Pure Maths

Question 1

Exercise
12A)
1 The diagram shows the curve with equation y = x2 - 2x.
a Copy and complete this table showing estimates for
the gradient of the curve.
b Write a hypothesis about the gradient of the curve at
the point where x = p.
c Test vour hypothesis by estimating the gradient of
the graph at the point (1.5, -0.75).

Answer 1

b) 2p-2

Question 2

F is the point with coordinates (3, 9) on the curve with equation y = x^2
a Find the gradients of the chords joining the point F to the points with coordinates:
i (4, 16)
i (3.5, 12.25)
iii (3.1, 9.61)
iv (3.01, 9.0601)
v (3 + h, (3 + h))
b What do you deduce about the gradient of the tangent at the point (3, 9)?

Answer 2

3)a)i) 7 (16-9/4-3)

Question 3

First principles? For x^2

Answer 3

Lim. f(x+h)-f(x). - (X+h)2 -x^2. - H@2+2xh
H-0 h. h. h.

Final answer is h+ 2x which equals to 2x

Question 4

Gradient formula?

Answer 4

Y2-Y1
X2-X1

Question 5

2y2-x3=0. What is dy/dx?

Answer 5

2y2=x3
Y^2=0.5x^3
Y=square root 0.5x^3
Dy/dx=1.5squareroot5x^2

Question 6

The line L is a tangent to the curve with equation
y = 4x2 + 1. L cuts the y-axis at (0, -8) and has a
positive gradient. Find the equation of L in the
form y = mx + c.

Answer 6

Y=4x^2+1
Dy/dx= 8x (gradient)
Tangent equation= y=(8x)x-8
4x^2+1=(8x)x-8
4x^2=9
X^2=9/4
X=3/2
Y=12x-8 (tangent equation)

Question 7

Least value in functions

Answer 7

Least value of y (involve differentiate to ensure equation is equal to 0 to find x and pluck that x value into the original equation)

Question 8

When sketching curve….

Answer 8

Differentiate to find turning points (make the differentiated equation equal to 0)
And differentiate again to find where the turning point is exactly (minimum, maximum, etc)
Consider x and y intercepts

Question 9

6 The function F is defined as F(x) = 12-2x + 2, for real numbers.
explain why f(x) > 0 for all values of x, and find the minimum value of f(x).

Answer 9

(X+a)^2 +b is greater than 0
(X+a)^2 is always greater
+b just makes it more positive

Question 10

a Prove that, if the values of a and c are r
of b so that f(x) = ax? + bx + c has distinct real
b Is it always possible to choose a value of b so that f(x) has equal roots? Explain your answer.

Answer 10

a F, for distinct real roots b2 −4ac>0
b2 >4ac for distinct real roots
If a > 0 and c > 0, or a < 0 and c < 0, choose b
such that b > square root 4ac
If a > 0 and c < 0, or a < 0 and c > 0, 4ac < 0, therefore 4ac < b2 for all b
b For equal roots, b2 − 4ac = 0 b2 = 4ac
If 4ac < 0, then there is no value for b to satisfy b2 = 4ac as b2 is always positive

Question 11

The diagram shows a section of a suspension bridge carryin a road over water. 0.00012x^2 +2200. Given that the towers at each end are 346 m tall, use your answer to part b to calculate the
length of the bridge to the nearest metre.

Answer 11

0.000 12x + 200 = 346
0.000 12x2 = 146 x2 = 146
0.000 12
146 x= ± 0.00012
So x = 1103 and x = −1103
c length = 1103 × 2 = 2206 m

Question 12

A fertiliser company uses a model to determine how the amount of fertiliser used, f kilograms
per hectare, affects the grain yield g, measured in tonnes per hectare.
g = 6 + 0.03f - 0.000 06f2
One farmer currently uses 20 kilograms of fertiliser per hectare. How much more fertiliser
would he need to use to increase his grain vield b 1 tonne ver hectare?

Answer 12

When f = 20, g = 6 + 0.03(20) − 0.00006(20)
= 6.576
For an extra tonne yield, g = 6.576 + 1
= 7.576 6 + 0.03f − 0.000 06f 2 = 7.576
1.576 − 0.03f + 0.000 06f 2 = 0
Using the formula, where a = 0.000 06, b = −0.03 and c = 1.576,
x=
−(−0.03) ± (−0.03)2 − 4(0.000 06)(1.576) 2(0.000 06)(20)
Using the formula, where a = −0.01, b = 0.975 and c = −16.5,
x = −0.975 ± 0.9752 − 4(−0.01)(−16.5) 2(−0.01)
x = 440.4 and x = 59.6 (to 1 d.p.) 59.6 − 20 = 39.6
39.6 kilograms per hectare

Question 13

The ratio of the lengths a: b in this line is the same as the ratio
of the lengths b:c. Find ratio

Answer 13

A(b+c) =. B
B. C
Cross mutliply that above
Use quadratic formula

Question 14

X= square root 1+….. find x

Answer 14

X= square root 1+x
X2= (1+x)
Find value using quadratic formula

Question 15

Find the shortest distance from line y = 2x + 4 is
parallel to the straight line I with equation 6x - 3y - 9 = 0.

