This question involves setting up and solving simultaneous equations in a and b
If g(3) = 20 then substitute x = 3 into g(x) and set this equal to 20

$\begin{array}{rcl} a \times 3 + b & = & 20 \\ 3 a + b & = & 20 \end{array}$

[1]

f^-1(33) = g(1)
Find the inverse function, f^-1(x) by first replacing f(x) with y

$y = 5 x + 3$

Without rearranging the equation, swap the x's and y's

$x = 5 y + 3$

Now make y the subject of this new equation (by subtracting 3 from both sides then dividing both sides by 5)

$\begin{array}{rcl} x - 3 & = & 5 y \\ \frac{x - 3}{5} & = & y \end{array}$

The function on the left-hand side is the inverse function, f^-1(x)
Replace y with f^-1(x)

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

[1]

Find f^-1(33) by substituting x = 33 into f^-1(x)

$\begin{array}{rcl} f^{- 1} (33) & = & \frac{33 - 3}{5} \\ = & \frac{30}{5} \\ = & 6 \end{array}$

Find g(1) by substituting 1 into g(x)

$g (1) = a + b$

[1]

f^-1(33) = g(1) where f^-1(33) = 6 and g(1) = a + b
Set 6 equal to a + b

$6 = a + b$

[1]

Write down the two simultaneous equations for a and b

$\begin{array}{rcl} 3 a + b & = & 20 \\ a + b & = & 6 \end{array}$

Solve to find a (for example, by subtracting the second equations from the first)

$\begin{array}{rcl} 2 a & = & 14 \\ a & = & \frac{14}{2} \\ a & = & 7 \end{array}$

Find b (for example, by substituting a = 7 into the second equation then solving for b)

$\begin{array}{rcl} 7 + b & = & 6 \\ b & = & - 1 \end{array}$

Write out the two solutions together

a = 7 and b = -1 [1]

Functions (CIE IGCSE Maths: Extended)

Exam Questions