Mathematics 1

37U4O5t42OCentre Number

Candidate Number

Surname

UNIVERSITY OF LONDONSCHOOL EXAMINATIONS BOARD

General Certificate of Education Examination

JUNE 1986 ADVANCED LEVEL

Mathematics L

One and a quarter hours

\-

INSTRUCTIONS TO CANDIDATES

USE AN HB PENCIL THROUGHOUT THE TEST

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SOBefore the test begins:l. Insert the information required in the spaces above.

2. Check that your answer sheet, which will be handed to you separately, is headed'Advanced level - 37ll4OSl420 Mathematics Paper l.'Take care that you do not crease or fold the answer sheet or make any marks onit other than those asked for in these instructions.

3. Insert the information required in the spaces provided on the answer sheet. When you have written yourCentre Number and Candidate Number in the boxes provided draw neat horizontal Iines with your HBpencil to join the dots under the appropriate numbers in the grids below the boxes. (You can see how to dothis in the sample shown below.)

The Test Number should then be inserted.If your Subject Code Number is 371 (Mathematics Syllabus B) your test number is 308.If your Subject Code Number is 405 (Pure Mathematics) your test number is 315.If your Subject Code Number is 420 (Pure Mathematics with Statistics) your test number is 316.

MAKE SURE THAT YOU HAVE MARKED THE RIGHT NUMBERS.

How to answer the test:

4. For each question there are five suggested answers, A, B, C, D and E. When you have selected your answerto the question, find the row on the answer sheet with the number of that question and draw a horizontalline to join the dots under the letter corresponding to the answer you have chosen.

For example, the answer C would be marked as shown ABCDEt 9t tHf tl t

5. Mark only one answer for each question. If you change your mind about an answer, rub out the first markcarefully, then mark your new answer.

6. There are 30 questions in this test and you are advised to answer all of them. You will score one mark foreach correct answer: no marks will be deducted for incorrect answers or omissions.

7. Do any necessary calculations and rough work in this booklet.

You must not take this booklet out of the examination room. All question booklets and answer sheets wiII becollecteil at the end of the test.

CENTRENUMB ER

CANDIOATENUMBER

00000Hl {t-tl t0 toooooI tf-tt tl 'tl t

1'ttltI tt-{t af-,ll t

1r111t-tl tt-.ll tl-t22222t it tl tt t{-t 22222f at tl ,lt-tt 222

lrfttt

Questions 1-20

SECTION I

(fwenty questions)

Directions. Each of these questions is followed by five suggested answers. Select the correct answer in each case and mark

the answer sheet aPProPriatelY.

1. Which one of the following is a factor of

x3-3x2*2x-6'!

A x-3

B x-2

C x-4

D x*3

E x*2

Given that | : c"* 1, then

0t-dx-A2x

B 2xe"+r

C 2x e2*

D (x2 + l)e"

E 2xe"

x*P 2 |G=W4A: i=1- i +2'

wherePisaconstant.

A P: 5

B P_ 3

C P- 2

D P:-3

E P:-5

4.(r-i)':A -2+2iB -4-2ic 4-2i

D 4+2i

E -2-2i

5. An equation of the straight line which is

perpendicular to 3x - 4y :5 and passes throughthe point (-2, 1) is

A 4x *.3Y: - l0

B 3x-4Y: 10

C 4x+3Y: -5D -3x + 4Y: l0

E 4x-3Y: 10

The equation of the given curve could be

A y- x(x-2)

B Y:-x(x-2)C y- x2(x+2)

D y:-x2(x-2)

E x21x -21

:J

V

| 7. logst:2.\./

v:A

B

C

D

E

Given that x2 + y2 : 4, then

4!-dx-

A 2x t2y

B J@-*',)

c-4v

10. The set {z : I z | ( I } can be represented in the Arganddiagram by the shaded region

I2x

Jxx2

2x

2x

B

!

Cvx

4-xv

o9.

D

U PQRSis a parallelogram.

