FEB 2016 SEC 4 ADDITIONAL MATHEMATICS CW & · 1 FEB EXAM 1 2016 SEC 4 ADDITIONAL MATHEMATICS CW & HW 1 Find the values of k for which the line y 6x 1 is a tangent to the curve y....

1

FEB

EXAM 1

2016 SEC 4 ADDITIONAL MATHEMATICS CW & HW

1 Find the values of k for which the line 16 xy is a tangent to the curve

22 7 xky . Find also the coordinates of the point at which this tangent touches

the curve. [5]

2

2

Given that q

2sin and both cos and tan are negative, find an expression, in

terms of q , for (i) tan , [2]

(ii) 2sin . [3]

3

3

(i) The graph of cxy 4 passes through the point (2, 3). Find the possible

values of the constant c .

[3]

(ii) Solve the inequality 1174 x . [2]

4

4 The height, h m, of a stone t seconds after it has been thrown vertically upwards

from ground level is given by 2btath , where a and b are constants.

t 2 4 6 8 10

h 14 -4 -45 -136 -250

The table shows experimental values of the variables t and h, but an error has

been made in recording one of the values of h.

(i) Express the given equation in the form suitable for drawing a straight line

graph and, using graph paper, draw the graph for the values given. [4]

Use the graph to (ii) correct the reading of h for which an error has been made, [2] (iii) estimate the value of a and b. [3] A second stone is thrown into the air from ground level. The height, h m, is directly

proportional to t and h = 60 m when t = 10 s. (iv) Draw a line on your graph to illustrate the motion of the second stone. [1] (v) Hence find the time when the two stones meet. [1]

5

6

5 Solve the equation 3loglog3

log 333 xx

. [5]

7

6 A cubic polynomial )(xf is such that the roots of 0)( xf are 3

1, 5 and -2 and it

gives a remainder of 24 when it is divided by )1( x . (i) Express )(xf as a cubic polynomial in x with integer coefficients. [4]

(ii) Hence, solve 0)2( xf . [2]

8

7 (i) Prove the identity xecx

x

x

x cos2

cos1

sin

sin

cos1

. [3]

(ii) Hence, find all the angles between 0 and 180 which satisfy the equation

70tan2cos1

2sin

2sin

2cos1

x

x

x

x.

[3]

9

8 A curve has the equation 23

4

x

xy .

(i) Find an expression for dx

dy and explain why the curve has no turning points. [3]

(ii) Find the gradient of the curve when y = 0. [2] (iii) Given that y is increasing at the rate of 0.28 units per second at the instant

when x = 2, find the rate of change of x at this instant. [3]

10

9 In the expansion of

9

2

2

px

xwhere p is a positive constant, the term independent

of x is 5376. (i) Show that p = 4. [4] (ii) With this value of p, find the coefficient of 9x in the expansion of

9

2 99 12

px x

x.

[4]

11

10 A line 022 yx cuts a curve yx

2

2

11 at two points A and B .

(i) Find the coordinates of A and B . [4]

(ii) Show that the perpendicular bisector of AB cuts the y -axis at

4

3,0 . [4]

12

EXAM 1 ANSWERS

1

.1,3

1 ispoint of sCoordinate

1,3

1

013

0169

meet, line and curvewhen

4

0)1)(7(436

04 curve, o tangent tis line

0167

716

2

2

2

2

22

22

yx

x

xx

k

k

acb

xxk

xkxy

M1 M1

A1 M1 A1 [5]

2

(i)

tantan

4

22q

4

22

q

M1

A1

(ii) sin2coscos2sin2sin

2sin cossin2

q

q

q

422

2

2

2 44

q

q

M1 M1 A1 [5]

13

3 (i) ycx 4 or ycx 4

324 c or 324 c

5c or 11

M1 A1, A1

(ii) 1174 x

117411 x

5.41 x

M1 A1 [5]

4 (i)

btat

h

btath

2

t 2 4 6 8 10

h 14 -4 -45 -136 -250

t

h

7 -1 -7.5 -17 -25

15

10

5

-5

-10

-15

-20

-10 10 20 30 40

B: (0.01, 13.68)

E: (2.47, 4.12)

Slope CD = -3.88

A: (5.89, -9.14)

E

B

B1 B1 for table of values B2 for straight line graph (i)

(ii)

54

9

h

t

h

B2

(iii) From graph, a = y-intercept = 14 b = gradient = 4 B1 M1, A1

(iv) Refer to line drawn above. B1

(v) t = 2.5 s B1

(i)

(iv)

14

5 Given 3loglog3

log 333 xx

3 3 33

33 3

3

3 3 3 3

3 3 3 3

3 3

3

1

33

log 3 log log log 3 [ 1]

log2log 3 log [ 1]

log 3

log 3 2log 3 log log

1log 3 log 3 log log [ 1]

2

1 log 2log

1log [ 1]

3

3 3 [ 1]

x x M

xx A

x x

x x M

x x

x A

x A

M1 A1

M1

M1

A1 [5]

