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Unit 1: Contemporary Issues in Health/Social > A-Level Math Exam Q Review > Flashcards

A-Level Math Exam Q Review Flashcards

1
Q

5x^2 + px - 8 = Q(x-1)^2 + r

A 
q = 5 
p = -10
r = -13
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2
Q

A rectangle R has length (1 + root 5 ) cm and area of root 80 cm^2
Calculate the width of R in the form p + q(root 5 )

A 
A = L x W 
W = A/L 
W = 5 - root 5
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3
Q

A:
2x + y = 1
x^2 - 4ky + 5k = 0
show that x^2 + 8kx + k = 0

B:
given that x^2 = 8kx + k = 0 has equal roots, find the solution of the simultaneous equation, k is a non zero constant

A 
k = 1/16
X = -1/4
y = 3/2
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4
Q

k^2 - 2k - 24 < 0, find the set of possible values of k

A

Critical values of k = 6, - 4

therefore -4 < k < 6

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5
Q

Curve C has the equation y= x^2 - 4x + 4 and line l is y = mx
a) in case m = 1, the curve and line intersect at point A & B. Find the midpoint of AB

b) find the non-zero value of m for which the line is a tangent to the curve and find the coordinates of the point where the tangent touches the curve

A

a) midpoint = (5/2, 5/2)
b) tangent = (-2, 16)
m = -8

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6
Q

A: (1, -6) , B: (7,2)

a) find the equation of the perpendicular bisector AB
b) a point c on the perpendicular bisector has coordinates (p, q) . the distance OC is 2 unit, write down 2 equations involving p & q and hence find the coordinates of the possible positions of c

A

a) y = 2x - 2
b) 1 : q = 2p - 2
2 : p^2 + q^2 = 4
(0, -2), (8/5, 6/5)

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7
Q

line l has the gradient -2 and passes via A: (3, 5). B is a point on the line such that the distant AB is 6 root 5. find the coordinates of the possible point B

A

B : (9, -7) , (-3, 17)

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8
Q

line l has the equation y = -1/2x + 6
C lies on l and has x coordinate equal to p , A : (2,5)
length AC = 5, show that p satisfies
p^2 - 4p - 16 = 0

A 
C : (p, -1/2p + 6) , A: (2, 5)
d = root (x2 - x1)^2 + (y2 - y1)^2 
5 = root (p - 2)^2 + (-1/2 p + 1)^2
25 = (p^2 - 4p + 4) + (1/4p^2 - p + 1)
100 = 4p^2 - 16p + 16 + p^2 - 4p + 4
5p^2 - 20p - 80 = 0
p^2 - 4p - 16 = 0
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9
Q

Transformation of Graph ( translation rule)

A 
Y = f( x - a) translates y = f(x) a units to the right 
Y = f(x + a) translates y = f(x) a units to the left 
Y = f(x) - a translates y = f(x) a units down
Y = f(x) + a translates y = f(x) a units up
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10
Q

Graph Transformation ( Reflection rule)

A

Y = - f(x) reflects y = f(x) in the x - axis

Y = f(-x) reflects y = f(x) in the y - axis

any variable point on the axis’s stays the same

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11
Q

Sketches of Graph Rules

A

Y = Af(x) stretches y = f(x) by a scale factor of A parallel to the y - axis

Y = f(Ax) stretches y = f(x) by a scale factor of 1/A parallel to the x - axis

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12
Q

Curve C has the equation y = f(x), x > 0 , f’(x) = 4x - 6root x + 8x^- 2
given that point p (4,1) lies on c
a) find f(x) and simplify your answer
b) find an equation normal to c at point p

A

a) 2x^2 - 4x^3/2 - 8x^- 1 + 3

b) 2x + 9y - 17 = 0

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13
Q

what does a positive skew in Blox plot look like?

A

Q3 - Q2 > Q2 - Q1

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14
Q

What does a negatively skewed distribution look like?

A

Q2 - Q1 > Q3 - Q2

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15
Q

what formula is used for lower & upper outliers?

