $2$ parallel to the $y$ -axis.

2019 H2 Maths Paper 1, Question 3

Part (i)

Part (ii)

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