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Review Circles with centre (0, 0)
● z = r, z , r, z ø r where r is a constant.
In this case, the length of the position vector representing the complex number, z, is the
constant value r. The argument of z can vary.
A point that moves so that is it always the same distance from a fixed point is a circle.
The fixed point is the centre of the circle. The fixed distance is the radius of the circle.
Recall that a position vector is simply a means of locating a point in space relative to an
origin, usually O.
For the complex number z = x + iy, where x and y are real, the locus of the point
P(x, y) that satisfies:
● z = r is the circumference of a circle with centre (0, 0) and radius r
● z , r is all points within a circle with centre (0, 0) and radius r, but the
circumference is not included
● z ø r is all points within a circle with centre (0, 0) and radius r or the
circumference of the circle.
WORKED EXAMPLE 11.14
On separate Argand diagrams, sketch these loci.
a z =5 b z ,3 c z ø4 289
Review Answer Im(z)
6
a z = 5 is the circumference of a circle, centre 5
(0, 0), radius 5. 4
3
It is only the circle, not the points within the 2
circle. 1
The circle is drawn as a solid line as it is –6 –5 –4 –3 –2 –1 O
included. –1
–2
–3 1 2 3 4 5 6 Re(z)
–4
Review –5
–6
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b z , 3 is a circle, centre (0, 0), radius 3. Im(z)
4
It is only the points within the circle.
3
The circumference of the circle is drawn as a dotted
line as it is not included. 2
1
–4 –3 –2 –1 O 1 2 3 4Re(z)
–1
Review –2
–3
–4
c z ø 4 is a circle, centre (0, 0), radius 4. Im(z)
5
It is all points within the circle and on the 4
circumference. 3
The circumference of the circle is drawn as a 2
solid line as it is included. 1
290
–5 –4 –3 –2 –1 O 1 2 3 4 5 Re(z)
–1
Review
–2
–3
–4
–5
Circles with centre (a, b) Im(z)
Review● z − z1 = r, z − z1 , r, z − z1 ø r where r is a constant and z1 is P(x, y)
the fixed point a + bi. Re(z)
Q(a, b)
We need to understand the vector representation of the complex
number z − z1. O
Let P(x, y) be the moving point that represents z = x + iy on an
Argand diagram.
Let Q(a, b) be the fixed point that represents z1 = a + bi on the
Argand diagram.
z = OP and z1 = OQ
Therefore, QP = −z1 + z
QP = z − z1
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We now interpret z − z1 = r as:
the length of the vector QP is r.
For the complex number z = x + iy, where x and y are real, and the fixed point
z1 = a + bi, where a and b are real, the locus of the point P(x, y) that satisfies
● z − z1 = r is the circumference of a circle with centre (a, b) and radius r
● z − z1 , r is all points within a circle with centre (a, b) and radius r, but the
circumference is not included
● z − z1 ø r is all points within a circle with centre (a, b) and radius r or the
circumference of the circle.
Review WORKED EXAMPLE 11.15
On separate Argand diagrams, sketch these loci.
a z − (2 + 4i) = 3 b z − 3i , 4 c z +5 ø5
Answer Im(z) 291
a z1 = 2 + 4i 7
Review z − (2 + 4i) = 3 is the circumference of a circle, centre 6
(2, 4), radius 3. 5
It is only the circle, not the points within the circle. 4
The circle is drawn as a solid line as it is included. 3
2
b z1 = 3i 1
z − 3i , 4 is a circle, centre (0, 3), radius 4.
It is only the points within the circle. –2 –1 O 1 2 3 4 5 6 Re(z)
The circumference of the circle is drawn as a –1
dotted line as it is not included. Im(z)
7
6
5
4
3
Review 2
1
–4 –3 –2 –1 O 1 2 3 4Re(z)
–1
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c z − z1 = z + 5 = z − (−5) Take care with signs.
z1 = −5 Im(z)
4
z + 5 ø 2 is a circle, centre (−5, 0), radius 2.
It is all points within the circle and on the circumference. 3
The circumference of the circle is drawn as a solid line as it is 2
included.
1
–7 –6 –5 –4 –3 –2 –1 O 1 2 Re(z)
–1
Review –2
–3
–4
Half-lines and part-lines Im(z)
● arg(z − z1) = α, where α is a fixed angle and z1 is the fixed point a + bi. α P(x, y)
With z − z1 defined as previously O Re(z)
z = OP, z1 = OQ and QP = z − z1
292
Q(a, b)
Review We now define arg(z − z1) as the angle made between the direction parallel to
the real axis and the direction of the vector QP.
Im(z)
Points below Q are not part of the locus as the angle made between these points
and the relevant horizontal line is π − α, not α . P(x, y)
β
● arg(z1 − z) = β, where β is a fixed angle and z1 is the fixed point a + bi.
With z1 − z = −(z − z1),
z = OP, z1 = OQ and PQ = z1 − z
Review Q(a, b) Re(z)
O
We now define arg(z1 − z) as the angle made between the direction parallel to
the real axis and the direction of the vector PQ.
Points above P are not part of the locus as the angle made between these points
and the relevant horizontal line is π − β, not β .
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WORKED EXAMPLE 11.16
On a single Argand diagram, sketch the loci z = 4 and arg(z + 2 + 3i) = π4 .
Show that there is only one complex number, z, that satisfies both loci.
Label this point as P on your diagram.
Answer Circle, centre (0, 0), radius 4
z =4
Review arg(z + 2 + 3i) = π
4
z − z1 = z + 2 + 3i
= z − (−2 − 3i)
z1 = −2 − 3i
Half-line from (−2, −3) at an angle of π radians.
4
Im(z) 293
6
5 P
4
Review 3
2
1
–5 –4 –3 –2 –1 O 1 2 3 4 5 6 7 8 Re(z)
–1
π –2
4 –3
–4
–5
–6
Perpendicular bisectors z − z1 ø z − z2 where z1 is the fixed point a + bi
Review ● z − z1 = z − z2 , z − z1 , z − z2 ,
and z2 is the fixed point c + di.
Let P(x, y) be the moving point that represents z = x + iy on an Argand diagram,
Q(a, b) be the fixed point that represents z1 = a + bi on the Argand diagram,
R(c, d ) be the fixed point that represents z2 = c + di on the Argand diagram,
then
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● when z − z1 = z − z2 , the point P(x, y) moves so that it is always the same
distance from Q as it is from R. The locus is, therefore, the perpendicular bisector
of the line QR.
● z − z1 , z − z2 is all points such that P is nearer to Q than to R. The
perpendicular bisector marks the boundary of this region but is not included.
● z − z1 ø z − z2 is all points such that P is nearer to Q than to R or equidistant
from Q and R. The perpendicular bisector marks the boundary of this region and
is included.
WORKED EXAMPLE 11.17
Review On an Argand diagram, sketch the locus z − 4 + i = z + 5i .
Find the Cartesian equation of this locus.
Answer
z − 4 + i = z − (4 − i) Write as z − z1 and find z1
z1 = 4 − i Write as z − z2 and find z2.
The locus is the perpendicular bisector of
z + 5i = z − (−5i)
z2 = −5i (4, −1) and ( 0, −5 ).
Midpoint of QP, (4, −1) and (0, −5),
294 is 4 + 0 , −1 + (−5) = (2, −3)
2 2
Review Gradient of QP = 1 and so the gradient of perpendicular
bisector = −1.
Cartesian equation is y + 3 = −1(x − 2)
y = −x − 1
Im(z)
3
2
1
–1 O 1 2 3 4 5 6 7 Re(z)
–1 P
–2
Review –3
–4
–5 Q
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EXERCISE 11E
1 Describe, in words, the locus represented by each of the following.
a arg(z − 2 + 3i) = π b z − 10 , z − 6i
12
5π
c z+6−i =7 d arg( z ) = 12
2 On different Argand diagrams, sketch the following loci. Shade any regions where needed.
a z − (2 + i) = z − (4 + 3i) b arg(z + 3i) = − π
c z + 4 − 5i ø 2 3
d z − 2i ø z − 4
Review PS 3 On a single Argand diagram, sketch the loci z = 4 and z + 2 = z − 4 . Hence determine complex numbers
that satisfy both loci, giving your answers in Cartesian form.
4 On an Argand diagram, sketch the locus z = z + 8i . Find the Cartesian equation of this locus.
5 Sketch the locus z − (3 − 6i) = 3 on an Argand diagram. Write down the Cartesian equation of this locus.
PS 6 Sketch the loci arg(z + 4 + 2i) = 5π and z − 5i = 4 on an Argand diagram. Determine whether or not there
6
is a complex number, z, that satisfies both loci.
PS 7 For complex numbers z satisfying z − 8 − 16i = 2 5 , find the least possible value of z and the greatest
possible value of z .
295
PS 8 By sketching the appropriate loci on an Argand diagram, find the value of z that satisfies z = 13 and
π
Review arg(z − 12) = 2 .
PS 9 On a single Argand diagram, sketch the loci z + 3 − 2i = 4 and arg(z + 1) = π4 . Hence determine the value
of z that satisfies both loci, giving your answer in Cartesian form.
10 The complex number z is represented by the point P(x, y) on an Argand diagram.
It is given that the locus of P is z − 5 − 5i = 5 .
a Write down the Cartesian equation of the locus of P.
b Sketch an Argand diagram to show the locus of P.
c Write down the greatest possible value of arg z and the least possible value of arg z.
Review
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Checklist of learning and understanding
i2 = −1
Cartesian form: x + iy where x and y are real values.
Arithmetic operations on z1 = a + bi and z2 = c + d i :
● Addition: z1 + z2 = (a + c) + (b + d )i
● Subtraction: z1 − z2 = (a − c) + (b − d )i
● Multiplication: z1z2 = (ac − bd ) + (ad + bc)i
● Division:
Review ● Modulus: z1 = (ac + bd ) + (bc − ad )i
z2 c2 − d2
z = x2 + y2
● Argument: Found using a diagram with tanθ = y
x
z1z2 = r1r2 = z1 z2 and arg( z1z2 ) = θ1 + θ2 = arg z1 + arg z2,
z1 = r1 = z1 and arg z1 = θ1 − θ2 = arg z1 − arg z2
z2 r2 z2 z2
Polar forms
● Modulus-argument form: r(cos θ + i sin θ )
296 ● Exponential form: reiθ
Review Roots of equations occur in complex conjugate pairs
● Quadratic equations have 2 real or 2 complex roots of the form x ± iy with y ≠ 0.
● Cubic equations have 3 real or 1 real and 2 complex roots of the form x ± iy with y ≠ 0.
● Quartic equations have 4 real or 2 real and 2 complex roots of the form x ± iy with y ≠ 0,
or 4 complex roots of the form x ± iy with y ≠ 0.
The cube roots of one are: z = 1, z = −1 + i 3 , z = −1 − i 3
2 2
Review
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END-OF-CHAPTER REVIEW EXERCISE 11
1 a Express 5 − 2i in the form x + iy where x and y are real numbers. [3]
1 + 3i [2]
b Solve w2 − 2w + 26 = 0. [3]
c On a sketch of an Argand diagram, shade the region whose points represent complex numbers [3]
satisfying the inequality z + 1 − 5i ø 2.
[5]
2 a The complex number z is defined as z = k − 6i, where k is a real value.
[2]
Find and simplify expressions, in terms of k, for zz* and z , giving your answers in the [4]
form x + iy where x and y are real. z*
Review 297
b The complex numbers u and w are defined as u = 4 cos 5π + i sin 5π and w = 2eiπ.
12 12 [2]
[3]
Find and simplify expressions for uw and u , giving your answers in the form reiθ , [3]
w [3]
where r . 0 and −π , θ ø π. [2]
[2]
3 The complex number w = 1 + 2i. [3]
[3]
a Represent w and w* by points P and Q on an Argand diagram with origin O and describe [3]
the polygon OPQ.
b Given also that u = −3 − i, write the complex number u in the form r( cosθ + i sinθ ),
where r . 0 and −π , θ ø π. w
4 z = 2 − 5i
a Find the real values x and y such that z* = (2x + 1) + (4x + y)i.
b On an Argand diagram, show the points A, B and C representing the complex
Review numbers z, z* and −z. What type of triangle is ABC ?
c Without using a calculator, express z*
−z
i in the form x + iy where x and y are real
ii in the form r(cosθ + i sinθ ), where r . 0 and −π , θ ø π.
5 z2 + (4 3 )z + 13 = 0
a Find the roots of this equation, giving your answers in the form x + iy where x and y are real.
b On an Argand diagram with origin O, show the position vectors OA and OB representing
the roots of the equation. Describe the geometrical relationship between OA and OB.
c Find the modulus and argument of each root.
6 z = 4 3 − 4i
a Find the exact values of the modulus and argument of z.
Review b Given that w = 2 2 cos π + i sin π , write z in the form reiθ , where r .0 and −π , θ ø π.
