June 2025 Paper 3 Q4

Notes

B1: For \(2x + k\)

M1: Correct order of operations for the composite function \(\mathrm{f}^{-1}\mathrm{g}\)

Alternative for first two marks in part (a)

B1: soi

M1: Condone sign errors only

M1*: Expanding and simplifying to obtain a three-term quadratic expression in \(x\) (allow sign errors only in the simplification and collection of terms) – all terms must be on the same side and the constant term must be a two-term linear expression in \(k\)

Must come from considering \(a(x^2 + kx + 5) + bk = 4 - 2kx\) where \(a, b \neq 0\)
Condone \(2x^2 + 4kx + (6 + k) \gt 0\) for M1 but cannot score the final A1 mark (as www)

M1dep*: Calculating correct discriminant for their 3TQ – allow \(4k^2\) recovered to \(16k^2\) for M1 only – do not award this mark if only embedded in quadratic formula

Condone \(= 0\), \(\lt 0\) etc. for this M1 mark

A1: AG www – sufficient working must be seen. Discriminant must have been set \(\gt 0\) before dividing by 8

A0 for setting discriminant \(= 0\) or \(\lt 0\) (oe wrong inequality) and later replacing with the correct inequality

Notes

M1: Correct method for solving their 3 term quadratic which would lead to the critical values of \(k\)

See appendix for solving 3TQ

B1: Correct set of values for \(k\) so must be strict inequalities – allow in terms of \(x\) and/or \(k\). Ignore any working seen

Condone poor set notation if intention is clear

B1FT: Correct union set notation for their values of \(k\) so must be in terms of \(k\) (and not \(x\)). Must be of the form \(\{k : k \lt k_1\} \cup \{k : k \gt k_2\}\) where \(k_1 \lt k_2\)
B0 for interval notation

Note that, for example, \(\left\{k \lt -\frac{3}{2}\right\} \cup \{k \gt 2\}\) and \(\left\{k : k \lt -\frac{3}{2} \cup k : k \gt 2\right\}\) etc. are B0

Note that for full marks to be awarded no incorrect working seen

Ignore inclusion of \(\in \mathbb{R}\) inside the \(\{\ \}\) brackets

Appendix: Rules for solving quadratics in question 4(b) ONLY

In question 4(b) candidates are required to solve a 3 term quadratic (3TQ) using DR – therefore we must see a correct, complete method for solving this quadratic – the correct answers do not imply the corresponding M mark, for example \(2x^2 - x - 6 = 0 \Rightarrow x = 2\) or \(x = -\frac{3}{2}\) is M0

Rules for factorising:
\(at^2 + bt + c \Rightarrow (mt + n)(pt + q)\) where \(a = mp\) and one of \(mq + np = b\) or \(c = nq\) (so when expanding their factorised expression it must give the correct quadratic term and one other term correct of the preceding 3TQ expression/equation)

e.g. in question 4b:
\(2x^2 - x - 6 = \left(x + \frac{3}{2}\right)(x - 2)\) is M0 (but the following B1 for the correct \(x \lt -\frac{3}{2}\) and \(x \gt 2\) can still be awarded)
\(2x^2 - x - 6 = (2x - 3)(x + 2)\) is M1 (when expanded the \(x^2\) and constant terms are correct)

Allow correct part factorisation for their 3TQ expression e.g. \(2x(x - 2) + 3(x - 2)\) scores M1

Rules for the formula:
Must apply the correct formula for their three-term quadratic (no errors even if correct formula is stated) – note that stating the formula (in terms of \(a\), \(b\) and \(c\)) followed immediately by the corresponding roots is M0 – we must see the formula being applied e.g. \(2x^2 - x - 6 = 0 \Rightarrow x = \dfrac{1 \pm \sqrt{1^2 - 4(2)(-6)}}{2(2)}\). Minimal acceptable working would be \(x = \dfrac{1 \pm \sqrt{49}}{4}\) (so must explicitly see the value of the discriminant) for M1

Rules for completing the square
The M1 is not awarded until correctly getting to the stage of \(x - \dfrac{1}{4} = \pm\sqrt{\dfrac{49}{16}}\) (must include \(\pm\) so implying two roots) with no errors (so consistent with applying the formula correctly)