Practice Paper 3 (DP IB Maths: AA HL)

Practice Paper Questions

1a
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1 mark

This question uses De Moivre’s theorem to derive an exact form for the value of sin pi over 5.

For a complex number with modulus r equals 1, De Moivre’s theorem is given by

 open parentheses cos space theta plus straight i space sin space theta close parentheses to the power of n equals cos space n theta plus straight i space sin space n theta

where n element of straight integer numbers to the power of plus and theta is measured in radians. 

Show that the theorem is true for n equals 1.

1b
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6 marks

Consider the case when  n equals 2.

(i)
Expand open parentheses cos space theta plus straight i space sin space theta close parentheses squared. 

(ii)
By equating real parts from both sides of De Moivre’s theorem, show that cos squared theta minus sin squared theta equals cos space 2 theta. 

(iii)
By equating imaginary parts, write down an identity for sin space 2 theta in terms of sin space theta and cos space theta.
1c
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7 marks

Consider the case when n equals 3 and let c equals cos space theta and s equals sin space theta.

 (i)     Expand open parentheses c plus straight i s close parentheses cubed

 (ii)
By considering real parts, write down an identity for cos space 3 theta in terms of sin space theta and cos space theta.

 (iii)
Use the Pythagorean identity cos squared theta plus sin squared theta equals 1 to rewrite the identity from part (c)(ii) in terms of cos space theta only, giving your answer in the form

p space cos cubed theta space minus q space cos space theta equals cos space 3 theta

        where p and q are integers to be found.

1d
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11 marks

The identity for sin space 5 theta is found by equating the imaginary parts of De Moivre’s theorem when n equals 5, then writing the result in terms of sin space theta spaceonly.  The identity is given by

sin space 5 theta equals 16 space sin to the power of 5 theta minus 20 space sin cubed theta plus 5 space sin space theta 

You may use this identity without proof for the rest of the question.

(i)
By substituting theta equals pi over 5 into both sides of the identity for sin space 5 theta, show that x equals sin space pi over 5 satisfies the polynomial equation 16 x to the power of 5 minus 20 x cubed plus 5 x equals 0.
 
(ii)
Showing clear algebraic working, solve the polynomial equation in part (d)(i), giving all your solutions as exact values.

(iii)
Using a sketch of y equals sin space x for 0 less than x less than pi over 2,  explain why 0 less than sin straight pi over 5 less than fraction numerator square root of 2 over denominator 2 end fraction.

 

(iv)
Justifying your choice of solution from part (d)(ii), prove that the exact value of sin straight pi over 5 is given by

sin pi over 5 equals 1 half square root of fraction numerator 5 minus square root of 5 over denominator 2 end fraction end root

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2a
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9 marks

This question explores the sequence of functions  bold italic f subscript bold n begin bold style stretchy left parenthesis x stretchy right parenthesis end style bold equals bold 1 bold minus bold italic x to the power of bold 2 bold plus bold italic x to the power of bold 4 bold. bold. bold. bold. bold plus begin bold style stretchy left parenthesis negative 1 stretchy right parenthesis end style to the power of bold n bold italic x to the power of bold 2 bold n end exponent  on the domain  bold minus bold 1 bold less than bold italic x bold less than bold 1  and uses them to find bounds on the value of pi.

Consider the sequence of functions given by

f subscript n open parentheses x close parentheses equals 1 minus x squared plus x to the power of 4... plus open parentheses negative 1 close parentheses to the power of n x to the power of 2 n end exponent 

where n element of straight integer numbers to the power of plus and  negative 1 less than x less than 1. 

The first three functions in the sequence are given below:                

f subscript 1 open parentheses x close parentheses equals 1 minus x squared      f subscript 2 open parentheses x close parentheses equals 1 minus x squared plus x to the power of 4      f subscript 3 open parentheses x close parentheses equals 1 minus x squared plus x to the power of 4 minus x to the power of 6            

(i)
Write down the function f subscript 4 open parentheses x close parentheses.

(ii)
Use your graphic display calculator to explore the stationary points on the graphs of  y equals f subscript n open parentheses x close parentheses over the domain negative 1 less than x less than 1. Hence copy and complete the following table:
 
n

Number of local maximum points

Number of local minimum points

1    
2    
3    
4    

(iii)
Use your table to predict the numbers of each type of stationary point that will occur on the graphs of  y equals f subscript n open parentheses x close parentheses for all odd values of n.

(iv)
Use f apostrophe subscript 2 open parentheses x close parentheses to find the exact coordinates of the stationary points for n equals 2 commastating clearly which coordinates correspond to which types of stationary point.

2b
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6 marks

As n rightwards arrow infinity, the graph of the limit of the sequence of functions f subscript n open parentheses x close parentheses is a smooth curve  y equals h open parentheses x close parentheses over the domain negative 1 less than x less than 1.

     (i)     By considering f subscript n open parentheses x close parenthesesas n rightwards arrow infinity as the infinite geometric series

1 minus x squared plus x to the power of 4 minus x to the power of 6 plus midline horizontal ellipsis

where negative 1 less than x less than 1 commause an appropriate series summation formula to show that 

h open parentheses x close parentheses equals fraction numerator 1 over denominator 1 plus x squared end fraction

(ii)
Use your graphic display calculator to sketch, on the same set of axes for negative 1 less than x less than 1,  the graphs of y equals f subscript 3 open parentheses x close parentheses comma space y equals f subscript 4 open parentheses x close parentheses and y equals h open parentheses x close parentheses.
2c
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3 marks

Show that the area under the curve y equals h open parentheses x close parentheses between x equals 0 and x equals 1 is equal to pi over 4square units.

2d
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8 marks
(i)
By considering the sketch from part (b)(ii), state with a reason which out of  y equals f subscript 3 open parentheses x close parentheses  or  y equals f subscript 4 open parentheses x close parentheses  would provide an underestimate of the area in part (c) when integrated between 0 and 1, and which would provide an overestimate.
 
(ii)
Calculate the values ofintegral subscript 0 superscript 1 f subscript 3 open parentheses x close parentheses space d x and integral subscript 0 superscript 1 f subscript 4 open parentheses x close parentheses space d x, showing your working and giving your answers as exact fractions.

 

(iii)
Hence use the results above to give a lower and upper bound on pi, giving your answer in the form a less than pi less than ba and b are correct to 2 decimal places.
2e
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4 marks

Starting from the result that

1 minus x squared plus x to the power of 4 minus x to the power of 6 plus midline horizontal ellipsis equals fraction numerator 1 over denominator 1 plus x squared end fraction comma 

derive the Maclaurin series for x(as given in the Formula Booklet). You may assume that an infinite series can be integrated term-by-term, although the value of any constant of integration will need to be found.

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