June 2025 Paper 2 Q5

Notes

Mark parts (a) and (b) together.

M1: Substitutes \(x = -3\) into \(x = \dfrac{t - 1}{2}\), attempts to find \(t\), and substitutes their \(t\) into \(y = 5(t + 2)^4\)
Condone slips. May be implied by substitution of \(t = -5\) into \(y\) or by 405 or by \(y = 5(-3)^4\)
Alternatively, they may substitute \(x = -3\) into their \(y = \mathrm{f}(x)\)

A1: cao Answer only scores full marks and ISW after seeing 405 e.g. \((-5,\ 405)\)
Accept \((-3,\ 405)\) or e.g. \(P = 405\)

Notes

Mark parts (a) and (b) together.

M1: Attempts to make \(t\) the subject of \(x = \dfrac{t - 1}{2}\) using the correct order of operations and substitutes into \(y = 5(t + 2)^4\) Condone slips e.g. dividing by 2 first: \(t = \dfrac{x}{2} + 1\) or missing the +2 in \(y = 5(t + 2)^4\)
Alt 1: Writes \(t = \left(\dfrac{y}{5}\right)^{\frac{1}{4}} - 2\), substitutes into \(x = \dfrac{t - 1}{2}\) and attempts to make \(y\) the subject using the correct order of operations, i.e., \(2x = \left(\dfrac{y}{5}\right)^{\frac{1}{4}} - 3 \Rightarrow (2x + 3)^4 = \dfrac{y}{5} \Rightarrow y = 5(2x + 3)^4\)
Alt 2: Makes \(t\) the subject and then substitutes into an expanded \(5(t + 2)^4\)
Do not be too concerned about their expansion, but it must contain \(t^4\) and a constant term.

A1: cao and ISW after a correct answer seen. May be scored for \(y = 5(2x + 1 + 2)^4\)
Allow e.g. \(y = 80x^4 + 480x^3 + 1080x^2 + 1080x + 405\) or e.g.
\(y = 5(2x + 1)^4 + 40(2x + 1)^3 + 120(2x + 1)^2 + 160(2x + 1) + 80\)
Their RHS may be unsimplified but do not allow if e.g. binomial coefficients are still present.
Do not accept \(\mathrm{f}(x) = \ldots\) It must be \(y = \ldots\)

Notes

Note: Differentiation seen in (a) or (b) can score marks if used in (c)

M1: Reduces the power of their \((\text{``}2x + 3\text{''})^n\) by one to \(Q(\text{``}2x + 3\text{''})^{n-1}\) where \(Q\) is a constant and could be 1. There should be no other terms using this method.
Alternatively, attempts to expand their \((\text{``}2x + 3\text{''})^n\) (may have been expanded in (b)) and reduces the power of \(x\) by one in at least one term.
They may use parametric differentiation, i.e., \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = a(t + 2)^3\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = b\) where \(a\) and \(b\) are constants, leading to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{a(t + 2)^3}{b}\) i.e. they must divide the correct way round.
Condone attempts at the chain rule that reach e.g. \(y^{\prime} = \ldots u^3\) but make a slip when substituting back in for \(x\) or \(t\), e.g., \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = (5x + 3)^3\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = (t + 2)^2\), provided the intention is clear.

dM1: Substitutes \(x = -3\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) which may be implied by their answer.
Using parametric differentiation, substitutes their \(t\) (found from an attempt at substituting \(x = -3\) into \(x = \dfrac{t - 1}{2}\) which may have been seen in (a)) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\).
Note for reference, if correct, the parametric differentiation is \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 20(t + 2)^3\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{1}{2}\) leading to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 40(t + 2)^3\)
They may substitute their value of \(t\) into \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) first before using the chain rule to reach \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) which is acceptable and implies the first M mark.

A1: \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) -1080\)
Correct answer only scores full marks.
May be seen labelled as \(m\) or something else.