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Date	November 2020	Marks available	2	Reference code	20N.1.SL.TZ0.S_10
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Show that	Question number	S_10	Adapted from	N/A

Question

The following diagram shows part of the graph of $f (x) = \frac{k}{x}$ , for $x > 0, k > 0$ .

Let $P (p, \frac{k}{p})$ be any point on the graph of $f$ . Line $L_{1}$ is the tangent to the graph of $f$ at $P$ .

Line $L_{1}$ intersects the $x$ -axis at point $A (2 p, 0)$ and the $y$ -axis at point B.

Find $f' (p)$ in terms of $k$ and $p$ .

[2]

a.i.

Show that the equation of $L_{1}$ is $k x + p^{2} y - 2 p k = 0$ .

[2]

a.ii.

Find the area of triangle $AOB$ in terms of $k$ .

[5]

The graph of $f$ is translated by $(\begin{matrix} 4 \\ 3 \end{matrix})$ to give the graph of $g$ .
In the following diagram:

point $Q$ lies on the graph of $g$
points $C$ , $D$ and $E$ lie on the vertical asymptote of $g$
points $D$ and $F$ lie on the horizontal asymptote of $g$
point $G$ lies on the $x$ -axis such that $FG$ is parallel to $DC$ .

Line $L_{2}$ is the tangent to the graph of $g$ at $Q$ , and passes through $E$ and $F$ .

Given that triangle $EDF$ and rectangle $CDFG$ have equal areas, find the gradient of $L_{2}$ in terms of $p$ .

[6]

Markscheme

$f' (x) = - k x^{- 2}$ (A1)

$f' (p) = - k p^{- 2} (= - \frac{k}{p^{2}})$ A1 N2

[2 marks]

a.i.

attempt to use point and gradient to find equation of $L_{1}$ M1

eg $y - \frac{k}{p} = - k p^{- 2} (x - p), \frac{k}{p} = - \frac{k}{p^{2}} (p) + b$

correct working leading to answer A1

eg $p^{2} y - k p = - k x + k p, y - \frac{k}{p} = - \frac{k}{p^{2}} x + \frac{k}{p}, y = - \frac{k}{p^{2}} x + \frac{2 k}{p}$

$k x + p^{2} y - 2 p k = 0$ AG N0

[2 marks]

a.ii.

METHOD 1 – area of a triangle

recognizing $x = 0$ at $B$ (M1)

correct working to find $y$ -coordinate of null (A1)

eg $p^{2} y - 2 p k = 0$

$y$ -coordinate of null at $y = \frac{2 k}{p}$ (may be seen in area formula) A1

correct substitution to find area of triangle (A1)

eg $\frac{1}{2} (2 p) (\frac{2 k}{p}), p \times (\frac{2 k}{p})$

area of triangle $AOB = 2 k$ A1 N3

METHOD 2 – integration

recognizing to integrate $L_{1}$ between $0$ and $2 p$ (M1)

eg $\int_{0}^{2 p} L_{1} d x, \int_{0}^{2 p} - \frac{k}{p^{2}} x + \frac{2 k}{p}$

correct integration of both terms A1

eg $- \frac{k x^{2}}{2 p^{2}} + \frac{2 k x}{p}, - \frac{k}{2 p^{2}} x^{2} + \frac{2 k}{p} x + c, {[- \frac{k}{2 p^{2}} x^{2} + \frac{2 k}{p} x]}_{0}^{2 p}$

substituting limits into their integrated function and subtracting (in either order) (M1)

eg $- \frac{k {(2 p)}^{2}}{2 p^{2}} + \frac{2 k (2 p)}{p} - (0), - \frac{4 k p^{2}}{2 p^{2}} + \frac{4 k p}{p}$

correct working (A1)

eg $- 2 k + 4 k$

area of triangle $AOB = 2 k$ A1 N3

[5 marks]

Note: In this question, the second M mark may be awarded independently of the other marks, so it is possible to award (M0)(A0)M1(A0)(A0)A0.

