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CSEC MATHEMATICS Past Paper Solution – May 2014 cxcDirect Institute

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** Please see the original past paper for the questions. Only the answers will be provided as per copyright obligations.

2 a) x –2 x+1 + = 3 4

**********************************************************

=

4x−8+3x +3 12

=

4x−8+3x +3 12

=

7x−5 12

(i)

x+4 =

x +10 2

(ii)

x −6 = 2x +9

Q1a. May 2014 Using a calculator: Q1a(i).

5.25÷0.015

= 350

Q1a(ii).

√(6.5025)

= 2.55

Q1a(iii).

3.142×2.2362

= 15.7

4( x−2)+3( x+1) 12

( 3 SF) 2 b)

Q1b(i) Cement

Sand

Gravel

Comments

Given

1

4

6

Mixing ratio

Q1b(i)

4

6 x 4 = 24

24 buckets of gravel needed

Q1b(ii)-a Q1b(ii)-b

20 4

=5

5

20 20

5 buckets of cement needed 6 x 5 = 30

30 buckets of gravel needed

2

2 c) (i)

3x + 5 = y

(ii)

3(4) + 5 = 17

(iii)

3x + 5 = 8

(1v)

y–5 3

so

x=1

=x

Q1.c (i) Total hire purchase price = deposit + total monthly payments

2 d) Simultaneous equations: ( solving by elimination)

Deposit = $350 Total monthly payments = 10 months x 120

By multiplying the second equation by 3, we get: = $1,200

Total hire purchase price = 350 + 1200

= $1,550

2x +3y = 9 9x – 3y = 24 By adding both equation we get: 11x = 33

so (x = 3)

Q1.c (ii) Amount saved if laptop is purchased for cash of $1,299 = 1,550 – 1,299

By substituting (x=3) into the first equation we get:

= $251

2(3)+3y = 9 so


-

Email: [email protected]

6

+ 3y = 9

website: www.cxcDirect.org

(giving y = 1)

2


Q3 a)

Q3b i(b) – To construct RX perpenducular to PQ R

U = {11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26} The set U represents integers between 11 and 26 INCLUSIVE.

6cm

(i) The number of members in set U are: n(U) = 16

60o P

p1

X

p2

Q

A = {12,14,16,18.20,22,24,26} (even #'s) B = {12,15,18,21,24} ( multiples of 3)

(ii) (iii)

p3

U A

B 14 16 20 22 26

15 21

12 18 24

1.

With centre R, and suitable compass separation, construct two arcs to cut PQ at p1 and p2.

2.

11, 13, 17, 19, 23, 25

With centre p1 and suitable compass separation, construct an arc below line PQ.

3.

With centre p2 and the same compass separation,

Q3 bi a) Construction details for Triangle PQR.

construct a second arc to intersect the first arc at

Note that the question require that you use compass, ruler and pencil ONLY to contruct the triangle.

perpendicular to line PQ

p3. The line that connects R and p3 is

R

4.

Draw RX which is perpendicular to line PQ.

p2 6cm

60o P

p1

Q

8cm

Steps: Draw a straight horizontal line PQ = 8 cm With centre P, and suitable compass separation, construct a wide arc above line PQ which cuts line PQ at p1. 3. With centre p1, and the same compass separation, draw a second arc to intersect the first arc at p2. 4. Draw line PR = 6cm which passes through points P and p2. 5. Draw line RQ to complete the triangle PQR 1. 2.


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3


Q4 (a ) - Map with scale 1:50,000 1.

1 cm on the map represents 50,000 cm on the island

2.

An area of 9

1cm

2.5×10 cm 3.

2

2

on the map represents on the island

If 1km = 100,000 cm, then 1cm on the map represents 0.5km on the island

4 (b ) LM represents a distance of 8cm on the map since (1cm = 0.5km) , then LM = 8×0.5 = 4km

4 (c) (i) The approximate number of squares representing the forest reserve as seen on the map is: number of full squares = 12 number of half-squares = 6 1 2

2 Total area of forest reserve = 12+(6× ) = 15cm

(ii) Now 1 cm on the map represents 0.5km ( shown earlier) so ⇒

1 cm

2

represents 0.5×0.5 = 0.25km 2

Area of forest reserve = represents 15×0.25 =


-

2

15cm 2 3.75km



4


Q5 (a) Enlargement with k = 2 and centre of origin is (0,0)

Q5b- Angle of Elevation T

80m 40 F

P

(i) The angle of elevation T from P is

0

(ii) now

tan 40

so

=

FP =

40

0

80 FP 80 tan 40

= 95.34m

Construction steps: 1) Finding Point A' of the image triangle 1. Using a ruler, draw a straight line from the centre of the enlargement O, and which passes through point A of the object triangle. 2. Now Point A' of the image and point A of the object are related such that: 3. 4.