12 12 w
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7 a Find the complex number w satisfying the equation w * −2 − 2i = 3iw. Give your answer [5]
in the form x + iy where x and y are real. [4]
b i On a single Argand diagram, sketch the loci z − 3 − 3i = 2 and arg(z − 3 − 3i) = π3 . [2]
ii Hence determine the value of z that satisfies both loci, giving your answer in the
form x + iy, where x and y are real.
Review 8 a (x + iy)2 = 7 − (6 2 )i [5]
Given that x and y are real numbers, find the values of x and the values of y. [1]
[4]
b i Show that z − 3 is a factor of 2z3 − 4z2 − 5z − 3.
ii Solve 2z3 − 4z2 − 5z − 3 = 0.
9 a Given that z1 = 5 − 3i and z1z2 = 21 + i, find z2, giving your answer in the form x + iy, [2]
where x and y are real. [3]
b Solve (3z + 1)3 = −27.
10 a It is given that w = 1 is a root of the equation f(w) = 2w4 + 5w3 − 2w2 + w − 6 = 0.
i Show that w + 3 is a factor of 2w4 + 5w3 − 2w2 + w − 6. [1]
ii Solve the equation. [4]
298 b i On an Argand diagram, sketch the locus z + 1 − i 3 = 1. [2]
ii Write down the minimum value of arg z. [1]
Review iii Find the maximum value of arg z. [2]
11 a i Given that z1 = − 3 + 7 i is a root of the equation z2 + pz + q = 0, where p and q
2 2
are real constants, find the value of p and the value of q.
[3]
ii Find z1 . [1]
b i Find the roots of the equation z3 + 1 = 0. [3]
ii On an Argand diagram, show the points A, B and C representing the roots of the equation. [2]
What type of triangle is ABC ?
12 z = 5 − i
a Show that z = 2 − 5 i. [3]
z* 3 3 [2]
[2]
b Find the value of z and arg z .
z* z* [4]
z [3]
Review c Find and simplify a quadratic equation with integer coefficients that has roots z* and its conjugate.
13 The complex number z is defined by z = k − 4i where k is an integer.
2k − i
7
a The imaginary part of z is Im z = 5 . Find the value of k.
b Find the argument of z.
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14 i Without using a calculator, solve the equation [4]
ii 3w + 2iw* = 17 + 8i, [5]
where w* denotes the complex conjugate of w. Give your answer in the form a + bi.
In an Argand diagram, the loci
arg(z − 2i) = 1 π and z−3 = z − 3i
6
intersect at the point P. Express the complex number represented by P in the form reiθ , giving
the exact value of θ and the value of r correct to 3 significant figures.
Review Cambridge International A Level Mathematics 9709 Paper 31 Q7 June 2013
15 Throughout this question the use of a calculator is not permitted.
i The complex numbers u and v satisfy the equations
u + 2v = 2i and iu + v = 3.
Solve the equations for u and v, giving both answers in the form x + iy, where x and y are real. [5]
[5]
ii On an Argand diagram, sketch the locus representing complex numbers z satisfying z + i = 1
3
and the locus representing complex numbers w satisfying arg(w − 2) = 4 π. Find the least value
of z − w for points on these loci.
Cambridge International A Level Mathematics 9709 Paper 31 Q8 November 2013
16 Throughout this question the use of a calculator is not permitted. 299
The complex numbers w and z satisfy the relation
Review w = z+i .
iz + 2
i Given that z = 1 + i, find w, giving your answer in the form x + iy, where x and y are real. [4]
[4]
ii Given instead that w = z and the real part of z is negative, find z, giving your answer in the
form x + iy, where x and y are real.
Cambridge International A Level Mathematics 9709 Paper 31 Q5 November 2014
Review
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CROSS-TOPIC REVIEW EXERCISE 24
P3 This exercise is for Pure Mathematics 3 students only.
1 Relative to the origin O, the position vectors of the points A and B are given by
−5 1
OA = 0 and OB = 7 .
3 2
a Find a vector equation of the line AB. [3]
Review b The line AB is perpendicular to the line L with vector equation:
4 m
r= 2 + µ 3 .
−3 9
i Find the value of m. [3]
ii Show that the line AB and the line L do not intersect. [4]
2 Given that y = 0 when x = 1, solve the differential equation
xy dy = y2 + 4, [6]
dx
300 obtaining an expression for y2 in terms of x.
Cambridge International A Level Mathematics 9709 Paper 31 Q5 June 2010
Review 3 The point P( p, q, −1) lies on the line L with vector equation [2]
r = i − j + 2k + λ(i + 8j + k).
a Find the value of each of the constants p and q. [2]
b The position vector of Q, relative to the origin O, is OQ = −10i + j − 5k. [5]
i Find the unit vector in the direction OQ.
ii Find angle POQ and hence find the exact area of triangle POQ.
4 The complex number z is given by
z = ( 3 ) + i.
i Find the modulus and argument of z. [2]
ii The complex conjugate of z is denoted by z*. Showing your working, express in the
Review form x + iy, where x and y are real,
a 2z + z*, b iz*. [4]
z
iii On a sketch of an Argand diagram with origin O, show the points A and B representing
the complex numbers z and iz* respectively. Prove that angle AOB = 1 π . [3]
6
Cambridge International A Level Mathematics 9709 Paper 31 Q6 November 2010
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5 The variables x and θ are related by the differential equation
sin 2θ dx = (x + 1) cos 2θ,
dθ
where 0 ,θ , 1 π. When θ = 1 π, x = 0. Solve the differential equation, obtaining an
2 12
expression for x in terms of θ, and simplifying your answer as far as possible.
[7]
Cambridge International A Level Mathematics 9709 Paper 31 Q4 November 2011
Review 6 PQRS is a parallelogram. The vertices, P, Q and R have position vectors, relative to an origin O,
1 3 2
OP = 4 , OQ = 2 and OR = −1 .
5 0 5
a Find OS. [2]
b Find the lengths of the sides of the parallelogram. [3]
c Find the interior angles of the parallelogram. [3]
[4]
7 The complex number u is defined by u = (1 + 2i)2 .
2+i
i Without using a calculator and showing your working, express u in the form x + iy,
where x and y are real. 301
ii Sketch an Argand diagram showing the locus of the complex number z such that z − u = u . [3]
Review Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2012
8 The complex number 2 + 2i is denoted by u.
i Find the modulus and argument of u. [2]
ii Sketch an Argand diagram showing the points representing the complex numbers 1, i and u. [4]
Shade the region whose points represent the complex numbers z which satisfy both
the inequalities | z − 1 | ø | z − i | and | z − u | ø 1.
iii Using your diagram, calculate the value of | z | for the point in this region for [3]
which arg z is least.
Cambridge International A Level Mathematics 9709 Paper 31 Q7 June 2010
Review 9 A certain substance is formed in a chemical reaction. The mass of substance formed t seconds [2]
after the start of the reaction is x grams. At any time the rate of formation of the substance [5]
is proportional to (20 − x). When t = 0, x = 0 and dx = 1. [2]
dt
i Show that x and t satisfy the differential equation
dx = 0.05(20 − x).
dt
ii Find, in any form, the solution of this differential equation.
iii Find x when t = 10, giving your answer correct to 1 decimal place.
iv State what happens to the value of x as t becomes very large. [1]
Cambridge International A Level Mathematics 9709 Paper 31 Q10 November 2010
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10 The line L1 has vector equation [2]
r = −2j + k + λ(i + 3k). [5]
The line L2 passes through the point P(−2, 1, −1) and is parallel to L1.
a Write down a vector equation of the line L2.
b Find the shortest distance from P to the line L1.
Review 11 In a certain country the government charges tax on each litre of petrol sold to motorists. The revenue
per year is R million dollars when the rate of tax is x dollars per litre. The variation of R with x is
modelled by the differential equation
dR = R 1 − 0.57 ,
dx x
where R and x are taken to be continuous variables. When x = 0.5, R = 16.8.
i Solve the differential equation and obtain an expression for R in terms of x. [6]
ii This model predicts that R cannot exceed a certain amount. Find this maximum value of R. [3]
Cambridge International A Level Mathematics 9709 Paper 31 Q7 November 2014
12 The complex number z is defined by z = 9 3 + 9i . Find, showing all your working,
3 − i
302 i an expression for z in the form reiθ, where r . 0 and −π , θ ø π, [5]
ii the two square roots of z, giving your answers in the form reiθ, where r . 0 and −π , θ ø π. [3]
Review Cambridge International A Level Mathematics 9709 Paper 31 Q5 June 2014
13 The variables x and y are related by the differential equation
dy = 6 ye3x .
dx 2 + e3x
Given that y = 36 when x = 0, find an expression for y in terms of x. [6]
Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2014
14 a Show that the straight line L with vector equation
1 8
r = 2 + λ −1
−10 3
Review intersects with the line through the points A and B with coordinates (0, 2, 7) and (7, 1, 27), [5]
respectively and find the position vector of this point of intersection.
b Find the acute angle between these two lines. [4]
15 The variables x and y satisfy the differential equation x dy = y(1 − 2x2 ), and it is given [6]
dx
that y = 2 when x = 1. Solve the differential equation and obtain an expression for y in
terms of x in a form not involving logarithms.
Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2016
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16 The complex number u is defined by u = 6 − 3i .
1 + 2i
1 π.
i Showing all your working, find the modulus of u and show that the argument of u is − [4]
ii For complex numbers z satisfying arg(z − u) = 1 π, find the least possible value of z . 2 [3]
4 [3]
iii For complex numbers z satisfying z − (1 + i)u = 1, find the greatest possible value of z .
Cambridge International A Level Mathematics 9709 Paper 31 Q8 June 2011
17 With respect to the origin O, the position vectors of the points A, B, C and D are given by
Review 2 3 1 2
1 m
OA = 5 , OB = 4 , OC = −5 and OD = −4 .
11 −2m
m +
a In the case where ABC is a right angle, find the possible values of the constant m. [3]
b In the case where D is the midpoint of the line BC, find the value of the constant m. [2]
c In the case where m = −5, find whether the lines AB and CD intersect. [5]
18 The complex number w is defined by w = 22 + 4i . [3]
(2 − i)2 [3]
i Without using a calculator, show that w = 2 + 4i.
ii It is given that p is a real number such that 1 π ø arg(w + p) ø 3 π. Find the set of
possible values of p. 4 4
303
Review iii The complex conjugate of w is denoted by w*. The complex numbers w and w* are [3]
represented in an Argand diagram by the points S and T respectively. Find, in the
form z − a = k, the equation of the circle passing through S, T and the origin.
Cambridge International A Level Mathematics 9709 Paper 31 Q8 June 2015
19 Given that y = 1 when x = 0, solve the differential equation
dy = 4x(3y2 + 10y + 3), [9]
dx
obtaining an expression for y in terms of x.
Cambridge International A Level Mathematics 9709 Paper 31 Q7 June 2015
20
Review 60º h
C
A tank containing water is in the form of a cone with vertex C . The axis is vertical and the
semi-vertical angle is 60°, as shown in the diagram. At time t = 0, the tank is full and the depth
of water is H . At this instant, a tap at C is opened and water begins to flow out. The volume
of water in the tank decreases at a rate proportional to h where h is the depth of water at time t.
The tank becomes empty when t = 60.
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i Show that h and t satisfy a differential equation of the form
dh = − − 3 ,
2
Ah
dt
where A is a positive constant. [4]
ii Solve the differential equation given in part i and obtain an expression for t in terms of h and H . [6]
Review iii Find the time at which the depth reaches 1 H . [1]
2
[The volume V of a cone of vertical height h and base radius r is given by V = 1 πr2h.]
3
Cambridge International A Level Mathematics 9709 Paper 31 Q10 November 2013
21 a Find, in the form r = a + λb, a vector equation of the line AB where the points have [3]
coordinates A(2, 5, 7) and B(9, −1, −2).
b Find the obtuse angle between the line AB and a line in the direction of i + 3j + 2k. [4]
22 The complex number 3 − i is denoted by u. Its complex conjugate is denoted by u* .
i On an Argand diagram with origin O, show the points A, B and C representing the
complex numbers u, u* and u * − u respectively. What type of quadrilateral is OABC ? [4]
ii Showing your working and without using a calculator, express u* in the form x + iy,
u
where x and y are real. [3]
304 iii By considering the argument of u*, prove that
u
tan−1 3 = 2 tan−1 1 . [3]
4 3
Review
Cambridge International A Level Mathematics 9709 Paper 31 Q9 November 2015
23 The complex number 1 + ( 2 )i is denoted by u. The polynomial x4 + x2 + 2x + 6 is denoted by p(x).
i Showing your working, verify that u is a root of the equation p(x) = 0, and write [4]
down a second complex root of the equation. [6]
ii Find the other two roots of the equation p(x) = 0.