---, ---+

PR + SQ:

---+A 2R^S

--+B 2SR

---+

C 2PS

---+

D 2SP

.---,

E 2PR

vTSE 8512797 Turn over

11. The complete set of real values of 0 for.o.6 : I is

A {?tn + rl6:neZ\ v {2nn - nl6:neZ}

B {nn + nl3:neZ}

C {2nn + nl3:neZ\

D {2nn * nl3:neZ} v {2nn - nl3:neZ}

E {nn + nl6:neZ} v {nn - fl6:neZ}

The complete solution set of

*r\'ois

A {x:x > 2}

B {x:-2<x<0}C {.r:x>2}v{x:-2<x<0}D {x:x< -2}u{x:x>2}E none ofthe above

13. The graph ofy: lx * 2l could be

v

A

12.

Y

C

D

-/

4

v/

17.14. A real root of the equation 2x3 + x2 - I : 0lies inthe interval

A (-2, -l)B (-1,0)

c (0,1)

D (1,2)

E (2,3)

The region shaded in the diagram is rotated through2n abottt the x-axis. The volume of the solid ofrevolution formed iS

A21B 72n

CTE

D 2nln2

E2n

16. The sum of the first six terms of the geometric seriesr+tl+2|+...isA 4i:3

Bsac t3*

D2qdE none ofthe above

The general solution of the differential equation

;*.I : 0, where x ) o, is, K being a constant,

A !:KxB l: Ke'

--2c t:i+xD y2:x2+K

E v:{

\r'

15.

\"

lrr18. I

Jo

A

B

C

D

E

xsinxd.r:

7E

1

0

-l_fE

The number of committees of 4 people that can beselected from 3 women and 3 men is

A6

B9

ct2D15

E81

The parametric equations of a curve are

x:secO*1, y:tan?-1.Points (x, y) of the curve satisfy

A x2 + y2 -2x *2y + l:0B x'- y'-2x -2y - l:0C x'- y'+2x +2y * l:0D x'- y'-2x -2y + l:AE x2 + y2 -2x -2y * l:0

19.

20.

\/

TSE 8512797 Turn over

Questions 2l-30

SECTION II(Ten questions)

Directions. For each of the following questions, ONE or MORE of the responses are correct. Decide which of theresponses is (are) correct. Then choose

Directions Summarized

A 123B t2C 23D 1

E 3

MARK ONE SPACE ONLY ON YOUR ANSWER SHEET FOR EACH QUESTION

A if 1,2 and 3 are correct

B ifonly I and2 are correct

C if only 2 and 3 are correct

D if only 1 is correct

E if only 3 is correct

The equation of a curve is y : -] ^.' x-zI y :2 is an asymptote

2 x:2 is an asymptote

3 The curve crosses the x-axis where x :0

I PQ:2J10 units

2 The centre of the circle which passes throughO,P and Q is (3, l)

3 The area of LPOQ is 6 square units

23. f : xt-* y2 * 2, xe R,

g: xt--------+2x - 1, xe R.

I The domain of fg is R

2 gf :xt--------+2x2 * 3, xeR

3 fg:gf

24. The equation of a circle is

x2+y2-8x-6y+16:0.

I The centre of the circle is the point (4, 3)

2 The circle touches the x-axis

3 The circle touches the y-axis

21.

P is the point (0, 2) and. Q is the point (6, 0).

25.

PQ is a chord of a circle with centre O and radius r.LPOO : 0 radians, where 0 < z.

1 Minor arc PQ: 7Q

2 Chord PQ :2r sin?

3 Area of minor sector POQ : 2r2 0

/8. z:-2-2i.

I lzl:zJ22 lz*l:zJ23 tan(argz): I

29. When expanded as a series of ascending powers of x,

l-3xffi: ao * a1x + a2x2 + ... .

I ar: 6

2 ar:1g

3 The expansion is valid only if t, I . I

30. The curve J, : .13 - 3x2 + 7

t has a stationary point at (0, 7)

2 has a point ofinflexion at (1, 5)

3 never has a gradient less than -3

STOPNow go back end chmk your work.

v

26. sin0 + 2cos0: rcos(0 - a),0<acnlZ.

1I Slnd:-

v)

2 cos* -- lv5

3 r:J5

where r>0 and

27 . The position vectors of the points P , Q wrlh respectto the origin O are (i - 3j), (2i + 5j) respectively.

