June 2022 Paper 3 Q10

AQACurrent spec13 marksDifferentiation Functions (including |mod|)

10 The function f is defined by

\[\mathrm{f}(x) = \frac{x^2 + 10}{2x + 5}\]

where f has its maximum possible domain.

The curve \(y = \mathrm{f}(x)\) intersects the line \(y = x\) at the points \(P\) and \(Q\) as shown below.

The curve y = f(x) with a vertical asymptote just left of the y-axis; the right-hand branch has a minimum at Q, just above and right of O, then rises gently; the left-hand branch has a maximum at P in the third quadrant then falls steeply; the line y = x passes through P and Q

(a) State the value of \(x\) which is not in the domain of f. [1 mark]

(b) Explain how you know that the function f is many-to-one. [2 marks]

(c)

(i) Show that the \(x\)-coordinates of \(P\) and \(Q\) satisfy the equation\[x^2 + 5x - 10 = 0\] [2 marks]

(ii) Hence, find the exact \(x\)-coordinate of \(P\) and the exact \(x\)-coordinate of \(Q\). [1 mark]

(d) Show that \(P\) and \(Q\) are stationary points of the curve.

Fully justify your answer. [5 marks]

(e) Using set notation, state the range of f. [2 marks]

Scheme	Marks	AO
States \(-2.5\) OE	B1	2.2a
	(1)

Mark scheme (b)

Scheme	Marks	AO
Explains that many-to-one function is when distinct values of \(x\) give the same value for \(y\)	E1	2.4
Uses the shape of the graph to justify their answer or gives an example of two \(x\) values eg \(\mathrm{f}(0) = \mathrm{f}(4)\) or states turning or minimum or maximum points indicate many-to-one	E1	2.4
	(2)

Typical solution

Many-to-one function is when two or more \(x\) values give the same \(y\) value.

This graph is many-to-one because you can draw a horizontal line and it will cross the graph twice.

Mark scheme (c)

Scheme	Marks	AO
(i) Equates \(x\) and \(\dfrac{x^2 + 10}{2x + 5}\)	M1	3.1a
Rearranges with at least one intermediate step to obtain quadratic equation AG Condone \(0 = x^2 + 5x - 10\)	R1	2.1
	(2)
(ii) Obtains \(\dfrac{-5 \pm \sqrt{65}}{2}\) Ignore any labels ISW	B1	1.1b
	(1)

Typical solution

(i)

\[x = \frac{x^2 + 10}{2x + 5}\]\[x(2x + 5) = x^2 + 10\]\[2x^2 + 5x = x^2 + 10\]\[x^2 + 5x - 10 = 0\]

(ii)

\[x = \frac{-5 \pm \sqrt{65}}{2}\]

Mark scheme (d)

Scheme	Marks	AO
Uses quotient rule to obtain an expression in the form of \(\dfrac{Ax(2x + 5) + B(x^2 + 10)}{(2x + 5)^2}\) or uses product rule to obtain an expression in the form of \(Cx(2x + 5)^{-1} + D(x^2 + 10)(2x + 5)^{-2}\) or uses implicit differentiation to obtain an equation of the form \(Ax\dfrac{\mathrm{d}y}{\mathrm{d}x} + By + C\dfrac{\mathrm{d}y}{\mathrm{d}x} = Dx\) \(A\), \(B\), \(C\) and \(D\) can be any values but not 0 Condone missing brackets	M1	3.1a
Obtains fully correct \(\mathrm{f}^{\prime}(x)\) or obtains \(2x\frac{\mathrm{d}y}{\mathrm{d}x} + 2y + 5\frac{\mathrm{d}y}{\mathrm{d}x} = 2x\) ACF May be unsimplified	A1	1.1b
Equates their \(\mathrm{f}^{\prime}(x)\) or their numerator of \(\mathrm{f}^{\prime}(x)\) to 0 or sets \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) = 0	M1	1.1a
Rearranges to obtain \(x^2 + 5x - 10 = 0\) or \(2x^2 + 10x - 20 = 0\) and links it to the equation in part c(i) or their answer to c(ii) or solves their quadratic \(\mathrm{f}^{\prime}(x) = 0\) correctly or deduces \(y = x\) and substitutes to get \(x = \dfrac{x^2 + 10}{2x + 5}\) then rearranges to get \(x^2 + 5x - 10 = 0\)	M1	1.1a
Completes a reasoned argument by using \(x = \dfrac{-5 \pm \sqrt{65}}{2}\) to conclude that \(P\) and \(Q\) are stationary points CSO Must have brackets correct throughout	R1	2.1
	(5)

Typical solution

\[\mathrm{f}^{\prime}(x) = \frac{2x(2x + 5) - 2(x^2 + 10)}{(2x + 5)^2}\]\[= \frac{2x^2 + 10x - 20}{(2x + 5)^2}\]\[\mathrm{f}^{\prime}(x) = 0 \Leftrightarrow 2x^2 + 10x - 20 = 0\]\[x^2 + 5x - 10 = 0\]

This is the same equation solved in part c(i) so \(P\) and \(Q\) must be stationary points.

Mark scheme (e)

Scheme	Marks	AO
Deduces critical regions from their answer to c(ii) condone strict inequalities or poor notation or decimal values	M1	2.2a
Writes correct range in correct set notation eg \(\left(-\infty, \dfrac{-5 - \sqrt{65}}{2}\right] \cup \left[\dfrac{-5 + \sqrt{65}}{2}, \infty\right)\) Accept other letters for \(x\) or using f(x) provided consistent throughout set Follow through their answer to c(ii)	A1F	2.5
	(2)
	(13 marks)

Typical solution

\[x \leqslant \frac{-5 - \sqrt{65}}{2} \quad \text{and} \quad x \geqslant \frac{-5 + \sqrt{65}}{2}\]\[\left\{x : x \leqslant \frac{-5 - \sqrt{65}}{2}\right\} \cup \left\{x : x \geqslant \frac{-5 + \sqrt{65}}{2}\right\}\]