Answer 15

Find line perpendicular to 2x+4
And find intersection between that line and both lines from question
And find length between the intersections

Question 16

What is a scalene triangle?

Answer 16

Triangle with different lengths

Question 17

2 The circle with centre Chas equation (x - 2)2 + (y - 1)2 = 10.
The tangents to the circle at points P and O meet at the point R with coordinates (6, -1).
a Show that CPRO is a square.

Answer 17

Find PR (not radius or length CR) by pythagoras
Identify equal lengths
And angles

Question 18

Area of kite

Answer 18

Times two lengths that are around perpendicular to one another

Question 19

Factorise ax3+bx+cx+d

Answer 19

Trial and error possibly (until we get the equation to equal to 0)

Question 20

Prove a set of coordinates lie on the circle

Answer 20

Create perpendicular bisectors
Find intersection
Find distance from centre to coordinates should be the same

Question 21

To find bearings

Answer 21

If obtuse - add 90 to the angle
Angle it makes with x axis - if it lies on negative x axis we subtract it from 180

Question 22

ABCD is a square. Angle CED is obtuse (40 degrees).
Find the area of the shaded triangle. CE IS 8

Answer 22

Use the sine rule to work out angle CED. sinE sinD
2ab cosD=5.9 +2.1 −4.2
2 2(5.9)(2.1)
e = d
2
2
sin E 10
sin 50° = 8
10 sin 50° 8
sin E =
E = 73.246 86° or 106.753 14°
2
The angle is obtuse so
Angle CED = 106.753 14°
Angle ECD = 180° − 50° − 106.753 14°
= 23.25°
Use trigonometry to work out the height of
triangle CDE.
sin 23.25° = height
The height of triangle ABE = 10 − 3.1575 = 6.84 cm
Area of triangle = 12 × 10 × 6.84 = 34.2
∴ Area of the shaded triangle is 34.2 cm2.

Question 23

sin x cos y + 3 cos x sin y = 2 sin x sin y - 4 cos x cos y,

Answer 23

Make that over cosxcosy and cancel some or divide to create tan

Question 24

28 A cylindrical biscuit tin has a close-fitting lid which overlaps the tin by 1 cm, as shown. The radii of the tin and the lid are both xem.
The tin and the lid are made from a thin sheet of metal of area 80t cm? and there is no wastage. The volume of the tin is Vem?.
a Show that V = (40* - 12 - *).
(5 marks)
Given that x can vary:
b use differentiation to find the positive value of x for
which V is stationary.
c Prove that this value of r gives a maximum value of V. (2 marks)
d Find this maximum value of V.
(3 marks)
(1 mark)
e Determine the percentage of the sheet metal used in the
lid when V is a maximum.

Answer 24

Let the height of the tin be h cm.
The area of the curved surface of the
tin = 2πxh cm2
The area of the base of the tin = πx2 cm2 The area of the curved surface of the
lid = 2πx cm2
The area of the top of the lid = πx2 cm2 Total area of sheet metal is 80π cm2. So 2πx2 + 2πx + 2πxh = 80π
h = 40 − x − x2
x
The volume, V, of the tin is given by
V = πx2h
= πx2 (40 − x − x2 )
x
= π(40x − x2 − x3)
b dV =π(40−2x−3x2) dx dV
Putting dx =0 40−2x−3x2 =0
(10−3x)(4+x)=0 Sox= 10 orx=− 4
3 Butxispositive,sox= 3
10
d2V
c dx2 = π(−2 − 6x)
Whenx=10 , dV =π(−2−20)<0 LetthelengthAB=CD=ymetres 3 dx2
So V is a maximum.

Question 25

The diagram shows an open tank for storing water,
ABCDEF. The sides ABFE and CDEF are rectangles.
The triangular ends ADE and BC are isosceles, and LAED = <BFC = 90°. The ends ADE and BCF are vertical and EF is horizontal.
Given that AD = * metres:
a show that the area of triangle ADE is 4 x2 m?
E
(3 marks)
Given also that the capacity of the container is 4000 m? and that the total area of the two triangular and two rectangular sides of the container is S m2.
b show that S =
+ 72
16000v2
It
(4 marks)
Given that x can vary:
c use calculus to find the minimum value of S.
d justify that the value of S you have found is a minimum.