A

Q1 - 1.5(IQR)

Q3 + 1.5(IQR)

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16
Q

19 employees take an aptitude test, the scores are 7, 18 (2), 24 (2), 26, 28, 32, 33 (3), 34, 35, 39, 40 (5)

a) illustrate the score in a stem and leaf diagram
b) find the median and IQR

An outlier is an observation whose value is less than Q1 - 1 x IQR

c) what are outliers
d) draw a box plot to illustrate employees scores

A

b) Median = 33
IQR = 16
C) 7 is the only outlier
D) L = 18 , H=40 , Q1 = 24, Q2 = 33, Q3 = 40, Outlier = 7

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17
Q

Data show that Penvile have Q1 = 31, Q2 = 39, Q3 = 55 whereas Greenslax have Q1 = 44, Q2 = 64, Q3 = 76
a) state the skewness of each distribution and justify your answer

A

Greenslax :
Q2 - Q1 = 20
Q3 - Q2 = 12
… SINCE Q2 - 21 > Q3 -Q2, its a negative skew

Penvile:
Q2 - Q1 = 8
Q3 - Q2 = 16
POSITIVE SKEW AS Q3 - Q2 > Q2 - Q1

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18
Q

45 students have Ex of 2497 and Ex^2 = 143 369

find the mean and standard deviation for the marks of these student

A

Mean = 2497/45
= 55.5 (3sf)
Standard deviation = square root ( (143 369/ 45 - (2497/45)^2 )
= 10. 3 (3sf)

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19
Q

what are other methods of measuring skewness

A

mean < median < mode = negative
mean > median > mode = positive

mean - median/standard deviation
if answer is + = positive skew
if answer is - = negative skew

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20
Q

the mean and standard deviation of all student were 55 and 10 respectively. examiners have decided that the total mark of all students should be scaled by subtracting 5 and then reducing the mark by a further 10%

find the mean and standard deviation of the scaled marks of students

A

new mean = ( 55 - 5) (0.9)
= 45
since standard deviation is only effective by the factor of “further 10%”
new standard deviation = 0.9 x 10
= 9

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21
Q

what is the equation for mean and standard deviation?

how else can we denote mean and standard deviation?

A

_ Mean = Ex/n
X Ex = the total value of data , n = sample size of data

_ Standard deviation = (square root Ex^2/ n - (EX/n)^2 )
ox

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22
Q 
For Abbey Hotel:            Balmoral Hotel:
mode 39                          50
mean 33.2                       52.5
stand  12.7                        12.2
skewedness - 0.454       0.203

Compare the 2 age distribution of the residents of each hotel (3)

A
  • Balmoral hotel is positively skewed vs Abbey is negatively skewed
  • Balmoral residents are generally older than Abbey residents
  • Balmoral ox < Abbey ox
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23
Q

when comparing and contrasting box plot diagrams, what are the things you should comment on and how do you determine the amount of comment given?

A
  • should give comments on IQR, Q2, Range, Skewness, Mean

- look at the marks, if its 2 mark then give 2 comments, etc..

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24
Q

a ball is projected upwards with a speed of 21ms^-1 from point A, which is 1.5m above the ground. After the projection the ball moves freely under gravity until it reaches the ground. Modelling the ball as a particle, find:

a) the greatest height above A reached by the ball
b) the speed of the ball as it touches the ground
c) the time between the instant the ball is projected and the instant when the ball reaches the ground

A

a) 22.5m
b) root 470.4 / 21.7 (3sf) ms^-1
c) 4.4 (1dp)

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25
Q

a ball of mass 0.3kg is released from rest at a point 2m above the ground. the ball moves freely under gravity. after striking the ground, the ball rebounds vertically and rise to a maximum height of 1.5m above the ground before falling to the ground again

a) find the speed of the ball at the instant before it strikes the ground
b) find the speed of the ball at the instant after it rebounds
c) find the magnitude of the impulse on the ball
d) sketch a velocity-time graph for the motion of the ball ( 1608 ESS)
e) find the time between the instant the ball is released and when it strikes the ground the 2nd time

A

a) 6. 26 ms^-1 (3sf)
b) 5.42ms^-1 (3sf)
c) 3.50 NS (3sf)
e) 1.75s

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26
Q

what is Impulse and its equation ?

What is its unit measures?