I Fa; : urr3

2 LPOQ is acute

-)?3 OM': ii + i, where M is the mid-point of PQ

Directions Summarized

A I 2 3

B 12C 23D I

E 3

vTSE 851279't

UNIVERSITY OF LONDON

GENERAL CERTIFICATE OF EDUCATION

ANSI.IERS TO MULTIPLE CI-OICE QUESTIONS

June 1986 A/L 371/405/420 Mathematics Papr 1

1A2B3A4E5C6E7C8C

11 D

L2C

13014C

15E

16D

t7E18A

2lc22A

23B

24825D

26E

27E

2BA

29C

304190

2089B

10D

Strnq 1996 Mu-tNPlq CkoicQ fta-t{^c<'^aLhcs

l. f-(3) . o ><-7 rr a Fcr-chrs @

2.. t = ga4z*l = e*t. e 4 > e.zr<e*' z z><ex.+tcL*

3. z-(x-+r-) -L><*t) - ?L+P -) P = 4+t = .f @

+ (t -2i +d') ( \-t) -z; Cr-t) ' -zi +zJL = -zd-z @

3 45 = 3><- 5 6rad;o^* , =/+ o4;raho^ oi linc. ertro^d,'c^,Ac^r

hD 0^ts l,tas OcaF{^'.-.} -+4 .zq,*o.:Kc>- .'s *f ftx'r^ 1t'llrx+C(-r.,) rrelslr fib (= -Or=*-, +C |= 1$ +<

7

6 .^rAeucaa f(lc) >o noh C

F(*) r.irq'EcrlQl a-3 ",< ir.tr€QS{S n,rb I or D

*t$ tecr+:c-r A |a*a.sr*trr'< ) aaat 6 (c*r.cc) 39q-,,t'.-t-sc-tacs(3 n& 4.,lxc.cfrqh'q . il*<sV Lo- g.

t. .2aL+3L '- I =) Lt<+ \H: o *+ $?r^ L o

.t-);e.satb zfti. 7 2. sA @

-). lquX, L t =3 O (atesa..cho'^ .,.! (€s)

9. A_{\Et.p==---_ !

O 1z\ < \ ie- x-L+..' . I (z--r't+ig)

rr. as9' La O : {- -! + Aia

t3. c.rrta-\r-(1 neb g. -[e-rr L : -1- I = | -z +zl -- o fir.4sk ,"" @

14. f (9 t FCt) -' z+t - 1 '- 2- s.g,^ c.lac.age har^^xr^ o, I g

rf, .,.se $ ir yz*rt \t * (11" a* , *ir !'r.t-. a* =F+, l.J, \r.-, . L:;.J,

,6. G?, 6'r.k Lot,q I ao+aco^ ra.hD Z/- sun^ J ftrrl. sir, l.<r^^q ,,s

l(l-(rr,)s) ;e-r r( esr\ q6s_

r_yl-- T+ ' ZT' A'12-

la' cciEic-9- rp-Qr,r-gs o *r\cl L cr,r*cr- - ? m--sk bz c or EC ucarkg 1 e

r'?. L4- 3 -lb alc

=) t^3 = -\n x + C l^y lnr<-+ lnkt,^3 = (rr 2& I = ?n @

7tl? \ ?4s\"1 >LA* uSq Oo-rtS ( r^c&z . tlv-(v&^')ulJ5.t+JS.

'--*- c.-os >< - i u)s aL Arc .f* @s ,L - Sia. -ll

=+IT@

cl4i:

C-)

ti'rf"U=tsC

zo ,.t+jt .. trrsecs)' + (C*o-,)t: t+ LSecO+'Secrg + ba,,^LE -ZFo^-O+t

rrot^) [ +- bo^rg = g-e.clO

te zr-L+=a : ( f ?-saco *Sc.tO - Zt-a,r€ p S.Q.cz O

= z-sc.czE +- 2-S4cO+t -Lha,\8

*t-5t = (+2-se.cB+Sece$ ( b,-,tt6+t -zta- d): |a 2-5eco + 2hcnB= 2zc *-3 +t

Zt l. a_s ?c-) aa I =n

c'€- 3-> I .'. 3"t r'g aLro'rfzanlcaQ

". sy rt^pkshe I .'s .^c-<s.srcc!

