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Date	November 2020	Marks available	3	Reference code	20N.1.AHL.TZ0.H_12
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Sketch and State	Question number	H_12	Adapted from	N/A

Question

Consider the function defined by $f (x) = \frac{k x - 5}{x - k}$ , where $x \in ℝ \ \{k\}$ and $k^{2} \neq 5$ .

Consider the case where $k = 3$ .

State the equation of the vertical asymptote on the graph of $y = f (x)$ .

[1]

State the equation of the horizontal asymptote on the graph of $y = f (x)$ .

[1]

Use an algebraic method to determine whether $f$ is a self-inverse function.

[4]

Sketch the graph of $y = f (x)$ , stating clearly the equations of any asymptotes and the coordinates of any points of intersections with the coordinate axes.

[3]

The region bounded by the $x$ -axis, the curve $y = f (x)$ , and the lines $x = 5$ and $x = 7$ is rotated through $2 π$ about the $x$ -axis. Find the volume of the solid generated, giving your answer in the form $π (a + b \ln 2)$ , where $a, b \in ℤ$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$x = k$ A1

[1 mark]

$y = k$ A1

[1 mark]

METHOD 1

$(f \circ f) (x) = \frac{k (\frac{k x - 5}{x - k}) - 5}{(\frac{k x - 5}{x - k}) - k}$ M1

$= \frac{k (k x - 5) - 5 (x - k)}{k x - 5 - k (x - k)}$ A1

$= \frac{k^{2} x - 5 k - 5 x + 5 k}{k x - 5 - k x + k^{2}}$

$= \frac{k^{2} x - 5 x}{k^{2} - 5}$ A1

$= \frac{x (k^{2} - 5)}{k^{2} - 5}$

$= x$

$(f \circ f) (x) = x$ , (hence $f$ is self-inverse) R1

Note: The statement $f (f (x)) = x$ could be seen anywhere in the candidate’s working to award R1.

METHOD 2

$f (x) = \frac{k x - 5}{x - k}$

$x = \frac{k y - 5}{y - k}$ M1

Note: Interchanging $x$ and $y$ can be done at any stage.

$x (y - k) = k y - 5$ A1

$x y - x k = k y - 5$

$x y - k y = x k - 5$

$y (x - k) = k x - 5$ A1

$y = f^{- 1} (x) = \frac{k x - 5}{x - k}$ (hence $f$ is self-inverse) R1

[4 marks]

attempt to draw both branches of a rectangular hyperbola M1

$x = 3$ and $y = 3$ A1

$(0, \frac{5}{3})$ and $(\frac{5}{3}, 0)$ A1

[3 marks]

METHOD 1

$volume = π \int_{5}^{7} {(\frac{3 x - 5}{x - 3})}^{2} d x$ (M1)

EITHER

attempt to express $\frac{3 x - 5}{x - 3}$ in the form $p + \frac{q}{x - 3}$ M1

$\frac{3 x - 5}{x - 3} = 3 + \frac{4}{x - 3}$ A1

attempt to expand ${(\frac{3 x - 5}{x - 3})}^{2}$ or ${(3 x - 5)}^{2}$ and divide out M1

${(\frac{3 x - 5}{x - 3})}^{2} = 9 + \frac{24 x - 56}{{(x - 3)}^{2}}$ A1

THEN

${(\frac{3 x - 5}{x - 3})}^{2} = 9 + \frac{24}{x - 3} + \frac{16}{{(x - 3)}^{2}}$ A1

$volume = π \int_{5}^{7} (9 + \frac{24}{x - 3} + \frac{16}{{(x - 3)}^{2}}) d x$

$= π {[9 x + 24 \ln (x - 3) - \frac{16}{x - 3}]}_{5}^{7}$ A1

$= π ⌊(63 + 24 \ln 4 - 4) - (45 + 24 \ln 2 - 8)⌋$

$= π (22 + 24 \ln 2)$ A1

METHOD 2

$volume = π \int_{5}^{7} {(\frac{3 x - 5}{x - 3})}^{2} d x$ (M1)

substituting $u = x - 3 \Rightarrow \frac{d u}{d x} = 1$ A1

$3 x - 5 = 3 (u + 3) - 5 = 3 u + 4$

$volume = π \int_{2}^{4} {(\frac{3 u + 4}{u})}^{2} d u$ M1

$= π \int_{2}^{4} 9 + \frac{16}{u^{2}} + \frac{24}{u} d u$ A1

$= π {[9 u - \frac{16}{u} + 24 \ln u]}_{2}^{4}$ A1

Note: Ignore absence of or incorrect limits seen up to this point.

$= π (22 + 24 \ln 2)$ A1

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.2—Functions, notation domain, range and inverse as reflection

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