OA' OA

(iii)

Angle of Elevation T from Q ( ∢Q ) T

= scale factor (k)

80m

(where scale factor is given as = 2) measure OA = 2.25cm Therefore: OA' = 2×OA = 4.5cm

20.6

F

95.34m

P

Similarly, find Points B' and C' of the image triangle, and construct the image triangle A'B'C'.

5a, (ii) From the diagram shown on the past paper,

Q

213.34m

Distance FQ

= 95.34 + 118

now

tan Q

=

80 213.34

so

∢Q

=

−1 0 tan (0.37498) = 20.6

where

∢Q

is the angle of elevation T from Q

4 5

118m

= 213.34m = 0.37498

The image triangle A"B"C" represents a translation of the object triangle ABC by a translation vector T, where : T=

A' ' 5 − −3

A 1 2

( ) ()


-

: giving T =

( ) 4 −5



5


Q6- Quadratic Function The graph of the quadratic funnction is shown on the past paper for y= x 2 .

N(3, 9)

Now if the lines are parallel their gradients are equal. so m = 2. Also if the line passes through the origin , then the intercept is equal to zero. ( so c = 0)

M(-1, 1)

a,(i)

(iii) Finding the equation of the parallel line that passes through the origin.

Giving equation of parallel line : y = 2x

Finding the coordinates of point M(-1, y) , and N(x, 9)

Q6- C – Draw a tangent at point ( 2,4)

2

now

y= x

so if

x = -1,

and when y = 9

2

then

y=

(−1)

=1

then

x=

±√ (9 )

=3

8

(2,4) 2 b) Finding Gradient of line MN now

x1 =−1 ,

Gradient m =

y 1=1 y2 – y2 x 2− x 1

=

x 2=3 9–1 = 3−(−1 )

and 8 4

y 2=9

=2

The gradient of the tangent at point (2,4) =

8 =4 2

(ii) Finding the equation of the line MN Now the general equation of the line is: y = mx + c From the graph it can be seen that the y-intercept = 3 ie ( c = 4) so the equation of line MN is y = 2x+3


-



6


Q7, a

Q8 – Number sequence

Num Tally books (x)

Freq (f)

f.x

1

11

2

2

2

111

3

6

3

1111 .

5

15

4

1111 1 6

24

5

1111 11

7

35

6

1111

4

24

7

111

3

21

sum

30

127

b)

c) i

Figure - 4

b) Fig (f)

Formula

Modal number of books in bag = 5

Total number of book in bag

Total number of dots Number (n)

1

5 x 2 -5

5

2

5 x 3 -5

10

3

5 x 4 -5

15

5

5 x 6 -5

25

6

5 x 7 -5

30

f

5 (f + 1) - 5

35

29

5(30) - 5

145

= 127

4 c) ii

mean number of books in bag =

127 30

= 4.23

d)

P ( X< 4) =

5+3+2 30

=

1 3

c) If (f) represents the figure number and n represents the number of dots: n

=

d) now if

n

= 145

then

145

= 5 (f +1) - 5

so

f+1

=

hence


-


5( f +1)– 5

Then:


f

145+5 5

= 30 - 1

= 30 = 29

7


SECTION II

9 (b)

Q9 a (i) Finding g(x) Given f(x) and g(x) as seen on the past paper: y=

(i) Now if

2x+7 x+1

the function will be undefined if the

denominator (x+1) is zero. That is... when

x = −1

(ii) Find: gf(5) First we must find f(5) where: Now

f (5)

gf(5)

(iii) Find:

f

−1

=

2( 5)+7 5+1

=

g(

=

4×

=

17 6

=

14

17 ) 6

17 +3 6

1 3

(a) Average speed after first 2 seconds :

( x)

Average speed =

Total Distance travelled in 2 sec time( 2 sec)

Now given f(x) as seen on the past paper, = Steps Action

= 40 m/s

Result

1

Replace f(x) with y

2

Exchange x and y

3

Make y the subject

y=

2x +7 x+1

x=

2y+7 y+1

(b) The speed of the ball after 3 second:

Mul. both sides by (y+1)

x(y+1) = 2y +7

expand the LHS

xy + x

Move 2y to LHS

xy - 2y + x = 7

Move x to RHS

xy - 2y = 7 – x

Factor y from the LHS

y(x -2) = 7 – x

Div both sides by (x -2) Replace y with


80 2

-

f −1 ( x )

= 2y +7

y = f −1 ( x) =

After 3 seconds the ball has reached its maximum height, and so its speed at that instant is the tangent to the curve at the turning point = 0 m/s

7−x x−2 7− x x− 2



8


Q10

From the diagram given on the past paper we can make 4 observations that will help us answer the questions:

#

Observations

Relevant Theorem

1

Isosceles Triangle BOE base angles are equal

2

Tangent BC

Angle in the alternate segment

3

Angle at the centre

= twice angle at circumference D

4

Cyclic quadrilateral

Opp. Angles are supplementary

so

∢ DBC = ∢ BED

But ⇒

∢ BED = 20 0 0 = 20 0 42

So

∢OED = 42 - 20 =

=

42

0

+ ∢OED + ∢OED 22

0

(iii) Finding angle BFE E

Finding angle BOE E

D

700 1400 O

200

F

1400 O

110 0

D

200

B B

First find angle BDE Observation #1 – Triangle BOE is isosceles so base angles are equal . Hence ∢OBE=∢OEB=20 0 and ∢ BOE=180 – 20 – 20 = 140 ( Finding

Observations #4 BDEF is a cyclic quadrilateral, so opposite angles are supplementary

∢' s in a triangle )

∢OED

Observation #2 – Angle between tangent BC and chord

BD = angle (BED) in the alternate segment.

so:

220

∢ BDE =

hence: so

D

200

+

∢ BDE

=

180

0

Observations #3 Now ∢ BOE=1400 is the angle subtended at the centre from chord BE, which is twice ∢ BDE subtended at the circumference. So:

E

∢ BFE

1 0 (140 ) 2

∢ BFE+70 ∢ BFE

O

0

0

=

70

= =

180 0 110

0

420

B

C 220


-



9


10b (i) Finding Bearing P from Q

(iii) Finding the distance PR ( use sine rule)

R

N

R 100km 100km

83.64km

Q 540

660 Q 540

80km

#

750

80km

P

P

The bearing P from Q is the sum of the angles 66 + 54 = 1200 now

sinP 100

=

Sin54 83.64

so

Sin P

=

100×Sin54 83.64

=

sin 0.9673

giving

P

−1

= 0.9673 =

0

75

(0 dp)

(ii) Finding the distance PR ( use cosine rule) R

100km

Q

83.64km

540

80km

P pr ⇒

pr

2

2

=

2

100 +80 −2×100×80 cos54

0

= 83.64 km


-



10


Q11

(iii) Finding the coordinates of the point R

( ) 7 2 p −1

given a Matrix =

This matrix will NOT have an Inverse if its determinant = 0 A Matrix whose determinat is zero is called a singular matrix. Now the determinant D of the above matrix is ; D = -7 – 2p so if D =0 Then 0 = −7 − 2p giving p = -3.5

Q11 b) The simultaneous equations as seen on the past paper may be expressed as:

(

4 −2 2 3

To find the coordinates of point R , we must first find the position vector O ⃗R Where that is :

O⃗ R O⃗ R

is the vector from O to Q and then Q to R ⃗ + Q⃗ = OQ R .....

) ( ) = (04) x y

Now we are given that

⃗ RQ

Q⃗ R

and what we need is the vector

Q11. C

⃗ OP

=

=

() 2 4

which is negative of

⃗ RQ

⇒

O⃗ R

=

⃗ OQ

=

() () ( )

=

⃗ - RQ

8 2 − 2 4

6 −2

so the coordinates of R is (6, -2)

()

2 a) O ⃗ P= 4

()

8 b) O ⃗ Q= 2

,

c) P ⃗ Q

is the vector going from P to Q

(iv) since

where but

⃗ PQ ⃗ PQ

⃗ + OQ ⃗ = PO ⃗ = −O P

Then the quadrilateral formed by the 4 vectors is a parallelogram

so

⃗ PQ

=

⃗ O⃗ R= P Q

and

⃗ RQ ⃗ O P=

⃗ −O ⃗ P + OQ

which may be written as ⃗ - OP

⃗ = OQ ⃗ PQ

=

----END---

() () ( ) 8 2

2 4

-


-

=

6 −2



11

cxcDirect-May2014

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