Practice Paper 3 | DP IB Maths: AA HL Practice Paper Questions 2021

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1 mark

This question uses De Moivre’s theorem to derive an exact form for the value of $\sin \frac{π}{5}$ .

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

${(\cos θ + i \sin θ)}^{n} = \cos n θ + i \sin n θ$

where $n \in ℤ^{+}$ and $θ$ is measured in radians.

Show that the theorem is true for $n = 1$ .

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6 marks

Consider the case when $n = 2 .$

(i)

Expand

{(\cos θ + i \sin θ)}^{2} .

(ii)

By equating real parts from both sides of De Moivre’s theorem, show that

\cos^{2} θ - \sin^{2} θ = \cos 2 θ .

(iii)

By equating imaginary parts, write down an identity for

\sin 2 θ

in terms of

\sin θ

and

\cos θ .

How did you do?

7 marks

Consider the case when $n = 3$ and let $c = \cos θ$ and $s = \sin θ$ .

(i) Expand

{(c + i s)}^{3}

(ii)

By considering real parts, write down an identity for

\cos 3 θ

in terms of

\sin θ

and

\cos θ

(iii)

Use the Pythagorean identity

\cos^{2} θ + \sin^{2} θ = 1

to rewrite the identity from part (c)(ii) in terms of

\cos θ

only, giving your answer in the form

$p \cos^{3} θ - q \cos θ = \cos 3 θ$

where $p$ and $q$ are integers to be found.

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11 marks

The identity for $\sin 5 θ$ is found by equating the imaginary parts of De Moivre’s theorem when $n = 5$ , then writing the result in terms of $\sin θ$ only. The identity is given by

$\sin 5 θ = 16 \sin^{5} θ - 20 \sin^{3} θ + 5 \sin θ$

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

(i)

By substituting

θ = \frac{π}{5}

into both sides of the identity for

\sin 5 θ

, show that

x = \sin \frac{π}{5}

satisfies the polynomial equation

16 x^{5} - 20 x^{3} + 5 x = 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 = \sin x

for

0 < x < \frac{π}{2}

, explain why

0 < \sin \frac{π}{5} < \frac{\sqrt{2}}{2} .

(iv)

Justifying your choice of solution from part (d)(ii), prove that the exact value of

\sin \frac{π}{5}

is given by

$\sin \frac{π}{5} = \frac{1}{2} \sqrt{\frac{5 - \sqrt{5}}{2}}$

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9 marks

This question explores the sequence of functions $f_{n} (x) = 1 - x^{2} + x^{4} . . . . + {(- 1)}^{n} x^{2 n}$ on the domain $- 1 < x < 1$ and uses them to find bounds on the value of $π$ .

Consider the sequence of functions given by

$f_{n} (x) = 1 - x^{2} + x^{4} . . . + {(- 1)}^{n} x^{2 n}$

where $n \in ℤ^{+}$ and $- 1 < x < 1 .$

The first three functions in the sequence are given below:

$f_{1} (x) = 1 - x^{2}$ $f_{2} (x) = 1 - x^{2} + x^{4}$ $f_{3} (x) = 1 - x^{2} + x^{4} - x^{6}$

(i)

Write down the function

f_{4} (x)

(ii)

Use your graphic display calculator to explore the stationary points on the graphs of

y = f_{n} (x)

over the domain

- 1 < x < 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 = f_{n} (x)

for all odd values of

n

(iv)

Use

f'_{2} (x)

to find the exact coordinates of the stationary points for

n = 2,

stating clearly which coordinates correspond to which types of stationary point.

How did you do?

6 marks

As $n \to \infty$ , the graph of the limit of the sequence of functions $f_{n} (x)$ is a smooth curve $y = h (x)$ over the domain $- 1 < x < 1 .$

(i) By considering $f_{n} (x)$ as $n \to \infty$ as the infinite geometric series

$1 - x^{2} + x^{4} - x^{6} + \dots$

where

- 1 < x < 1,

use an appropriate series summation formula to show that

$h (x) = \frac{1}{1 + x^{2}}$ .

(ii)

Use your graphic display calculator to sketch, on the same set of axes for

- 1 < x < 1

, the graphs of

y = f_{3} (x), y = f_{4} (x)

and

y = h (x) .

How did you do?

3 marks

Show that the area under the curve $y = h (x)$ between $x = 0$ and $x = 1$ is equal to $\frac{π}{4}$ square units.

How did you do?

8 marks

(i)

By considering the sketch from part (b)(ii), state with a reason which out of

y = f_{3} (x)

y = f_{4} (x)

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 of

\int_{0}^{1} f_{3} (x) d x

and

\int_{0}^{1} f_{4} (x) 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

π

, giving your answer in the form

a < π < b

a

and

b

are correct to 2 decimal places.

How did you do?

4 marks

Starting from the result that

$1 - x^{2} + x^{4} - x^{6} + \dots = \frac{1}{1 + x^{2}},$

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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Practice Paper 3 (DP IB Maths: AA HL)

Practice Paper Questions