A2 June 2024 Paper 2 Q16

AQACurrent spec9 marksGraphs & Inequalities Integration

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

\[\mathrm{f}(x) = \frac{ax + 5}{x + b}\]

where \(a\) and \(b\) are constants.

The graph of \(y = \mathrm{f}(x)\) has asymptotes \(x = -2\) and \(y = 3\)

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

(b) The diagram shows the graph of \(y = \mathrm{f}(x)\) and its asymptotes.

The shaded region \(R\) is enclosed by the graph of \(y = \mathrm{f}(x)\), the \(x\)-axis and the \(y\)-axis.

Graph of y = f(x) with dashed asymptotes x = −2 and y = 3; the left branch lies above y = 3, and the right branch rises from below the x-axis near x = −2, crosses the x-axis between −2 and O and approaches y = 3; the region R between this branch, the x-axis and the y-axis is shaded

(i) The shaded region \(R\) is rotated through 360° about the \(x\)-axis to form a solid.

Find the volume of this solid.

Give your answer to three significant figures. [3 marks]

(ii) The shaded region \(R\) is rotated through 360° about the \(y\)-axis to form a solid.

Find the volume of this solid.

Give your answer to three significant figures. [4 marks]

Scheme	Marks	AO
Deduces that \(a = 3\)	B1	2.2a
Deduces that \(b = 2\)	B1	2.2a
	(2)

Mark scheme (b)

Scheme	Marks	AO
(i) Deduces that \(x\)-intercept is \(-\dfrac{5}{3}\). PI by correct answer.	B1	2.2a
(i) Uses \(\pi\displaystyle\int y^2\,\mathrm{d}x\) Condone missing \(\mathrm{d}x\) and missing/incorrect limits. PI by correct answer.	M1	1.1a
(i) Obtains AWRT 21.2	A1	1.1b
	(3)
(ii) Deduces that \(y\)-intercept \(= 2.5\) PI correct answer.	B1	2.2a
(ii) Deduces an expression for \(x\) in terms of \(y\) PI correct answer.	M1	2.2a
(ii) Uses \(\pi\displaystyle\int x^2\,\mathrm{d}y\) Condone missing \(\mathrm{d}y\), use of \(\mathrm{d}x\) and missing/incorrect limits. PI by correct answer.	M1	1.1a
(ii) Obtains AWRT 14.1	A1	1.1b
	(4)
	(9 marks)

Typical solution

(i)

\[\mathrm{f}(x) = \frac{3x + 5}{x + 2}\]\[x\text{-intercept} = -\frac{5}{3}\]\[V = \pi\int_{-\frac{5}{3}}^{0} \left(\frac{3x + 5}{x + 2}\right)^2 \mathrm{d}x\]\[= 21.2040\ldots\]\[= 21.2\]

(ii)

\[y\text{-intercept} = 2.5\]\[y = \frac{3x + 5}{x + 2}\]\[yx + 2y = 3x + 5\]\[x(y - 3) = 5 - 2y\]\[x = \frac{5 - 2y}{y - 3}\]\[V = \pi\int_{0}^{2.5} \left(\frac{5 - 2y}{y - 3}\right)^2 \mathrm{d}y\]\[= 14.1360\ldots\]\[= 14.1\]