6

(i) Let 253

1)(

xxxaxf

1033

1 2

xxxa

3

109

3

10 23 xxxa

Since 24)1( f ,

243

109

3

101

a

248 a 3a

3

109

3

103)( 23 xxxxf

103103 23 xxx

M1 M1

M1

A1

(ii) 0)2( xf

022523

123

xxx

0433

53

xxx

,3x 3

5, 4

M1

A1 [6]

15

7

(i)

xx

xxx

x

x

x

x

cos1)(sin

sincoscos21

cos1

sin

sin

cos1 22

xx

x

cos1)(sin

1cos21

xx

x

cos1)(sin

cos12

ecxcos2

M1 M1

A1

(ii)

70tan2cos1

2sin

2sin

2cos1

x

x

x

x

70tan2cos2 xec

70tan

22sin x

3.133,7.462x

7.66,4.23x

M1 M1

A1 [6]

8

(i) 23

4

x

xy

2

2

2

23

14

23

12323

23

34123

x

x

xx

x

xx

dx

dy

Since 0dx

dy, ∴ the curve has no turning point.

M1

A1

B1

(ii) when y = 0,

4

23

40

x

x

x

∴ gradient of the curve 14

1

14

142

M1 A1

16

8

(iii)

when x = 2, 32

7

8

142

dx

dy

dt

dx

dx

dy

dt

dy

∴

sunits

dt

dx

/28.1

32

728.0

M1

M1

A1 [8]

9

(i)

M1

A1

M1

AG1

(ii)

B1

M1 M1 A1 [8]

9th 9 21 term

2

rr

r

kr C x

x

9 18 2

9 18 2

9 18 3

18 3 0

9 6 6

7 6

6 6

6

6

6

2

2

2 [M1]

3 18

6 [ 1]

2 5376 [M1]

84 2 5376

5376

84 2

4096

4 [1]

rr r

r

r r r r

r

r r r

r

r

C x p x

C p x x

C p x

x x

r

r A

T C p

p

p

p

p

9

92 9 2 1 94

9 1 2 9 12

x x x x xx

9 6 32 1 9 9 2 1

4 3

9

9 9 9

9 9

9

9

For 2 , term in 2

672 [B1]

term in 672 9 5376 [M1]

48384 672

47712

coefficient of 47712 [A1]

x x x T C x x

x

x x x

x x

x

x

17

10

[A1]

[M1]

[M1]

[A1]

[M1]

[M1]

[AG1]

[M1]

[8]

1

FEB

EXAM 2

2016 SEC 4 ADDITIONAL MATHEMATICS CW & HW

1 Given that 2 1 42 8 2 2 ,x x x find the exact value(s) of x. [5]

2

2 The equation of a circle C is 04422 xyx .

(i) Find the centre and radius of the circle. [3] (ii) The circle C is reflected in the y -axis to obtain the circle D . Write down

the equation of the circle D . [1]

(iii) The two circles intersect at points P and Q , as shown in the diagram

above. Find the length of line segment PQ . [4] (iv) Two line segments are then drawn from P to meet each of the circle at X

and Y respectively. Find the coordinates of X and Y , given that PX and PY are diameters of the circles C and D respectively. [6]

x

y

Circle C Circle D

P

Q X Y

3

4

3 (i) State the amplitude of the curve

xy

2

1sin31 . [1]

(ii) Sketch the graph of

xy

2

1sin31 for the interval 20 x . [2]

(iii) By drawing a suitable line on the same diagram in part (ii), determine how

many solutions there are of the equation

xx

2

1sin6210 for the

interval 20 x . [3]

5

4 (i) Solve the equation xxx 3sin2sin8sin for 1800 x . [4]

(ii) Find all angle(s) between 0 and 2

for which 5cos6sin4 22 . [3]

6

5 A particle moves in a straight line so that its velocity, v m/s, is given by tt 216 2 , where t is the time in seconds after passing O. Find an expression in terms of t for

(i) its acceleration, [1] (ii) its displacement from O. [1] Calculate (iii) the value of t at which the particle passes through O again, [2] (iv) the minimum velocity of the particle, [3] (v) the total distance traveled by the particle in the interval t = 0 and t = 10. [5]

7

8

6 The diagram shows a rectangle PQRS . PQT is an

isosceles triangle with PT = 14 cm and PTQ = 2 radians. TV is parallel to PS , and

TV = 7 cm.

(i) Show that the perimeter, W cm, of the rectangle PQRS is given by

14cos28sin56 .

[3]

(ii) Express W in the form ba sin , where 0a and 900 . [2]

(iii) Find the value of for which W = 73 cm. [2]

2

7 cm

P Q

S R

T

V

14 cm

9

7 The equation 052 2 xx has roots and and the equation 042 pxx

has roots

k and

k. Find the value of k and of p.

[8]

10

8

The diagram shows two circles with centres O and P intersecting at two points A

and B. P is a point on the circumference of the circle with centre O. A straight line through B meets the circles at D and C. The line CP meets AB at E and CP produced meets AD at M. Prove that

(i) APAC = AECP, [4] (ii) CM is perpendicular to AD. [5]

A

B

O P

M E

C

D

11

12

9

(i) Given that 1221

2

b

a, where a and b are integers, find, without using

a calculator, the value of a and of b.

