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IBMaths: AA HLDPExam Questions6. Extended Questions6.2 Extended Questions (Section B, HL)

Extended Questions (Section B, HL) (DP IB Maths: AA HL): Exam Questions

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1a
Sme Calculator2 marks

The function f is defined by f open parentheses x close parentheses equals fraction numerator 2 x minus 1 over denominator x squared plus 3 x minus 4 end fraction comma for x element of straight real numbers comma space space x not equal to m comma space space x not equal to n. 

Find the values of m and n.

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1b
Sme Calculator3 marks

Find an expression for f apostrophe open parentheses x close parentheses.

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1c
Sme Calculator2 marks

The graph of  y equals f open parentheses x close parentheses has exactly one point of inflection.

Find the x-coordinate of the point of inflection.

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1d
Sme Calculator4 marks

Sketch the graph of y equals f open parentheses x close parentheses for negative 6 less or equal than x less or equal than 6,  showing the coordinates of any axis intercepts and local maxima and local minima, and giving the equations of any asymptotes.

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1e
Sme Calculator3 marks

The function g is defined by gopen parentheses x close parentheses equals fraction numerator x squared plus 3 x minus 4 over denominator 2 x minus 1 end fraction,  for x element of straight real numbers comma space x not equal to 1 half.

Find the equation of the oblique asymptote of the graph of y equalsg open parentheses x close parentheses.

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1f
Sme Calculator4 marks

By considering the graph of y equals f open parentheses x close parentheses minus g open parentheses x close parentheses comma or otherwise, solve g open parentheses x close parentheses less than f open parentheses x close parentheses for x element of straight real numbers.

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2a
Sme Calculator3 marks

The function f has a derivative given by f apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator 3 x open parentheses k minus x close parentheses end fraction comma space x element of straight real numbers comma space x not equal to 0 comma where k is a positive constant.

The expression for f apostrophe open parentheses x close parentheses can be written in the form fraction numerator a over denominator 3 x end fraction plus fraction numerator b over denominator k minus x end fraction  where p comma space q element of straight real numbers. Find a  and b in terms of k.

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2b
Sme Calculator3 marks

Hence find an expression for f open parentheses x close parentheses.

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2c
Sme Calculator7 marks

R is the population of rabbits on an island. The rate of change of the population can be modelled by the differential equation fraction numerator d R over denominator d t end fraction equals fraction numerator 3 R open parentheses k minus R close parentheses over denominator 4 k end fraction,  where t is the time measured in years, t greater or equal than 0,  and k is the maximum population that the island can support. 

The initial population of the rabbits is 20. 

By solving the differential equation, show that  R equals fraction numerator 20 k e to the power of 3 over 4 t end exponent over denominator k minus 20 plus 20 e to the power of 3 over 4 t end exponent end fraction

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2d
Sme Calculator3 marks

After two years, the population of rabbits has risen to 70.

Find k.

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2e
Sme Calculator2 marks

Find the value of t at which the population of rabbits is growing at its fastest rate.

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3a
Sme Calculator6 marks

A particle is moving in a vertical line and its acceleration, in ms to the power of negative 2 end exponent , at time t seconds, t greater or equal than 0 is given by a equals negative fraction numerator 1 minus v over denominator 2 end fraction comma where v is the velocity in meters per second and v less than 1.

The particle starts at a fixed origin O with initial velocity v subscript o space ms to the power of negative 1 end exponent.

By solving a suitable differential equation, show that the particle’s velocity at time t is given by v open parentheses t close parentheses equals 1 minus e to the power of negative t over 2 end exponent open parentheses 1 minus v subscript o close parentheses.

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3b
Sme Calculator4 marks

The particle moves down in the negative direction, until its displacement relative to the origin reaches a minimum. Then the particle changes direction and starts moving up, in a positive direction. 

