Pure Maths Flashcards

1
Q

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).

A

b) 2p-2

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2
Q

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)?

A

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

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3
Q

First principles? For x^2

A

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

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4
Q

Gradient formula?

A

Y2-Y1
X2-X1

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5
Q

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

A

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

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6
Q

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.

A

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)

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7
Q

Least value in functions

A

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)

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8
Q

When sketching curve….

A

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

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9
Q

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).

A

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

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10
Q

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.

A

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

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11
Q

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.

A

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

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12
Q

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?

A

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

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13
Q

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

A

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

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14
Q

X= square root 1+….. find x

A

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

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15
Q

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

A

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

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16
Q

What is a scalene triangle?

A

Triangle with different lengths

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17
Q

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.

A

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

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18
Q

Area of kite

A

Times two lengths that are around perpendicular to one another

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19
Q

Factorise ax3+bx+cx+d

A

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

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20
Q

Prove a set of coordinates lie on the circle

A

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

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21
Q

To find bearings

A

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

22
Q

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

A

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.

23
Q

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

A

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

24
Q

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.

A

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.

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 =
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.
26
2^x+2+5
Treat x-2 as translation by +2 in x direction And +5 as translation by +5 in y direction
27
AE^bx+C
C is the asymptote A is how much to multiply the 1 by A+C is equal to y intercept
28
E^3x
3E^3x
29
Logarithm rule
Loga(BC) is logaB + logaC Loga(A)=1 Ine^3x=3x Loga(b/c) is logab - logac Loga(b^2) is 2 logab
30
Given that a = 2i + 5j and b = 3i - j, find: a \if a + Ab is parallel to the vector i
The values of j should equal to 0
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.
X^2+y^2=13 Simultaneous equations for OB
32
Triangle has OA and m as midpoint. Find vector expressivion for OB
OA +AB OM+MB
33
Prove circle lies in circle
Consider centre for each and the radius
34
Prove that if ab is irrational then one of a and b is irrational
A=d/e B=c/f Ab= dc/ef If ab is irrational at least one of the factors has to be irrational
35
Prove a surd is irrational
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
36
Prove there is no least positive number
N=a/b M/a/2b M is less than N No least positive rational
37
To find number of jumps
Difference between last and first number Over the Difference between 2 and 1 integer
38
When suggesting a recurrance relationship
Try multiply by the term and add to get to the next term 2n+1 may be written as un+2
39
To prove a vector passes through an origin
Find the points it is in at A and B Check to see if these points are parallel If yes it passes through origin
40
For prove of odd numbers
Use 2k+1 as odd numbers not n+1
41
Find the range in e^2x-3 = f(x)
F(x) is greater than -3
42
Cot 90
Equal to 1/tan90 Equal to 1/1/0 Equal to 0
43
When drawing inverse of sinx cosx and/or tanx
Rotate sinx/cosx/tanx by x=0 and then rotate anticlockwise by 90 degrees
44
2sinAcosB (from identities of sin(A+B) and sin(A-B)
Sin(A+B) + sin(A-B)
45
2cosAsinB (from identities of sin(A+B) and sin (A-B))
Sin(A+B)-sin(A-B)
46
2cosAcosB (from identities of cos(A+B) and cos(A-B))
cos(A+B)+cos(A-B)
47
2sinAsinB (from identities of cos(A+B) and cos(A-B))
Cos(A-B)-cos(A+B)
48
Sin P + sin Q
2sinP+Q/2xcosP-Q/2
49
Sin P - sin Q
2sinP-Q/2xcosP+Q/2
50
CosP+cosQ
ii 2cosAcosB≡cos(A+B)+cos(A−B)⇒ cosP+cosQ≡2cos P+Q /2cos P−Q/2
51
CosP-cosQ
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
52