Cambridge International A Level Mathematics 9709 Paper 31 Q9 November 2012
Review
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24 G F
N h cm
D E
C
B
kj M
4 cm
Oi 10 cm A
Review The diagram shows a cuboid OABCDEFG with a horizontal base OABC. The cuboid has a
length OA of 10 cm, a width AB of 4 cm and a height BF of h cm. The point M is the
midpoint of CB and the point N is the point on DG such that ON = j + 3k. The unit
vectors i, j and k are parallel to OA, OC and OD, respectively.
a Write down the value of h. [1]
[2]
b Find the unit vector in the direction ON . [4]
[3]
c Find angle NME .
d Find, in the form r = a + λb , a vector equation of the line MF .
25 The variables x and y are related by the differential equation
dy = 6xe3x . 305
dx y2
Review It is given that y = 2 when x = 0. Solve the differential equation and hence find the value of y when [8]
x = 0.5, giving your answer correct to 2 decimal places.
Cambridge International A Level Mathematics 9709 Paper 31 Q7 June 2012
26 i Showing all your working and without the use of a calculator, find the square roots of the complex [5]
number 7 − (6 2 )i. Give your answers in the form x + iy, where x and y are real and exact.
[4]
ii a On an Argand diagram, sketch the loci of points representing complex numbers w [2]
and z such that w − 1 − 2i = 1 and arg(z − 1) = 3 π.
4
b Calculate the least value of w − z for points on these loci.
Cambridge International A Level Mathematics 9709 Paper 31 Q10 June 2016
−2 1
27 The line L1 has vector equation r = 3 + λ −1 .
0 2
Review −4 −1
The line L2 has vector equation r = 5 + µ 0 .
m 1
a In the case where m = 2, show that L1 and L2 do not intersect. [4]
b Find the value of m in the case where L1 and L2 intersect. [2]
c For your value of m from part b, find the acute angle between L1 and L2. [4]
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28 The number of birds of a certain species in a forested region is recorded over several years.
At time t years, the number of birds is N, where N is treated as a continuous variable.
The variation in the number of birds is modelled by
dN = N(1800 − N ) .
dt 3600
It is given that N = 300 when t = 0.
Review i Find an expression for N in terms of t. [9]
ii According to the model, how many birds will there be after a long time? [1]
Cambridge International A Level Mathematics 9709 Paper 31 Q10 June 2011
29 i Showing your working, find the square roots of the complex number 1 − (2 6 )i . Give your [5]
answers in the form x + iy, where x and y are exact.
ii On a sketch of an Argand diagram, shade the region whose points represent the complex [4]
numbers z which satisfy the inequality z − 3i ø 2. Find the greatest value of arg z for
points in this region.
Cambridge International A Level Mathematics 9709 Paper 31 Q10 November 2011
30 Liquid is flowing into a small tank which has a leak. Initially the tank is empty and t minutes later,
the volume of liquid in the tank is V cm3. The liquid is flowing into the tank at a constant rate of
306 80 cm3 per minute. Because of the leak, liquid is being lost from the tank at a rate which,
at any instant, is equal to kV cm3 per minute where k is a positive constant.
Review i Write down a differential equation describing this situation and solve it to show that
V = 1 (80 − 80e−kt ). [7]
k
ii It is observed that V = 500 when t = 15, so that k satisfies the equation k = 4 − 4e−15k .
25
Use an iterative formula, based on this equation, to find the value of k correct to 2 significant figures.
Use an initial value of k = 0.1 and show the result of each iteration to 4 significant figures. [3]
iii Determine how much liquid there is in the tank 20 minutes after the liquid started flowing, and state
what happens to the volume of liquid in the tank after a long time. [2]
Cambridge International A Level Mathematics 9709 Paper 31 Q10 June 2013
Review
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PURE MATHEMATICS 2 PRACTICE EXAM-STYLE PAPER
Time allowed is 1 hour 15 minutes (50 marks)
1 By writing cot x as cos x show that d (cot x) = −cosec2 x. [3]
sin x dx [4]
[5]
2 Solve the inequality 3x − 1 ù 2x .
[4]
3 Solve the equation 32x − 3x+1 = 10 giving the value of x correct to 3 significant figures. [2]
4 The polynomial x3 + 8x2 + px − 25 leaves a remainder of R when divided by x − 1 and a [3]
remainder of −R when divided by x + 2. [3]
Review a Find the value of p. 307
b Hence, find the remainder when the polynomial is divided by x + 3. [4]
[4]
5 The sequence of values given by the formula [3]
[4]
xn+1 = 8xn2 , [1]
3 sec xn [5]
[5]
with initial value x1 = 1, converges to α.
a Use this formula to calculate α correct to 2 decimal places, showing the result of each
iteration to 4 decimal places.
b State an equation satisfied by α and hence find the value of α correct to 7 decimal places.
6 The parametric equations of a curve are x = ln(2t + 1), y = t − e2t.
a Find an expression for dy in terms of t.
dx
Review b Find the equation of the normal to the curve at the point where t = 0.
7 a Express 8 sinθ + 6 cosθ in the form R sin(θ + α ), where R . 0 and 0° , α , 90°, giving
the value of α correct to 2 decimal places.
b Hence solve the equation 8 sinθ + 6 cosθ = 7 giving all solutions in the interval 0° , θ , 360°.
c Write down the greatest value of 8 sinθ + 6 cosθ + 3 as θ varies.
8 a Show that cos 3x ≡ 4 cos3 x − 3 cos x.
∫b Hence show that π cos3 x dx = 2 .
2 3
0
Review
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PURE MATHEMATICS 3 PRACTICE EXAM-STYLE PAPER
Time allowed is 1 hour 50 minutes (75 marks) [3]
1 Use logarithms to solve the equation e2x = 2x+5, giving your answers correct to 3 decimal places. [5]
∫2 5 = 25 ( ln 25 − 1). [5]
Show that dx [5]
x ln x
04
3 The polynomial 2x3 − 9x2 + ax + b, where a and b are constants, is denoted by f(x). It is given
that x − 4 is a factor of f(x) , and that when f(x) is divided by (x − 1) the remainder is −12.
Find the value of a and the value of b.
Review 4 Solve the equation 2 sin(x − 60°) = 3 cos x for −180° ø x ø 180°.
5 y
P
Ox
Q
The diagram shows the curve y2 = 4x2 − x4 + 5. The curve is symmetrical about both axes.
The point P is one of the curve’s maximum points and the point Q is one of the curve’s minimum
308 points. Find the exact distance between the points P and Q. [7]
= 7x2 − 5x + 27
(x − 2) x2 + 5
( )6f(x)
Review a Express f(x) in partial fractions. [5]
b Hence obtain the expansion of f(x) in ascending powers of x, up to and including the term in x2. [5]
7y
O πx
2
The diagram shows the curve y = x sin 2x for 0 ø x ø π .
2
( )a dy x2 d2 y − 2x dy + 2 1+ 2x2 y = 0.
Find dx and show that dx2 dx [5]
[5]
b Find the area of the region enclosed by this part of the curve and the x-axis.
[5]
Review 8 a The complex numbers u and w are such that: [4]
3u − iw = 15
u + w = 5 + 10i.
Without using a calculator, solve these equations to find u and w, giving each answer in
the form x + iy, where x and y are real.
b The complex number z is defined by z = 2 e− π i. On an Argand diagram, show the
4
points A, B and C representing the complex numbers z, z* and z2, respectively and
find the length of the longest side of triangle ABC .
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9 a Given the vectors 5i + 7j + pk and i − 2j + pk are perpendicular, find the possible values of [3]
the constant p. [2]
b The line L1 passes through the point (5, 0, 2) and is parallel to the vector i − 6j + k. [4]
i Write down a vector equation of the line L1.
ii The line L2 has vector equation: [6]
[2]
r = 4j + 3k + µ( 3i + 9j + k ) [3]
[1]
Review Show that L1 and L2 do not intersect.
10 A liquid is heated so that its temperature, x °C , at time t seconds satisfies the differential equation
dx = α (100 − x)
dt
where α is a positive constant. The temperature at t = 0 is 25 °C.
a Show that ln 75 x = αt.
100 −
b It is given that x = 500α when t = 2. Show that α = 0.2 − 0.15e−2α.
c Use an iterative formula based on the equation in part b to find the value of α correct to 2 significant
figures. Use a starting value of 0.1 and show the result of each iteration to 4 significant figures.
d Find the temperature of the liquid when t = 30.
309
Review
Review
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Answers 8 x = ± 3
4
1 Algebra 9 x = 7, y = 5
Prerequisite knowledge Exercise 1B
1 a 357 b 381 1 a ∨-shaped graph, vertex = (−2, 0), y-intercept = 2
c 133 remainder 27 x+2 if x ù −2
−(x + 2) if x , −2
2 Straight line, gradient 2, crossing axes at 5 , 0 y =
and (0, −5). 2
b ∨-shaped graph, vertex = (3, 0), y-intercept = 3
ReviewExercise 1A b −2, 3 c −6, 22 y = x−3 if x ù3
e −20, 40 3 3−x if x ,3
1 a −1, 5
2 f7
c ∨-shaped graph, vertex = (10, 0), y-intercept = 5
d −15, 3
2 a 9 , 11 b −1, 3 c 7 , 23 1 x − 5 if x ù 10
73 e2 46 2
y =
d 1, 3 f −7, 3 1x
5 − 2 if x , 10
3 a −1, − 1 b −3, 3 c 2, 4 2ax 0 1 2 3 4 5 6
3 5 f 0, 12 y5 43 2 3 45
d −4, − 6 e −7, 1 13
5
310 4 a ±3 b ∨-shaped graph, vertex = (3, 2), y-intercept = 5
Review b −1, 2, 1 ( −1 − 33 ) , 1 ( 33 − 1) c 3
22 Translation 2
c −2, ±1
3 a −1 b Translation 5
d 3 − 1, 1 + 5 c Translation 2 −2
Reflection in x-axis,
e 1, 2, 1 − 3, 1 + 3
f 0, 2, 6 0
translation 2
5 a x = 0, y = 4 or x = − 8, y = 16
33 d Stretch, stretch factor 2, with y = 0 invariant,
b x = −1, y = 3 or x = − 5 , y = 15 0 −2
22 translation −3 . 1
6 x = 4 or x = 6 e Reflection in x-axis, translation .