A

impulse = change in momentum
(Mass) (v) - (mass) (u)
Newtons second

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27
Q

a circle c has centre (-1, 7) and passes via point (0,0). find an equation for C (4m)

A 
(x - x1) + ( y - y1) = r^2 
... ( x + 1) + ( y - 7) = R^2
R^2 = (-1 - 0 )^2 + ( 7 + 0)^2 
       = 50 
... ( x + 1 ) + ( y - 7 ) = 50
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28
Q

The points of A & B has ( -2 , 11 ) & ( 8 , 1 ) respectively. Given that AB is a diameter of circle C
a) show that centre C has coordinates ( 3 , 6 )
b) find an equation for C
c) find an equation of the tangent to C at point ( 10 , 7 ) in y = mx + c
5 min

A

a) Centre C = the midpoint of AB
… 1/2 ( -2 + 8) , 1/2 ( 11 + 1)

b) ( x - 3) + (y - 6) = R^2
R^2 = (3 + 2)^2 + ( 6 - 11)^2
R^2 = 50

C) M1 = 7 - 6 / 10 - 3
        = 1/7 
    M2 = - 7 
   y - 7 = - 7 ( x - 10 )
   y = -7x + 77
29
Q

circle C has centre A with coordinates ( 7, 5) . Line l with equation
y = 2x + 1 is tangent to C at P.
a) show that an equation of line PA is 2y + x = 17
b) find an equation for C
c) the line with equation y = 2x + k, k doesn’t = 1, is also tangent to C
find the value of K

A

a) M1 = 2 , M2 = -1/2
y - 5 = -1/2 ( x - 7 )
2y - 10 = - x + 7
2y + x = 17

b) ( x - 7 ) + ( y - 5 ) = R^2
   Use Pythagoras theorem = R^2
  P = Instaneous equation 
  2( 2x + 1) + x = 17 
  ... X = 3 
     Y =  2(3) + 1 
 ... R^2 = 20 
( x - 7) + ( y - 5) = 20
c) the displacement from P to A = the displacement from A to p'
  PA = ( 4 , -2 ) as a vector 
.... P' = (11, 3) coordinates 
   3 = 2(11) + k
.... K = -19
30
Q

Circle C has the equation x^2 - 4x + y^2 + 10y = k

State the possible range of K

A

(X - 2)^2 + ( y+ 5) = 29 + k
R^2 = 29 + K , > 0
… k > -29

31
Q

The line Joining the points ( -1, 4) & ( 3, 6) is a diameter of the circle C
Find an equation for C? 6m

A

Equation for C : (x - 1)^2 + (y - 5)^2 = 5
Centre C = The Midpoint of AB
R^2 = the use of Pythagoras theorem
= ( -2)^2 + ( - 1)^2

32
Q

A circle C has the equation x^2 - 20x + y^2 - 24y + 195 = 0
The centre of C is at point M
a) find the coordinates of M
b) find the radius of C

N is a point with coordinates (25, 32)
c) find the length of the line MN

Tangent to C at Point P on the circle passes via N
d) find the length of the line NP

A

a) M: ( 10, 12)
b) R = 7
c) MP = square root ( 15)^2 + (20)^2 )
= 25
d) NP = square root ( (25)^2 - (7)^2 )
= 24

33
Q

A circle with centre T and radius R has equation
x^2 + y^2 - 20x - 16y + 139 = 0
a) find the coordinates of the centre and show that radius = 5

the line line has equation x = 13 and crosses circle C at P and Q
b) find the y coordinates of P & Q

given that to 3dp the angle PTQ is 1.855 radians#
c) find the perimeter of the sector PTQ

A

a) T: ( 10 , 8) / r = square root of 25
b) P = 12 & Q = 4
P & Q are both points of intersection thus simultaneous equation must be used - e.g. - ( 13 - 10)^2 + (y - 8) = 25

c) both side has radius of 5, therefore 5 + 5 = 10
Arc Length PQ = theta/ 2pie x 2pie(r)
= 1.855/ 2pie x 2pie(5)
= 9, 275
Perimeter = 19.275

