C3 June 2010 Q6

EdexcelOld spec10 marksFunctions (including |mod|)Sketching & Transforming Graphs

Figure 2: U-shaped curve y = f(x) through (0, 5) with minimum point A(3, -4) — Figure 2

Figure 2 shows a sketch of the curve with the equation \(y = \mathrm{f}(x)\), \(x \in \mathbb{R}\).
The curve has a turning point at \(A(3, -4)\) and also passes through the point \((0, 5)\).

(a) Write down the coordinates of the point to which \(A\) is transformed on the curve with equation

(i) \(y = \left|\mathrm{f}(x)\right|\),

(ii) \(y = 2\mathrm{f}\left(\tfrac{1}{2}x\right)\).

(4)

(b) Sketch the curve with equation\[y = \mathrm{f}(\left|x\right|)\]On your sketch show the coordinates of all turning points and the coordinates of the point at which the curve cuts the \(y\)-axis. (3)

The curve with equation \(y = \mathrm{f}(x)\) is a translation of the curve with equation \(y = x^2\).

(d) Explain why the function f does not have an inverse. (1)

Scheme	Marks
(i) \((3, 4)\)	B1 B1
(ii) \((6, -8)\)	B1 B1
	(4)

Sketch of y = f(|x|): W shape with cusp at (0, 5) and minimum points (-3, -4) and (3, -4) — Mark scheme (b)

Scheme	Marks
	B1 B1 B1
	(3)

Mark scheme (c)

Scheme	Marks
\(\mathrm{f}(x) = (x - 3)^2 - 4\) or \(\mathrm{f}(x) = x^2 - 6x + 5\)	M1A1
	(2)

Notes

(c) M1: Either states \(\mathrm{f}(x)\) in the form \((x \pm \alpha)^2 \pm \beta\); \(\alpha, \beta \ne 0\)
Or uses a complete method on \(\mathrm{f}(x) = x^2 + ax + b\), with \(\mathrm{f}(0) = 5\) and \(\mathrm{f}(3) = -4\) to find both \(a\) and \(b\).

A1: Either \((x - 3)^2 - 4\) or \(x^2 - 6x + 5\)

Mark scheme (d)

Scheme	Marks
Either: The function f is a many-one {mapping}. Or: The function f is not a one-one {mapping}.	B1
	(1)
	(10 marks)

Notes

(d) B1: Or: The inverse is a one-many {mapping and not a function}.
Or: Because \(\mathrm{f}(0) = 5\) and also \(\mathrm{f}(6) = 5\).
Or: One \(y\)-coordinate has 2 corresponding \(x\)-coordinates {and therefore cannot have an inverse}.