(i)
If the initial velocity of the particle is negative 3 space ms to the power of negative 1 end exponent, find the time at which the minimum displacement of the particle from the origin occurs, giving your answer in exact form.

(ii)
If T is the time in seconds when the displacement reaches its smallest value, show that T equals 2 space ln open parentheses 1 minus v subscript o close parentheses.

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3c
Sme Calculator5 marks
(i)
Find a general expression for the displacement, in terms of t and v subscript o.

(ii)
Combine this general expression with the result from part (b)(ii) to find an expression for the minimum displacement of the particle in terms of v subscript o.

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3d
Sme Calculator5 marks

Let v open parentheses T minus k close parentheses represent the particle’s velocity k seconds before the minimum displacement and v open parentheses T plus k close parentheses the particle’s velocity k seconds after the minimum displacement. 

(i)
Show that v open parentheses T minus k close parentheses equals 1 minus e to the power of begin inline style k over 2 end style end exponent.

(ii)
Given that v open parentheses T plus k close parentheses equals 1 minus e to the power of negative k over 2 end exponent comma show that v open parentheses T minus k close parentheses plus v open parentheses T plus k close parentheses greater or equal than 0.

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4a
Sme Calculator5 marks

The diagram below shows the graph of f open parentheses x close parentheses equals arctan open parentheses x close parentheses comma space x element of straight real numbers. The graph has rotational symmetry of order 2 about the origin.

mi-q12a-ib-aa-hl-pp1-set-c-maths-dig

A different function, g, is described by gopen parentheses x close parentheses equals negative arctan open parentheses x minus 1 close parentheses comma space x element of straight real numbers.

(i)
Describe the sequence of transformations that transforms f open parentheses x close parenthesesto gopen parentheses x close parentheses.

(ii)
Sketch the graph of  gopen parentheses x close parentheses on the axes above.

(iii)
Using your answers to parts (i) and (ii) to help you, describe the relationship between integral subscript 0 superscript 1 arctan open parentheses x close parentheses d xand integral subscript 0 superscript 1 minus arctan open parentheses x minus 1 close parentheses d x. .

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4b
Sme Calculator6 marks
(i)
Prove that  arctan space p minus arctan space q equals arctan open parentheses fraction numerator p minus q over denominator 1 plus p q end fraction close parentheses.

(ii)      Show that arctan open parentheses fraction numerator 1 over denominator x squared minus x plus 1 end fraction close parenthesescan be written as arctan open parentheses x close parentheses minus arctan open parentheses x minus 1 close parentheses.

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4c
Sme Calculator7 marks

Using the results from parts (a) and (b), evaluate integral subscript 0 superscript 1 arctan open parentheses fraction numerator 1 over denominator x squared minus x plus 1 end fraction close parentheses d x commaleaving your answer in exact form.

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5a
Sme Calculator6 marks

Paola is modelling a small vase from her house for her maths project. To model the edge of the vase in cross-section, she decides to use a function f of the form

f open parentheses x close parentheses equals fraction numerator q straight e to the power of x over 2 end exponent over denominator 2 plus straight e to the power of x end fraction

 

where x element of straight real numbers comma space x greater or equal than 0 and q element of straight real numbers to the power of plus. 

The function and the vase are represented in the diagrams below.

mi-q11a-ib-aa-sl-pp2-set-c-maths-dig1

mi-q11a-ib-aa-sl-pp2-set-c-maths-dig2



The vertical height of the vase, OB, is measured along the x-axis. The radius of the vase’s opening is OA, and its base radius is BC. 

To model the vase, she will rotate by 2 pi radians about the x-axis the region enclosed by the graph of y equals f open parentheses x close parentheses ,  the x-axis, the y-axis, and the line x equals ln space 43 . 

Show that the volume of the solid of revolution thus formed is fraction numerator 14 q squared straight pi over denominator 45 end fraction units cubed.

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5b
Sme Calculator2 marks

The volume of the actual vase is 100 cm cubed.

Use this information to find the value of q.

