June 2018 Paper 1 Q14

Notes

M1: Uses \(\cos 2t = 1 - 2\sin^2 t\) in an attempt to eliminate \(t\)

A1*: Proceeds to \(y = 6 - (x - 3)^2\) without any errors
Allow a proof where they start with \(y = 6 - (x - 3)^2\) and substitute the parametric coordinates. M1 is scored for a correct \(\cos 2t = 1 - 2\sin^2 t\) but A1 is only scored when both sides are seen to be the same AND a comment is made, hence proven, or similar .

Notes

(b) M1: For sketching a \(\cap\) parabola with a maximum in quadrant one. It does not need to be symmetrical

A1: For sketching a \(\cap\) parabola with a maximum in quadrant one and with end coordinates of \((1, 2)\) and \((5, 2)\)

B1: Any suitable explanation as to why \(C\) does not include all points of \(y = 6 - (x - 3)^2,\ \ x \in \mathbb{R}\)
This should include a reference to the limits on sin or cos with a link to a restriction on \(x\) or \(y\).
For example
‘As \(-1 \leqslant \sin t \leqslant 1\) then \(1 \leqslant x \leqslant 5\)’ Condone in words ‘\(x\) lies between 1 and 5’ and strict inequalities
‘As \(\sin t \leqslant 1\) then \(x \leqslant 5\)’ Condone in words ‘\(x\) is less than 5’
‘As \(-1 \leqslant \cos(2t) \leqslant 1\) then \(2 \leqslant y \leqslant 6\)’ Condone in words ‘\(y\) lies between 2 and 6’
Withhold if the statement is incorrect Eg "because the domain is \(2 \leqslant x \leqslant 5\)"
Do not allow a statement on the top limit of \(y\) as this is the same for both curves

Notes

B1: Deduces either

• the correct that the lower value of \(k = 7\) This can be found by substituting into \((5, 2)\)
  \(x + y = k \Rightarrow k = 7\) or substituting \(x = 5\) into \(x^2 - 7x + (k + 3) = 0 \Rightarrow 25 - 35 + k + 3 = 0 \Rightarrow k = 7\)
• or deduces that \(k \lt \dfrac{37}{4}\) This may be awarded from later work

M1: For an attempt at the upper value for \(k\).
Finds where \(x + y = k\) meets \(y = 6 - (x - 3)^2\) once by using an appropriate method.
Eg. Sets \(k - x = 6 - (x - 3)^2\) and proceeds to a 3TQ

A1: Correct 3TQ \(x^2 - 7x + (k + 3) = 0\) The = 0 may be implied by subsequent work

M1: Uses the "discriminant" condition. Accept use of \(b^2 = 4ac\) oe or \(b^2 \ldots 4ac\) where ... is any inequality leading to a critical value for \(k\). Eg. one root \(\Rightarrow 49 - 4 \times 1 \times (k + 3) = 0 \Rightarrow k = \dfrac{37}{4}\)

A1: Range of values for \(k = \left\{k : 7 \leqslant k \lt \dfrac{37}{4}\right\}\) Accept \(k \in \left[7, \dfrac{37}{4}\right)\) or exact equivalent

Alternative (c)