IB DP Maths Topic 2.6 Logarithmic functions and their graphs: x↦logax SL Paper 2

 

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Question

Let \(f(x) = {\log _3}\frac{x}{2} + {\log _3}16 – {\log _3}4\) , for \(x > 0\) .

Show that \(f(x) = {\log _3}2x\) .

[2]
a.

Find the value of \(f(0.5)\) and of \(f(4.5)\) .

[3]
b.

The function f can also be written in the form \(f(x) = \frac{{\ln ax}}{{\ln b}}\) .

(i)     Write down the value of a and of b .

(ii)    Hence on graph paper, sketch the graph of f , for \( – 5 \le x \le 5\) , \( – 5 \le y \le 5\) , using a scale of 1 cm to 1 unit on each axis.

(iii)   Write down the equation of the asymptote.

[6]
c(i), (ii) and (iii).

Write down the value of \({f^{ – 1}}(0)\) .

[1]
d.

The point A lies on the graph of f . At A, \(x = 4.5\) .

On your diagram, sketch the graph of \({f^{ – 1}}\) , noting clearly the image of point A.

[4]
e.
Answer/Explanation

Markscheme

combining 2 terms     (A1)

e.g. \({\log _3}8x – {\log _3}4\) , \({\log _3}\frac{1}{2}x + {\log _3}4\)

expression which clearly leads to answer given     A1

e.g. \({\log _3}\frac{{8x}}{4}\) , \({\log _3}\frac{{4x}}{2}\)

\(f(x) = {\log _3}2x\)     AG     N0

[2 marks]

a.

attempt to substitute either value into f     (M1)

e.g. \({\log _3}1\) , \({\log _3}9\)

\(f(0.5) = 0\) , \(f(4.5) = 2\)     A1A1     N3

[3 marks]

b.

(i) \(a = 2\) , \(b = 3\)     A1A1     N1N1

(ii)
     A1A1A1     N3

Note: Award A1 for sketch approximately through \((0.5 \pm 0.1{\text{, }}0 \pm 0.1)\) , A1 for approximately correct shape, A1 for sketch asymptotic to the y-axis.

(iii) \(x = 0\) (must be an equation)     A1     N1

[6 marks]

c(i), (ii) and (iii).

\({f^{ – 1}}(0) = 0.5\)     A1     N1

[1 mark]

d.


     A1A1A1A1     N4

Note: Award A1 for sketch approximately through \((0 \pm 0.1{\text{, }}0.5 \pm 0.1)\) , A1 for approximately correct shape of the graph reflected over \(y = x\) , A1 for sketch asymptotic to x-axis, A1 for point \((2 \pm 0.1{\text{, }}4.5 \pm 0.1)\) clearly marked and on curve.

[4 marks]

e.

Question

Let \(f(x) = 2\ln (x – 3)\), for \(x > 3\). The following diagram shows part of the graph of \(f\).

Find the equation of the vertical asymptote to the graph of \(f\).

[2]
a.

Find the \(x\)-intercept of the graph of \(f\).

[2]
b.

The region enclosed by the graph of \(f\), the \(x\)-axis and the line \(x = 10\) is rotated \(360\)° about the \(x\)-axis. Find the volume of the solid formed.

[3]
c.
Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\;\;\;\)horizontal translation \(3\) units to the right

\(x = 3\) (must be an equation)     A1     N2

[2 marks]

a.

valid approach     (M1)

eg\(\;\;\;f(x) = 0,{\text{ }}{e^0} = x – 3\)

\(4,{\text{ }}x = 4,{\text{ }}(4,{\text{ }}0)\)     A1     N2

[2 marks]

b.

attempt to substitute either their correct limits or the function into formula involving \({f^2}\)      (M1)

eg\(\;\;\;\int_4^{10} {{f^2},{\text{ }}\pi \int {{{\left( {2\ln (x – 3)} \right)}^2}{\text{d}}x} } \)

\(141.537\)

volume = \(142\)     A2     N3

[3 marks]

Total [7 marks]

c.

Question

The price of a used car depends partly on the distance it has travelled. The following table shows the distance and the price for seven cars on 1 January 2010.

M16/5/MATME/SP2/ENG/TZ2/08

The relationship between \(x\) and \(y\) can be modelled by the regression equation \(y = ax + b\).

On 1 January 2010, Lina buys a car which has travelled \(11\,000{\text{ km}}\).

The price of a car decreases by 5% each year.

Lina will sell her car when its price reaches \(10\,000\) dollars.

(i)     Find the correlation coefficient.

(ii)     Write down the value of \(a\) and of \(b\).

[4]
a.

Use the regression equation to estimate the price of Lina’s car, giving your answer to the nearest 100 dollars.

[3]
b.

Calculate the price of Lina’s car after 6 years.

[4]
c.

Find the year when Lina sells her car.

[4]
d.
Answer/Explanation

Markscheme

Note:     There may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working.

(i)     valid approach (M1)

eg\(\,\,\,\,\,\)correct value for \(r\) (or for \(a\) or \(b\) seen in (ii))

\( – 0.994347\)

\(r =  – 0.994\)     A1     N2

(ii)     \( – 1.58095,{\text{ }}33480.3\)

\(a =  – 1.58,{\text{ }}b = 33500\)     A1A1     N2

[4 marks]

a.

Note:     There may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working.

correct substitution into their regression equation

eg\(\,\,\,\,\,\)\( – 1.58095(11000){\text{ }} + 33480.3\)     (A1)

\(16\,089.85{\text{ }}(16\,120{\text{ from 3sf}})\)    (A1)

\({\text{price}} = 16\,100{\text{ }}({\text{dollars}})\) (must be rounded to the nearest 100 dollars)     A1     N3

[3 marks]

b.

Note:     There may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working.