34
Q

a circle has centre C (-2 , 4) and radius 5. the tangent to the circle at point P (-5 , 8) has equation 3x - 4y + 47 = 0 and T ( 3, 14) lies on the tangent line
a) find the area of the triangle CPT

A

A) AREA = 25 UNIT^2
1/2 (10) (5) = area
sketch and find lengthen CP = 10

35
Q

Figure 3 shows a sketch of the circle C with centre N and equation
( x - 2)^2 + (y + 1)^2 = 169/4. The chord AB of C is parallel to the x-axis, lies below the x-axis, and is of length 12 units

a) find the coordinates of A and B
b) show that angle ANB = 134.8 to the nearest 0.1 degree

the tangents to C at points A and B meets at the point P
c) find the length AP giving your answer to 3sf

A

a) A: ( -4, -7/2) , B: (8. -7/2)
N cuts the chord in half, … 2 - 6 & 2 + 6
Y- axis = substitution of either -4 or 8 into equation C
since it’s below the x-axis = negative - 7/2

b) A = cos^-1 (b^2 + c^2 - a^2 / 2bc)
c) AP = 13/2 x tan 67.380 …
= 15. 6

36
Q

circle C has equation x^2 + y^2 - 6x + 4y = 12
P & Q has points (-1 , 1) & (7 , -5) respectively and is a diameter
Point R lies on the positive y-axis and the angle PRQ = 90
A) find the coordinates of R

A 
If y lies on the positive y-axis, x = 0
0^2 + y^2 - 6(0) + 4y = 12
y^2 + 4y - 12 = 0
 y = 2 , -6
since y > 0 , y = 2
... R: (0 , 2)
37
Q

Point P (-3 , 2) , Q (9 , 10) and R (a , 4) lies on circle C.

a) Find the value A ?
b) Find the equation of circle C?

A 
a) Gradient PQ = 2/3 
Gradient QR = -3/2 
Gradient = dy / dx
-3/2 = 10 - 4 / 9 - a
-3 (9 - a) = 12
-27 + 3a = 12
 3a = 39
  a = 13

b) (x - 5)^2 + (y - 3)^2 = 65

38
Q

the positive constant a is such that: the integral
2a (2x^3 - 5x^2 + 2 ) dx = 0
divide
a x^2
show that 3a^3 - 5a^2 + 2 = 0
show that a = 1 is a root of 3a^3 - 5a^2 + 2 = 0
find other possible values of a in surd form

A 
f(a) = 3a^3 - 5a^2 + 2 
f (1) = 3(1)^3 - 5(a)^2 + 2
      = 0 
hence a = 1 is a root of f(a)
( a - 1 ) ( 3a^2 - 2a - 2)
hence other possible values of a is 1 & 1 + root 7 / 3 
  A can't be 1 - root 7 / 3 as A > 0
39
Q

solve the equation:

4cos^3 theta - 7cos theta - 3 = 0 for 0 < theta < 2pie in the exact form

A 
cos theta = X 
4x^3 - 7x - 3 = 0
X = 3/2 , -1/2, -1
cos theta = 3/2 , - 1/2 , - 1
theta = inverse cos - 1/2
         = 120, 240 degrees
theta = inverse cos - 1
          = 180 degrees
cos theta = 3/2 has no solution 
hence theta = 2/3 pie , pie , 4/3 pie
40
Q

2 cubic polynomials are defined by:
f(x) = x^3 + ( a - 3)x + 2b , g(x) = 3x^3 + x^2 + 5ax + 4b

Given that f(x) & g(x) have a common factor of ( x - 2) show that
a = - 4 and find the value of b

factorize f(x) fully & show that f(x) & g(x) have 2 common factor

A

b = 3

x - 2 ) ( x + 3) is a common factor of f(x) & g(x

41
Q

find the set of values of x for which

both 3x - 7 > 3 - x & x^2 - 9x < ( or equal to) 36

A

10/4 < x < (equal to) 12

42
Q

a rectangular tile has length 4x cm and width (x + 3) cm. The area of the rectangle is less than 112 cm^2.
Determine the set of possible values of x