55
7 a x = ±2, x = ±3 f Stretch, stretch factor 2, with y = 0 invariant,
by 0
reflection in x-axis, translation 5
6
Review 4 a ∨-shaped graph, vertex = (−1, 2)
b ∨-shaped graph, vertex = (5, −2)
–6 –4 –2 O 2 4 6x c ∧-shaped graph, vertex = (0, 2)
d ∨-shaped graph, vertex = (0, −3)
c x=0 e ∧-shaped graph, vertex = (−2, 1)
f ∧-shaped graph, vertex = (0, 5)
5 3 ø f(x) ø 14
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6 a, b y 3 a −2 ø x ø 1 b −2 , x , 6
8
− 5
c x ø 3 or x ù3
6 4 a 7 ø x ø 3 b x . 1
(0, 5) 3 3
4 c xù1
5 a x , − 4 or x . 4 b x,2 or x . 8
3 3
2
(2, 1) c x ù 1 d 3,x,7
2
–2 –1 O 1 2 3 4 5 6 x 9
e x ø 2 or x ù 8 f 1 , x , 5
c x = 1 or x = 5
Review − 5 5
6 x , 4 or x . 4
7 a, b y
4
Exercise 1D
3 b x2 − 5x + 7
2 (0, 2) 1 a x2 + 3x − 1 d x2 − 2x + 5
c 3x2 − 4x + 2
1 (2, 0) e −5x2 + 3x − 4
–3 –2 –1 O 1 2 3 4 5 6 7 8x f −6x3 − 6x2 − 6x − 19
c x = ±1
8a y 2 a Quotient = x2 + x + 4, remainder = −8
6
5 b Quotient = 6x2 + 19x + 38, remainder = 70
4
3 c Quotient = 4x2 + 5 x + 1, remainder = 5 311
2
1 24 4
d Quotient = −2x2 + 3x + 9, remainder = −36
Review e Quotient = x + 3, remainder = −10x − 4
f Quotient = 5x2 − 7, remainder = 15 − 13x
–3 –2 –1 O 1 2 3x 3 a Proof b Proof
b x=±2 4 a Proof
Exercise 1C b (x − 2)(2x − 3)(x + 7)
1 3,x,7 5 a Proof b ± 1, − 4
23
2a y 6 a Proof b Proof
5
4 y = |2x – 1| 7 −3, 1 , 5
2
3 y = 4 – |x – 1|
Review 2 Exercise 1E
1 1 Proof
–4 –3 –2 –1–1O 1 2 3 4 5 x 2 Proof
b x , − 2 or x . 2 3 a = −4
3 4 a = 1−b
3
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Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity
5 a = −2, b = 1 4 ±5, ± 3
6 a p = −1, q = −6 b Proof 5 a a=6
7 −4, 0, 1 b (3x + 1)(2x + 3)(x − 4)
8 a p = −7, q = −6 6 a (3x − 1)(2x + 3)(x − 5)
b (x + 1)(x + 2)(x − 3) and (x + 1)(x + 3)(x + 4)
b ± 1 , ±5
9 a p = −19, q = 30 3
7 a = −10, b = 8
b (x − 1)(x + 2)(x − 5)(x + 3) 8 a Quotient = x − 3, remainder = 2x − 6
Review10 a ±2, 5 b −7, − 1, 3 b Proof
c − 5 , 2, 3 d −4, −2, 1 9 a Quotient = 4x2 + 4x − 3, remainder = 5
2 3
b − 3, −1, 1, 1
e −3, ±2, 1 f − 1 , 1, 4 22
2
11 Proof
10 a k = 7
12 0 , k , 5 b 1 ( −7 − 41 ) , 1 ( −7 + 41 ) , 1 ( 7 − 53 ) , +
222
Exercise 1F b8 c4 d − 31 − 3 , 1 ( 7 + 53 )
b b=8 4 2
1 a6 11 a a = 6
2 a a=5 b (2x − 1)(x + 3)(x − 2)(x + 1)
312 3 a = 2, b = 2 12 a a = 7 b −5, 2 , 2
3
Review4 a = 4, b = 0 13 a 33
b Quotient = 2x + 13, remainder = 41x − 15
5 a a = 6, b = −14
b (3x − 4)(2x + 1)(x + 2) 14 a a = 3, b = −10 b 3x − 4
15 a a = −21 b − 5, − 1, 2
6 a p = 14 b 57
32
7 a a = 5, b = −15 16 a a = −3, b = −11
b (2x − 1)(x + 2)(x − 3)
b 2, −7 + 53 , −7 − 53
22 17 a a = 2, b = −16 b −18
8 a k=5 b −36
2
18 a Quotient = 5x2 − 8x + 9
9 2550
b Proof
10 a = 2, b = −4, c = −2. 19 a k = −29 b −2, 1 , 9
c ±3, ± 1 4
2
ReviewEnd-of-chapter review exercise 1
1 − 4, 2 20 a a = 3, b = 50
37
b Quotient = 2x − 3, remainder = 56 − 4x
− 4
2 x ø 5 or x ù 2 b −2, 3, 5
21 a a = −11, b = 30 2
5
3 1 , x , 3 22 i x + 2, 3x + 4
ii Proof
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23 i −16 ii − 1 , 3, 3 e −2 f3 g −1 h −4
22 2 2 3
24 i a = 2, b = −6
ii (2x + 3)(x − 1)(x − 3) 6 a3 b1 c5 d −4
3 2
h9
e −6 f5 g2 2
2 3
2 Logarithmic and exponential
functions 7 f −1(x) = 3 + 2x−1
Prerequisite knowledge 8 a3 b −3, 1
2
1 a1 b4 c1 9 log4 3, log2 2, log3 4, log2 3, log3 9, log3 20, log2 8
25 c 8x15
Review b 1 5 Exercise 2C b log6 5 c log5 2
21 27 e log2 18 f log4 8
5 −7 5 x2 1 a log2 77
d log3 2 b 2 log6 10 c 2 + log2 3
3 a 2x 2
2 a3
Exercise 2A
1 a 2 = log10 100 b x = log10 200 3 a log5 4 or 2 log5 2
c x = log10 0.05 b 2.40
b x = 101.2 b log3 4 or 2 log3 2
2 a 1.72 b 575
c −0.319 4 23, 2−2, − 3 b7 d −3
c 1.5 2 4 2
3 a 10 000 = 104 f 0.5
c x = 10−0.6 5 a3 c −1
4 a 75.9 6 y = x−2 313
x2
c 0.0398
= 1
Review 5 a2 b −4 7 z 1 − 3y3
d1 e 2 1 8 a 5y b 2+y
3 3
c 3 y−3 d3
6 f −1(x) = log10 (x + 3) 2 y
7 10 10 9 a 3+x b 1−y
2
Exercise 2B c 4x+y
d 2x − 1 y − 1
1 a 2 = log5 25 b 4 = log2 16 10 a 3 2
c −5 = log3 1 d −10 = log2 1 c 16 b −9
243 1024 2
11 x = 2 log3 2 − 2 log2 5,
e x = log8 15 f y = logx 6 y = log2 5 − 2 log3 2, d5
z = log2 5 − log3 2 3
g b = loga c h 5y = logx 7
Review 2 a 23 = 8 b 34 = 81 c 80 = 1 Exercise 2D
f 24 = y
1 1 1 a 10
2 a 41 c − 46
d 162 = 4 e 83 = 2 9
2
g a0 = 1 h xy = 5 3 a5 c −3
10
3 a8 b9 c1 d7 b 3.5 d 54
b9 c 3, 6
4 a4 b 11 c3 d −3 d3
5 a3 b2 c1 7 2
b 10
2 d1
4
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Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity
4 a8 b 12 c 5 d4 Exercise 2F
3
5 a 8, 32 1 a x , log 5 b x ù log 7
c 1 , 25 b 0.1, 1000 log 2 log 5
3125
d 2 2, 2 c x , log 3 d x . log 0.3
6 a x = 3, y = 27 2 log 2 log 0.8
b x = 1 , y = 1 3
2
c x = 4, y = −12 2 a x .5 − 1
d x = 625 , y = 25 log 8
42
7 log10 x = 3, log10 y = −2 b x 1 log 20 −
2 log 3 5
Review .
Exercise 2E b 5.13 c 0.946 c xø 1 log 3
e 3.86 f 1.71 2 2 − 1
1 a 1.80 h 0.682 i −0.756
d 3.64 k −0.443 l −15.4 log 5
g 0.397
j 6.76 b 1.26 c 1.83 log 4
f 0.535 − 7
2 a Proof b 1.11 d x .3 log 5
e 3.21
3 a −0.322 6
d 1.03
0 or x log 2
314 3 x , .
log 5
4 2, 3
4 Proof
5 a 61 b1
Review 5 a 0.431, 0.683 b 0, 2.32 Exercise 2G
c 0, 1.77 d 0.792, 1.58
1 a 20.1
6 2, 2.58 c 2.23 b 14.9
d 0.135
7 a 2.81
c 0.431, 1.29 b 1.46 2 a 1.10 b 0.336
d 0.792, 0.161 c −0.105 d −1.90
8 a 1.58
c 1.58, 2.32 b −0.792, 0.792 3 a2 b3 c 30 d2
d 1.37
9 a 1.77
10 0, 200.8 b 0.510 c 1.98 4 a5 b 15 c4 d1
c 1.08 3
11 a 1.29, 1.66 5 a 2.89 b 1.61
d 0.834 d 2.89
g ±1.89
b 1.63 c 1.91, 2.91 6 a ln13 b 1 ln 7
12 a 13.5 e −0.515 f 1.87 c 1 (1 + ln 6) 3
h ±2.81 2
13 0, 1, 2 d 2[ln(4) − 3]
b 1.45
Review 7 a x . ln10
b x ø 1 (2 + ln 35)
5
c x , − 1 (3 + ln 5)
2
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8 a 148 b 0.0183 4 a log10 y = 3 x + 2
c 405 d −0.432 2
b y = 100 × 3 x
9 a 1.22 b 5.70 c 1.41 10 2
d 0.690 e 1.16 f 1.08
5 a ln y = 3 ln x − 2
10 a y = 1 b y = x3
x−2 e2
b y = 2 6 Gradient = ln 3 y-intercept = 0, ln 3
ex2 − 1 , 2 ln5
ln 5
11 a ln 3 b ln 2, ln 3 7 a ln m
d ln 7
4
Review c ln 5
2
3
ln x − 2
12 f −1(x) = 5
2
13 0.151
14 a x = e, y = 5 1
e
O 10 20 30 40 50 t
b x = 2 ln 2, y = − 1 ln 2
5 b m0 = 50, k = 0.02
c 35 days
15 − 1 , x ø 3 8 a ln(T − 25) = −nt + ln k
2
b k = 45, n = 0.08
Exercise 2H c i 70 °C 315
1 a ln y = ax + b, Y = ln y, X = x, m = a, c = b ii 34 minutes
iii 25 °C
Review b log y = ax − b, Y = log y, X = x, m = a, =
m = a, c = −b
c ln y = −b ln x + ln a, Y = ln y, X = ln x, = End-of-chapter review exercise 2
= m = −b, c = ln a log 7
log 2
d ln y = x ln b + ln a, Y = ln y, X = x, l 1 x .
m = ln b, c = ln a q2
3+q
e x2 = −by + ln a, Y = x2, X = y, m = −b, = 2 p =
m −b, c = ln a ,3 x , − log 3
f ln y = − a ln x + ln 8 , Y = ln y, X = ln x, m = 2
bb
log 8
= n , m = − a , c = ln8
Review bb 4 20.1
5 −1, 2 log 2 − log 3
g ln x = − y ln a + ln b, Y = ln x, X = y, = l
log 2
= , m = − ln a, c = ln b 64
h ln y = −bx + ln a, Y = ln y, X = x, =
3
= m = −b, c = ln a 7 K = 7.39, m = 1.37
2 a = 66, n = −0.53 8 i Proof ii 0, 1.58
9 9.83
3 k = 9.5, n = 0.42
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10 i (x + 2)(4x + 3)(3x − 2) Exercise 3A b2 c −1
ii 3y = 2 , y = −0.369 3
3 1 a2
11 3.81 d2 e2 f −3
12 22.281 g −2 h −3
3 b1
13 0.438 c2
2 a2 3 f −2
14 a 2 ln 3 e −2
ln 5 d −1
3 3
b 1, − 2 ln 3 h2
ln5 g1
Review 3 3
15 K = 1.73, b = 1.65
3 a 70.5°, 289.5° b 51.3°, 231.3°
3 Trigonometry c 199.5°, 340.5° d 41.4°, 318.6°
Prerequisite knowledge 4 a π , 5π bπ
66 d 2.76, 5.90
1 ay
c 0.464, 3.61
3
5 a 28.2°, 61.8° b 37.8°, 142.2°
2 c 22.5°, 112.5° d 24.1°, 155.9°
1
O 6 a 60°, 180° b 35.9°, 84.1°
–1
90 180 270 360 x c 2.82, 5.96
d −2.28, −1.44, 0.865, 1.71
316 b y
Review O 90 180 270 360 x 7 a −150°, −30°, 30°, 150°
b −109.5°, −70.5°, 70.5°, 109.5°
cy c −112.6°, 112.6°
d −180°, 0°, 180°
1 e −135°, 45°
f −60°, 60°
O 90 180 270 360 x
8 a 48.2°, 180°, 311.8°
–1 b 31.0°, 153.4°, 211.0°, 333.4°
c 19.5°, 160.5°, 203.6°, 336.4°
2a3 b− 2 c− 3 d 60°, 180°, 300°
2 2 3 e 107.6°, 252.4°
f 27.2°, 152.8°
Review3 a 31.0°, 211.0° 9 a 41.8°, 138.2°
b 13.3°, 22.5°, 103.3°, 112.5°
b 30°, 150°, 270° c 45°, 60°, 120°, 135°
d 97.2°, 172.8°
10 a 2π , π, 4π
33
b 7π , 11π
66
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11 a i y –π2 π –32π 2π x vi –π π –3π 2π x
–π π –3π 2π x 22
6 22 y
4 –π π –3π 2π x 3
2 22 2
O 1
–2 –π π –3π 2π x O
