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Name:

SECOI{DARY THRSE EXPRESS

Classes: Sec 303, 304, 305, 306

Additional materials: Writing paper

CHIJKATONG CONYENT

1VtrD.YEAR EXAMINATION 2013

ADDITIONAL MATHEMATICS

Class RegistrationNumber

4047

Duration: 2 hours

27

CHIJ Katono Convent Mid-Year Exam 201 3 Add Math Sec 3E

Mathematical Formulae

1. ALGEBRAQuadratic Equation

For the equation ax2 + bx + c : 0

-b4u' -*t"2a

Binomial expansion

(a+bl = s,.(;),^,r-(;),,-,a,- .[:),"-,b, +...+b,,

where ra is a positive integer and tl = A#,1r = "q#dREAD THESE INSTRUCTIONS FIRST

Write your name, class and registration number in tle spaces provided at the top of this page and onall the work you hand in. Write in dark blue or black pen. You may use a soft pencil for anydiagrams or graphs. Do not use staples, paper clips, highlighters, glue or corection fluid.

Answer all questions.

Write your answers on the separate writing paper provided.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place ia the case ofangles in degrees, unless a different level of accuracy is specified in the question. The use of a

scientifrc calculator is expected, where appropriate. You are rerninded of the need for clearpresentation in your answers.

At the end ofthe examination, fasten all your work securely together.

The number ofmarks is given in brackets [ ] at the end of each question or part question. The totalnumber of marks for this paper is 80.

2. TRIGONOMETRY

Identtties

Formulaefor AABC

sin2A+cos2A=1

sec2A = l+tan2A

cosec2A=1+cot2Asin(n t e) = sinAcosB I cosAsin8

cos(R t B) = cosAcosB msinAsinB

tan(nte)=3!418!ql mtanAtanB

sin2A = 2sinAcosA

cos2A = cos2A - sinzA =2cos2A-'l=1-2sin2A

.r,nza= 2tan41-tan2A

sinA + sinB = zsinl(a+ a)cos.1(,A - e)2" 2'sinA - sinB = zcos](,n * a)sinltn- a)

cosA + cos8 = zcosl(A * a)co=1(,e - a)2" 2'cosA - cosB = -zsin1(.a + A)sinl(.a - a)2" 2'

abcsinA sinB sinCa2 = b2 + c2 -2bc cosA

A =1ab sinC2

2

question paper consists of 4 printed pages (including this cover page).

[Tum over

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CHIJ Katono Convent Mid-Year Exam 20'13 Add Math Sec 3E

1. The simultaneous equatiors mx+ny =|andm2x-n2y = -10 have a solution, = 2

andy=3. Findthepossiblevaluesoftheconstantszandn. t5]

2. Solve 3,[4-3a =y122. t3]

( a\ t3. A rectanglehas length l, -i]metres and breaa*t (s +./E)*"t es. Find\ ''tt )

(a) the perimeter ofthe rectangle, t3](b) the area of the rectanglg expressing your answers in the form a + bJi , wherc

a and b are integers. t4l

4. Using a suitable substitution or otherwise, find the values ofx such that

5x+t *52-x _126. t4l

5. Solutions to this question by accurate drawing will not be accepted.

R(4,'1)

P(1c

QQ,1)

or

The diagram above, which is not drawn to scale, shows a triangle PQR in which P is&e point (1, 3), Q is (7, 1) and R is (4, 7). The point S which lies on PQ is the foot ofthe perpendicular line from R.(a) Find the equation ofthe line PQ and ofthe line RS. i61(b) Find the coordinates ofS. tzl(c) A point T is located such that the quadrilateral PTQR is akite. Determine the

coordinates ofT. 121

29

8.

CHIJ Katong Convent Mid-Year Exam 2013 Add Math Sec 3E

Find the range of values of .r which satisfy

(a) (x+2)z > 2x+7 ,

(b) x2-8<5r+6<8

9. Show that the equation xz + la = 5 - 2k has real and distinct roots for all real values

ofk. 141

10.