A 
A = L x W 
4x(x + 3) < 112
4x^2 + 12x - 112 < 0
Critical Values = 4 , - 7
hence from the graph :  -7 <  x < 4
however since x > 0 because can't have negative length :
0 < x < 4
43
Q

a rectangular tile of length 4y cm & width (y + 3) cm has a rectangle length of 2y cm & width y cm removed from one corner
given that the perimeter of this tile is between 20 cm & 54 cm

Determine the set of possible values of y

A 
perimeter = 4y + 3 + 2y + y + 2y +( y + 3)
                = 10 y + 6
20 < 10 y + 6 < 54
14 < 10 y < 48
1 . 4 < y < 4 . 8
44
Q

a) sketch the curve y = - x^2 - x + 12 showing coordinates of all intercepts
b) solve the inequality 12 - x - x^2 > 0

c) find the coordinates of the points of intersection of the curve
y = - x^2 - x + 12 & line 3x + y = 4

A

a) shape: upside down smile , axes: ( 3 , 0) , (-4 , 0) , ( 0 , 12)
b) from the graph: the curve > 0 if x lies between - 4 < x < 3
c) ( 4 , - 8) , ( -2 , 10)

45
Q

find the set of values of x for which both
A) 3(x - 2) < 8 - 2x & (2x - 7) ( 1 + x) < 0
B) 4x - 3 > 7 - x & 2x^2 - 5x - 12 < 0

A 
A) = -1 < x < 14/5
B) = 2 < x < 4
46
Q

the equation kx^2 + 4x + (5 - k) = 0 where k is a constant has 2 different solution for x

a) show that k satisfies k^2 - 5k + 4 > 0
b) find the set of possible values of k

A

a) b^2 - 4ac > 0
(4)^2 - 4(k)(5 - k) > 0
16 - 20k + 4k^2 > 0
k^2 - 5k +4 > 0

b) k < 1 U k > 4

47
Q

for the events A & B: P(A’ n B) = 0.22 , P( A’ n B’) = 0.18
A) find P( A)
B) Find P ( A U B)

Given P(A | B) = 0.6
C) Find P(A n B)
D) determine whether or not A & B are independent

A 
A) P( A) = 1 - ( 0.22 + 0.18)
             = 0.6 
B) P(A U B) = 1 - 0.18 
                  = 0.82
C) P ( A | B) = P ( A n B) / P (B)
     0.6  = x /x + 0.22
     0.6x + 0.132 = x
        0.132 = 0.4x
        0.33  = x

D) P( A n B) = P(A) x P(B)
0.33 = 0.6 x 0.55
0.33 = 0.33
therefore A & B are independent

48
Q

what is the equation P ( A u B) ?

A

P ( A u B) = P(A) + P (B) - P( A n B)

49
Q

what is the equation of P( X | Y) ?

A

P( X | Y) = P( A’ n B’) / P (B’)

50
Q

a lorry is moving along a straight horizontal road with constant acceleration. the lorry passes point A with speed u ms^-1, (u < 34), and 10 second later passes point B with speed 34 ms^-1 . Given that
AB = 240m
a) find the value of u
b) the time taken for the lorry to move from A to midpoint of AB

A

a) u = 14 ms^- 1
b) t = 6 seconds
S = 120 U = 14 V = x A = x T = ?
We must find A since it is always constant
A : V = U + AT
34 = 14 + 10a
20 = 10a
2 = a
S = ut + 1/2 at^2
120 = 14t + 1/2 (2) (t)^2
t^2 + 14t - 120 = 0
t = 6, -20
t > 0 , t = 6

51
Q

3 posts P , Q , R, are fixed in that order at the side of a straight horizontal road. PQ = 45m & QR = 120m. The car is moving along the road with constant acceleration a ms^-2. The speed of the car as it passes P is u ms^-1. The car passes Q 2s after passing P & passes R 4s after Q.
Find the value of a & u ? 5min

A 
PQ: 
S= 45 U = U V= X A= A T= 2
S  = UT + 1/2 AT^2
45 = 2U + 1/2 (A) (2)^2
45 = 2U + 2A
PR:
S= 165 U= U V= X A= A T= 6
S = UT + 1/2 AT^2
165 = 6U + 18A
55 = 2U + 6A
10 = 4A
2.5 = A 
55 = 2U + 6(2.5)
20 = U
52
Q

i and j are horizontal unit vectors due east and north respectfully
Particle P moves with constant acceleration. At t=0, the particle is at O and is moving with velocity ( 2i - 3j) ms^-1. At t = 2, P is at A with position vector ( 7i - 10j) m.