METHOD 1

valid approach     (M1)

eg\(\,\,\,\,\,\)\(P \times {({\text{rate}})^t}\)

\({\text{rate}} = 0.95\) (may be seen in their expression)     (A1)

correct expression     (A1)

eg\(\,\,\,\,\,\)\(16100 \times {0.95^6}\)

\(11\,834.97\)

\(11\,800{\text{ }}({\text{dollars}})\)    A1     N2

METHOD 2

attempt to find all six terms     (M1)

eg\(\,\,\,\,\,\)\(\left( {\left( {(16\,100 \times 0.95) \times 0.95} \right) \ldots } \right) \times 0.95\), table of values

5 correct values (accept values that round correctly to the nearest dollar)

\(15\,295,{\text{ }}14\,530,{\text{ }}13\,804,{\text{ }}13\,114,{\text{ }}12\,458\)    A2

\(11\,835\)

\(11\,800{\text{ }}({\text{dollars}})\)     A1     N2

[4 marks]

c.

Note:     There may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working.

METHOD 1

correct equation     (A1)

eg\(\,\,\,\,\,\)\(16\,100 \times {0.95^x}{\text{ = }}10\,000\)

valid attempt to solve     (M1)

eg\(\,\,\,\,\,\)M16/5/MATME/SP2/ENG/TZ2/08.d/M, using logs

9.28453     (A1)

year 2019     A1     N2

METHOD 2

valid approach using table of values     (M1)

both crossover values (accept values that round correctly to the nearest dollar)     A2

eg\(\,\,\,\,\,\)\({\text{P}} = 10\,147{\text{ }}({\text{1 Jan 2019}}),{\text{ P}} = 9\,639.7{\text{ }}({\text{1 Jan 2020}})\)

year 2019     A1     N2

[4 marks]

d.

Question

Let \(f(x) = \ln x\) and \(g(x) = 3 + \ln \left( {\frac{x}{2}} \right)\), for \(x > 0\).

The graph of \(g\) can be obtained from the graph of \(f\) by two transformations:

\[\begin{array}{*{20}{l}} {{\text{a horizontal stretch of scale factor }}q{\text{ followed by}}} \\ {{\text{a translation of }}\left( {\begin{array}{*{20}{c}} h \\ k \end{array}} \right).} \end{array}\]

Let \(h(x) = g(x) \times \cos (0.1x)\), for \(0 < x < 4\). The following diagram shows the graph of \(h\) and the line \(y = x\).

M17/5/MATME/SP2/ENG/TZ1/10.b.c

The graph of \(h\) intersects the graph of \({h^{ – 1}}\) at two points. These points have \(x\) coordinates 0.111 and 3.31 correct to three significant figures.

Write down the value of \(q\);

[1]
a.i.

Write down the value of \(h\);

[1]
a.ii.

Write down the value of \(k\).

[1]
a.iii.

Find \(\int_{0.111}^{3.31} {\left( {h(x) – x} \right){\text{d}}x} \).

[2]
b.i.

Hence, find the area of the region enclosed by the graphs of \(h\) and \({h^{ – 1}}\).

[3]
b.ii.

Let \(d\) be the vertical distance from a point on the graph of \(h\) to the line \(y = x\). There is a point \({\text{P}}(a,{\text{ }}b)\) on the graph of \(h\) where \(d\) is a maximum.

Find the coordinates of P, where \(0.111 < a < 3.31\).

[7]
c.
Answer/Explanation

Markscheme

\(q = 2\)     A1     N1

Note:     Accept \(q = 1\), \(h = 0\), and \(k = 3 – \ln (2)\), 2.31 as candidate may have rewritten \(g(x)\) as equal to \(3 + \ln (x) – \ln (2)\).

[1 mark]

a.i.

\(h = 0\)     A1     N1

Note:     Accept \(q = 1\), \(h = 0\), and \(k = 3 – \ln (2)\), 2.31 as candidate may have rewritten \(g(x)\) as equal to \(3 + \ln (x) – \ln (2)\).

[1 mark]

a.ii.

\(k = 3\)     A1     N1

Note:     Accept \(q = 1\), \(h = 0\), and \(k = 3 – \ln (2)\), 2.31 as candidate may have rewritten \(g(x)\) as equal to \(3 + \ln (x) – \ln (2)\).

[1 mark]

a.iii.

2.72409

2.72     A2     N2

[2 marks]

b.i.

recognizing area between \(y = x\) and \(h\) equals 2.72     (M1)

eg\(\,\,\,\,\,\)M17/5/MATME/SP2/ENG/TZ1/10.b.ii/M

recognizing graphs of \(h\) and \({h^{ – 1}}\) are reflections of each other in \(y = x\)     (M1)

eg\(\,\,\,\,\,\)area between \(y = x\) and \(h\) equals between \(y = x\) and \({h^{ – 1}}\)

\(2 \times 2.72\int_{0.111}^{3.31} {\left( {x – {h^{ – 1}}(x)} \right){\text{d}}x = 2.72} \)

5.44819

5.45     A1     N3

[??? marks]

b.ii.

valid attempt to find \(d\)     (M1)

eg\(\,\,\,\,\,\)difference in \(y\)-coordinates, \(d = h(x) – x\)

correct expression for \(d\)     (A1)

eg\(\,\,\,\,\,\)\(\left( {\ln \frac{1}{2}x + 3} \right)(\cos 0.1x) – x\)

valid approach to find when \(d\) is a maximum     (M1)

eg\(\,\,\,\,\,\)max on sketch of \(d\), attempt to solve \(d’ = 0\)

0.973679

\(x = 0.974\)     A2     N4 

substituting their \(x\) value into \(h(x)\)     (M1)

2.26938

\(y = 2.27\)     A1     N2

[7 marks]

c.
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