[4]

(ii) Given that ln x2y = a and ln y

x = b, express ln

2x

y in terms of a and b. [5]

13

14

10 The diagram shows part of the graph of the curves

2sin

xy and

12

cos3

xy for x0 .

(i) Show that the point of intersection of the 2 curves is

5.0,

3

. [4]

(ii) Given that the curve 12

cos3

xy cuts the x -axis at the point P. Show

that the x -coordinate of P is approximately 1.91. [2]

(iii) The line 3

x is drawn to divide the area enclosed by the 2 curves and the

x -axis into 2 regions, A and B . A student claimed that the regions A and B

are of the same size by just looking at it. Determine if this claim is true, explaining your argument clearly. [5]

2sin

xy

12

cos3

xy

x

y

P 0

15

16

EXAM 2 ANSWERS

1 2 1 42 8 2 2x x x

Let 2 .xy

2 4

2

2 8 2

2 17 8 0

2 1 8 0

y y y

y y

y y

1

2y or 8y

12 2x or 32 2x

1x or 3x

M1

M1 for sub. 2xy

M1 for factorisation

A1, A1 [5]

2

[M1]

[A1]

[A1]

04022222 yxyx

Centre of circle = (2, 0)

Radius = 42022

= 228

[B1]

[M1]

[A1]

OR

[B1]

17

2

18

2

[A1]

[M1]

[M1]

[A1]

[14]

19

3 (i) Amplitude = 3 B1

(ii)

x

y

-2

-1

0

1

2

3

4

5

B1 for ½ cycle of sine curve B1 for correct position

(iii)

xx

2

1sin6210

x

x

2

1sin31

25 Draw

25

xy

No. of solutions = 2

M1 for getting eqn of line M1 for correct line drawn A1 [6]

(ii)

(iii)

20

4 (i) xxx 3sin2sin8sin

xxxxx 3sin282

1sin28

2

1cos2

xxx 3sin3sin5cos2

015cos23sin xx

03sin x or 5.05cos x 540,360,1803x 780,660,420,300,605x

180,120,60,0x 156,132,84,60,12x

180,156,132,120,84,60,12,0x

M1

M1 M1

A1

(ii) 5cos6sin4 22

4

5.0sin

1sin2

5sin66sin4

5sin16sin4

2

22

22

M1 M1 A1 [7]

5 (i)

2112

216 2

ta

ttv

B1 (ii)

2

212

0,0,0when

2

212

216

23

23

2

tts

cst

ct

ts

ttv

B1

(iii)

st

tt

tt

s

4

15

4

21

02

212

02

212

again O passed particle when 0

2

23

M1 A1

21

5 (iv)

smv

st

t

/8

318

4

721

4

76

4

7

12

21

02112

0a velocity,min

2

M1 M1 for finding t A1

(v)

m

ms

s

ms

t

t

75.1035)28

742(950 distance Total

950)100(2

21)10(2

10 t tos2

7 tfrom travelleddistance

8

742

2

7

2

21

2

72

2

7 t to0 t from travelleddistance

2

7,0

07-2t3t

021t-6

0,when v

3

23

2

M1

M1 M1

M1 A1 [12]

22

6 (i)

14cos28sin56

cos1472sin282

cos147

sin28sin142

W

W

PS

PQ

M1

M1

M1

AG

(ii)

57.2656

28tan

61.622856

1

22

a

146.26sin6.62 P

A2

(iii) 731457.26sin61.62

61.62

5957.26sin

45.7057.26

9.43 , 83.0º

M1

A1 [7]

7

5.2

5.0

441

1600

21

191

21

4025.5)5.2(4

25.55.225.0

2

4

2

2

222

22

p

pkkk

k

kkk

B1 B1 M1 M1 A1 A1

M1 A1 [8]

(shown)

23

8 (i) 1. Join PB

2. PA=PB ( radii of circle with centre P)

3. PBEPAE (base angles, isos )

4. PCAPBE ( s in the same seg)

5. PCAPAE

6. CPAAPE (common angle)

7. APE is similar to CPA (AA)

8. AC

AE

CP

AP

9. CPAEACAP (proven)

M1

M1

M1

A1

(ii) 1. ADBAPB 2 ( at centre = 2 at circumference)

2. 180PBEPAEAPB ( s sum of )

3. 18022 PAEADB

4. 90PAEADB

5. BCEPAE ( s in the same seg)

6. 90BCEADB

7. 180CMDBCEADB

8. 90CMD CM is perpendicular to AD .

