Higher June 2018 Paper 6 Q17
17 Here is a function.
(a) The output of function A is \(x\).
Write an algebraic expression, in terms of \(x\), for the input of function A. [2]
(b) A number, \(k\), is put into function A.
The output is also \(k\).
The output is also \(k\).
Find the value of \(k\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{x}{5} - 14\) oe | 2 | M1 for \(\dfrac{x}{5}\) If 0 scored then SC1 for \(\dfrac{x - 14}{5}\) oe | Condone use of another letter for M1 max Must use \(x\) in SC1 0 for \(x - 14 \div 5\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| −17.5 or \(-\dfrac{35}{2}\) oe nfww | 3 | M1 for 5(‘\(k\)’ + 14) = ‘\(k\)’ or ‘\(k\)’ = \(\dfrac{k}{5} - 14\) M1FT for 4’\(k\)’ = −70 or better or re-arrangement of their comparable f(\(k\)) = g(\(k\)) equation into the form \(ak = b\). M1FT solving their \(ak = b\) | eg \(5k + 14 = k\) becomes \(4k = -14\) and then \(k = -3.5\) scores M0 M1FT M1FT \(k + 70 = k\) is not comparable Answers may be in decimal or fractional form but fractions equating to integers should be simplified |
| Alternative (FT as above): M1 for ‘\(k\)’ = \(\dfrac{k}{5} - 14\) M1FT for \(\dfrac{4k}{5}\) = −14 or better M1FT solving their \(ak = b\) Trials or no working: SC3 for −17.5 | |||