Higher June 2025 Paper 4 Q11
11 Function A and function B are shown below.
When the input for both machines is 11, the output of Function A is equal to the output of Function B.
When the input for both machines is 29, the output of Function A is twice the output of Function B.
Find the value of \(a\) and the value of \(b\). [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(a =\)] 9 [\(b =\)] 7 | 5 | B2 for \(a(11 - 5) = 3(11 + b)\) oe and \(a(29 - 5) = 2 \times 3(29 + b)\) oe | e.g. \(6a = 3(11 + b)\) or better e.g. \(24a = 6(29 + b)\) or better |
| or B1 for any two of the expressions below \(a(11 - 5)\), \(3(11 + b)\), \(a(29 - 5)\), \(3(29 + b)\) oe | e.g \(6a\), \(24a\), \(33 + 3b\), \(87 + 3b\) | ||
| M1 for equating coefficients of one variable in their original linear equations or making \(a\) or \(b\) the subject of one of their original equations e.g. \(24a = 12(11 + b)\) and \(24a = 6(29 + b)\) or \(b = 2a - 11\) | Note : if they write two correct equations and manipulate them incorrectly e.g. expanding brackets incorrectly then they can score B2 M0 M1 | ||
| M1 for correctly eliminating one variable in their linear equations with one coefficient of one variable equal e.g. \(6(29 + b) = 12(11 + b)\) oe or \(24a = 6(29 + 2a - 11)\) | B2 M2 for a correct linear equation in \(a\) or \(b\) e.g. \(2a - 11 = 4a - 29\) oe or better Allow one arithmetical error in one of the processes For substitution award second M1 at the point of substitution | ||
| If 0 scored SC1 for \(a(x - 5)\) or \(3(x + b)\) allowing any letter for \(x\) | Trials : M1 for each attempt using values for \(a\) and \(b\) and checking both functions with 11 and 29 to a maximum of M2 | ||