Home
Subjects
Textbook solutions
Create
Study sets, textbooks, questions
Log in
Sign up
Upgrade to remove ads
Only $35.99/year
mock January maths pure p2
STUDY
Flashcards
Learn
Write
Spell
Test
PLAY
Match
Gravity
Terms in this set (11)
a set of 500 toy cubes are all given a number from 1 to 500. they are numbered in ascending order of the side length such that cube number 1 is the smallest and cube number 500 is the largest
given that:
- the side length of each cube forms an arithmetic progression
- the side length of cube 1 is 16mm
- the side length of cube 21 is 24
a) find the difference between the heights of any two consecutively numbered cubes in the set
John forms a tower by stacking all 500 blocks
b) find the height of the tower
a) Un = a + (n-1)d
U21 = 16 + (21 -1) d = 24
16 + 20d =24
20 d = 8
d = 0.4
b) sn = 1/2 [2a + (n-1)d]
S500 = 1/2 [2(16) + (500-1) 0.4]
= 57900
a (x) = 7 -2x^2
b (x) = 3x/(5x-1)
a) write down the range of a(x)
b) find the value of ba(1.8)
c) find the inverse function of b(x)
a) a(x) <= 7
b) a(1.8) = 7 - 2(1.8)^2
= 0.52
b(0.52) = 3(0.52)/5(0.52)-1 = 0.975
c) x = 3y/5y-1
x(5y-1)=3y
5xy -x = 3y
5xy -3y =x
y(5x-3) = x
y = x/(5x-3)
solve log3(12x+5) -log3(1-3x) = 2
log3(12x+5/1-3x)=2
3^2 = 12x+5/1-3x
9(1-3x) = 12x+5
9 - 27x = 12x + 5
4 = 39x
x=4/39
given that x is small and measured in radians, show that
4sinx/2 + 3 cos^2x = ax^2 + bx +c
finding the values of the integers a,b,c
sin x/2 = x/2
cos^2x = 1 -sin^2x
= 1 - x^2
4(x/2)+3(1-x^2)
= 2x + 3 -3x^2
a=-3
b=2
c=3
a curve has equation
f(t) = 5t^4 - 24t^3 + 42t^2 - 32t + 11
a) find f'(t) and f"(t)
b) i) verify that curve has stationary point at t=1
f'(t) = 20 t^3 - 72t^2 +84t -32
f"(t) = 60t^2 + 144t + 84
i) f'(1) = 20 (1)^3 - 72(1)^2 +84(1) -32 = 0 therefore stationary point
here is a shape made from UQT, a sector of a circle with centre Q and radius 2x, and two congruent sectors QRS and PQV with centres Q and radii x where PQR is a straight line. the angle UQT is o radians
a) write down the size of angle VPQ in terms of alpha
b) show that the are is 1/2x^2(3alpha + pi)
c) find the perimeter
a) o + alpha + o = pi
2o+ alpha = pi
o = pi-alpha/2
pi-alpha/2 < VQP
b) area of sector 1 = 1/2 (2x^2) alpha
area of sector 2 = 1/2x^2 (pi-alpha/2)
total area = 2x^2+ x^2 (pi-alpha /2)
= x^2 (2alpha + pi-alpha/2)
= x^2 (2alpha + 1/2 pi - 1/2 alpha)
= x^2 (3/2alpha +1/2 pi)
= 1/2 x^2 (3alpha + pi)
c) 2x+ 2x(pi-alpha/2) + 2x + 2 xalpha
= 2 x ( 2 + 1/2 alpha + pi/2)
h(x) = x^3 - 10x^2 +27x -23
the point a (5, -13) lies on the curve
the line l is a tangent to the curve at the point a
a) find the equation for the line l in the form y =mx +c
b) verify that the curve meets line L again on the x-axis
a) h'(x) = 3x^2 -20x + 27
h'(5) = 3(5)^2 - 20(5) + 27 =2
y = 2x + c
-13 = 2(5) + c
c= -23
y = 2x - 23
b) when x=0
y = 2(0) -23
(0,-23)
h'(0) = 3(0)^2 - 20(0) + 27 = -23
(0,23)
ax^3 +bxy + 3y^2 = 26
a) find an expression for dy/dx
