June 2024 Paper 1 Q8

Notes

M1: Correct method, e.g. attempts to find g(2) \(\left(= \dfrac{5}{4-9}\right)\) and substitutes its value into f to achieve a value.
Alternatively, attempts \(\mathrm{fg}(x) = 4 - 3\left(\dfrac{5}{2x-9}\right)^2\), condoning slips, and substitutes \(x = 2\) to achieve a value.

A1: Correct answer only. If gf(2) is also attempted, then mark the final attempt which is the most complete.

Notes

M1: Eliminates the fraction and puts the \(xy\) term (or \(x\) term) onto one side of the equation.
Alternatively swaps \(x\)’s and \(y\)’ s, eliminates the fraction and puts the \(xy\) term (or \(y\) term) onto one side of the equation. Condone minor slips in rearranging e.g. \(-9y\) instead of \(+9y\)

A1: Correct expression for the inverse, \(x\) in terms of \(y\) or \(y\) in terms of \(x\). Need not be simplified.

Note that \(y = \dfrac{5}{2x-9} \Rightarrow 2x - 9 = \dfrac{5}{y} \Rightarrow 2x = \dfrac{5}{y} + 9\) is M1 and \(\Rightarrow x = \dfrac{\frac{5}{y} + 9}{2}\) is A1

A1: Fully correct notation for the inverse including its domain and including the e.g. \(\mathrm{g}^{-1} =\).
Condone \(x \neq 0\) without \(x \in \mathbb{R}\)  Need not be simplified.
Do not be too worried about \(\mathrm{g}^{-1}\) looking a bit like \(y^{-1}\) due to poor handwriting but if it is clearly \(y^{-1}\) then withhold this mark.

Accept e.g. \(\mathrm{g}^{-1}(x) = \frac{5+9x}{2x}\ x \in \mathbb{R}, x \neq 0\) or \(\mathrm{g}^{-1}(x) = \dfrac{5}{2x} + \dfrac{9}{2}\ \ x \neq 0\) or \(\mathrm{g}^{-1} = \dfrac{\frac{5}{x} + 9}{2}\ \ x \neq 0\)

Ignore any reference to the range of g.

Notes

(c)(i)

M1: Correct method. Attempts to substitute f into g, condoning slips, e.g. missing the 3.

A1: Correct simplified fraction. Ignore any reference to domains. Do not isw.
There is no need to include the \(\mathrm{gf}(x) =\)
If \(\mathrm{fg}(x)\) is also attempted, then mark the final attempt which is the most complete.

(ii)

B1: Deduces the correct range. May be scored even if \(\mathrm{gf}(x)\) is incorrect (but not a follow through).
Allow e.g. \(-5 \leqslant y \lt 0,\ y \in [-5, 0)\), \([-5, 0)\)
Do not allow e.g. \(-5 \leqslant x \lt 0,\ y \in (-5, 0)\), \(-5 \leqslant f(x) \lt 0\), \(-5 \leqslant g(x) \lt 0\)

Notes

M1: Sets \(\mathrm{f}(x) = \mathrm{h}(x)\) and attempts to collect terms to obtain a 3TQ = 0
The = 0 may be implied by use of the discriminant. Condone copying slips in \(\mathrm{f}(x)\) and \(\mathrm{h}(x)\).

dM1: Recognises the need to use "\(b^2 - 4ac \ldots 0\)" on their 3TQ and uses this to establish a value or range of values for \(k\). Allow for an attempt to solve \(b^2 - 4ac \ldots 0\) or \(b^2 \ldots 4ac\), which must be in terms of \(k\) only, where … is an equality or any inequality.

(Alt 1) Attempts to complete the square for their 3TQ (usual rules) and uses its minimum value set … 0 to establish a value or range of values for \(k\). Their expression for the minimum value must be in terms of \(k\) only. Condone any equality or inequality when comparing their minimum value to 0.
e.g. \(5x^2 - 6x + k - 4 \rightarrow 5\left(x - \dfrac{3}{5}\right)^2 - \dfrac{29}{5} + k \rightarrow \text{``}-\dfrac{29}{5} + k\text{''} \gt 0 \Rightarrow k \gt \ldots\) scores dM1

(Alt 2) Differentiates their 3TQ with respect to \(x\) to achieve a linear expression, sets = 0 (which may be implied), solves for \(x\) and substitutes \(x\) back into their 3TQ set … 0 to establish a value or range of values for \(k\). Here … can be any equality or inequality.
e.g. \(5x^2 - 6x + k - 4 \rightarrow 10x - 6 \rightarrow x = 0.6 \Rightarrow 5(0.6)^2 - 6(0.6) + k - 4 \gt 0 \Rightarrow k \gt \ldots\) scores dM1

A1: Deduces the correct range for \(k\), e.g. \(k \gt \dfrac{29}{5}\) o.e. Must be in terms of \(k\) and not e.g. \(x\)
Accept e.g. \(k \in (5.8, \infty)\) or just \(\left(\dfrac{29}{5}, \infty\right)\) but not e.g. \(k \geqslant \dfrac{29}{5}\) or \(x \gt \dfrac{29}{5}\) or \([5.8, \infty)\)