>Lrz r-S c1n aSg.*fhl'e, U-dX aJ +r-4 2 U € oo L l-r*€

.;ho-n >L'-O L(=C: S.> 3 t's brqQ @,)

-v zz Po = fir;T- -- fr; = J+-; . z-,f,o r ,/g. oJec-- aF ba'an6te POGz = i-6 - L = (

"t*a '-,a-iltr Zt'/

z iF ce^-bre c.S e.'r.la is (3,9 IVS rcr-clt'r.r.s is 5o6u (r<--f)t * (j -t)t = ,o il trta- au*qhcvl

x-'- o, \ , L 5-.'bs so .locs (G, o\ ' /

L? r. [ft & 3 -) g R Ao F [r* S] 1/

t. 5t t hrsb *L+L v|.a,, Z(rt1+z) -l L*t+33. F3 = F(z*-,) I (.-r-r'-9L ts z + 3F Jv

L+ cot4a{e- =.t* J" + 2gx- t z FX + c 2t) (ca^I< (1,-F) )

,ti6ftrc^h*viag 2zt-+r*?*-s-"yn =o' 1{["s'e) z $''2']L

rr - 9-L*- ushqn za = zl- 43- . O t6 * Uz_ ?Z _ 6: +t6 =6-d* z)-6 crr,4

B= o salrtsltes \a^-a'g so 2- t's Er4e 5: - 65 ' o

7- a -L"t

-

25-a

utrLo^ q a a.)

dtas(S gcocl'r"4alr

a4L+9atd€s ndr

atues

.'s ,1f'rui(e u)r-r,^

- E*c - l8+t ( = O

-ra-Fts$ $La'5

a61, hPtcjr 3 4|a't

9:3 / ltd\^)

}/-z - 8z< +- I

6raA.:a* /t6)

*g)Ltt*

'-o ?a-: o

3x

@ | is krre'q

G.fPa. of gecbcxf

o?Q. i s !'s asc€-(e s Z F si,\O (Zx)'\z

sc\ ol)9d.,. & +

betn ol = L1

r'- fs

-)Po=

( (9-:

-.)-of

I -2.r- 2i I

aJS z- =

51'^ a/ : J-,lg'

<AT D? !4, zvS'

J;ft r'a.eorr4clr

71. tx

3 o.T.t1

i*8-; I t'o) ..

AH

c\e-o.tt1 .^,g(c is a.J\r +cL\k€ L

= oP +LPQ : ^:'-=i

*- 1c f raJ

r'rarrrf i'{t l,

\( cr8j)3 Frr,te.

23.

l2-'( l

4ta

3DU2d>L

pr"".,, tc'\

a4 -- 3rt- - 6r-d*

= 5 ' = z.fz_ 7_y'

-r35" Eon. -t3S" = |

-z- -- -z -z i

q/

L9. 1r -a*;6t+l*;-l

'( a. 1 : t8

3"r-t - 6-t

PrJ\* t^rAo^ Y 46F ,-".Rc,.io-. olc- {a 10 ;L

.}.!:l 3= S 2-y' d-rt'

r^:!e.n r4-aO 4 --o to t&.ra iS a- i?a*io.l.c2E.!u )-..,- lt z o 3: -1

6*-6 :o =) *=l

(arruir{ 1;n Valha crF taUS oc.q.JrJ -,)fora >L 1t

cLaL

;k-,@

lr-zr.O = )-?Q^L

r

( 3 .'s usnrna )

:) Pe :

LcosD; C c{rs(O-".) = C[*tgc.:so( +-si^$(

| = r SC.r y' (g (.e.4r,ctrt^3 a.r-qgt6^'e^"tr J r.^^ 1}\2 = f c.ssol (.) ( tt .o_r O)

;ai'iYe LiSl-eJ\ ?t : o

( (entrl^ ar.S ",.rc i5 r O )

P O,g '- L_rL D

(2] )

@

I z+g'

zL r aa