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Date	May 2022	Marks available	10	Reference code	22M.1.SL.TZ2.8
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Find	Question number	8	Adapted from	N/A

Question

Consider the functions $f (x) = \frac{1}{x - 4} + 1$ , for $x \neq 4$ , and $g (x) = x - 3$ for $x \in ℝ$ .

The following diagram shows the graphs of $f$ and $g$ .

The graphs of $f$ and $g$ intersect at points $A$ and $B$ . The coordinates of $A$ are $(3, 0)$ .

In the following diagram, the shaded region is enclosed by the graph of $f$ , the graph of $g$ , the $x$ -axis, and the line $x = k$ , where $k \in ℤ$ .

The area of the shaded region can be written as $\ln (p) + 8$ , where $p \in ℤ$ .

Find the coordinates of $B$ .

[5]

Find the value of $k$ and the value of $p$ .

[10]

Markscheme

$\frac{1}{x - 4} + 1 = x - 3$ (M1)

$x^{2} - 8 x + 15 = 0$ OR ${(x - 4)}^{2} = 1$ (A1)

valid attempt to solve their quadratic (M1)

$(x - 3) (x - 5) = 0$ OR $x = \frac{8 \pm \sqrt{8^{2} - 4 (1) (15)}}{2 (1)}$ OR $(x - 4) = \pm 1$

$x = 5 (x = 3, x = 5)$ (may be seen in answer) A1

$B (5, 2)$ (accept $x = 5, y = 2$ ) A1

[5 marks]

recognizing two correct regions from $x = 3$ to $x = 5$ and from $x = 5$ to $x = k$ (R1)

triangle $+ \int_{5}^{k} f (x) d x$ OR $\int_{3}^{5} g (x) d x + \int_{5}^{k} f (x) d x$ OR $\int_{3}^{5} (x - 3) d x + \int_{5}^{k} (\frac{1}{x - 4} + 1) d x$

area of triangle is $2$ OR $\frac{2 \cdot 2}{2}$ OR $(\frac{5^{2}}{2} - 3 (5)) - (\frac{3^{2}}{2} - 3 (3))$ (A1)

correct integration (A1)(A1)

$\int (\frac{1}{x - 4} + 1) d x = \ln (x - 4) + x (+ C)$

Note: Award A1 for $\ln (x - 4)$ and A1 for $x$ .
Note: The first three A marks may be awarded independently of the R mark.

substitution of their limits (for $x$ ) into their integrated function (in terms of $x$ ) (M1)

$\ln (k - 4) + k - (\ln 1 + 5)$

${[\ln (x - 4) + x]}_{5}^{k} = \ln (k - 4) + k - 5$ A1

adding their two areas (in terms of $k$ ) and equating to $\ln p + 8$ (M1)

$2 + \ln (k - 4) + k - 5 = \ln p + 8$

equating their non-log terms to $8$ (equation must be in terms of $k$ ) (M1)

$k - 3 = 8$

$k = 11$ A1

$11 - 4 = p$

$p = 7$ A1

[10 marks]

Examiners report

Nearly all candidates knew to set up an equation with $f (x) = g (x)$ in order to find the intersection of the two graphs, and most were able to solve the resulting quadratic equation. Candidates were not as successful in part (b), however. While some candidates recognized that there were two regions to be added together, very few were able to determine the correct boundaries of these regions, with many candidates integrating one or both functions from $x = 3$ to $x = k$ . While a good number of candidates were able to correctly integrate the function(s), without the correct bounds the values of $k$ and $p$ were unattainable.

[N/A]

Syllabus sections

Topic 1—Number and algebra » SL 1.5—Intro to logs

Topic 1—Number and algebra » SL 1.7—Laws of exponents and logs

Topic 5 —Calculus » SL 5.10—Indefinite integration, reverse chain, by substitution

Topic 5 —Calculus » SL 5.11—Definite integrals, areas under curve onto x-axis and areas between curves

Topic 1—Number and algebra

Topic 5 —Calculus

Question

Markscheme

Examiners report

Syllabus sections

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