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5c
Sme Calculator4 marks

Find the cross-sectional radius of the vase

(i)     at its base,

(ii)    at its widest point.

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5d
Sme Calculator4 marks

Paola wants to investigate how the cross-sectional radius of the vase changes.

Sketch a graph of the derivative of f, and use it to find the value of x at which the cross-sectional radius of the vase is decreasing most rapidly.

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1a
Sme Calculator3 marks

The points A(2, 3, 0), B(-2, 4, 1), C(1, -1, 3) and D(5, -2, 2) lie on the plane capital pi subscript 1 and form a parallelogram, where AB and CD are one pair of parallel edges and BC and AD are the other pair of parallel edges. Each unit on the coordinate grid is equivalent to 1 cm in length.

Find the vector product of AB with rightwards arrow on top and AC with rightwards arrow on top.

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1b
Sme Calculator2 marks

Hence, or otherwise, find the Cartesian equation of the plane capital pi subscript 1.

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1c
Sme Calculator4 marks

A second plane capital pi subscript 2 contains the point with position vector open parentheses table row 5 row cell negative 3 end cell row 5 end table close parentheses and also the line L, which has vector equation bold space bold italic r equals space open parentheses table row 6 row 1 row 2 end table close parentheses plus lambda open parentheses table row 4 row cell negative 1 end cell row cell negative 1 end cell end table close parentheses.

Show that capital pi subscript 1 and capital pi subscript 2are parallel.

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1d
Sme Calculator3 marks

A parallelepiped is a 3D object made up of six faces that are parallelograms lying in pairs of parallel planes.  EFGH is a parallelogram on capital pi subscript 2 that is congruent to ABCD, and points A, B, C and D on capital pi subscript 1 are joined to points E, F, G and H respectively on capital pi subscript 2 to form a parallelepiped.

Given that the coordinates of E are (3, 6, 0), find the coordinates of point H.

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1e
Sme Calculator5 marks

The volume of a parallelepiped can be found using the formula open vertical bar open parentheses bold italic a cross times bold italic b close parentheses. bold italic c close vertical bar where bold italic a comma space bold italic b and bold italic c  are vectors corresponding to three edges meeting at a single vertex of the parallelepiped.

Show that the volume of the parallelepiped ABCDEFGH is 40 cm3. 

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2a
Sme Calculator1 mark

A function g is defined by g open parentheses x close parentheses equals arccos open parentheses fraction numerator x squared minus 1 over denominator x squared plus 1 end fraction close parentheses comma space x element of straight real numbers. 

Show that g is an even function.

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2b
Sme Calculator2 marks

By considering the limit of g as x tends to infinity, show that the graph of  y equals g open parentheses x close parentheses has a horizontal asymptote and state its equation.

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2c
Sme Calculator9 marks
(i)
Show that  g to the power of apostrophe open parentheses x close parentheses equals fraction numerator negative 2 x over denominator open parentheses square root of x squared end root close parentheses open parentheses x squared plus 1 close parentheses end fraction for x element of straight real numbers comma space x greater or equal than 0.

 

(ii)
Considering the fact that square root of x squared end root equals open vertical bar x close vertical bar commaand also the expression for g to the power of apostrophe open parentheses x close parentheses above, show that g is increasing for x less than 0.

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2d
Sme Calculator5 marks

A new function, h, is created by restricting the domain of g, such that h open parentheses x close parentheses equals arccos open parentheses fraction numerator x squared minus 1 over denominator x squared plus 1 end fraction close parentheses comma space x element of straight real numbers comma space x greater or equal than 0.,  ,  .

Find an expression for h to the power of negative 1 end exponent open parentheses x close parentheses, carefully considering the range of h in determining your final answer.

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2e
Sme Calculator2 marks

State the domain of h to the power of negative 1 end exponent open parentheses x close parentheses.