M1 M1 M1 M1

A1 [9]

9

(i) Given 12

21

2

b

a

2)1()12(2

21)2(22

)21)(12(2

bba

bba

ba

Hence by comparing coefficients, we obtain

12 ba -------- (1)

11 b -------- (2)

Solving, we obtain b = - 2 and a = - 3

M1 M1 A1, A1

24

OR

2 2 12 1 [ 1]

1 2 2 1

2 2 22 1

2 1

1 2 3 0 [ 1]

ab M

a ab

a b a A

Hence by comparing coefficients, we obtain

01ba -------- (1)

03 a -------- (2) [A1]

Solving, we obtain b = - 2 and a = - 3 [A1]

M1 M1 A1, A1

9 (ii) ln x2y = a

ln x2 + ln y = a

2 ln x + ln y = a ------ (1)

ln x

y = b

ln x – ln y = b ------- (2)

Solving simultaneously eqns (1) and (2),

ln x = 1

3(a+b),

ln y = 1

23

a b

ln 2x

y = ln y – ln x 2 = ln y – 2 ln x

= 1

23

a b – 2 [1

3(a+b) ]

= ba 43

1

M1

M1

A1 M1

A1 [9]

25

10

26

10

The claim is not true as area of region A area of region B. [A1] [11]

2

[M1]

[A1]

[A1]

04022222 yxyx

Centre of circle = (2, 0)

Radius = 42022

= 228

[B1]

[M1]

[A1]

OR

[B1]

FEB

EXAM 3

2016 SEC 4 ADDITIONAL MATHEMATICS CW & HW

1 Find the range of values of k if 32 xy intersects 42 kyx at two distinct points. [5]

2

2 The roots of the equation 0235 2 xx are and .

(a) State the value of and . [2]

(b) Find the quadratic equation in x whose roots are 2 and 2 . [5]

3

3 A wire is bent to form a rectangle of area 253 cm2. If the breadth of the rectangle is

12 cm, find the length of the wire, giving your answer in surds. [4]

4

4 Express 2

2

)1(3

4116

x

xx in partial fractions. [5]

5

5 Solve the simultaneous equations [8]

.9813

,2loglog

32

3

24

yx

yx

6

6 Solve the equation 01437163 23 xxx . Hence solve 01437163 23 xxx . [7]

7

7 Write down and simplify, in ascending powers of x, the first three terms of the expansion of

(a)

5

21

x (b) 523 x

Hence, or otherwise, obtain the first three terms of the expansion of

5

2

23

x

x and use it

to estimate the value of 594.2 correct to 3 decimal places. [8]

8

8 A circle with centre C passes through points A 7,1 and B 8,0 .

(i) Explain why the perpendicular bisector of AB will pass through C. [1]

(ii) Given further that the line 22 xy passes through the centre of the circle,

show that the coordinates of C is 4,3 . [4]

(iii) Hence find the equation of the circle. [2]

9

y

x

4

2

4

3

1

0

1

cbxay sin

2

9 The following diagram shows the graph of cbxay sin for x0 .

(a) State the possible values of a and the value of b and of c. [2]

(b) Copy the graph on your answer sheet. By sketching an additional graph on the same

axes and domain, find the number of solutions of 01cos2sin cxbxa . [4]

10

10 Answer the whole of this question on a sheet of graph paper.

The table below shows the experimental values of two variables x and y.

x 1 2 3 4 5

y 1.19 0.87 0.69 0.57 0.48

It is known that x and y are related by an equation of the form nx

my

where m and n are

unknown constants. Draw a graph of y against xy and use the graph to estimate

(i) the value of m and of n,

(ii) the value of x when 75.0y . [8]

11

12

EXAM 3 ANSWERS

1

)2(4

)1(32

32

2

kyx

xy

xy

Sub (1) into (2):

0342

432

4)32(

2

2

2

kkxx

kkxx

xkx

For 2 distinct roots,

1or4

0)1)(4(

043

012164

0)34)(1(4)2(

04

2

2

2

2

kk

kk

kk

kk

kk

acb

M1

M1

M1

M1

A1

5

2(a)

2

5

2

3

2

3

5,3,2

0235 2

a

c

a

b

cba

xx

B1

B1

2

2(b)