–4 22 –1
–6 –2
–3
ii y
Review b x = π , x = 5π , x = 9π , x = 13π
O 88 8 8
iii y 12 a Proof b Proof
6 c Proof d Proof
4
2 13 a Proof b Proof c Proof
O
–2 d Proof e Proof f Proof
–4
–6 g Proof h Proof
iv y 14 a 48.2°, 180° b 45°, 63.4°, 161.6°
6 Exercise 3B
4
2 1 3 cos x − 1 sin x 317
O 22
–2
–4 2 a1 b2 c1
–6 2 2
Review
vy d3 e1 f3
2 3
6
3 a 6+ 2 b 2+ 3 c 2− 6
4 4 4
e 6− 2
2 d −2 + 3 4 f 6− 2
4
O h 6+ 2
g −2 + 3 4
4 4+3 3
10
5 Proof
Review 6 a 33 b 16 c − 33
65 65 56
7 a 77 b 36 c 13
85 85 84
–π π –3π 2π x
2 2
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8 t−2 8 1 (1 + cos 4x)
2t + 1 2
91 b2 9 a Proof
2 b sin 3x ≡ 3sin x − 4 sin3 x
b −0.2
10 a 1 10 35.3°, 60°, 120°, 144.7°
3 b −30°, 150°
b 70.9°, 250.9°
11 a 3 d 150°, 330° 11 a Proof b4
2 b 18.4°, 116.6° 12 a Proof b − 2π , − π , π , 2π
d 35.0°
12 a Proof f 74.1° 13 a Proof 3 33 3
b − 5π , − π , π, 5π
13 a 19.1°, 199.1°
6 66 6
Review c 5.9°, 185.9°
14 a 38.4°, 111.6° 14 a Proof
c 16.0° b sin 3θ ≡ 3sinθ − 4 sin3 θ, cos 2θ ≡ 1 − 2 sin2 θ
e 18.4°, 26.6°
c Proof
15 22.5°, 112.5° d 5 −1
4
16 Proof 15 2π , θ , 4π
17 p2 + q2 − 2 3 3
2 16 210° ø θ ø 330°
Exercise 3C 17 Proof
318 1 a sin 56° b cos 68° c tan 34° 18 a Proof
2 a 24 b 0° , x , 45° or 120° , x , 135°
25 b −7
25 19 45° , θ , 135° or 225° , θ , 315°
Review c − 24 d − 44 Exercise 3D
7 117
1 a Proof
3 a 336 b − 336 d Proof b Proof c Proof
625 527 g Proof e Proof f Proof
h Proof
c7 d 24 2 a Proof c Proof
25 7 d Proof b Proof f Proof
g Proof e Proof
4 a − 24 b 336 h Proof
25 625 3 a Proof
b Proof
c 24 d2 4 Proof
7
5 Proof
51
3 6 Proof
6 a 14.5°, 90°, 165.5°, 270°
b 60°, 300°
Review c 48.6°, 131.4°, 270°
7 a 30°, 150° Exercise 3E b 64.1°, 172°
b 33.6°, 180° b 12.6°, 234.8°
c 30°, 150° 1 a 17 sin(θ − 28.07°) b 38.2°, 197.9°
d 33.2°, 90°, 146.8°
e 39.2°, 90°, 140.8° 2 a 13 cos(θ + 56.31°)
f 24.9°, 98.8°
3 a 17 sin(θ − 28.07°)
c 34
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4 a 2 13 sin(θ −56.31°) b 80.9° 5 41.8°, 138.2°, 194.5°, 345.5°
c 49, −3
6 a Proof b 30°, 330°
5 a 5 sin(θ + 53.13°) b 103.3°, 330.4° 7 a Proof
b Proof b 61.3°, 118.7°, 241.3°, 298.7°
c −2
6 a 2 cos θ − π 8 i Proof ii 18.4°, 26.6°
3
7 a 4 5 sin(2θ + 26.57°) 9 a Proof b 35.8°, 125.8°
b 66.9°, 176.5°, 246.9°, 356.5°
10 i R = 10, α = 18.43°
Review c1 b 70.5°, 180° ii 34.6°, 163.8°, 214.6°, 343.8°
8
11 a Proof
8 a 3 cos(θ + 54.74°)
bi 2
c1 2
6
ii 15°, 75°
cos θ π b 23π
9 a 2 + 4 12 12 a Proof
b 2 13 cos(θ − 56.31°)
c − 2økø 2 b 32.3°, 290.8°
c 52
10 a 10 sin(θ − 71.57°) 13 i R = 2 13, α = 56.31°
c 1 + 10, 80.8° ii 80.9°, 211.7°
11 a 3 cos(θ − 41.81°) b 41.8° iii 60, 8
c 302.6°
319
12 a Proof b 5 sin(θ + 53.13°) 14 i Proof ii Proof
c 115.3°, 318.4° iii 0.322, 0.799, − 1.12
Review
13 a 3 sin(θ + 60°) b 84.7°, 335.3° 15 a Proof
14 a 10 + 5, 10 − 5 b 60°, 104.5°, 255.5°, 300°
16 a Proof b 21.8°, 161.6°
b 18.4°, 45°, 198.4°, 225° 17 a Proof b 5 sin(2x + 36.87°)
End-of-chapter review exercise 3 c 71.6°, 161.6°
1y Cross-topic review exercise 1
1 2 , x , 4
5
2 0.631, −0.369
3 3 i x=7 ii 0.222
6 ii −0.569
O 45 90 135 180 x
–3 4 i −2, 2
5
Review
5 A = 8.5, b = 1.6
6 i 1.77 ii ±1.77
2 131.8° 7 i n = 1.50, C = 6.00
ii n ln x + ln y = lnC is linear in ln y and ln x.
3 − 15
8 8 −0.405, 1.39
4 0°, 131.8°, 228.2°, 360°
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9 θ = 135°, φ = 63.4° or θ = 53.1°, φ = 161.6° b 4
10 i − 2 , 4 ii −0.161 (1 − 2x )23
35 ii 3 , − 1 , 3
3 x − 5y = 16
11 i a = −16 22
4 (0, 2) maximum, (2, −2) minimum
12 i a = −4, b = 6
ii quotient = 2x − 4, remainder = −2 Exercise 4A
13 i R = 10, α = 71.57° 1 a (6x − 2)(x − 2)4 b 5(2x + 1)2(8x + 1)
ii 61.2°, 10.4°
c 3x + 4 d 3x + 9
2 x+2 2 x+5
Review14 i Proof
ii (x − 2)(4x + 1)2 e x2(7x − 3)
iii 2 2x − 1
15 i a = 6, b = −3 f (13x2 + 2)(x2 + 2)2
ii (x + 1)(4x + 1)(2x − 1) 2x
g (x − 3)(x + 2)4(7x − 11)
16 i a = −17, b = 12 h 2(2x − 1)4(3x + 4)3(27x + 14)
ii x = −3, x = 4, x =1 i 2(3x2 + 1)(15x2 − 30x + 1)
3 2 −1.5
17 i a = 1, b = −10
ii quotient = x − 1, x = 1, x = 2, x = −4 3 16x + y = 32
320 18 i a = 19, b = −36 45
ii (x + 2)(x + 3)(5x − 6), 0.113
5 −1, 3 , 3
Review19 i quotient = x2 + 2x + 1, remainder = 5x + 2 5
ii p = 7, q = 4
iii x = −1 6 −1
3
7a y
20 i 2 ii − 9 8
3 20
21 i 29 sin(2θ + 21.80°)
ii 13.1°, 55.1°, 193.1°, 235.1° (1, 3) y = (x – 1)2(5 – 2x) + 3
(2, 4)
iii 1
116 O x
22 i a = 2, b = −5
ii a Proof b2
b 109.5°
Exercise 4B
Review4 Differentiation 1 a − (x 11 b 1
− 4)2 (2 − x)2
Prerequisite knowledge c 2(x2 − x + 3) d 11
(2x − 1)2 (2 − 5x)2
1 a 15x2 + 6 + 1
x3 x
− 2(8x + 1) − 20x3
b 5 x4 − 4x − 1 e (x + 4)3 f (x2 − 1)3
2 2x2
2 a 12(3x − 5)3
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g 13x2 + 30x − 35 4 a xex + ex b 3x2e3x + 2xe3x
(x2 + 2x + 5)2
c e−2x (5 − 10x) d ex (2x + 1)
+ 4)(2x2 + 12x − x
h − 2(x (x2 + 1)4 1)
21 e e6x (6x − 1) f − e−2x (4x + 1)
4 x2 2x x
3 (−6, −7), (1, 0) g 3ex h 3xe3x + 3e6x + e3x
(ex + 2)2
4 (2, 1), (8, −5) i 2x2ex + 5xex + 2xe2x − ex − 2
(ex + 2)2
Review 5 y = 9x − 4 5 −4
9
6 a −5x − 1 b x+4 6 −1, − 1
2 x (5x − 1)2 e
3
(2x + 3)2 7 y = 3x + 3, (−1, 0)
c − x(x2 + 1) d 5(x − 1)2(5x + 13)
3 3 8 (3, −e3 ) minimum
(x2 − 1)2 2(x + 2)2
73 9 (1, e2 ) minimum
8 3y = x + 7 10 a x = 0 minimum, x = 2 maximum
b Proof
9 a −3, 1, 5
b 9, − 1 , − 5 11 x = 1 − 1 , x = 1 + 1 321
33 22
Review Exercise 4C b −4e−4x 12 1 , 2
d −15e−5x 2
1 a 5e5x f 2e2x−7
h 2 + 3e x 13 Proof
c 12e6x
2x 14 3 + 3 ln 3
x j 6e3x
l 10xex2 − 10
e 2e2
x
g 2xex2 −3
i 5 ex + 2e−2x Exercise 4D
2
1 a1 b1
k 3e2x − e−2x x x
2a y c 2 d 2x
2x + 1 x2 + 1
y=1
O2 e 4 f 1
2x − 1 2(x − 3)
Review y = 1 – e2–x g 5 h 3− 1
x+3 x
1 – e2 i 5 − 2 j 1
2x − x ln x
1
b y = −x + 2 k 1 l 1 + 5x
3 0.0283 grams per year x(5x + ln x)
x ( x −2)
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2 Possible justification: ln 3x = ln 3 + ln x and g x 1 3 − x 1 4 − 1
ln 7x = ln 7 + ln x − + x −1
d (ln 3x) = d (ln 3 + ln x) = 0 + 1 and h − x 2 1 − 1 2
dx dx x + x−
d (ln 7x) = d (ln 7 + ln x) = 0 + 1 i x 1 2 + 2 − 1 − x 1 5
+ 2x − 1 x +
dx dx x
3 a 1 + ln x b 2x2(1 + 3 ln x) 11 a 4x b 9x2 + 2
2x2 − 1 3x3 + 2x
c 2x 1 + ln(2x + 1) d 3(1 + ln 2x) c 2x − 4
2x + 1 − ln5x (x + 1)(x − 5)
Review e 1 + ln(ln x) f x2 12 −5
ln x
g − 2 Exercise 4E
x(ln x)2
1 a cos x
3x − (3x − 2)ln(3x − 2) c −2 sin x − sec2 x b 2 cos x − 3sin x
h x2(3x − 2) d 6 cos 2x
2(4x − 1) − 4(2x + 1)ln(2x + 1) e 20 sec2 5x
i (2x + 1)(4x − 1)2
f −2(3sin 3x + 2 cos 2x)
4 ay x = 3 y = 1n(2x – 3) g 3sec2(3x + 2) h 2 cos 2x + π
2 3
2
322 i −6 sin 3x − π
1 6
2 a 3sin2 x cos x b −15 sin 6x
Review O c 2 sin x(1 + cos x) d 4 sin x (3 − cos x)3
1 2 3 4 5x
e 12 sin2 2x + π cos 2x + π
–1 6 6
–2 f −12 cos3x sin x + 8 tan 2x − π sec 2 2x − π
4 4
b2
7 3 a x cos x + sin x
5 −8 b 5(cos 3x − 3x sin 3x)
6 2 + 4 ln10, 3 + 2 ln10 c x2 sec2 x + 2x tan x
d cos2 2x(cos 2x − 6x sin 2x)
7 1 , − 1 , minimum e 15 tan 3x sec 3x f sec x(x tan x + 1)
e 2e
x sec2 x − tan x 1 + 2 cos x
Review8 e, 1 , maximum g x2
e h ( 2 + cos x )2
(3x − 1) cos x − 3sin x
9 y = 5x − 5 i (3x − 1)2
j −6 cot 2x cosec3 2x
10 a 5 b −3
2(5x − 1) 3x + 2 k 3(1 − 2x cot 2x) cosec 2x
c 1 + 5 d 2 3 − 1 l 2
x x +1 2x + x −1 sin 2x − 1
e 3 1 − 2 f 1 + x 1 2 − x 1 4 4 a cos x esin x b −2 sin 2xecos 2x
3x − x x − +
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c 3sec2 3 xetan 3x e 18x2 y2 dy + 12xy3 f 2y dy + x dy + y
dx dx dx
d (cos x + sin x)e(sin x−cos x)
g 3x2 − 7x dy − 7y + 3y2 dy
e (cos x − sin x)ex dx dx
f (2 cos 2x + sin 2x)ex h x cos y dy + sin y − y sin x + cos x dy
dx dx
g ex (cos x − 3sin x) i x3 dy + 3x2 ln y
y dx
h x2 (3 − x sin x)ecos x
j −2x sin 2y dy + cos 2y
i −tan x dx
Review j x cot x + ln(sin x) k 5 dy + ex cos y dy + ex sin y
dx dx
k − 2(sin 2x + cos 2x)
e2x +1 l −2x sin y ecos y dy + 2ecos y
(1 − 2x) sin 2x + 2x cos 2x dx
l e2x 2 a − 3x2 + 2 y b 5 − 2xy
2x + 3y2 x2 + 2y
51
4x + 5y d y(2 − ln y)
6 2 3−6 c − 5x + 2y x
7 π, π
e − 2 y(ex y2 + 1) f − y x
63 3ex y2 + 2 2xy2 +
8 Proof
g y4 h y(5xy + 1)
9 a tan x sec x 2 − 3xy3 x(2 − 5xy)
c −cosec2x
b −cot x cosec x 3 − 11 323
10 Proof 25
11 y = −13.3x + 12.9
43
Review 5 y= 1 x−9
44
12 0.464, 2.03 6 y= 3x+5 b (−1, 5), (1, −5)
88
13 x = π , maximum
4 7 a Proof
8 a (4, 18), (4, −2) b Proof
c y = 4x + 2
14 x = π
8 9 a Proof b (1, 2)
