Practice Paper 1 (DP IB Maths: AA HL)

Practice Paper Questions

1
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4 marks

Prove that the square of an odd number is always odd.

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2a
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2 marks

Show that the equation 2 space sin squared space x space space plus space 3 space cos space x space equals space 0 can be written in the form space a space cos squared space x space plus space b space cos space x space plus space c space equals space 0 space, where a comma space b and c are integers to be found.

2b
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3 marks

Hence, or otherwise, solve the equation 2 space sin squared space x space plus space 3 space cos space x space equals 0 for negative 180 degree space less or equal than space x space less or equal than space 180 degree.

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3
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5 marks

In the expansion of left parenthesis x space plus space h right parenthesis to the power of 5 comma where h space element of space straight real numbers , the coefficient of the term in x cubed is 320.

Find the possible values of h.

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4a
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4 marks

The diagram below shows part of the graph of y space equals space straight f left parenthesis x right parenthesis, where straight f left parenthesis x right parenthesis is the function defined by

straight f left parenthesis x right parenthesis space equals space left parenthesis x squared long dash space 1 right parenthesis space In left parenthesis x space plus space 3 right parenthesis comma space x space greater than space long dash 3

q4a-practice-paper1-setc-ib-dp-aa-hl

Points A comma space B and C are the three places where the graph intercepts the x-axis.

Find straight f apostrophe space left parenthesis x right parenthesis.

4b
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2 marks

Show that the coordinates of point A are (-2, 0).

4c
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3 marks

Find the equation of the tangent to the curve at point A.

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5
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4 marks

The points A, B, C and form the vertices of a parallelogram with position vectors bold italic a comma bold italic b comma bold italic c bold spaceand bold italic d respectively.

Show that the area of the parallelogram is open vertical bar bold italic a cross times bold italic b plus bold italic b cross times bold italic d plus bold italic d cross times bold italic a close vertical bar.

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6
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7 marks

The following triangle shows triangle ABC, with AB = 3a, BC = a and AC = 7.

q6-practice-paper1-setc-ib-dp-aa-hl 

Given that cos space straight A straight B with hat on top straight C space equals 1 half , find the area of the triangle. Give your answer in the form fraction numerator p square root of 3 over denominator r end fractionwhere p comma q element of straight real numbers.

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7a
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2 marks

 alphaandspace beta are non-real roots of the equation x squared plus 3 k x plus 2 k plus 1 equals 0,  where k greater than 0 is a constant. 

Find alpha plus beta and alpha beta, in terms of k.

7b
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2 marks

Given that alpha squared plus beta squared equals 3, show that left parenthesis alpha plus beta right parenthesis squared equals 4 k plus 5.

7c
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3 marks
Hence find the value of k .

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8a
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4 marks

Two lines, l subscript 1  and  l subscript 2, are parallel and their vector equations are given below:

l subscript 1 colon r subscript 1 equals open parentheses table row 2 row 1 row 10 end table close parentheses plus lambda open parentheses table row 4 row cell negative 2 end cell row 2 end table close parentheses

l subscript 2 colon r subscript 2 equals open parentheses table row cell negative 3 end cell row cell negative 1 end cell row 0 end table close parentheses plus mu open parentheses table row cell negative 2 end cell row p row q end table close parentheses

          (i)     State the values of p  and q.

 

(ii)   Show that l subscript 1 and l subscript 2 are not collinear.

8b
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5 marks
Find the minimum distance between l subscript 1and l subscript 2.

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9
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7 marks

Use the substitution  u equals cos space x to find integral fraction numerator sin space x space cos space x over denominator cos squared x plus 3 cos space x minus 4 end fraction d x.

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10a
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4 marks

Consider the function f defined by space f space left parenthesis x right parenthesis space equals space 3 space sin space x minus 3 , for 0 less or equal than x less or equal than 3 straight pi
The following diagram shows the graph of y space equals f open parentheses x close parenthesesq10a-practice-paper1-setc-ib-dp-aa-hl

The graph of f touches the x-axis at point A as shown. Point B is a local minimum of f
.
The shaded region is the area between the graph of y space equals space f space left parenthesis x right parenthesis and the x-axis, between the
points A and B.

 Find the x-coordinates of A and B.

10b
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5 marks

Show that the area of the shaded region is 3straight pi units2 .

10c
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2 marks

The right cone in the diagram below has a curved surface area of twice the shaded area in
the previous part of the question.
The cone has a slant height of 3, base radius r, and height h.

q10c-practice-paper1-setc-ib-dp-aa-hl

Find the value of r.

10d
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4 marks

Hence find the volume of the cone.

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11a
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6 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 begin inline style t over 2 end style end exponent open parentheses 1 minus v subscript o close parentheses.
11b
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4 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.
11c
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5 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.
11d
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5 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 begin inline style k over 2 end style 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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12a
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5 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. .

12b
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6 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.
12c
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7 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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