11. Acnrveisdefinedas y=3x2 -6x+5.(a) Express y=3v2 -6x+5 inthe form y = o(r-h)z +k,wherea,hatdkarc

constants. t21

(b) Sketch the gr aph of y = 3262 - 6x + 5 . t21

(c) Thecurveisalwaysabovetheline y = loc+2'Fndtherangeofvalues oft' t3l

12. Solution to this question by accurate drawing will not be accepted.

The diagram, which is not drawn to scale, shows a AIDC in which the pointl (8, 8)and the point B lies on they-axis. The point C lies on the perpendicular bisector ofl.Band the equation of the line 8C is -y = 8, - 4. Find

(a) the coordinates ofB,(b) the equation ofthe perpendicular bisector oflB,(c) thecoordinates of C.

The point D is such thatlCBD is a rhombus. Find(d) the coordinates ofD,(e) the area of rhombus ICBD.

-------,----END OF PAPER------------

The roots ofthe quadratic equation 9x2 -76x + 4 = 0 are az and B'? . Find

(a) a quadratic equation whose r*t" "." 4 *a

f .

(b) two distinct quadratic equations whose roots are a and B if ap < 0 .

t3l

t4l

t6l

tsI

lilt5ll2l

tzlt3l

R)

6. Find the value ofk given that^k+1.kJ .O-aa- =96 L4l

x

7. At rhe beginning of 1981, the population, P, of a certain species of insects after a

period of r years is given by p = 3gggg"-0'025n -

(r) Find the population ofthe insects(i) atthebeginning, tll(ii) atthebeginningoflgSg,correctto3significantfigures. 12)

(b) Skelch ilre graph ofP against n. l2l

3

[Turn over

B

4

r

II

I

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3E AM MYE 2013 Solutions

mX+nY:8 (t)

m2 x - n'y = -10 ---------- (2)

Sub x : 2, y :3 into (l) and (2)

2m+3n =8 (3)

2m2 -3n' = -10---------- (4)M1

,(r*)' -3n2:-to

(e+ - +e" +9n')- 6n2 : -20

From (3),

Sub (s) into (4)

3n' -48n+84 = 0

n'-l6n+28=0

(n -ru\n-z)= o

n =14 or

m=-17

.'.n=14, m=-17

8-3nm =_ ____________ (5)

2

M1

M1

M1

n=2

m=1

or n=2,n'l=l A1

2

3-t[4-i: x+22

9@ -3x) = (x +72)2

36-27x:12 +44x+484 M1

30

x2 +7lx+ 448 = 0

)

)

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(x+7)(x+64)=0 M1

.'. x = -7 or x = -64(rejected, because doesn't satisfy original eqn) A1

3

(a) ,l(,-#).F..^)]

=r[t-f..n]

J'1"o):16

.'.Perimeter of the rectangle is 16 m.

(b) #)b .6)

= 15 +:.r/B- -

rsJi *zJaJ, -20-BJ,la

\]L

tJi-s J,=-Y- J' J'

=7 -4J,'.Area of the rectangle t (z - +Ji) ^'

sJ, - q +z(z)

J'

M1

M1

A1

(,

20 qJs:-:Jz Jz

ML

M1

M1

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4.

5'*l + s2-* =126

5x.5+L=D65x

Let u :5x

5u2 -126u +25 = O

(s" -t\" - 25)= s

(sr-t)=s

5.

(a)

M1

M1

M1

or (" -zs)= o

Iu=-5

5r = 5-l

x=-l

Sub x=1, y =3intotheequation.

^l<---L-J

10

3

.'.Eqn of line PQ: y = -1, * I33

21 IGradient PO= - '---

t-7 3

Let equation of the line bey = -1x+cJaJ

u =25

5x -')5-

x=2 A1

GradientRs: -r*(-i)=,

M1

M1

A1

M1

M1

A1

Let equation of the line be y = 3x + d

Sub x =4, /:Tintotheequation.

7 =3(4)+ du --J.'. Eqn of line RS: y = 3x - 5

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(b)

6.

