AS June 2021 Paper 1 Q16

AQACurrent spec8 marksGraphs & Inequalities

16 Curve \(C\) has equation \(y = \dfrac{ax}{x + b}\) where \(a\) and \(b\) are constants.
The equations of the asymptotes to \(C\) are \(x = -2\) and \(y = 3\)

Graph of y = ax/(x + b) with a dashed vertical asymptote left of the y-axis and a dashed horizontal asymptote above the x-axis; the left branch lies above the horizontal asymptote and rises steeply towards the vertical asymptote; the right branch rises from below through O towards the horizontal asymptote

(a) Write down the value of \(a\) and the value of \(b\) [2 marks]

(b) The gradient of \(C\) at the origin is \(\dfrac{3}{2}\)

With reference to the graph, explain why there is exactly one root of the equation

\[\frac{ax}{x + b} = \frac{3x}{2}\]

[2 marks]

[4 marks]

Mark scheme (a)

Scheme	Marks	AO
Obtains \(a = 3\) Accept \(y = \dfrac{3x}{x + b}\) with any non-zero value of \(b\)	B1	2.2a
Obtains \(b = 2\) Accept \(y = \dfrac{ax}{x + 2}\) with any non-zero value of \(a\)	B1	2.2a
	(2)

Typical solution

\[y = \frac{3x}{x + 2}\]\[a = 3 \quad \text{and} \quad b = 2\]

Mark scheme (b)

Scheme	Marks	AO
Explains that \(y = \frac{3x}{2}\) is a tangent to the curve. Condone ‘tangent’ only.	B1	2.4
Explains that a tangent to a hyperbola meets the curve exactly once. Accept ‘conic’ instead of ‘hyperbola’. Condone no conclusion given.	B1	2.4
	(2)

Typical solution

\(y = \frac{3x}{2}\) is a tangent to the hyperbola.

Any tangent to a hyperbola only intersects once.

So there is exactly one root of the equation.

Mark scheme (c)

Scheme	Marks	AO
Forms a simplified quadratic equation from \(ax - (1 - x)(x + b) = 0\) Accept any inequality sign instead of = PI by \(x^2 + 4x - 2\) or \(-x^2 - 4x + 2\) PI by \(-2 + \sqrt{6}\) or \(-2 - \sqrt{6}\) Or forms a simplified cubic equation from \(ax(x + 2) - (1 - x)(x + b)(x + 2) = 0\)	M1	1.1a
Identifies \(-2 + \sqrt{6}\) or \(-2 - \sqrt{6}\) as a critical value. Follow through their \(a\) and \(b\)	A1F	1.1b
Identifies \(-2\) as a critical value.	B1	1.1b
Obtains the correct set of values of \(x\)	A1	3.2a
	(4)
	(8 marks)

Typical solution

\[\frac{3x}{x + 2} = 1 - x\]\[3x = (1 - x)(x + 2)\]\[3x = -x^2 - x + 2\]\[x^2 + 4x - 2 = 0\]\[x = \frac{-4 \pm \sqrt{16 + 4 \times 2}}{2}\]\[x = -2 \pm \sqrt{6}\]\[x \leqslant -2 - \sqrt{6},\]\[-2 \lt x \leqslant -2 + \sqrt{6}\]