Answer 25

29 a
Let the equal sides of ∆ADE be a metres.
Using Pythagoras’ theorem, a2 +a2 =x2
2a2 =x2 a2 = x22
Areaof∆ADE= 1 ×base×height 2
= 1×a×a 2
= x2 m2
Areaoftwotriangularends and rectangular ends

Areaoftworectangularsides =2×ay=2ay
SoS=x22 +2x22y=x22+xy2
But capacity of storage tank = 1/4 x2 × y
So 1/4 x2y = 4000
y= 16000 x2
Substituting for y in equation for S gives: S= x2/2 +16000 root2/x (2 =root2)
2x
c dS=x−160002
dx x2 Putting dSx = 0
x=16000 2 x2
x3 = 16 000 2
x = 20 2 = 28.28 (4 s.f.)
Whenx=20 2,
S = 400 + 800 = 1200
d d2S =1+ 32000 2 dx2 x3
Whenx=20 2, d2S =3>0,sovalueisaminimum.

Question 26

2^x+2+5

Answer 26

Treat x-2 as translation by +2 in x direction
And +5 as translation by +5 in y direction

Question 27

AE^bx+C

Answer 27

C is the asymptote
A is how much to multiply the 1 by
A+C is equal to y intercept

Question 28

E^3x

Answer 28

3E^3x

Question 29

Logarithm rule

Answer 29

Loga(BC) is logaB + logaC
Loga(A)=1
Ine^3x=3x
Loga(b/c) is logab - logac
Loga(b^2) is 2 logab

Question 30

Given that a = 2i + 5j and b = 3i - j, find:
a \if a + Ab is parallel to the vector i

Answer 30

The values of j should equal to 0

Question 31

The point B lies on the line with equation 2y = 12 - 3x. Given that |OB| = V13,
find possible expressions for OB in the form pi + ai.

Answer 31

X^2+y^2=13
Simultaneous equations for OB

Question 32

Triangle has OA and m as midpoint. Find vector expressivion for OB

Answer 32

OA +AB
OM+MB

Question 33

Prove circle lies in circle

Answer 33

Consider centre for each and the radius

Question 34

Prove that if ab is irrational then one of a and b is irrational

Answer 34

A=d/e
B=c/f
Ab= dc/ef
If ab is irrational at least one of the factors has to be irrational

Question 35

Prove a surd is irrational

Answer 35

N=a@2/b@2
Nb@2 is a@2
Multiple of n
A=nc
A@2=nc@2
Multiple of n
Irrational - both a and b have common factors

Question 36

Prove there is no least positive number

Answer 36

N=a/b
M/a/2b
M is less than N
No least positive rational

Question 37

To find number of jumps

Answer 37

Difference between last and first number
Over the
Difference between 2 and 1 integer

Question 38

When suggesting a recurrance relationship

Answer 38

Try multiply by the term and add to get to the next term
2n+1 may be written as un+2

Question 39

To prove a vector passes through an origin

Answer 39

Find the points it is in at A and B
Check to see if these points are parallel
If yes it passes through origin

Question 40

For prove of odd numbers

Answer 40

Use 2k+1 as odd numbers not n+1

Question 41

Find the range in e^2x-3 = f(x)

Answer 41

F(x) is greater than -3

Question 42

Cot 90

Answer 42

Equal to 1/tan90
Equal to 1/1/0
Equal to 0

Question 43

When drawing inverse of sinx cosx and/or tanx

Answer 43

Rotate sinx/cosx/tanx by x=0 and then rotate anticlockwise by 90 degrees

Question 44

2sinAcosB (from identities of sin(A+B) and sin(A-B)

Answer 44

Sin(A+B) + sin(A-B)

Question 45

2cosAsinB (from identities of sin(A+B) and sin (A-B))

Answer 45

Sin(A+B)-sin(A-B)

Question 46

2cosAcosB (from identities of cos(A+B) and cos(A-B))

Answer 46

cos(A+B)+cos(A-B)

Question 47

2sinAsinB (from identities of cos(A+B) and cos(A-B))

Answer 47

Cos(A-B)-cos(A+B)

Question 48

Sin P + sin Q

Answer 48

2sinP+Q/2xcosP-Q/2

Question 49

Sin P - sin Q

Answer 49

2sinP-Q/2xcosP+Q/2

Question 50

CosP+cosQ

Answer 50

ii 2cosAcosB≡cos(A+B)+cos(A−B)⇒ cosP+cosQ≡2cos P+Q /2cos P−Q/2

Question 51

CosP-cosQ

Answer 51

2sinAsinB≡cos(A−B)−cos(A+B) ⇒ cosQ−cosP≡2sin P+Q sin P−Q 22
⇒ cosP−cosQ≡−2sin P+Q /2sin P−Q/2