SPM Practice (Short Question)

Posted on April 18, 2020 by user

Question 1:
Given the function g : x → 3x – 2, find
(a) the value of x when g(x) maps onto itself,
(b) the value of k such that g(2 – k) = 4k.

Solution:
(a)

$\begin{array}{l} g (x) = x \\ 3 x - 2 = x \\ 3 x - x = 2 \\ 2 x = 2 \\ x = 1 \end{array}$

(b)

$\begin{array}{l} g (x) = 3 x - 2 \\ g (2 - k) = 4 k \\ 3 (2 - k) - 2 = 4 k \\ 6 - 3 k - 2 = 4 k \\ - 7 k = - 4 \\ k = \frac{4}{7} \end{array}$

Question 2:
Given the functions f : x → px + 1, g : x → 3x – 5 and fg(x) = 3px + q.
Express p in terms of q.

Solution:

$\begin{array}{l} f (x) = p x + 1, g (x) = 3 x - 5 \\ f g (x) = p (3 x - 5) + 1 \\ = 3 p x - 5 p + 1 \\ Given f g (x) = 3 p x + q \\ 3 p x - 5 p + 1 = 3 p x + q \\ - 5 p + 1 = q \\ - 5 p = q - 1 \\ 5 p = 1 - q \\ p = \frac{1 - q}{5} \end{array}$

Question 3:
Given the functions h : x → 3x + 1, and gh : x → 9x² + 6x – 4, find
(a) h^-1 (x),
(b) g(x).

Solution:
(a)

$\begin{array}{l} Let h^{- 1} (x) = y, \\ thus h (y) = x \\ 3 y + 1 = x \\ 3 y = x - 1 \\ y = \frac{x - 1}{3} \\ ∴ h^{- 1} (x) = \frac{x - 1}{3} \\ h^{- 1} : x \mapsto \frac{x - 1}{3} \end{array}$

(b)

$\begin{array}{l} g [h (x)] = 9 x^{2} + 6 x - 4 \\ g (3 x + 1) = 9 x^{2} + 6 x - 4 \\ Let y = 3 x + 1 \\ thus x = \frac{y - 1}{3} \\ g (y) = 9 {(\frac{y - 1}{3})}^{2} + 6 (\frac{y - 1}{3}) - 4 \\ = \frac{9 {(y - 1)}^{2}}{9} + 2 (y - 1) - 4 \\ = y^{2} - 2 y + 1 + 2 y - 2 - 4 \\ = y^{2} - 5 \\ ∴ g (x) = x^{2} - 5 \end{array}$