0125308

08

125

4

15

8

125

2

5

4

15

2

3

2

5

)(

2

2

3

3

3322

22

xx

xx

M1

A1

M1

A1

A1

5

13

3

cm227

12

2510323

12

12

12

253rectangle ofLength

cm 2216

228

212227

Perimeter wireofLength

M1

A1

M1

A1

4

4

22

2

22

22

2

1

3

13

12

)1(3

4116

3

12

3

7(2),into2Sub

242:)2()1(

)2(3

793342224,2When

)1(3

59334,0When

334116,1When

3)1(3)1(34116

11)1(3

4116

xxx

xx

BA

AA

BABAx

BABAx

CCx

CxBxAxx

x

C

x

BA

x

xx

M1

M1

M1

M1

A1

5

5

)4(67

3

333,(2)From

)3(16

16

4log

4loglog

2log2

log,(1)From

)2(9813

)1(2loglog

67

634

2

2

22

2

22

22

32

3

24

yx

yx

y

x

y

x

yx

yx

yx

y

yx

yx

M1

M1

M1

M1

M1

8

14

2

1and4

42

167),4(into

2

1Sub

2

1)(

8

7

0)12)(78(

07616

6716

,(4)into(3)Sub

2

2

yx

xy

yorNAy

yy

yy

yy

M1

A1

A1

6

Let 1437163)( 23 xxxxf

7or3

1or2

7or3

1or2

014)(37)(16)(3

01437163

01437163

7or3

1or2

)1)(13)(2(0

)7223)(2()(

22

442

2737

,oftcoefficiencomparingBy

)73)(2(1437163

).(offactorais)2(

014)2(37)2(16)2(3)2(

23

23

23

2

223

23

xxx

xxx

xxx

xxx

xxx

xxx

xxx

xxxxf

b

b

b

x

bxxxxxx

xfx

f

M1

M1

M1

M1

A1

M1

A1

7

15

7

375.219

...)1.0(2

675)1.0(

2

40524394.2

),1( into1.0ngsubstitutiBy

)1(...2

675

2

405243

23

...2

12152025

2

12151080810243

...1080810243...2

5

2

51

232

1

232

12

3

...1080810243

...2)3(2)3(323)(

...2

5

2

51

...22

12

1)(

25

2

5

2

222

22

5

5

55

2

2

23

2

54

1

555

2

2

2

5

1

5

5

x

xxxx

xxxxx

xxxx

xx

xx

xx

xx

xCxCxb

xx

xC

xC

xa

M1

A1

M1

A1

M1

A1

M1

A1

8

8(i)

By the property of circle, the perpendicular bisector of chord passes through

centre of circle. Therefore, the perpendicular bisector of chord AB passes

through C.

B1

1

8(ii)

473(1),into3Sub

3

93

227(2),into(1)Sub

)2(22

)1(7

72

1

2

15,

2

15and

2

1When

1 AB ofbisector lar perpendicu ofGradient

1)1(0

78 ofGradient

2

15,

2

1

2

87,

2

01 ofpoint Mid

yx

x

x

xx

xy

xy

ccyx

AB

AB

The coordinates of C are 4,3 .

M1

M1

M1

A1

4

16

8(iii)

Radius = units5)48()03( 22

Equation of circle:

08625)4()3(

5)4()3(

2222

222

yyxxoryx

yx

M1

A1

2

9(a)

Possible values of 3or3 a

Value of 2b and 2c

B0.5,B0.5

B0.5,B0.5

2

9(b)

1cos2sin

01cos2sin

xcbxa

cxbxa

Award 2 marks for correct graph drawn (with labels).

Number of solutions = 3

M1

B2

B1

4

y

x

4

2

4

3

1

0

1

2

cbxay sin

1cos2 xy

3

17

10

(i)

(ii)

n

mxy

ny

mxyny

mnyxy

nx

my

1

1 2 3

0.5

1

1.5

2

xy

y

Award 1 point for each of the following:

Correct labelling of the graph, x-axis and y-axis

Appropriate use of scale and plotting of all points

Drawing of best fit line

gradient n

1 73.1578.0

1 n

n 05.0

y-intercept 85.1 20.385.1 mn

m 05.0

From the graph, 95.1xy when 75.0y .

05.060.2

95.175.0

x

x

xy 1.19 1.74 2.07 2.28 2.40

y 1.19 0.87 0.69 0.57 0.48

M1

M1

M3

A1

A1

A1

8

FEB

EXAM 4

2016 SEC 4 ADDITIONAL MATHEMATICS CW & HW

1. Find the coordinates of the points of intersection A and B of the line 2 2 0x y and

the curve 1 2 1

2x y . Hence show that the distance AB is 3 5 units. [5]

2. Given that the expression 12532 23 xxx is exactly divisible by

42 2 pxx .

(a) Find the value of p. [3]

(b) Hence solve the equation )3)(12(12532 23 xxxxx . [5]

3. Given the quadratic equation 2 3 3 8 0x kx k , where k is a real number,

(a) if the equation has equal real roots, find the values of k, [4]

(b) if the equation has complex roots, find the range of values of k. [2]

4. (a) Solve the equation 2 1 12 7(2 ) 36x x , leaving your answers in two decimal

places. [4]

(b) Without using a calculator, find the value of p for which

4 5(log 25)(log ) 3p . [3]

(c) Simplify 1

27 73 2

. [3]

5. (a) Express 3 2

2

2 3 9 2

4

x x x

x

in partial fractions. [5]

(b) Prove the identity 21 cos

cos cot1 cos

xec x x

x

[4]

6. (a) Sketch the following graphs cos 1y x and sin 2y x on the same axes, for

0 360x . [3]

(b) Hence, find the number of solutions for the following equation [1]

sin2 cos 1x x .

7. (a) Given that 4 x 6, find the value of x for which

3 tan x + cot x = 5 cosec x. [5]

(b) Express 9sin3 7cos3 in the form sin 3R where R is a positive

number and α is acute. Hence, or otherwise, solve the equation

9sin3 7cos3 3 for 0° ≤ ≤ 180°. [6]

EXAM 4 ANSWERS

1. 2x + y + 2 = 0 -----

1 2 1

2x y -----

: y = – 2x – 2 -----

: 2y + 4x = xy -----

Sub into ,

2(– 2x – 2) + 4x = x(– 2x – 2) [M1]

– 4x – 4 + 4x = – 2x2 – 2x

2x2 + 2x – 4 = 0 [M1]

(2x + 4)(x – 1) = 0

x = – 2 or 1

y = 2 or – 4

(x, y) = (– 2, 2), (1, – 4) [A2]

