maths OCR Core 2 MEI January 2010 4752 - Chatterton Tuition · OCR (MEI) Mathematics Advanced Subsidiary GCE Core 2 (4752) January 2010 0775 950 1629 Page 3 Question 3 Draw a right

OCR (MEI) Mathematics Advanced Subsidiary GCE

Core 2 (4752) January 2010

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QUALIFICATIONS, QUALIFICATIONS BY TYPE, AS/A LEVEL GCE, MATHEMATICS

(MEI), VIEW ALL DOCUMENTS, PAST PAPERS – JANUARY SERIES 2010,

QUESTION PAPER UNIT 4752/01 – CORE MATHEMATICS 2

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Section A

Question 1

a) To integrate you first add a power to the x term and then divide by the new

power. However you also need to make sure all the x terms are on the top line

to a negative power if need be. As we don’t have limits we do need to add a

constant c

� (x- 3x-2) dx

= ��x

2 -

��x

-1 + c

= ��x

2 + 3x-1 + c




Question 2

i) the pattern is repeating every 4 terms

4th

term is 3

8th

term is 3

12th

term is 3

52nd term is 3

u52 = 3

u53 = 1

u54 = 3

u55 = 5

55th

term is 5

ii) the sum of the first 4 terms is 1 + 3 + 5 + 3 = 12

there are 13 sets of these

13 x 12 = 156

S55 = 156 + u53 + u54 + u55 = 156 + 1 + 3 + 5 = 165




Question 3

Draw a right angled triangle and label the sides (opp, adj and hyp) from the

point of view of the angle θ. We know sinθ = √�� , so label the opp √2 and the

hyp as 3.

from pythagorus we can work out the length of the adjacent side

adj2 = 3

2 – (√2)

2 = 9 – 2 = 7

adj = √7

we know tanθ = ��

tanθ = √�√�

3

√2 opp

θ

hyp

adj




Question 4

Area of sector = ��

�

= �� .�

� = 8.45

multiply both sides by 2

r2 x 0.4 = 16.9

divide both sides by 0.4

r2 = 42.25

square root both sides

r = 6.5 cm




Question 5

i) y = f(2x) this is a stretch (as we are multiplying or dividing rather than adding

or subtracting)

The change is inside the brackets so affecting the x values

When the x values are affected they are affected in the opposite way to what

you think

This is a stretch of scale factor ½ parallel to the x axis (all the x values are half

what they were, and all the y values are unchanged)

Points P, Q and R will now be

(-��, 2), (0, 4) and (2, 2)




ii) y = ��f(x) this is a stretch (as we are multiplying or dividing rather than adding

or subtracting)

The change is outside the brackets so affecting the y values

This is a stretch of scale factor �� parallel to the y axis (all the y values are

��

times what they were, and all the x values are unchanged)

Points P, Q and R will now be

(-1, ��), (0, 1) and (4,

��)




Question 6

i) we are given that u1 = 5 and un+1 = un + 4

u2 = 5 + 4 = 9

u3 = 9 + 4 = 13

etc

so we see we have an arithmetic sequence with the first term a as 5 and the

common difference d of 4

we want the 51st

term

un = � + (n – 1)d

substitute a = 5, d = 4, n = 51

u51 = 5 + 4(51 – 1) = 5 + 200 = 205

ii) the first term is 5

a = 5

the common ratio is 2/5 = 0.4

r = 0.4

sum to infinity = �

�� = �

��.� = ��




Question 7

i)

we can use the sine rule ��.� =

�� !".�

multiply both sides by 5.6

sinθ = �.� � �� !

".� = 0.6544

take the inverse sin of both sides (sin-1)

θ = sin-1(0.6544) = 40.9⁰

A

8.4

5.6

D

B

θ

79⁰




ii) we know angle BDC is 101⁰ (180 – 79)

this time we can use the cosine rule

a2 = b

2 + c

2 – (2bccosA)

Where a = x, b = 7.8, c = 5.6 and A = 101⁰

x2 = 7.8

2 + 5.6

2 - (2 x 7.8 x 5.6 x cos 101⁰) = 60.84 + 31.36 - -16.66907

x2 =108.86907

square root both sides

x = 10.434 = 10.4 (3 significant figures)

D

7.8

5.6 C

B

x

101⁰




Question 8

to find the gradient of a curve we differentiate the curve

to then find the gradient at a particular point we substitute in the x value at

that point

to differentiate you multiply by the power of x and then reduce the power by 1

y = 6x1/2

# $ = 3x

-1/2

when x = 16 # $ =

�√�� =

��

We need to find the y coordinate that goes with the x coordinate of 16

When x = 16, y = 6 x √16 = 6 x 4 = 24

we have the gradient �� and a point that it goes through (16, 24) so we can work

out the equation of the tangent using

y – y1 = m(x – x1)

y – 24 = �� (x – 16)

multiply both sides by 4

4y – 96 = 3x - 48

Subtract 4y from both sides

-96 = 3x – 4y – 48

Add 96 to both sides

3x – 4y + 48 = 0




Question 9

i)

ii) take logs of both sides

log1032x+1

= log1010 = 1

bring down the power to the front

(2x + 1)log103 = 1

Divide both sides by log103

2x + 1 = 1/log103 = 2.095903

Subtract 1 from both sides

2x = 1.095903

Divide both sides by 2

x = 0.55 (2 decimal places)




Section B

Question 10

i) turning points are stationary points which are minimum or maximum and we

find them by differentiating and setting the differential equal to 0

to differentiate you multiply by the power of x and then reduce the power by 1

# $ = 3x

2 - 6x – 9

3x2 - 6x – 9= 0

Divide through by 3

x2 - 2x – 3= 0

See if it will factorise

See if we can find two numbers that multiply to make - 3 and add to make -2

The two numbers are -3 and +1

(x – 3)(x + 1) = 0

x – 3 = 0

x = 3

or

x + 1 = 0

x = -1




We need to investigate which of these is the maximum and which is the

minimum.