15 x = π , minimum 10 − 44
12 21
16 x = π maximum, x = 5π minimum, x = 7π 11 (6, −3), (−2, 0)
6 6 6
maximum, 11π minimum
x = 6 12 5x − 8y = 1
Review 17 0.452 13 e−1
Exercise 4F 14 (−4, −8) minimum, (4, 8) maximum
1 a 5y4 dy b 3x2 + 4y dy Exercise 4G 2 − 2 sin 2θ
dx dx cos 2θ
1 a1
c 10x + 1 dy d cos y dy 3t
y dx dx
b
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sin 2θ d 4 cos 2θ 12 i Proof ii (5.47, 0.693)
c 3sec2 θ 13 i Proof ii k = 5, c = 68
1 − cos 2θ
sinθ
e − sec2 θ sin 2θ
f sinθ − 2 sin 2θ
2 sec2 θ
g sin 2θ h −e2t − 1 5 Integration
i 1 t2e−t (t + 3) j 2et(t + 3) Prerequisite knowledge
2 l4
5(1 − t) 1 a 6 cos 2x + 5 sin x b 5e5x−2
t2 t
k c 2
2x + 1
−3, 1
Review2 3 2 Proof
3 33 Exercise 5A
4 (2, 4) 1 a 1 e2x + c b − 1 e−4x + c c 2e3x + c
2 4
5 x+y=2
6 a Proof b e4, 4 − 1 d 1x + c e −2e−x + c
7 a Proof e2
8e 2
b (1, 2) f 1 e2x+4 + c g 1 e3x−1 + c
2 3
8 a Proof b 6, minimum
h −2e2−3x + c i 1 e8x−3 + c
9 y=− 1 x+4 4
324 2 2 a x − e−x + c
c x − e2x + e4x + c b 5 ex(e3x + 8) + c
10 a Proof b (1, 2) c Proof 4 4
e 1 e−2x (ex − 4) + c
Review11 a Proof b Proof 2 d x − 2e−2x + c
f x − 2e−2x + 4e−x + c
End-of-chapter review exercise 4 3 a 1 (e6 − 1) b 1 (e2 − 1)
3 4
1 i Proof ii (1, 6)
ii 2 c 15 d 1 e(e6 − 1)
2 i4 8 2
25
3 i Proof ii Proof e 4e − 4 f 1 (−3 + 4e + e2 )
iii (e−6, 4e−2 + 3) e 2
4 i5 ii −3 g 1 (−17 + 6e2 + 8e3 + 3e4 )
12
5 i 5x + 4y − 6 = 0 ii Proof h − 19 − 25 + 2e2 i 9− 1 − 2
2 2e2 2 2e4 e2
6 i Proof ii Proof iii (−3, −2) 4 y = 3e2x − 2e−x + 1
7 y = 8.66x − 2.53 5 y = 5e−2x + 2x − 2
Review8 i Proof ii 0.294, 1.865 6 1 4 − 1 + e
ii x = 1 2 e3
iii − 13
9 2 3
9 5x − 9y + 22 = 0 7 4e2 − 4
10 i − 1 8 a Proof b 1 + 2e3
3
1 − x (x + 1)2 9 a 7 − 2e−2a − 5e−a b7
11 dy = −cost
dx
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10 6 − 7 ln 2 6 a y = 5 − 2 cos 2x − π
11 a 1 2
b Proof
b x + 2y = 10 − 2 3 + π
3
Exercise 5B 7 π+ 3
1 a 6 ln x + c b 1 ln x + c 2
2
c 1 ln(3x + 1) + c 8 21
3 d 3 ln(2x − 5) + c 4
e − 5 ln(2 − 3x) + c f 3 ln(5x − 1) + c 9 a 1 ( 3 − 1) b Proof
3 10 2
2 a ln 7 b 1 ln 3 10 a 1 ( 3 − 2 ) b Proof
2 2 2
Review
c 3 ln 13 d 1 ln17 Exercise 5D
23 29
e − 3 ln 7 f −ln 9 1 a 3 x + 3 sin 2x + c
23 24
3 a 12 + 5 ln14 b ln 27
35 25 b 2x + 2 sin x + c
c 4 + ln81 c 1 x − 1 sin 6x + c
2 12
4 a A=2 b Proof
d 2 tan x − 2x + c
5 a Quotient = 3x + 10, remainder = 50
e 2 tan 3x − 6x + c
b Proof f 3 x + 1 sin 2x + 1 sin 4x + c
6 y = x2 + 3 ln(x + e) − 3 ln 2e 84 32
325
7 k = 4e2 − 3 π− 3 b 2 3−π
2 a 68 6
8 (2 ln 2, 2 ln 2 − 3)
Review c 3+π d 2+π
86 16
Exercise 5C e 1 (2π − 3 3 ) f 1 (3 3 − π)
6 6
1 a − 1 cos 3x + c b 1 sin 4x + c
3 4 3 a 1 (2π − 5 3 ) b 1 (6 + 2π + 3 3 )
c −2 cos x + c d − 3 cos 2x + c 24 24
2 2 c 1 (2 + 5π) d 5π − 9 − 3
5 1 8
e 3 sin 3x + c f 2 tan 2x + c 34
g − 2 sin(1 − 5x) + c h − 3 cos(2x + 1) + c 1 ( 2π + 3) f 53
5 2 e 24 11 4
2
i 5 tan(5x − 2) + c 4 a Proof b Proof
2a3 b1 5 a Proof b Proof
8
c3 d2 6 a Proof b Proof
2
Review 7 a Proof b Proof
e 1− 3 f 5π − 4
4 2 8 a Proof b Proof
3 a x cos x b 1 ( 3π − 3) 9 π 2 + 5π
4 y = x + 3 cos 2x − π 6 4
24 10 a Proof c Proof
5 y = 3sin 2x + 2 cos x − 2x + π − 3
b Proof
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Exercise 5E 14 i Proof ii x sin x iii 1
15 i Proof ii Proof
1 a 5.22 b 17.09 c 0.92
d 5.61 e 0.40 f 7.68
2 1.55 6 Numerical solutions of equations
3 a 6.76 Prerequisite knowledge b −48.7505 …
b Over-estimate since the top edges of the strips 1 a 35 d 5.405 465
all lie above the curve 27 b x= y+7
y c 0.223130 3
Review 2 a x = y − 13
4
c x = 5 1+ y
27
3 Proof
Oπ x 4 x = 2, x = 5
2π
5a y
4 1.77, under-estimate since top edges of the strips
all lie below the curve 1
Ox
5 4.07, over-estimate since top edges of the strips
all lie above the curve
End-of-chapter review exercise 5
326 1 Proof
2 12 + ln 81
25
Review
3 1.81, under-estimate since top edges of the strips by
all lie below the curve
41
12
5 Proof O 2π x
–2
6 a 1 (19 − 15e−2k − 4e−3k )
6 –4
b 19
6
7 Proof cy
8 a A = −20 b Proof
9 i Proof ii Proof –2 O 2 4 6x
Review10 a 12ex + 4 e3x + c b Proof –16
3 ii 5 π + 1
11 i Proof 84
12 i Proof ii Proof
iii − 1 cot x + c
2
13 i 1 tan 2x + 1 x + 1 sin 4x + c
2 28
ii ln(16e20 )
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y
20
y = x3 + 5x2
10
–5 O x
y = 5 – 2x
0.5
Review O x
by 3 points of intersection, so 3 roots.
O1 2 x b Let f(x) = x3 + 5x2 + 2x − 5 = 0 then
f(0) = 03 + 5(0)2 + 2(0) − 5 = −5 and
f(2) = 23 + 5(2)2 + 2(2) − 5 = 27. Change of
sign indicates presence of root.
3a y
y = x3
cy 1 x 327
–1 O 1
1
Review O 2π x y = 1 – 5x
–1
One point of intersection so one solution only
of x3 + 5x − 1 = 0.
Exercise 6A y y = x² b Let f(x) = x3 + 5x − 1 then
f(0.1) = 0.13 + 5(0.1) − 1 = −0.499 and
1a f(0.5) = 0.53 + 5(0.5) − 1 = 1.625. Change of
sign indicates presence of root.
4a y y = 3x – 4
y = ln(x + 1)
–1 O 4 x
–4 3
Review 1 y = √1 +x
–1 O x
b 2 points of intersection so 2 roots. b ln(x + 1) = 3x − 4 in 2 places and so
ln(x + 1) − 3x + 4 = 0 has 2 roots.
c Let f(x) = x2 − 1 + x = 0 then
f(−1) = (−1)2 − 1 + (−1) = 1 and
f(0) = (0)2 − 1 + 0 = −1. Change of sign
indicates presence of root.
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Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity
5a y One point of intersection for −0.5 ø x ø 0.5
y = ex and so only one root in this domain.
y=x+6
8a y
6
1 y = sin x –π Ox
–6 O –2π
x
y = 2x + 3
Review Graphs intersect at 2 points, so 2 roots. One point of intersection for −2π , x , − π
2
b Let f(x) = ex − x − 6 then sin x
f(2.0) = e2 − 2 − 6 = −0.610 … and and so only one root of 1= 2x + 3 on this
f(2.1) = e2.1 − 2.1 − 6 = 0.0661… Change of
sign indicates presence of root. domain. Also, should x be less than −2π or
should x be greater than − π , the line and
6 a Let f(x) = (x + 2)e5x − 1 = 0
then f(0) = (0 + 2)e0 − 1 = 1 and 2
f(−0.2) = (−0.2 + 2)e−1 − 1 = −0.337 …
Change of sign indicates presence of root. curve will not intersect again and so this is the
b y y = e5x only point of intersection of y = sin x and
y = 1. 2x + 3
b Let f(x) = sin x −1= 0 then
2x + 3
328 f(−2) = sin(−2) −1= −0.09070 … and
1 −4 + 3
1 –2 O x f(−1.9) = sin(−1.9) − 1 = 0.1828 … Change of
y= −3.8 + 3
Review
x+2 sign indicates presence of root.
9a y
y = x3 + 4x
Graphs intersect at 1 point, so 1 root only. 4
7 a Let f(x) = cos−1 2x − 1 + x = 0 then
–5 O 5x
f(0.4) = cos−1(0.8) − 1 + 0.4 = 0.0435… and y = 7x + 4
f(0.5) = cos−1 1 − 1 + 0.5 = −0.5. Change of
sign indicates presence of root. One point of intersection for 0 ø x ø 5 and so
by only one root.
π b Let f(x) = x3 − 3x − 4 then
f(2) = 23 − 3(2) − 4 = −2 and
Review 3π/4 f(3) = 33 − 3(3) − 4 = 14
π/2
Change of sign indicates presence of root.
π/4 y=1–x
–1 –0.5 O 0.5 1 x
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10 a y 13 a y
y = 2x 4 y=x+4 O y = cosec x
–4 x
1 y = sin x
O π 2π x
Two points of intersection, so two roots. 2 points of intersection for 0 , x , 2π and so
Review b Let f(x) = 2x − x − 4 = 0 then 2 roots.
f(2.7) = 22.7 − 2.7 − 4 = −0.2019 … and
f(2.8) = 22.8 − 2.8 − 4 = 0.1644 … Change of b Let f(x) = cosec x − sin x
sign indicates presence of root.
f 3π = cosec 3π − sin 3π = 0
2 2 2
3π = 4.71 correct to 3 significant figures.
11 a y 2
y = cot x
14 a f(x) = 20x3 + 8x2 − 7x − 3 and so
y = x2 f(0.5) = 20(0.5)3 + 8(0.5)2 − 7(0.5) − 3 = −2
and f(1) = 20 + 8 − 7 − 3 = 18. Change of sign
indicates presence of root.
O πx by 329
2
y = (5x – 3)(2x + 1)2
Review One point of intersection so one root. Ox
b Let f(x) = cot x − x2 then
–3
f(0.8) = cot 0.8 − 0.82 = 0.3312 … and
f(1) = cot1 − 12 = −0.3579 … Change of sign c Proof
indicates presence of root. 4 πr3
12 a y 15 a 800 = 3 + πr2(20)
2
y=x
b f(r) = πr3 + 30πr2 − 1200 = 0 and
Review y = tan 2x f(3) = −266.946 … and f(4) = 509.026 …
Change of sign indicates presence of root.
O π 2π x Or a suitable pair of graphs drawn.
3 points of intersection for 0 , x , 2π and so Exercise 6B
3 roots. 1 a 1.1338, 1.1085, 1.1276, 1.1133, 1.1240
b 1.12, f(1.115) = (1.115)3 + 5(1.115) − 7
b Let f(x) = x − tan 2x = 0 then = −0.0388 … , f(1.125) = (1.125)3 + 5(1.125) − 7
f(2.1) = 2.1 − tan 4.2 = 0.3222… and = 0.0488 … Change of sign indicates presence
f(2.2) = 2.2 − tan 4.4 = −0.8963… Change of of root.
sign indicates presence of root.
