October 2021 Paper 3 Q4

OCR ACurrent spec5 marksFunctions (including |mod|)Sketching & Transforming Graphs

(a) Sketch, on a single diagram, the following graphs.

\(y = |x - 1|\)
\(y = \dfrac{k}{x}\), where \(k\) is a negative constant

[2]

(b) Hence explain why the equation \(x|x - 1| = k\) has exactly one real root for any negative value of \(k\). [1]

Sketch: V-shaped graph y = |x − 1| with vertex on the x-axis at x = 1; hyperbola y = k/x with negative k, one branch in the second quadrant and one in the fourth quadrant, not meeting the axes — Mark scheme (a)

Scheme	Marks	AO
	B1 B1	1.1 1.1
	[2]

Mark scheme (b)

Scheme	Marks	AO
The graphs in (a) intersect at only point (for any negative values of \(k\)) and therefore \(\|x - 1\| = \dfrac{k}{x} \Rightarrow x\|x - 1\| = k\) has exactly one real root	B1	2.4
	[1]

Notes

B1: Dependent on both marks in (a) – must mention that the solution of the equation \(x|x - 1| = k\) corresponds to where the two graphs in (a) intersect (so just stating that the graphs in (a) intersect at only one point is B0)

Mark scheme (c)

Scheme	Marks	AO
\(x\|x - 1\| = -6 \Rightarrow x(1 - x) = -6\) \(x^2 - x - 6 = 0\)	M1	3.1a
\(x = -2\)	A1	2.2a
	[2]

Notes

M1: Uses graph and sets up quadratic (oe) – allow if \(x^2 - x + 6 = 0\) stated as well (but M0 if this is the only quadratic (oe) considered)
Or setting up a four-term quartic from \((x - 1)^2 = 36x^{-2}\)

A1: BC \(x = -2\) only (www)

SC If no marks awarded, then B1 for \(x = -2\) only and then B1 for explicitly showing that \(-2\lvert -2 - 1\rvert = -2(3) = -6\)