(a) Find an algebraic expression for the output of the inverse of function A when the input is \(x\). [2]
(b) Here is a composite function C.
Find the value \(x\) when \(z = 4x\). [5]
Mark scheme (a)
Answer
Marks
Part marks and guidance
\(\dfrac{x + 2}{3}\) or \(\dfrac{x}{3} + \dfrac{2}{3}\) final answer
2
M1 for \(y + 2 = 3x\) or \(x + 2 = 3y\) or for \(x = 3y - 2\) or for \([x =]\ \dfrac{y + 2}{3}\)
If 0 scored, SC1 for answer \(y = \dfrac{x}{2} - 7\)
For 2 marks, condone answer \(y = \dfrac{x + 2}{3}\) Allow M1 for correct reverse flowchart with arrows reversed ← ÷ 3 ← + 2 ←
Mark scheme (b)
Answer
Marks
Part marks and guidance
−5
5
For method marks, condone inclusion of multiplication signs Method must be seen in working space for part (b) If brackets omitted then allow recovery for method
M3 for \(2(3x - 2 + 7) = 4x\) oe or \(\dfrac{2x - 7 + 2}{3} = x\) oe
or M2 for \(2(3x - 2 + 7)\) oe seen or \(\dfrac{2x - 7 + 2}{3}\) oe seen
or M1 for \(3x - 2\) or \(3x + 5\) oe seen or \(2x - 7\) or \(2x - 5\) oe seen
For M3 eg \(2x - 7 = 3x - 2\)
M1dep for correct rearrangement of their eqn with at least 2 terms in \(x\) to \(ax + b = 0\) or better
M1dep on at least M1 earned previously e.g. \(2(3x - 2) + 7\) [\(= 4x\)] scores M1 and then can earn M1 dep if they then correctly rearrange to \(ax = b\)