show that the magnitude of the acceleration of P is 2.5 ms^-2

A

s = (7i - 10j) u= ( 2i - 3j) v= x a= ? t= 2
s = ut + 1/2 at^2
7i - 10 j = (2i - 3j) (2) + 1/2 (a) (2)^2
7i - 10 j = 4i - 6j + 2a
3i - 4j = 2a
1.5i - 2j = a

a | = square root ( 1.5)^2 + (-2)^2
= 2.5 as required

53
Q

find and simplify the first 3 terms in the expansion of ( 2 + 5x)^6, in ascending powers of x

in the expansion of (3 + cx)^2 (2 + 5x)^6, the coefficient of x is 4416. find the value of c

in the expansion of ( 3 + bx)^5 the coefficient of x^2 is twice the coefficient of x . find the value B

A

64 + 960x + 6000x^2 …
C = - 11
B = 3

54
Q

given that the vector (40 , 4 ) = 40! /(4!) (b!)

What’s the value B?

A

ncr = ( n / r) = n! / r! ( n - r)!

hence, B = 36

55
Q

What are all the binomial equations?

A

(1 + a)^n = 1 + na + n(n-1)/ 2! (a^2) + n(n-1)(n-2)/3! ( a^3) + ………

(a + b)^n = nCo (a^n) b^0 + nC1 (a^n-1)(b) + nC2 (a^n-2)(b^2) + ……

56
Q

in the binomial expansion of ( 1 + ax)^40 , the coefficients of x^4 & x^5 are p and q respectively. Find the value of q/p

A

x^4 = p , x^ 5 = q
X^4 :
(40C4) (1^36)(x^4) = 91390
P = 91390

x^5 :
(40C5) ( 1^35) (X^5) = 658008
Q = 658008

Q/P = 658008 / 91390
= 36 / 5

57
Q

use your expansion of (1 + x/4)^8 to estimate the value of 1.025^8
giving your answer to 4 dp

use your expansion of ( 1 + x/2)^10 to estimate the value of 1.005^10 giving your answer to 5 dp

A

( 1 + x/4)^8 = 1 + 2x + 7/4 x^2 + 7/8 x^3
x = 0.1
(1.025)^8 = 1.2184 4 dp

(1 + x/2)^10 = 1 + 5x + 45/4 x^2 + 15x^3
x = 0.01
(1.005)^10 = 1. 05114 to 5 dp

58
Q 
if x is so small, so that x^2 and higher powers can be ignored, 
show that ( 1 + x) (1 - 2x)^5 is roughly 1 - 9x
A

( 1 + x) ( 1 - 10x + 40x^2 - 80x^3)
= 1 - 10x + …. + x + ….
= 1 - 9x

59
Q

find the binomial expansion of (3 + 2x)^5 + ( 3 - 2x)^5

A

= 486 + 2160x^2 + 480x^4

60
Q 
use the binomial expansion in ascending powers of x to show that :
square root ( 4 - x) = 2 - 1/4 x + kx^2 + .....
where k is a rational constant to be found

a student attempts to substitute x = 1 into both side of this equation to find an approximate value for root 3.
state giving reason if the expansion is valid for this value of x

A 
square root ( 4 - x) = (4 - x)^1/2 
( 4 ( 1 - 1/4x ) )^1/2 
2 ( 1 + (1/2)(-1/4x) + (1/2)(1/2 - 1) (-1/4x)^2 / 2! )
2 - 1/4 x - 1/64 x^2 + ....
hence k = - 1/64

generally the expansion is valid if - 1 < a < 1
hence valid if this expansion - 1 < -1/4 x < 1
- 4 < x < 4
therefore, when x = 1 , it’s valid

61
Q

summarize binomial validity key point

A
  • the binomial expansion ( 1 + a)^n is Only valid for all positive integer
    values of n But if n is negative or a rational value Then it is only valid
    for -1 < a < 1
62
Q