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3a
Sme Calculator2 marks

The function f is defined by f open parentheses x close parentheses equals fraction numerator 4 x plus 3 over denominator 9 x squared minus 4 end fraction,  for x element of straight real numbers comma space x not equal to p comma space x not equal to q. 

Given that p less than q ,  find the value of p and the value of q.

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3b
Sme Calculator3 marks

Find an expression for f to the power of apostrophe open parentheses x close parentheses.

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3c
Sme Calculator2 marks

The graph of y equals f open parentheses x close parentheses has exactly one point of inflection. 

Find the x-coordinate of the point of inflection.

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3d
Sme Calculator5 marks

Sketch the graph of y equals f open parentheses x close parentheses for negative 3 less or equal than x less or equal than 3 commashowing the values of any axes intercepts, the coordinates of any local maxima and local minima, and giving the equations of any asymptotes.

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3e
Sme Calculator4 marks

The function g is defined by g open parentheses x close parentheses equals fraction numerator 9 x squared minus 4 over denominator 4 x plus 3 end fraction comma for x element of straight real numbers comma space x not equal to negative 3 over 4. 

Find the equations of all the asymptotes on the graph of  y equals g open parentheses x close parentheses.

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3f
Sme Calculator4 marks

By considering the graph of y equals f open parentheses x close parentheses minus g open parentheses x close parentheses comma or otherwise, solve f open parentheses x close parentheses less than g open parentheses x close parentheses for x element of straight real numbers.

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4a
Sme Calculator3 marks

The derivative of the function f is given by f apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator x open parentheses k minus x close parentheses end fraction comma space x element of straight real numbers comma space x not equal to 0 comma space x not equal to k comma where k greater than 0 is a real constant. 

By finding appropriate constants a and b in terms of k, show that the expression for f apostrophe open parentheses x close parenthesescan be written in the form a over x plus fraction numerator b over denominator k minus x end fraction comma where a comma b element of straight real numbers.

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4b
Sme Calculator3 marks

Hence find an expression for f open parentheses x close parentheses.

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4c
Sme Calculator8 marks

Consider a population of lizards, P, which has an initial size of 800. The rate of change of the population can be modelled by the differential equation fraction numerator d P over denominator d t end fraction equals fraction numerator P open parentheses k minus P close parentheses over denominator 25 k end fraction,  where t is the time measured in years, t greater or equal than 0 comma and k is the maximum sustainable population. 

By solving the differential equation, show that

P equals fraction numerator 800 k over denominator open parentheses k minus 800 close parentheses e to the power of negative t over 25 end exponent plus 800 end fraction

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4d
Sme Calculator3 marks

At t equals 12 the lizard population has reduced in size to three fourths of its original value. 

Find the value of k, giving your answer correct to four significant figures.

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4e
Sme Calculator3 marks

Find the value of t when the population is decreasing at a rate of 16 lizards per year.

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5a
Sme Calculator3 marks

A mathematical function f is defined by f open parentheses x close parentheses equals x e to the power of 2 x end exponent.

Show that f apostrophe apostrophe open parentheses x close parentheses equals open parentheses 4 x plus 4 close parentheses e to the power of 2 x end exponent.

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5b
Sme Calculator7 marks

Prove by mathematical induction that if f open parentheses x close parentheses equals x e to the power of 2 x end exponent then f to the power of open parentheses n close parentheses end exponent open parentheses x close parentheses equals open parentheses 2 to the power of n x plus n 2 to the power of n minus 1 end exponent close parentheses e to the power of 2 x end exponent.

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5c
Sme Calculator7 marks

Let  g open parentheses x close parentheses equals ln open parentheses 1 plus m x close parentheses comma space m element of Z to the power of plus. 

Consider the function h defined by h open parentheses x close parentheses equals f open parentheses x close parentheses cross times g open parentheses x close parentheses.

Given that the term in x to the power of 4 of the Maclaurin series for h open parentheses x close parentheses has coefficient 6, find the value of m.

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