Higher November 2023 Paper 6 Q7
7 Here are two functions.
| Function A: | input → \(\times k\) → \(+ 21\) → output |
| Function B: | input → \(- 1.5\) → \(\times k\) → output |
5 is input into Function A.
5 is also input into Function B.
The output of Function A is 10 times the output of Function B.
Work out the value of \(k\).
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| [0].7 oe with correct working | 5 | By Equation: M3 for \(5k + 21\) and \(10(5 - 1.5)k\) oe or M1 for \(5k + 21\) M1 for \((5 - 1.5)k\) or \(3.5k\) AND M1FT for \(30k = 21\) | “correct working” requires evidence of at least M3 \(5k + 21\) [M1] = \(26k\) [isw, but ruled out of M3 and M1FT if \(26k\) used] \(5 - 1.5k = 3.5k\) [M1 bod] FT their linear equation in the form \(ak + b = ck\) | ||||||||||||||||||||||||||||||||||||||||||||||
| If 0 or 1 scored, instead award SC2 for answer [0].7 with no working or insufficient working If 0 scored, instead award SC1 for \(-\frac{140}{31}\) or −4.516… to −4.52 as final answer | SC1 is from ×10 of wrong function | ||||||||||||||||||||||||||||||||||||||||||||||||
| By Trials: M4 for \(5k + 21\) and either \(10(5 - 1.5)k\) oe or \((5 - 1.5)k\) oe both correctly evaluated with \(k = 0.7\) or M3 for \(5k + 21\) and either \(10(5 - 1.5)k\) oe or \((5 - 1.5)k\) oe both correctly evaluated in one trial with consistent \(k\) or M1 for \(5k + 21\) correctly evaluated in one trial M1 for either \(10(5 - 1.5)k\) oe or \((5 - 1.5)k\) oe correctly evaluated in one trial SC award marks as above |
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