given that the point Q (-1, -4) lies on the curve and the equation of the normal to the curve at point Q s 19X + 26Y + 123 = 0
b) find the values of a and b
a) 3ax^2 + b(xdy/dx + y) + 6y dy/dx = 0
3ax^2 + bx dydx + 6y dy/dx + by = 0
dy/dx (6y + bx) = -3ax^2 - by
dy/dx = -3ax^2 - by / bx + 6y
b) a9-1)^3 + b(-1)(-4) + 3 (-4)^2 = 26
-a + 4b + 48 =26
-a + 4b = -22
19X + 26Y +123 = 0
26Y = -19X - 123
Y = -19/26 X - 123/26
V=kx^n
a) show that this can be written as
log10V = nlog10x + log10 a
graph with points (-0.7,0),(0.21,0.45)
b) find the complete equation for the model
a) log 10 V = log 10 k (x^n)
log 10 V = log 10 k + log 10 (x^n)
log 10 V = log 10 k + n log 10 x
b) log 10 V - 0 = (0.45-0/0.21+7) ( log 10 x + 0.7)
log 10 V = 45/91( log 10 x + 0.7)
log 10 V = 45/91 log 10 x + 9/26
n= 45/91 = 0.495
log 10 K = 9/26
k = 10 ^ 9/26 = 2.22
V = 2.22 x ^ 0.495
y = |2x - 3a|
a) sketch the graph with the equation y = g(x) where
g(x) = a - |2x - 3a|
b) find in terms of a the set of values for which
a - |2x - 3a| > x -a
c) find in terms of a the coordinates of the minimum point of
y = 3 -5g(1/2x)
a) when 2x = 3a
y = |3a -3a| = 0
x = 3a/2
when x = 0
y = |-3a|
y = 3a
a - |2x - 3a| = 0
2x -3a = a
2x = 4a
x = 2a
a - |2x - 3a| = 0
-(2x-3a) = a
-2x + 3a = a
-2x = -2a
x=a
coordinates = (0, -2a), (3a/2, a), (2a, 0)
b) a> x -a + | 2x -3a|
2a - x > |2X - 3a|
(2a - x)^2 > (2X - 3a)^2
4a^2 - 4ax + x^2 > 4x^2 - 12ax + 9a^2
0> 3x^2 - 8ax + 5a^2
3x^2 - 3ax - 5ax + 5a^2
3x(x-a) - 5a (x-a)
(3x -5a)(x-a)
x = 5a/3 or a
c)1. (3a/2, -5a)
2. (3a, -5a)
3. (3a, 3-5a)
a) show that 2cosx -sin x can be written in the form
Rcos (x+a) R>0 0<a<pi/2
b) y = 3 + 4cos (0.5s) - 2sin (0.5s)
using the model find the maximum height of p above the water level and the value of s when this max height first occurs
a) Rcosxcosa - Rsinxsina
Rcos a =2
-Rsin a = -1
tan a = 1/2
a = 0.464
r^2 (cos ^2 a + sin^2 a) = 5
r = root 5
2cosx - sinx = root 5 cos (x + 0.464)
b) first double the first equation
4cos x - 2sinx = root 5 cos (x + 0.464)
let 0.5s = x
4cos(0.5s) - 2sin(0.5s) = 2 root 5 (0.5s + 0.5s + 0.464)
3 + 4 cos (0.5s) - 2sin (0.5s) = 3 + 2 root 5 (cos0.5s + 0.464)
max value of y = 3 + 2 root 5
= 7.47 m
3 + 2 root 5 =3 + 2 root 5 cos (0.5s + 0.464)
1 = cos (0.5s + 0.464)
let o = 0.5s + 0.464
cos o = 1
o = -2pi, 0, 2pi
s = 2 (2pi - 0.464)
= 11.6
c) 3 + 4cos (0.5s) - 2sin (0.5s) = 0
3 + 2 root 5 cos (0.5s + 0.464) = 0
cos ( 0.5s + 0.464) = -3/2 root 5
0.5s + 0.464 = 2.306
= 3.68
0.5s + 0.464 = 3.97
= 7.02
7.02 - 3.68 = 3.34
Sets with similar terms
Math Placement Exam Practice Problems Georgia Stat…
73 terms
Algebra 1 Review problems
14 terms
Midpoints #1
24 terms
Abeka 8th Grade Pre-Algebra Math Test 12 (final ex…
50 terms
Other sets by this creator
synthesis all organic reactions
37 terms
mock January maths pure p1
4 terms
blood clotting
7 terms
maths stats
59 terms
Other Quizlet sets
Chapter 2: Exchange Rate Regimes
11 terms
AP UNIT 9: Suliraning Teritoryal at Hangganan
19 terms
AP Psych Unit 1 test
40 terms
FL RE - overview
49 terms