recognizing use of transformation (M1)

eg area of triangle $AOB$ = area of triangle $DEF$ , $g (x) = \frac{k}{x - 4} + 3,$ gradient of $L_{2} =$ gradient of $L_{1}$ , $D (4, 3), 2p+4,$ one correct shift

correct working (A1)

eg area of triangle $DEF = 2 k, CD = 3, DF = 2 p, CG = 2 p, E (4, \frac{2 k}{p} + 3), F (2 p + 4, 3), Q (p + 4, \frac{k}{p} + 3),$

gradient of $L_{2} = - \frac{k}{p^{2}}, g' (x) = - \frac{k}{{(x - 4)}^{2}},$ area of rectangle $CDFG = 2 k$

valid approach (M1)

eg $\frac{ED \times DF}{2} = CD \times DF, 2 p \cdot 3 = 2 k, ED = 2 CD, \int_{4}^{2 p + 4} L_{2} d x = 4 k$

correct working (A1)

eg $ED = 6, E (4, 9), k = 3 p, gradient = \frac{3 - (\frac{2 k}{p} + 3)}{(2 p + 4) - 4}, \frac{- 6}{(\frac{2 k}{3})}, - \frac{9}{k}$

correct expression for gradient (in terms of $p$ ) (A1)

eg $\frac{- 6}{2 p}, \frac{9 - 3}{4 - (2 p + 4)}, - \frac{3 p}{p^{2}}, \frac{3 - (\frac{2 (3 p)}{p} + 3)}{(2 p + 4) - 4}, - \frac{9}{3 p}$

gradient of $L_{2}$ is $- \frac{3}{p} (= - 3 p^{- 1})$ A1 N3

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.3—Differentiating polynomials, n E Z

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18N.1.SL.TZ0.S_10c:
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22M.1.SL.TZ2.7d:
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19M.1.AHL.TZ1.H_7:
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17N.1.SL.TZ0.S_8d:
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17M.2.AHL.TZ1.H_12g.ii:
Hence, show that there are no solutions to $({g^{ - 1}})'(x) = 0$ .
16N.1.SL.TZ0.T_14a:
Find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ .
18M.1.AHL.TZ2.H_6b:
Hence, or otherwise, find $\int\limits_0^\pi {{{\text{e}}^{ - x}}\,{\text{sin}}\,x\,{\text{d}}x}$ .
18N.2.SL.TZ0.T_6g:
Use your answer to part (f) to show that the width of Nanako’s bag is 12 cm.
19M.1.SL.TZ1.T_15c:
Find the value of $r$ when the area of the curved surface is maximized.
18N.2.SL.TZ0.T_6d:
Write down and simplify an expression in x and y for the area of cloth, A, used to make Nanako’s bag.
19M.1.SL.TZ2.T_15b:
Differentiate $P\left( x \right)$ .
20N.1.SL.TZ0.T_13c:
Find the value of $k$ .
17M.2.SL.TZ2.T_6a:
Write down the $y$ -intercept of the graph.
17M.2.SL.TZ2.T_6c.i:
Show that $a = 8$ .
17M.2.SL.TZ2.T_6b:
Find $f'(x)$ .
20N.1.SL.TZ0.T_13a:
Write down $f' (x)$ .
19M.2.SL.TZ2.T_5h:
Determine the range of $f\left( x \right)$ for $p$ ≤ $x$ ≤ $q$ .
17N.2.SL.TZ0.T_5e:
Write down the coordinates of the point of intersection.
17M.2.SL.TZ2.T_6g:
The equation $f(x) = m$ , where $m \in \mathbb{R}$ , has four solutions. Find the possible values of $m$ .
16N.2.SL.TZ0.T_6d:
Show that $A = \pi {r^2} + \frac{{1\,000\,000}}{r}$ .
16N.2.SL.TZ0.T_6g:
Find the value of this minimum area.
19M.1.SL.TZ1.T_15b:
Find $\frac{{{\text{d}}A}}{{{\text{d}}r}}$ .
18N.2.SL.TZ0.T_6e:
Use your answers to parts (c) and (d) to show that

$A = 3{x^2} + \frac{{10368}}{x}$ .
16N.2.SL.TZ0.T_6a:
Write down a formula for $A$ , the surface area to be coated.
16N.2.SL.TZ0.T_6h:
Find the least number of cans of water-resistant material that will coat the area in part (g).
17N.2.SL.TZ0.T_5d:
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17M.2.SL.TZ2.T_6c.ii:
Find $f(2)$ .
17M.2.SL.TZ2.T_6d.ii:
Write down the intervals where the gradient of the graph of $y = f(x)$ is positive.
17M.2.SL.TZ2.T_6e:
Write down the range of $f(x)$ .
18N.2.SL.TZ0.T_6b:
Calculate the volume, in cm³, of the bag.
18N.2.SL.TZ0.T_6c:
Use this value to write down, and simplify, the equation in x and y for the volume of Nanako’s bag.
18N.2.SL.TZ0.T_6f:
Find $\frac{{{\text{d}}A}}{{{\text{d}}x}}$ .

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