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2 a f(1) = ln(2) + 2 − 4 = −1.306 … 11 a y
f(2) = ln(3) + 6 − 4 = 3.0986 … Change of
sign indicates presence of root. ( )6 32– x 1
–
b 1.535 5
y=
4A
3
3 a Proof 2
b 0.7231, 0.6142, 0.6584, 0.6387, 0.6472, 0 1
, 0.6435, 0.6451, 0.6444, 0.6447 so 0.64
–4 –3 –2 –1–1O 1 2 3 4 5 6 7 x
4 a Proof b 0.5382 –2
–3
5 a Proof b Proof y = x –4
Review π → 0.234 48, 0.240 34, 0.24169, 0.242 00, 0 –5
15
c → ln(xn + 1)
ln(1.5)
00, 0.242 07 → 0.242 b e.g. xn =
or π → 0.24656, 0.24311, 0.242 33, 0.24215, 0
12 → c A(3.94, 3.94) → OA = 5.6
15, 0.24210 → 0.242 12 a 33 − 3r b Proof
or 0.2 → 0.232 24, 0.23982, 0.24157, 0 c Pro→of
.24157, 0.24197, 0.242 06, 0.242 09 → 0.242 d 8 → 8.058 69, 8.032 00, 8.04419, 8.038 63,
6 a 0.6325, 0.8345, 0.7416, 0.7885, 0.7658, 0 8.04117, 8.040 01, 8.040 54, 8.040 30, 8.040 41,
.7658, 0.7771, 0.7716, 0.7743, (0.7730, …) → 0.77 (8.040 36, 8.040 38), … → 8.040
330 b x3 + 5x2 + 2x − 5 = 0 e The radius of the cone that would give a
container of the required volume
7 a Proof b xn+1 = 4 1 + xn
Review c 1.2574, 1.2258, 1.2214, 1.2208, 1.2208… → 1.22 13 e.g xn+1 = xn3 + 1 → α = 0.39,
7
= sin−1 1 = 1
8 a xn+1 xn2 or xn+1 sin xn xn+1 = 3 7xn2 − 1 → β = 6.98
b x = sin−1 1 → 0.4605 …, which leads to 14 a Proof b Proof
x2
c 0.9082 … , 0.9015 … , 0.9069 … , 0.9026 … ,
inverse sine of a value . 1.
0.9061 … , 0.9032 … , 0.9055 … , 0.9037 … ,
c x = 1 → 1.00125, 1.089 70, 1.06210, 1 0.9052 … , 0.9040 … , 0.9049 … , → 0.90
sin x
d e.g. xn +1 = tan−1 1 → 0.9063 … ,
10, 1.070 04, 1.067 69, 1.068 38, (1.06818, 1.068 24, 1 sin xn
24, 1.068 22) → 1.068 0.9038 … , 0.9048 … , 0.9044 … ,
9 a 2.0794, 2.1192, 2.1390, 2.1489, 2.1538, 2.1563, 0.9045 … , 0.9045 … , → 0.90
2.1563, 2.1575, 2.1581 → 2.16 Exercise 6C
Review b x2 = ex − 4 1 a Proof
10 a Proof b e.g. xn+1 = 3 + xn with x1 = 1.5 → 1.574 681, 1
2 e2xn
b 1 → 1.5, 1.2603, 1.3713, 1.3186, 1.3434, 1.3317, 1
, 1.3372, 1.3346, 1.3358, 1.3352, … → 1.34 , 1.567 522, 1.568…184, 1.568122, 1
, 1.568128 → 1.5681
or 1.4 → 1.3053, 1.3497, 1.3287, 1.3386, 1 …
.3386, 1.3339, 1.3361, 1.3351, 1.3356, … → 1.34
or 1.2 → 1.4007, 1.3050, 1.3499, 1.3286, 1 …
.3286, 1.3386, 1.3339, 1.3361, 1.3351, 1.3356, … → 1.34
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2 a Proof 6 a Proof b Proof
b e.g. 1.5 → 1.432164, 1.407 497, 1.398 602, c e.g. 1.5 → 1.3083, 1.3689, 1.3495, 1.3557, 1
1.395 404, 1.394 256, 1.39384→4, 1.393696, .3557, 1.3537, 1.3543, (1.3541), … → 1.35
1.393643 → 1.394
d −1.35
c (8, −6) 7 a ln(3x + 5) + e6x a = 0.6
3 a Proof 3 3 0
b e.g. x1 = π → 1, 0.9093, 0.9695, 0.9330, 0.9567, 0 b 0.2 → 0.164 717 … , 0.165896 … , 0 …
4 … , 0.165857 … , (0.1→65858 …), → 0.166
.9567, 0.9419, 0.9514, 0.9454, 0.9493 → 0.95 8 a Proof
Review 4 a Proof b Proof b 1 → 1.06366, 1.072 02, 1.07311, 1.07324, 1 →
24, 1.07327 → 1.073 →
c 0.3 → 0.257 594, 0.273→768, 0.267 462, ( →
462, (0.269 900, 0.268 954, 0.269 321, 0
321, 0.269179) → 0.27 9 a Proof
d 0.007 30 correct to 3 significant figures
b f(θ ) = θ − cos−1 3π = 0 → f(0.8) = 0 − −1
32θ
End-of-chapter review exercise 6 −1 ) = 0.8 − cos−1 3π = −0.11882 … , f(1.2) =1 −
) = 1.2 − cos−1 32(0.…8)
1 a 1.5 → 1.5397, 1.5546, 1.5606, 1.5632, 1.5643, 1
, 1.5647, 1.5649, 1.5650, 1.5650, … → 1.57 … 3π = 0.1475 … Change of
32(1.2)
6 1 x 4+ 1
b x = 7 x + x3 → 7x = 6 x3 → = sign indicates presence of root.
c 1 → 0.99715, 0.996 22, 0.995 92, 0.99582, 0
331
→ 7x4 = 6x4 + 6 → x4 = 6 → α = 4 6 82, 0.995 79, 0.995 78, 0.995 78, … → 0.996
→
Review 2 a 1, 2 b Proof
c e.g. 1.5 → 2.1985, 1.7717, 2.0039, 1.8688, 1 10 a e.g. f(x) = ex−2 − sin x →= 0 → f(0.155) =
1.9445, (1.9011, 1.9256, 1.9117, 1.9196) → 1.9 e0.155−2 − sin 0.155 = 0.00364 …
f(0.165) = e0.165−2 − sin 0.165 = −0.00463 …
3 a Proof Change of sign indicates presence of a root.
b f(x) = e2x+1 − 14 + x3 = 0 → f(0.5) = e2 + b e.g. ex−2 = sin x (take logs to base e) →
…
3→ ) = e2(0.5)+1 − 14 + (0.5)3 = −6.4859 … → f(1) = 2
x − 2 = ln(sin x) → x = 2 + ln(sin x) → q
→ = e2(1)+1 − 14 + (1)3 = 7.08553 … change of sign q = 2 + ln(sin q) when x = q. q
…
indicates presence of root.
c Proof c e.g. 2 → 1.9049, 1.9431, 1.9290, 1.9344,
d e.g. xn+1 = ln(14 − xn3 ) − 1 with (1.9324, 1.9332, 1.9329, 1.9330, 1.9329,
2 1.9329) → 1.93
x1 = 0.75 → 0.804 230, 0.800 597, 0.800 858, 0 11 a Proof
Review 858, 0.800 839, 0.800 840, … → 0.8008
4 a 1 → 1.1825, 1.1692, 1.1662, 1.1657, 1 b e.g. 1.2 → 1.294 39 … , 1.234 93 … , 1.27370 … ,
, 1.1656, … → 1.17
… 1.248 93 … , 1.264 98 … , 1.254 67 … ,
b Proof
1.26133 … , 1.257 04 … , (1.25981… ,
1.258 02 … , 1.25917 … , 1.258 43 … ,
5 a Proof b 3 and 4 1.258 91… , …) → 1.26
c 3 → 3.098 61, 3.130 95, 3.14134, 3.144 65, 3 …
65, 3.145 70, 3.146 04, 3.14614, 3.14618, … → 3.146
12 a Proof
d x = 5y = 3.146, y = log5 3.146 = 0.71 b Proof
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c e.g. f(x) = sec x − π − x π + x 9 i dy = e−2x (sec2 x − 2 tan x) = e−2x (1 − tan x)2
2 4 dx
f(−0.215)= sec(−0.215) −ππ2 + 0.215 π − 0.215 ii e−2x . 0 and (1 − tan x)2 ù 0
4
iii x = 1 π
= 0.00495... 4
− π π 10 i Proof
2 4
f(−0.205)= sec(−0.205) + 0.205 − 0.205 ii a 2 2 b3
11 (−3a, −a)
= −0.00928...
Change of sign indicates presence of root. 12 i Proof ii Proof iii 1.54
d 1 → 0.4102, 0.6822, 0.6936, 0.6913, (0.6918, 13 i Proof
Review ii y = − 1 x + 1
0.6917, 0.6918, …) → 0.69 14 i − 11 3 22
2
13 a Proof ii − 5
b 0.2 → 0.30889, 0.314 70, 0.315 09, 15 i Proof 6
09, 0.315 11 → 0.315
→ ii Proof
→
iii Proof iv Proof
16 i Proof
14 a [x ln x − ]x a = 5 17 i a = 9 ii (2, 3) iii − 3
1 18 i Proof 8
b 5 → 5.592 01, 5.572 41, 5.572 39, 5 ii 8e − 14
, 5.572 39 → 5.572
ii (−2, − 1), (0, 1.44)
332 Cross-topic review exercise 2 19 i Proof ii 0.678
20 i a x − 3e−2x + c
1 Proof
b 3 sin 2x + 3x + c
Review2 π,π 42
63
ii 4.84
3 i Proof
ii (ln 3, −2) 21 i y
4 i Proof ii 2x − 5y + 8 = 0
5 a 0.11 Ox
b x + 4e−x − 2e−2x + c
ii Proof iii Proof iv 1.26
6i y 22 i y
Ox y = cosec x
Review ii Proof iii 4.84 O y = x(π – x)
7 i Proof ii Proof πx
8 i Proof ii Proof
iii a 0.66
ii 1.854
b 2.48
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23 i 5 Exercise 7B
2
b (−5.15, −7.97) 1 a x 4 3 + x 2 2 b x 5 4 − 3
ii a Proof + − − 2x
24 i Proof c 7 − 6
x −1 3x +
ii a 1.11, 2.03 1
b 13
2 d 1− 1
2(x − 3) 2(3x − 5)
25 i Proof e 2 + 3 − 4
ii a −0.572, 0.572 x x −1 2x + 1
Review b 3 π+ 1 f 2 3 − x 2 2 + x 1 3
32 4 2x + + −
26 i Proof ii Proof 2 − 4
+ + 2)2
27 i 0.362 or 20.7°, 1.147 or 65.7° 2 a x 2 (x
ii 33 b 3 + 4 − 2
2x + +1 + 1)2
28 i Proof ii Proof iii 2.728 1 x (x
29 i Proof
ii a = 2, b = −1 c − 2 + 3 − (x 1
x x −1 − 1)2
iii 4 − 3 2 d 5 + 4 − 1 1)2
2x − 2x + +
3 1 (2x
7 Further algebra e 3 2) − 3 2) + 4(x 3 2)2
16(x + 16(x − −
Prerequisite knowledge
f − 2 2) + 2 1) + 7 333
1 a A = 6, B = −3, C = −9 9(x + 9(x − 3(x − 1)2
b A = −1, B = − 5 , C = 4
Review 3 3 a 2− 3
x x2 + 1
2 a 1 + 14x + 84x2 3 2
b 243 − 810x + 1080x2 b 2x + 1 + x2 + 5
3 Quotient = x − 5, remainder = −11 c 1 5 − 2 1
3x + 2x2 +
d 7 5 − 3 5
2x2 + 3x −
Exercise 7A 4 a 2 + 3 − x 2 2
x −1 +
7 7
1 a 4 + 20 5 b 2 − 3 2 b 1+ 4(x − 2) − 4(x + 2)
2x − 3x +
1 7 2 3x −1
c 2x2 − x + 2 − 2(2x + 1) c −4 + x − 4 + x2 +1
d x + 2 − 4x + 7 3 d 2 + 7 + 1 1 − (x 2
x2 + 2x + 2x x− − 1)2
Review e 7x + 2 + 30x + 11 f x2 + 1− 6 5 A = 2, B = −3, C = 4, D = −1
x2 −5 x2 + 1
2 A = 1, B = 4, C = 12, D = 29 6 a (2x − 1)(x − 2)(x + 1)
3 A = 1, B = −1, C = 6, D = −6, E = 5 b 5 − x 3 2 + 1
2x − 1 − x +1
4 A = 2, B = 3, C = −1, D = 6 7 a (2x + 1)(x − 3)2
b 2 1 − x 1 3 + (x 3
2x + − − 3)2
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Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity
8 a a = −4, b = −3 8 a = 2, b = −5
b (2x + 1)(x + 3)(x − 1) 9 a a=−1
2
c x 6 3 − 32 + 10
+ 2x + 1 x −1 b 1 + 3 x + 3 x2 + 5 x3 + 15 x4
1 1 2 2 4 16
9 a x − x + 2
10 a = −9, a = 5
3 1 1 3n2 + 5
b 2 − (n + 1) − n+2 or 2(n + 1)(n + 2) 11 a a = 8, n = −3
c3 b −5120x3
2
Review10 Telescoping series, Sn = 1 1 − 1 + 1 , Exercise 7D
2 2 n+1 +
n 2 1 − 1 x + 3 x2 − 1 x3,
S∞ = 1 1 a 4 4 16 8 x ,2
4
1 + 2 x + 4 x2 + 8 x3, 5
b 5 25 125 625 x ,2
Exercise 7C 3 − 1 x − 1 x2 − 1 x3,
1 a 1 − 2x + 3x2 − 4x3, x , 1 c 6 216 3888 x ,9
b 1 − 3x + 9x2 − 27x3, x 1 d 2 + 1 x − 1 x2 + 5 x3, x 8
,3 4 32 768 ,3
c 1 + 8x + 40x2 + 160x3, x 1 e 4 + 4 x + 8 x2 + 40 x3, x ,3
,2 27 27 81 729
d 1− 3 x+ 3 x2 − 5 x3, x ,2 f − 1 − 16 x − 84 x2 − 64 x3, x 5
2 2 4 ,
125 625 3125 3125 2
334
1 1 1
e 1+ x − 2 x2 + 2 x3, x ,2 2 a 1 − 1 x2 + 3 x4, x , 2
4 4 16
Review f 1− x − x2 − 5 x3, x , 1
3 3
b 2 − 1 x2 − 1 x4, x, 8
g 2 + 16x + 96x2 + 512x3, x , 1 4 32 3
4
9 3 − 15 3 x2 + 15 3 x4,
h 1 − 9x + 42x2 − 152x3, x , 1 c 28 x, 3
2
i 1 + 2x + 5 x2 + 4x3, x , 1 3 2 − 25 x + 23 x2 + 67 x3
2 2
6 72 2592
2 a 1 − 3x2 + 6x4, x , 1 4 a 1 + 1 x + 1 x2, 1 − 6x + 27x2
24 8
b 1− 2 x2 − 4 x 4 , 1
3 9 x , 2 1 − 11 x + 97 x2, 1
,
1 − 6x2 + 6x4, x 1 b 24 8 x 3
,2
c
3 2 + 3x + 5x2 + 15 x3 + 75 x4 5 a = 2, b = 75 , 125 x3
24 16 8
4 a Yes, (3x − 1)−2 = (−1)−2 (1 − 3x)−2 = (1 − 3x)−2 6 a = −2
Review b No, 2x − 1 = −1 1 − 2x and −1 is not a 7 a 1− 2 + 4 − 8 b Proof
real number. x x2 x3
5 −1 − 6x − 24x2 − 80x3 c x − x2 + x3 − x4
2 4 8 16
6 Proof d 2 , 1 gives x , − 2 or x . 2, x , 1 gives
x 2
−2 , x , 2. The two ranges do not overlap.
