A2 June 2020 Paper 1 Q9

AQACurrent spec13 marksGraphs & Inequalities

9 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = \frac{x(x + 3)}{x + 4} \qquad (x \in \mathbb{R},\ x \neq -4)\]

(a) Find the interval \((a, b)\) in which \(\mathrm{f}(x)\) does not take any values.

Fully justify your answer. [5 marks]

(b) Find the coordinates of the two stationary points of the graph of \(y = \mathrm{f}(x)\) [2 marks]

(d) Sketch the graph of \(y = \mathrm{f}(x)\) on the axes below. [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Forms an equation \(\dfrac{x(x + 3)}{x + 4} = {}\)‘\(k\)’ or Differentiates using quotient rule.	M1	3.1a
Rearranges their equation into a quadratic in \(x\). or Obtains the correct \(\mathrm{f}^{\prime}(x) = \dfrac{(x + 4) \times (2x + 3) - (x^2 + 3x) \times 1}{(x + 4)^2}\).	M1	1.1a
Explains that the discriminant of this quadratic \(\lt 0\) or Explains that because there is a vertical asymptote the minimum is higher up the graph than the maximum and the two turning points lie on different branches of the graph.	E1	2.4
Forms a quadratic equation or inequality in ‘\(k\)’ from their discriminant. or Equates their \(\mathrm{f}^{\prime}(x)\) or its numerator to 0 and solves	M1	1.1a
Completes a rigorous argument to show that \(\mathrm{f}(x)\) does not take any values in the interval \((-9, -1)\). Condone \(-9 \lt k \lt -1\)	R1	2.1

Typical solution

Assume \(y = k\) meets \(y = \dfrac{x(x + 3)}{x + 4}\)

Then

\[\begin{aligned} k(x + 4) &= x(x + 3) \\ kx + 4k &= x^2 + 3x \\ x^2 + (3 - k)x - 4k &= 0 \end{aligned}\]\[\Delta \lt 0 \therefore (3 - k)^2 - 4(1)(-4k) \lt 0\]\[\begin{aligned} k^2 + 10k + 9 &\lt 0 \\ (k + 9)(k + 1) &\lt 0 \\ -9 \lt k &\lt -1 \end{aligned}\]

So \(\mathrm{f}(x)\) does not take any values in the interval \((-9, -1)\)

Mark scheme (b)

Scheme	Marks	AO
Substitutes their \(-9\) or \(-1\) into \(y = f(x)\) and forms a quadratic in \(x\) or Differentiates using the quotient rule and equates their \(\mathrm{f}^{\prime}(x)\) or its numerator to 0	M1	1.1a
Finds the coordinates of both stationary points	A1F	1.1b

Typical solution

Stationary points \((-6, -9)\) and \((-2, -1)\)

Mark scheme (c)

Scheme	Marks	AO
Divides the numerator by \(x + 4\) and obtains \(\mathrm{f}(x) = x + \cdots\)	M1	3.1a
Obtains the correct equation of the asymptote \(y = x - 1\)	A1	1.1b

Typical solution

\[\begin{aligned} \mathrm{f}(x) = \frac{x^2 + 3x}{x + 4} &= \frac{(x + 4)(x - 1) + 4}{x + 4} \\ &= x - 1 + \frac{4}{x + 4} \end{aligned}\]

Asymptote is \(y = x - 1\)

Sketch of y = f(x) with asymptotes x = −4 and y = x − 1: the right branch comes down beside x = −4, crosses the x-axis at −3, has a minimum at (−2, −1), passes through O and approaches y = x − 1; the left branch rises to a maximum at (−6, −9) and falls beside x = −4 — Mark scheme (d)

Scheme	Marks	AO
Draws a curve asymptotic to \(x = -4\) or their \(y = x - 1\)	B1F	1.1b
Draws a curve with two branches asymptotic to their asymptotes	B1F	1.1b
Draws one branch of the curve in the correct position	M1	1.1b
Deduces the correct curve and draws a fully correct sketch, with roots at \(-3\) and \(0\) FT their oblique asymptote	A1F	2.2a
	(13 marks)