Determine the validity range for binomial expansions of:

a) root ( 1 - 2x)
b) 3 / 2 - x
c) cube root ( 8 - 3x ) / (3 - x)^2

A

a) = - 1/2 < x < 1/2
b) = -2 < x < 2 or can be written as | x | < 2
c) validity | x | < 8 / 3

63
Q

show that the equation tan 2x = 5sin 2x

can be written in the form ( 1 - 5cos 2x) sin 2x = 0

A 
tan 2x = 5sin 2x
sin 2x / cos 2x = 5sin 2x
sin 2x = 5sin 2x cos 2x
sin 2x - 5sin 2x cos 2x = 0
sin 2x( 1 - 5cos 2x) = 0
( 1 - 5cos 2x) sin 2x = 0
64
Q

A frequency distribution is shown by :
class interval : 1 - 10 | 11 - 20 | 21 - 30 | 31 - 40 | 41-50
frequency : 10 | 20 | 30 | 24 | 16
C.F : 10 | 30 | 60 | 84 | 100

use interpolation to estimate the value of 30th & 70th percentile
Hence estimate the 30% - 70% inter-percentile range

A

30th : 100 x 0.3 = 30
p30 - 10.5 / 10 = 20 / 20
20p30 - 210 = 200
p30 = 20.5

70th: 100 x 0.7 = 70
p70 - 30.5 / 10 = 10 / 24
24p70 - 732 = 100
p70 = 34. 67

inter-percentile range = 34.7 - 20.5
= 14.2

65
Q

Curve C has equation y = x^2 - 2x - 24x^1/2 , x > 0

verify that C has a stationary point when x = 4

A 
dy /dx = 2x - 2 - 12x^- 1/2
dy / dx = 2(4) - 2 - 12(4)^ - 1/2
            = 8 - 2 - 6
             = 0
hence C is at a stationary point when x = 4
66
Q

Curve C has equation y = (x + 3) (x - 8) / x , x > 0

a) find dy / dx
b) find an equation of the tangent to c at the point where x = 2

A

a) dy / dx = 1 + 24x^-2
b) M = 7 & y = - 15
y + 15 = 7( x - 2)

67
Q

Curve C has equation: y = ( x + 1) ( x + 3)^2 . Point A with x- coordinate -5 lies on C

a) find the equation of the tangent to c at A
b) another point B also lies on c. the tangents to c at A & B are parallel. find the x-coordinate of B

A 
a) y = ( -5 + 1) ( -5 + 3)^2
      = - 16
   A : ( -5 , - 16)
   dy /dx = 3x^2 + 14x + 15
              = 20 
    y - y1 = m ( x - x1)
    y + 15 = 20 ( x + 5 )
b) If A & B are parallel, then M = 20 
    20 = 3x^2 + 14x + 15 
    3x^2 + 14x - 5 = 0
     hence x = 1/3 , -5
      since x > 0 , x-coordinate of B = 1/3
68
Q

A rectangular box has sides measuring x cm, (x + 3) cm, 2x cm.
a) write down an expression for the volume of the box
b) Given that the volume of the box is 980 cm^3
show that x^3 + 3x^2 - 490 = 0
c) show that x = 7 is a solution to this equation
d) prove that the equation has no other real solution

A

a) V = l x w x h
V = (x) (x + 3) (2x)
V = 2x^3 + 6x^2
V = 2x^2 ( x + 3)

b) 980 = 2x^3 + 6x^2
2x^3 + 6x^2 - 980 = 0
x^3 + 3x^2 - 490 = 0

c) f(x) = x^3 + 3x^2 - 490
f ( 7) = 7^3 + 3(7)^2 - 490
= 0
since x - 7 ) is a factor & x = 7 is a solution to f(x)

d) f(x) = (x - 7) ( x^2 + 10x + 70)
discriminant is negative - 180, hence have no real solution and the equation has no more real solution

69
Q

solve these equations to 3sf:

a) 4^x = 23
b) 7^2x+1 = 1000
c) 10^x = 6^x+2

A

a) x = 2.26
b) x = 1. 27
c) x = 7. 02

Unit 1: Contemporary Issues in Health/Social (3 decks)

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