Pure Maths Flashcards
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).
b) 2p-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)?
3)a)i) 7 (16-9/4-3)
First principles? For x^2
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
Gradient formula?
Y2-Y1
X2-X1
2y2-x3=0. What is dy/dx?
2y2=x3
Y^2=0.5x^3
Y=square root 0.5x^3
Dy/dx=1.5squareroot5x^2
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.
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)
Least value in functions
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)
When sketching curve….
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
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).
(X+a)^2 +b is greater than 0
(X+a)^2 is always greater
+b just makes it more positive
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 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
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.
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
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?
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
The ratio of the lengths a: b in this line is the same as the ratio
of the lengths b:c. Find ratio
A(b+c) =. B
B. C
Cross mutliply that above
Use quadratic formula
X= square root 1+….. find x
X= square root 1+x
X2= (1+x)
Find value using quadratic formula
Find the shortest distance from line y = 2x + 4 is
parallel to the straight line I with equation 6x - 3y - 9 = 0.
Find line perpendicular to 2x+4
And find intersection between that line and both lines from question
And find length between the intersections
What is a scalene triangle?
Triangle with different lengths
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.
Find PR (not radius or length CR) by pythagoras
Identify equal lengths
And angles
Area of kite
Times two lengths that are around perpendicular to one another
Factorise ax3+bx+cx+d
Trial and error possibly (until we get the equation to equal to 0)
Prove a set of coordinates lie on the circle
Create perpendicular bisectors
Find intersection
Find distance from centre to coordinates should be the same
To find bearings
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
ABCD is a square. Angle CED is obtuse (40 degrees).
Find the area of the shaded triangle. CE IS 8
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.
sin x cos y + 3 cos x sin y = 2 sin x sin y - 4 cos x cos y,
Make that over cosxcosy and cancel some or divide to create tan
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.
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.
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.
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.
2^x+2+5
Treat x-2 as translation by +2 in x direction
And +5 as translation by +5 in y direction
AE^bx+C
C is the asymptote
A is how much to multiply the 1 by
A+C is equal to y intercept
E^3x
3E^3x
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
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
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
Triangle has OA and m as midpoint. Find vector expressivion for OB
OA +AB
OM+MB
Prove circle lies in circle
Consider centre for each and the radius
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
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
Prove there is no least positive number
N=a/b
M/a/2b
M is less than N
No least positive rational
To find number of jumps
Difference between last and first number
Over the
Difference between 2 and 1 integer
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
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
For prove of odd numbers
Use 2k+1 as odd numbers not n+1
Find the range in e^2x-3 = f(x)
F(x) is greater than -3
Cot 90
Equal to 1/tan90
Equal to 1/1/0
Equal to 0
When drawing inverse of sinx cosx and/or tanx
Rotate sinx/cosx/tanx by x=0 and then rotate anticlockwise by 90 degrees
2sinAcosB (from identities of sin(A+B) and sin(A-B)
Sin(A+B) + sin(A-B)
2cosAsinB (from identities of sin(A+B) and sin (A-B))
Sin(A+B)-sin(A-B)
2cosAcosB (from identities of cos(A+B) and cos(A-B))
cos(A+B)+cos(A-B)
2sinAsinB (from identities of cos(A+B) and cos(A-B))
Cos(A-B)-cos(A+B)
Sin P + sin Q
2sinP+Q/2xcosP-Q/2
Sin P - sin Q
2sinP-Q/2xcosP+Q/2
CosP+cosQ
ii 2cosAcosB≡cos(A+B)+cos(A−B)⇒ cosP+cosQ≡2cos P+Q /2cos P−Q/2
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
