June 2025 Paper 1 Q7

Notes

M1: Either

• \(x \leqslant -1\) o.e. e.g. \(-1 \geqslant x\)
• \(2 \leqslant x \leqslant 5\) o.e.

but condone use of strict inequalities anywhere for this mark.
e.g. \(2 \lt x \lt 5\) or \(2 \lt x \leqslant 5\) or \(2 \leqslant x \lt 5\) May also write e.g. \(x \lt 5\) and \(x \gt 2\) which scores M1 but not "\(x \lt 5\) or \(x \gt 2\)"

Allow interval notation such as e.g. \([2, 5]\) or \((-\infty, -1]\) or condone e.g. \((2, 5)\)

Ignore incorrect inequality statements not related to the one which is valid.
e.g. “\(2 \leqslant x \lt 5\) and \(x \gt -1\)” which scores M1 for the first inequality.

A1: Requires \(\{\ \}\) and \(\cup\)

\(\{x : x \leqslant -1\} \cup \{x : 2 \leqslant x \leqslant 5\}\) or \(\{x \mid x \leqslant -1\} \cup \{x \mid 2 \leqslant x \leqslant 5\}\) either way round

but condone \(\{x \leqslant -1\} \cup \{2 \leqslant x \leqslant 5\}\), \(\{x \leqslant -1 \cup 2 \leqslant x \leqslant 5\}\).

Allow e.g. \(\{x : x \leqslant -1\} \cup \{x : 2 \leqslant x \cap x \leqslant 5\}\)

Use of \(\cap\) to join the two separate regions is A0
It is acceptable (but not required) to mention \(\mathbb{R}\)
e.g. \(\{x : x \in \mathbb{R}, x \leqslant -1\} \cup \{x : x \in \mathbb{R}, 2 \leqslant x \leqslant 5\}\)

Condone use of a lower limit written as e.g. \(\{x : -\infty \leqslant x \leqslant -1\} \cup \{x : 2 \leqslant x \leqslant 5\}\)

Notes

Note a correct equation written down scores all 3 marks.
A correct expression but missing e.g. \(y = \ldots\) or \(\mathrm{f}(x) = \ldots\) scores M1dM1A0

M1: Forms the equation of the form \((y =)\ \alpha(x+1)^2(x-5)^2\). Condone \(\alpha = 1\)
Award for sight of \(\alpha(x+1)^2(x-5)^2\) even with \(\alpha = 1\) i.e. \((x+1)^2(x-5)^2\)

dM1: Substitutes \((0,\ -75)\) into the form \(y = \alpha(x+1)^2(x-5)^2\) and attempts to find the value for \(\alpha\). It is dependent on the previous method mark.

A1: \(y = -3(x+1)^2(x-5)^2\) o.e. (e.g. \(y = -3x^4 + 24x^3 - 18x^2 - 120x - 75\))
isw after a correct answer. Condone \(\mathrm{f}(x) = -3(x+1)^2(x-5)^2\) but not \(C = -3(x+1)^2(x-5)^2\)
A correct equation scores all 3 marks. Allow if seen in (c)
isw if they attempt to multiply out.

Alternative I part (b):

Using the form \(y = ax^4 + bx^3 + cx^2 + dx + e\), then setting up and solving simultaneous equations.
There are various versions of this but can be marked similarly.

M1: Sets \(e\) equal to \(-75\) (may just be seen in their equation) and forms three correct different equations in \(a\), \(b\), \(c\) and \(d\) which may be unsimplified.
Note that the form \(y = ax^4 + bx^3 + cx^2 + dx + e\) is M0 until e is set equal to \(-75\)
There are 5 equations that can be formed, only 3 are necessary for this mark.
Do not condone slips.

dM1: Forms four correct different equations and solves to find values for \(a\), \(b\), \(c\) and \(d\). You do not need to be concerned by the process of solving. A calculator can be used to solve the equations.

A1: \(y = -3x^4 + 24x^3 - 18x^2 - 120x - 75\) o.e. isw if they attempt to factorise but withhold this mark if they e.g. divide all terms by 3.
Condone \(\mathrm{f}(x) = \ldots\) but not \(C = \ldots\)
A correct equation scores all 3 marks. Allow if seen in (c)

Alternative II part (b): Uses the form \(y = (x+1)(x-5)(ax^2 + bx + c)\)

M1: Substitutes \(x = 0,\ y = -75\)   \(-75 = -5c \Rightarrow c = 15\), multiplies out, differentiates
\(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = (2x - 4)(ax^2 + bx + 15) + (x^2 - 4x - 5)(2ax + b)\)
and forms a correct equation in \(a\) and \(b\) which may be unsimplified.

dM1: Forms two correct different equations and solves to find values for \(a\) and \(b\). You do not need to be concerned by the process of solving. A calculator can be used to solve the equations.

A1: \(y = (x+1)(x-5)(-3x^2 + 12x + 15)\) o.e. isw if they attempt to multiply out or factorise
Condone \(\mathrm{f}(x) = \ldots\) but not \(C = \ldots\) but withhold this mark if they e.g. divide all terms by 3. A correct equation scores all 3 marks. Allow if seen in (c)

Alternative III part (b): Uses \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \beta(x+1)(x-2)(x-5)\) (\(\beta\) may be 1) and integrates.

M1: Integrates \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) \beta(x+1)(x-2)(x-5)\) to \((y =)\ \beta\left(\dfrac{1}{4}x^4 - 2x^3 + \dfrac{3}{2}x^2 + 10x + k\right)\) and forms one correct equation using either \((0,\ -75)\): \(-75 = \beta k\) (allow \(-75 = k\))
\((-1,\ 0)\): \(0 = \beta\left(\dfrac{1}{4} + 2 + \dfrac{3}{2} - 10 + k\right)\)   \((5,\ 0)\): \(0 = \beta\left(\dfrac{625}{4} - 250 + \dfrac{75}{2} + 50 + k\right)\)

dM1: Forms a different equation using one of \((0,\ -75),\ (-1,\ 0),\ (5,\ 0)\) and solves to find values for \(\beta\) and \(k\). You do not need to be concerned by the process of solving. A calculator can be used to solve the equations.

A1: \(y = -12\left(\dfrac{1}{4}x^4 - 2x^3 + \dfrac{3}{2}x^2 + 10x + \dfrac{25}{4}\right)\) o.e. isw if they attempt to multiply out or factorise
Condone \(\mathrm{f}(x) = \ldots\) but not \(C = \ldots\) but withhold this mark if they e.g. divide all terms by 3. A correct equation scores all 3 marks. Allow if seen in (c)

Notes

M1: Substitutes \(x = 2\) into their \(y = -3(x+1)^2(x-5)^2\) (must be a quartic in any form) and proceeds to find a value for \(y\). Sight of their \(\pm y\) (or \(\pm 243\)) scores M1.
You may need to check this on your calculator if only a value is seen.

A1ft: \(0 \lt k \lt 243\) o.e. ft on their negative \(y\) value at \(x = 2\).
Allow use of set notation, interval notation and allow e.g. \(k \lt 243,\ k \gt 0\) but do not allow OR or \(\cup\). Do not accept \(0 \leqslant k \leqslant 243\) o.e.
If there are multiple attempts at describing the region, mark what appears to be their final answer.
This mark can only be scored if they have a negative quartic graph function
i.e. \(\alpha \lt 0\) for their \(y = \alpha(x+1)^2(x-5)^2\) or \(a \lt 0\) for their \(y = ax^4 + bx^3 + cx^2 + dx + e\)