We do this by differentiating again. If the second differential comes out as

positive then the point is a minimum, if it comes out negative then it is a

maximum

�# $� = 6x - 6

When x = -1 �# $� = (6 x -1) – 6 = -6 – 6 = -12 (negative so a maximum)

When x = 3 �# $� = (6 x 3) – 6 = 18 – 6 = +12 (positive so a minimum)

Maximum at x = -1

Minimum at x = 3




ii)the curve crosses the x axis when y = 0

so set

x3 - 3x

2 – 9x = 0

We are told to give the answer in exact form so unlikely that it would factorise

completely. However we can factorise out the x to give

x(x2 – 3x – 9) = 0

one solution is x = 0

to get the other solutions we need to solve the quadratic

x2 – 3x – 9 = 0

we could complete the square or use the quadratic formula. I shall complete

the square:

(x - ��)

2 -

!� - 9 = (x -

��)

2 -

!� -

�� = (x -

��)

2 -

�� = 0

Add �� to both sides

(x - ��)2 =

��

Square root both sides

x - �� =' (��

�

add �� to both sides

x = �� ' (��

� = �� '

√��

we have 3 x coordinates

x = 0 , �� ' √��

� (-1.85, 0, 4.85)




iii)




Question 11

i) The trapezium rule is given in the formulae booklet

First set up a table for each of the terms

n 0 1 2 3 4 5 6 7 8

xn 0 1.5 3 4.5 6 7.5 9 10.5 12

yn 2.3 2.7 3.3 4.0 4.8 5.2 5.2 4.4 2.0

��h((y0 + yn) + 2(y1 + y2 + … + yn))

h = 1.5 = strip width

�� x 1.5 x ((2.3 + 2.0) + 2(2.7 + 3.3+ 4.0 + 4.8 + 5.2 + 5.2 + 4.4))

0.75 x ((4.3) + 2(29.6))

0.75 x (4.3 + 59.2) = 47.625 = 47.6 m2 to 3 significant figures

ii) in order to get a lower bound we always need to take the lower height of

the two.

All the rectangles have the same width so area =

1.5 x (2.3 + 2.7 + 3.3 + 4.0 + 4.8 + 5.2 + 4.4 + 2) = 1.5 x 28.7

Lower bound = 43.05 m2




iii) we can integrate the function to get the area under the curve

the limits should be 0 and 12

� �� (-0.013x

3 + 0.16x

2 – 0.082x + 2.4) dx

to integrate you first add a power to the x term and then divide by the new

power

as we have limits we don’t need to worry about adding a constant c

*��.��$+� , �.��$-

� . �.�"�$�� , 2.40112

0 =

put in the limits

47.664 – 0 = 47.664 = 47.7 to 3 significant figures

iv) when x = 7.5 the model would give a height of

y = (-0.013 x 7.53) + (0.16 x 7.5

2) – (0.082 x 7.5) + 2.4

y = 5.300625 = 5.30 to 3 significant figures

this compares with 5.2m from the diagram (it is therefore an overestimate of

about 2%)




Question 12

i) take logs of both sides

log10P = log10(a x 10bt

) = log10a + log1010bt

bring the power to the front

log10P = log10a + btlog1010

log1010 = 1

log10P = log10a + bt

this can be compared to y = mx + c

where y = log10P, x = t, m = b, c = log10a

so if we plot the points with y as log10P and x as t we should get a straight line

the intercept will be log10a

the gradient will be b




ii)

Year 1955 1965 1975 1985 1995 2005

t 10 20 30 40 50 60

P 131 161 209 277 372 492

log10P 2.12 2.21 2.32 2.44 2.57 2.69




iv)

from the graph we have c (intercept) = 1.96

1.96 = log10a

So

101.96

= a

a = 91.2

Again from the graph:

Gradient is �22 �� # 3�456� �22 �� $ 3�456�

Gradient = �.�"��.!�

�� = 0.012 = m

So m = b

b = 0.012

putting these values back in to P = a x 10bt

we have

P = 91.2 x 100.012t

iv) in 2050, t = 105

substitute t = 105 into the equation in part iv)

P = 91.2 x 100.012 x 105 = 1660 million

This will not be very reliable though because we are extrapolating from the

data – we are assuming that the relationship will continue to hold as t gets

bigger and bigger. This is unlikely to be true since the model predicts the

population will eventually become infinite.




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maths OCR Core 2 MEI January 2010 4752 - Chatterton Tuition · OCR (MEI) Mathematics Advanced Subsidiary GCE Core 2 (4752) January 2010 0775 950 1629 Page 3 Question 3 Draw a right