7 k = 177
2
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Exercise 7E 12 2 − 9 − 23
5(2x − 5(x +
2 − 3 1) 2)
1− 1 + 2x
1 a x 13 k = 16
b −1 + 8x − 10x2 + 26x3 14 i 1 + 3 + 12
+ − − 3)2
2 1 3 x 1 x 3 (x
1 − 3x 1− x (1 − x)2
2 a + − ii 4 − 4 x + 4 x2
39 3
b x + 10x2 + 43x3
1 3x − 1
5 + 3x − 1 15 i − x − 2 + x2 + 3
1− x 1 + 2x2
3 a
Review b 4 + 8x + 7x2 − x3 ii 1 + 5 x + 17 x2
6 4 72
4 a 1+ 6 + 2 5 16 i − 1 + 2x + 1
1 + 2x − 3x 1+ x 1 + 2x2
b 237 ii 3x − 3x2 − 3x3
8
5 a x 3 4 − x 3 3 8 Further calculus
− +
b − 7 + 7 x − 91 x2 Prerequisite knowledge
4 48 576
1 a 3 cos 3x b 2xex2 +1
5 1 3 d 2 sec2 2x + 5 sin x
6 a x+ 1 + x − 2 − (x − 2)2 c 5
5x − 3
b 15 − 6x + 69 x2
4 16 2 a 1 e5x+1 + c b x − 1 sin 4x + c 335
5 28
End-of-chapter review exercise 7
c 5 ln(3x − 2) + c d 1 tan 3x + c
Review 1 1 + 8x + 40x2 + 160x3 3 3
2 1 − 2x − 4x2 − 40 x3 3 a 3 3) + 1 1) b 7x 2 − 3
3 2(x + 2(x − x2 + x
3 2 − 7x + 18x2 c 5 − 5 + (x 5 d 3 − x 1 1 + x 1 2
x x −1 − 1)2 + −
4 1 + 2x − 3 x2 Exercise 8A
2
1 a 2 b 5
5 1 − 3 x + 27 x2 − 135 x3 4x2 + 1 25x2 + 1
28 16 c 3 d 1
x2 + 9 x2 − 2x + 2
6 5 + 5 x + 15 x2
22 8 e 2x f 2
x4 + 1 5x2 + 2x + 1
7 1 + 5 x + 75 x2 x
2 16 256 2 a x2 + 1 + tan−1 x
Review 8 A = − 8 , B = −4, C = 16 b 2x − (4x2 + 1) tan−1 2x
33 x2 (4x2 + 1)
9 A = 5, B = 3, C = −2 c ex 1 + tan−1 x
x2 +
2 5x −3 1
x x2 +1
10 + 3 x − 4y = 2 − π
11 A = 3, B = −1, C = 2, D = −2 4 3+ π
5
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Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity
Exercise 8B e 20 f 5 (π + 2)
9 32
1 a 1 tan−1 x b 1 tan−1 x
3 3 4 4 g 1 (2π − 3 3 ) hπ
6 2
c 1 tan−1(2x) 1 tan−1 3x
2 d 12 4 5π
e 3 tan−1 2 3x f 6 tan−1 6 x 64
6 3 6 2 15
2 aπ bπ c 6π 7 π
12 4 9 3
3 2π2 8 6e − 6
Review e
Exercise 8C Exercise 8E
1 a 2 ln(x3 − 1) + c b ln(1 + sin x) + c 1 a 1 ln 2 x x + c
c 2 ln(x2 − 5x + 1) + c d ln(sin x) + c 2 −
e − 1 ln(2 − x2 ) + c f ln(1 + tan x) + c b 3 ln(x + 2) − ln(1 − x) + c
2
c 2 ln(2x2 − 9x − 5)
2 a 1 ln 3 b ln5
2 2 d ln(x − 3) + 2 tan−1 x + c
e 5 ln(x + 1) + ln(2 − x) + 2 + c
c 1 ln 5 d 1 ln 7
2 2 2 2 2−x
f 2x − 3 ln(x − 1) + ln(x + 4) + c
336 e 1 ln 2 f ln 3
2
2 a ln 100
Review3 Proof 27 b ln10
4 Proof c ln 9 3 d 7 (π − 4 ln 3)
16 10
5 p = e2 − 1 e π + ln 2 f 1 (π + ln 4)
4 4
Exercise 8D g 1 + ln 4 h 2 ln 9 − 3
6 42
1 x2 − 3 + c i ln 9 − 1
23
− 3x + 2 + − 1 (1 − 3 + b 4 − ln 12 c 2 − ln 2
6(x + 2)3 3 a 1 + ln 16 5
2 a c b 2x2 )2 c 5
6
4 Proof
c sin6 x + c d 2 (ex + 3 + c 5 Proof
6 6 Proof
3 2)2
e x + 1 − 1 ln(5x + 1) + c
5 5
Review Exercise 8F
5 3
f 2 2 1 a 3xex − 3ex + c
(3x − 1)2 + (3x − 1)2 + c
45 27
3π b x sin x + cos x + c
4
c 1 x2 ln 2x − 1 x2 + c
4 a2 b 18 ln 3 − 16 24
3 d 22
d 1 sin 2x − 1 x cos 2x + c
c 3 ln( 3 + 2 2 ) 4 2
2
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e 1 x4 ln x − 1 x4 + c 12 i −1 − 1 ii 4 ln 2 − 15
4 16 3e 16
e 3,
f 2 x ln x − 4 x + c ii p = 3.40
13 i x = 3 2
2 a 1 (π − 2) Cross-topic review exercise 3
18
1 1 + 15x + 135x2 + 945x3
b ln16 − 1 ln 2 − 3
24
c 1 (e2 + 1) d ln 27 − 2 2 3π
4 9
e 5− 2 f 8 3 4 − 6x + 6x2 − 5x3
9 9e3 25
Review 3 a 2(ln 2 − 1)2 b 1 (π2 − 8)
32
4 1 + 1 x + 3 x2
c π2 − 4 d 1 e2 2 8 64
4
5 π−2
e2 f 1 (1 + eπ ) 6 1 + 3 x + 3 x2 + 15 x3
2 2 8 16
4 a 1 (1 + 2e3 )
9 bπ 7 Proof
c1 d 2 − 26 8 i 1 + 2x + 6x2 ii 5
2 e4 9 i a= 2
5 1 π(e4 − 5) 2 ii 1 − 2x + 3 x2 337
4 10 i Proof 2
8 + 2 ln 1
11 Proof ii 2
Review End-of-chapter review exercise 8 12 a Proof b Proof
1 1 − 1 e−1 13 a Proof b 6.56
42
2 4(ln 4 − 1)
3 14 14 i Proof ii 2e2 − 10
9
15 i 2 3 x + 4 4x ii Proof
4 i Proof − + x2
ii Proof
16 i Proof ii 1.94
1π − 3
5 i Proof ii 3 2 17 i 1 + 3 − 1
− 2(1 + 2x) 2(1 + 2x)2
3 x
ii 15 ln5 − 4
6 i Proof ii 4 − 8 x + 1 x2
3 9 27
1 − 1
7 i x +1 x + 3 ii Proof 18 i 2 − 1 + 3
1− − − x)2
iii Proof x 2 x (2
Review 8 a 3x + 1 tan 2x + c ii 9 + 5 x + 39 x2
2 4 2 16
3 − 6x + 1
1π 1 ln 1 19 a x − 1 2x2 − 1
8 2 2
b 3 − b −2 + 3x − x2 + 9x3
9 i Proof ii Proof 20 a A = 1 , B = 4, C = 2, D = − 15
ii 24 − 8e 22
10 i 5 − 1
b Proof
11 Proof 1 − 1 ii 1 (2e3 + 1)
e 2e 9
21 i ,
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Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity
22 i Proof ii 11 Exercise 9B
96
1 a 1 (−3j + 7k) b λ=3
23 i 5 cos(θ − 0.6435) 58 b ON = 6i + 4j + 2k
ii a 1.80, 5.77 2 a d=2
b 2 tan(θ − 0.6435) + c
c 2 i+ 2 j+ 1k
9 Vectors 333
−6 2 −8
Prerequisite knowledge
3 ai AD = d−a = −6 − 2 = −8
1 36.7° (correct to 1 decimal place) 0 0 0
Review2 a y = 3 x−1 b y=− 2 x−6 = 128 = 8 2,
2 3
13 2 11
3 a 4 cm b 61
2 = b−a = − =
AB 5 2 3
4 0 4
4 (2, 12)
= 112 + 32 + 42 = 146,
Exercise 9A 5 13 −8
AB = 5 BC = 5 BC = c−b = −3 − 5 = −8
−3 2 4 4 0
1 a
b AC = 10 = 8 2,
−1
338 5 −6 11
= −7 DC = c−d = −3 − −6 = 3
3 4 0 4
2 a EF
Review b DF − DE = 3 − 10 = −7 = 112 + 32 + 42 = 146
5 2 3
ii Opposite sides are parallel and equal in
3 QR = PR − PQ = PR + QP = QP + PR QED length.
4 a XY = b − a and BC = 2b − 2a = 2(b − a) 1 2 1 11 7.5
BC is a scalar multiple of XY , therefore BC is 2 = 2 2 3.5
parallel to XY . b i OM = OA + AB 0 + 3 = 2
4
b k=1 so M (7.5, 3.5, 2)
2
ii OP = OB + 1 BD
12 3 3
5 a 2 b 4 13 −6 13
0 2 −6 5
= 5 + 1 − 4
4 3 0
6 a i q+s−p ii p − 2q − r − s
Review b For example, angle AHC = 45° (Interior angle of 13 1 −19 1 20
a regular octagon = 135° and angle GHC = 90°) 5 3 3 4
and the exterior angle (at A) is 45° and so the line = 4 + −11 = 8
segments AB and HC are parallel. −4
k = 1+ 2 so P 20 , 4, 8
3 3 3
7 Proof
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