2 2( 2 1) (2 4) 45 3 5AB [B1]

2(a) )3)(42(12532 223 xpxxxxx

Coeff of x2 : 3 = 2(3) +p(1) [M1]

p = – 3 [A1]

2(b) )3)(12(12532 23 xxxxx

2x3 + 3x2 – 5x + 12 – (2x + 1)(x + 3) = 0

(2x2 – 3x + 4)(x + 3) – (2x + 1)(x + 3) = 0

(x + 3)[2x2 – 3x + 4 – 2x – 1] = 0 [M1]

(x + 3)(2x2 – 5x + 3) = 0

(x + 3)(x – 1)(2x – 3) = 0 [M1]

x = – 3, 1, 1.5 [A3]

3(a) 2 3 3 8 0x kx k

x2 – 3kx + (3k+ 8) = 0

if equation has equal real roots, b2 – 4ac = 0 [M1]

(–3k)2 – 4(1)(3k + 8) = 0

9k2 – 12k – 32 = 0 [M1]

(3k – 8)(3k + 4) = 0

4 8 or

3 3k k [A2]

3(b) if complex roots, b2 – 4ac < 0 [M1]

4 8

3 3x [A1]

4(a) 2 1 12 7(2 ) 36x x

22x. 21 = 7(2x. 2) + 36 [M1]

Let y = 2x.

2y2 – 14y – 36 = 0

(2y + 4)(y – 9) = 0 [M1]

y = – 2 or 9

2x = – 2 or 2x = 9 [M1]

(rejected) x = lg 9 / lg 2

x = 3.17 (2 dec pl) [A1]

4(b) 4 5(log 25)(log ) 3p

log 25 log

3log 4 log5

p [M1]

2log5 log

32log 2 log5

p

log p = 3 log 2 [M1]

p = 23 = 8 [A1]

4(c) 1

27 73 2

= 1 2 3

3 3 72 3 2 3

[M1]

= 2 3

3 3 74 3

[M1]

= 9 2 3 [A1]

5(a)

3 2

2

2 3 9 2

4

x x x

x

= 2

102 3

4

xx

x

= 10

2 3( 2)( 2)

xx

x x

[M1]

Let 10

( 2)( 2) 2 2

x A B

x x x x

[M1]

x – 10 = A(x – 2) + B(x + 2)

when x = 2, – 8 = 4B

B = – 2 [A1]

when x = – 2, – 12 = – 4A

A = 3 [A1]

3 2

2

2 3 9 2

4

x x x

x

= 3 2

2 32 2

xx x

[A1]

5(b) 21 cos

cos cot1 cos

xec x x

x

RHS =

21 cos

sin sin

x

x x

[M1]

=

21 cos

sin

x

x

=

2

2

1 cos

sin

x

x

[M1]

=

2(1 cos )

1 cos 1 cos

x

x x

[M1]

= 1 cos

1 cos

x

x

= RHS [A1]

6(a)

6(b) From the graph, 2 solutions [B1]

7(a) 3 tan x + cot x = 5 cosec x

sin cos 5

3cos sin sin

x x

x x x [M1]

3sin2 x + cos2 x = 5cos x

3(1 – cos2 x) + cos2 x = 5cos x

2cos2 x + 5cos x – 3 = 0 [M1]

(2cos x – 1)(cos x + 3) = 0

cos x = 0.5 or cos x = – 3 [M2]

basic = 1.05 (rejected)

Soln in 1st and 4th quadrant,

x = 1.05, 5.23 [A1]

(rejected)

Correct Graph – M2

Correct Range and Axis – M1

7(b) 9sin3 7cos3 = sin 3R

2 29 7 11.402R [M1]

7

tan9

037.87 [M1]

9sin3 7cos3 = 011.402sin 3 37.87 [A1]

011.402sin 3 37.87 = 3

0sin(3 37.87 ) 0.263 [M1]

3 - 37.870 = 15.250, 164.750,

375.250, 524.750

735.250, 884.750

3 = 53.120, 202.620, 413.120, 562.620,

773.120, 922.620

= 17.70, 67.50, 137.70 (1 dec place) [A2]

1

FEB

EXAM 5

2016 SEC 4 ADDITIONAL MATHEMATICS CW & HW

1. (a) (i) Expand

9

14

x

up to the first 3 terms. [2]

(ii) Hence, given that 9

2 28 2 3 1 8 .......4

xx x hx kx

,

find the values of h and k. [3]

(b) Evaluate the coefficient of x7 in the binomial expansion of

14

2

2

1

xx . [3]

2

2. (a) If α and β are the roots of the equation 22 9 0x hx and β = 2α, calculate

the values of h. [3]

(b) Given that α and β are the roots of the equation 23 5 1 0x x , form another

equation whose roots are 2

1

and

2

1

. [4]

3

3. The line 7 34y x cuts the circle 2 2 2 4 20 0x y x y at two points, A and B.

Find

(a) the coordinates of A and B, [5]

(b) the equation of the perpendicular bisector of AB and show that it passes

through the centre of the circle. [6]

4

4. The diagram shows a prism such that each cross-section is a quadrant of a circle of

radius x cm, with angle at the centre equal to 90. The cross-sections are OAB and

PDC where A, B, C, D lie on the curved surface of the prism and the vertical line OP

is the intersection of the vertical plane faces OADP and OBCP. The cross-sections are

horizontal and y cm apart.