110v - --x+-,-^

JJ

!:3x-5

Sub (l) into (2)

rl0-_x+--3x-5a-J7

*:2!, v =2!22

(l)

(2)

(c) Midpoint of.RZ =(r)r))

Let coordinate of f be (p,q)

(ry,ry)=?+,,:)

." r(t,-z)

-k+3 .kJ ,'o =96gk+l

'(':,':)

2k .3:96

2k =32

k=5

M1

A1

M1

A1

lk .2..1 .6k _ ,ugn .g

tgk .l _ru9K

M1

M1

M1

A1

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7.

(aXD When n:0, P = 30000

(ii) When n: 8, P = 30000a-o'o2s(8)

=24600

P

(b)

8

(a)

(x +2)2 > 2x +7

x2 +4x+4-2x-7 >O

x2 +2x-3>0

(x+3)(x-l)> 0

.'.x<-3 or x>l

(b)

x'-8 <5x+6

u

AL - intercept

41 - shape

No axes : deduct 1 mark

M1

A1

M1

A1

n0

M1

M1

A1

and 5x+6<8

*2 -5*-14<o

(x -t\x+ z)< o

-2<x<7

2.;s(-5

M1

M1

'. -2. * <?5

A1

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9.

x'+ls+2k-5=0

D: k2 -4(2k-5)

: k2 -Bk+20

: (k-4)2 -16+20

: (k-4)2 +4

New equation:

Product of roots: o'B' =19

New equation

I I B2+a2SUmOt'rOOtSi ^ +-=-d, B, O, B,

76499

=L =194

lllProduct ofroots: - X - =-----;------a' p' a'B'

M1

M1

A1

9

4

For all real values of k,(k - 4)' > O, (k - 4)2 + 4 > 0 .

Since D>0 for all real values of k, the equation has real and distinct roots.

A1

10

(a) Sum ofroot", o2 * B' =+ B1

B1

M1

AL

A1

A1.16 9y.__y*_=0

44

4x2 -76x *9 = 0

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(b) New equation

o'B' =t

oB ='f,

ap:?rgei))aB=-a

t^J

M1

A1or

(o*B)'=a'+82+2aB

=19*r( -?)e [ 3)

=649

88.'.d+B-- or a+lJ:--tataJJ

.'.New equation is x2 -lr-1= o or

3x2 -Bx -2 = 0

(a) !=3x2 -6x+5

=3(x2 -r.*]l

M1

A1

.82x'+-x---035

3x2 +8x -2 = O

M1

A1

A1

A1 No axes : deduct l mark

A1

1.L.

=r[G -t)2 -,'.;]

: 3(, - t)2 +z

(b) Min point (1,2)

When r : 0, y: 5. .'.y-intercept: (0, 5)

Shape: smile

)

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(c) 3x2 -6x+5>lac+2

3x2 -6x-ta+3 > 0

D<0

e6-D2-4(3X3)<o

k2 +l2k <0

k(k +12) <0

.'.-12<k<0

tZ. (a) Sub x : 0 into | = 8x - 4

(b)

v:-4.'. r(0,-+)

Gradient of AB = s-(-+)8-0

J

M1

M1

A1

2

B1

M1

M1

M1

Gradient of,perpendicular bisect o, = -?3

8+0 8-4Midpoint of AB =

22

= (+,2)

Let equation ofperpendicularbisector be y = -l* * r.3

Sub x = 4, ! : 2intothe equation.

)2=-1(4\+c

a"J

t4L --

J

.'.Equation of perpendicular bisect ot y = -?,.+JJ

M1

A1

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3B

(c) !:8x-4 ------------- (l)

214t=-=x+ _ ------(2)-33

Sub (1) into (2)

Bx-4=-2 **14aaJJ

x=1, !=4

.'. c(t,+)

(d) Midpoint of AB: Midpoint of CD

Let coordinates of D b" (p,q).

M1

(+,2)=l+ p 4+q

22

A1

M1

A1

(e)

:.p-7,q:o

.'- o(t,o)

Area of rho mbus ACBD :!2

ls l o 781

lr 4 -4 o 8l

= )lt r- 4 + o + s6)- (a + o - zs + o)]

M1

M1

A1

= l(toa)2'

.):52 unrt-Danya

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