June 2022 Paper 1 Q3

OCR ACurrent spec7 marksQuadratics Sketching & Transforming Graphs

(a) In this question you must show detailed reasoning.
Find the coordinates of the points of intersection of the curves with equations \(y = x^2 - 2x + 1\) and \(y = -x^2 + 6x - 5\). [4]

(b) The diagram shows the curves \(y = x^2 - 2x + 1\) and \(y = -x^2 + 6x - 5\).
This diagram is repeated in the Printed Answer Booklet.

The U-shaped curve y = x squared minus 2x plus 1 touching the x-axis at x = 1, and the n-shaped curve y = minus x squared plus 6x minus 5 crossing the x-axis at x = 1 and further right; the curves meet at x = 1 and again near the top of the n-shaped curve

On the diagram in the Printed Answer Booklet, draw the line \(y = 2x - 2\). [2]

(c) Show on your diagram in the Printed Answer Booklet the region of the \(x\)-\(y\) plane within which all three of the following inequalities are satisfied. \[y \geqslant x^2 - 2x + 1 \qquad y \leqslant -x^2 + 6x - 5 \qquad y \leqslant 2x - 2\] You should indicate the region for which all the inequalities hold by labelling the region \(R\). [1]

Mark scheme (a)

Scheme	Marks	AO
DR \(2x^2 - 8x + 6 = 0\) \(x^2 - 4x + 3 = 0\)	M1	1.1
\((x - 1)(x - 3) = 0\)	M1	1.1
\(x = 1\), \(x = 3\)	A1	1.1
\((1, 0)\) and \((3, 4)\)	A1	1.1
	[4]

Notes

M1: Equate, and rearrange to three term quadratic
Attempt to gather like terms, but not necessarily on same side of equation
Condone no ‘= 0’

M1: Attempt to solve quadratic
If factorising then expansion should give \(x^2\) and one other term correct
Quadratic formula should be correct – allow one slip when substituting as long as general formula already seen as correct
Completing the square needs to go as far as \(x - p = \pm\sqrt{q}\)

A1: Obtain both correct \(x\) values
Or one correct \((x, y)\) coordinate following a correct factorisation oe

A1: Obtain both correct pairs of coordinates
Allow as eg \(x = 1\), \(y = 0\) as long as pairings are clear

SC If no method shown for solving quadratic then allow
M1 for obtaining 3 term quadratic
A1 for \(x = 1\), \(x = 3\)
A1 for \((1, 0)\) and \((3, 4)\)

SC If no method at all shown then allow B1 for both \((1, 0)\) and \((3, 4)\)

The two curves with the straight line y = 2x minus 2 drawn through both points where the curves meet, crossing the negative y-axis — Mark scheme (b)

The two curves and the line, with the region R shaded: between the line y = 2x minus 2 and the U-shaped curve, from x = 1 to x = 3 — Mark scheme (c)

Scheme	Marks	AO
Correct region labelled with \(R\), or otherwise clearly identified	B1FT	2.2a
	[1]