(a) Given that the volume of the prism is 20 cm3, express y in terms of x.

Hence show that the total surface area, A cm2, of the prism is given by

2 40 4

2

xA

x

. [4]

(b) Find the value of x for which A has a stationary value, [2]

(c) Find the stationary value of A, [1]

(d) Determine if the stationary value of A is a maximum or a minimum. [2]

(e) Given also that the total surface area, A, is increasing at a constant rate of

3 cm2 s-1, find the rate at which x is changing when x 4. [3]

y cm

O A

B

C

D P

x cm

5

6

5. (a) Solve the following equations for 0 x 360:

(i) 3cos 2sin2 0x x , [3]

(ii) 2sec 4 tan 6x x . [4]

(b) Prove the identity

sin 7 sin 5

tancos 7 cos5

x xx

x x

. [3]

7

6. Differentiate the following with respect to x :

(a) (3 2x3)10 [2]

(b) 122 xex [3]

(c)

4

32ln

x

x, leaving your answer to the simplest form. [4]

8

7. Variables x and y are related by the equation x

bxay , where a and b are

constants. The table below shows measured values of x and y.

x 1 2 3 4 5 6

y 5.7 5.6 5.9 6.2 6.6 6.9

(a) On a graph paper, plot xy against x, using a scale 2 cm to represent 1 unit

on the x axis and 1 cm to represent 1 unit on the xy axis. Draw a straight

line graph to represent the equation x

bxay . [3]

(b) Use your graph to estimate the value of a and of b. [3]

(c) On the same diagram, draw the line representing the equation x

xy

3 and

hence find the value for which a

bx

3. [2]

9

10

8. A particle moves along a straight line so that its displacement, s metres, from a fixed

point P is given by 3 214 15

3s t t t , where t is the time in seconds after passing P.

Find the

(a) initial velocity and acceleration of the particle, [4]

(b) minimum velocity, [2]

(c) range of values of t for which the velocity is negative. [2]

11

9. The diagram shows the curve 1

42

x

xy crosses the x-axis at P.

(a) Find the coordinates of P. [1]

(b) Find the equation of normal to the curve at P. [4]

P

R

S

y

x

Q

12

10. The solution to this question by accurate scale drawing will not be accepted.

The diagram shows a rectangle ABCD. The coordinates of A and D are A(4, 1) and

D(16, 5) and the equation of AC is y = x – 3.

Find

(a) the equation of CD, [2]

(b) the coordinates of C and B, [3]

(c) the length of AB, [1]

(d) the area of the rectangle ABCD. [2]

y

x A(4, 1)

D(16, 5)

C

B

O

13

EXAM 5 ANSWERS

1(a) 9

2 28 2 3 1 8 .......4

xx x hx kx

(i)

9 29 91 1

4 4 4

x x x

+ ……. [M2]

(ii) 2

2 9 98 2 3 1 ...

4 4

x xx x

=

22 29

8 18 18 2 3 ...2

xx x x x

=

2218 16 ...

2

xx [M2]

21

16 and 2

h k [A1]

1(b)

14

2

2

1

xx

General term or 1th

r term of the expansion

= 14

14 2 1

2

rr

rC xx

= 14 28 31

2

r

r

rC x

[M1]

For the term in 7x ,

7 28 3rx x

7 = 28 – 3r

r = 7 [M1]

Coeff of

7

7 14

7

1 1326

2 16x C

[A1]

2(a) 22 9 0x hx

2

h

9

2 [M1]

Given β = 2α

2α2 = 9

2 α =

3

2 [M1]

2 6 9h [A1]

14

2(b) 23 5 1 0x x

5

3

1

3 [M1]

Sum of new roots

=

2 2

22 2

1 1

=

2

2

2

[M1]

= 31

Product of new roots

=

2

19

[M1]

New equation is 2 31 9 0x x [A1]

3(a) 7 34y x ---

2 2 2 4 20 0x y x y ---

Sub y = 34 – 7x into eqn : [M1]

22 34 7 2 4 34 7 20 0x x x x

2 21156 476 49 2 136 28 20 0x x x x x

250 450 1000 0x x

2 9 20 0x x [M1]

5 4 0x x

x = 4 or 5

y = 6 or – 1 [A2]

A(4, 6) and B(5, – 1) [A1]

3(b) Gradient of line AB = 1 6

75 4

Gradient of line = 1

7 [M1]

Midpoint of AB = 4 5 6 1 9 5

, ,2 2 2 2

[M1]

Eqn of bisector of AB is

5 1 9

2 7 2y x

15

7 13y x [M1]

2 2 2 4 20 0x y x y ---

Centre of Circle is 2g = – 2, – g = 1

2f = – 4, – f = 2

[M2]

(1, 2)

Sub (1, 2) into eqn :

LHS = RHS [A1]

