ii)

Find $gf (x)$ by substituting the function $f (x)$ into the function $g (x)$ .

$\begin{array}{rcl} gf (x) & = & {(f (x))}^{2} + f (x) \\ = & {(4 - 3 x)}^{2} + (4 - 3 x) \end{array}$

Expand ${(4 - 3 x)}^{2}$ .

$\begin{array}{rcl} (4 - 3 x) (4 - 3 x) & = & (4) (4) + (4) (- 3 x) + (4) (- 3 x) + (- 3 x) (- 3 x) \\ = & 16 - 12 x - 12 x + 9 x^{2} \\ = & 16 - 24 x + 9 x^{2} \end{array}$

[1]

Substitute into $gf (x)$ .

$\begin{array}{rcl} gf (x) & = & (16 - 24 x + 9 x^{2}) + (4 - 3 x) \\ = & 20 - 27 x + 9 x^{2} \end{array}$

Find $9 g (x)$ by multiplying the function $g (x)$ by 9.

$\begin{array}{rcl} 9 g (x) & = & 9 (x^{2} + x) \\ = & 9 x^{2} + 9 x \end{array}$

[1]

Substitute into $gf (x) - 9 g (x)$ .

$\begin{array}{rcl} gf (x) - 9 g (x) & = & (20 - 27 x + 9 x^{2}) - (9 x^{2} + 9 x) \end{array}$

[1]

Expand and simplify.

$\begin{array}{rcl} gf (x) - 9 g (x) & = & 20 - 27 x + 9 x^{2} - 9 x^{2} - 9 x \\ = & 20 - 27 x - 9 x + 9 x^{2} - 9 x^{2} \\ = & 20 - 36 x \end{array}$

$4 (5 - 9 x)$ [1]
The answer 20 - 36x is also accepted.

Functions (Cambridge O Level Maths)

Exam Questions