4(a) volume of prism = 20π cm3

base area of prism

= 21

4x [M1]

volume of prism = 21

4x y = 20π

2

80y

x [A1]

Curved surface area ABCD = 1

24

xy

= 2

1 802

4x

x

= 40

x

cm2

Surface Area of OADP and OBCP

= 2xy = 160

xcm2

Surface Area of quadrant OAB and PCD

=

2

2

xcm2

Total Surface Area of Prism, A

=

2

2

x+

40

x

+ 2xy

=

2 40 160

2

x

x x

A =

2 40( 4)

2

x

x

cm2 (shown)

16

4(b) For stationary value, 0dA

dx

2

40 4dAx

dx x

= 0 [M1]

2

40 4x

x

3

40 4x

x = 4.50 cm [A1]

4(c) Stationary value of A = 95.3 cm2 [A1]

4(d)

2

40 4dAx

dx x

2

2 3

80 4d A

dx x

[M1]

At A = 95.3 cm2, 2

2 3

80 4d A

dx x

>0

minimum point. [A1]

4(e) Given 2 13 cm

dAs

dt

dA dA dt

dx dt dx [M1]

when x = 4,

2

40 4dAx

dx x

= – 5.288

5.288

1.7623

dt

dx

[M1]

0.567dx

dt cm/s [A1]

17

5(a)(i) 3cos 2sin2 0x x

3cos 4sin cos 0x x x

cos (3 4sin ) 0x x [M1]

cos 0x or 3 4sin 0x

3

sin4

x

90 , 270x 228.6 , 311.4x

[A2]

5(a)(ii) 2sec 4 tan 6x x

21 tan 4 tan 6x x [M1]

2tan 4 tan 5 0x x

tan 5 tan 1 0x x [M1]

tan x = 5 or tan x = – 1

basic = 78.69 basic = 45

x = 135 ,78.7 ,315 ,258.7 [A2]

5(b) sin 7 sin 5

tancos 7 cos5

x xx

x x

LHS = sin 7 sin 5

cos 7 cos5

x x

x x

=

7 5 7 52cos sin

2 27 5 7 5

2cos cos2 2

x x x x

x x x x

[M2]

= sin

tancos

xx RHS

x [A1]

6(a) 10

33 2d

xdx

= 9

3 210 3 2 6x x [M1]

= 9

2 360 3 2x x [A1]

6(b) 2 2 1xdx e

dx

= 2 2 1 2 1 2x xd d

x e e xdx dx

[M1]

= 2 2 1 2 12 2x xx e e x [M1]

= 2 12 ( 1) xx x e [A1]

18

6(c) 2 3

In4

d x

dx x

= [ln 2 3 ln 4 ]d

x xdx

[M1]

= 2 1

2 3 4x x

[M2]

=

11

2 3 4x x

[A1]

7(a)

x

bxay

y x ax b

Y = mX + C

y x 5.70 7.92 10.22 12.4 14.76 16.90

x 1 2 3 4 5 6

y 5.7 5.6 5.9 6.2 6.6 6.9

[M1]

Correct Graph – B1

Correct Points – B1

7(b) a = gradient = 16.9 5.7

2.246 1

[B2]

b = y-intercept = 3.45 [B1]

19

7(c) Draw the line Y = 3X [B1]

a

bx

3

3x – ax = b

3x = ax + b

Points of intersection of the line Y = 3X and Y = aX + b

(4.6, 13.8) [A1]

8. 3 21

4 153

s t t t

(a) initial velocity =

0t

ds

dt

= 2 8 15t t [M1]

= 15 m/s [A1]

initial acceleration =

2

2

0t

d s

dt

= 2 8t [M1]

= – 8 m/s2 [A1]

20

8(b) Minimum velocity

2

20

d s

dt

2t – 8 = 0

t = 4 s [M1]

min velocity = 2 8 15t t = – 1 m/s [A1]

8(c) 2 8 15t t < 0

5 3 0t t [M1]

3 5t [A1]

9. 1

42

x

xy

(a) when y = 0, x = – 2 P(– 2, 0) [B1]

(b) Gradient at point P

= 2 4

1

d x

dx x

=

2

1 2 4 2 4 1

1

d dx x x x

dx dx

x

=

2

6

1x

[M1]

= 2

3 [M1]

Normal gradient = 3

2 [M1]

Eqn of normal at point P:

3 3

2 32 2

y x x [A1]

10(a) Gradient of AD = 5 1 1

16 4 3

Gradient of line to AD = – 3 [M1]

Eqn of CD,

5 3( 16)y x

3 53y x [A1]

21

10(b) 3 53y x ---

y = x – 3 ---

x – 3 = – 3x + 53

4x = 56

x = 14

y = 11

C(14, 11) [A1]

Midpoint of AC and BD is the same.

16 5 4 14 1 11

, ,2 2 2 2

x y

[M1]

x = 2, y = 7

B(2, 7) [A1]

10(c) Length of AB

= 2 2

4 2 1 7 2 10 6.32 units

[B1]

10(d) Area of ABCD

= 4 2 14 16 41

1 7 11 5 12 [M1]

= 80

= 80 units2 [A1]