This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **ﬁiﬁﬁéﬂﬁi HONG KONG EXAMINATIONS AUTHORITY ~J‘LAE5’Fﬁiﬁ'Iiﬂﬁf’9‘
HONG KONG CERTIFICATE OF EDUCATION EXAMINATION I983 III in £5! E ADDITIONAL MATHEMATICS
E‘II =3 -- PAPER | :4~Mf'7‘£ Two hours
J;-’FAB§E~I<5}E-I—B#E +5? 8.30 a.m.—10.30 a.m.
*‘ﬁiﬁﬁ‘ﬁmﬁ RIF-2% This paper must be answered in English Answer ALL questions in Sectian A and any THREE
questions from Section B. All working must be cleariy shown. SECTION A (40 marks)
Answer ALL questions in this section. 1. Determine the range of values of A for which the equation
x"+4x+2+ A(2x+ I) = 0 has no real roots.
(5 marks) 2. Given that a, b, c are in arithmetic progression and the pusitive numbers x, y, z are in
geometric progression; prnve that (b—€)logx + (c-Iz)Iogy +(I1—b)Iogz ‘—‘ 0.
(6 marks) 83-CE»ADD MATHS MEI—1 3. Figure 1 show: an isosceles triangle ABC with A
BC = 21 and AB = AC. The perimeter of
the triangle is 2 metres. The triangle is revolved
about BC so as to form a solid consisting of
two cones with a common base of radius AD.
Express the volume of this solid in terms of
x. Hence ﬁnd the value of x for which this volume is a maximum,
I ( 6 marks) x D x
Figure l 4. Expand (l + ax)‘(l .. 4x)j in ascending powers of I up to and including the term containing x1. Given that the coefﬁcient of x is zero, evaluate the coefﬁcient of x’.
(7 marks) 5. Solve the inequality lx(x - 2)] < l . (8 marks) 6. The complex number 2 satisﬁes the condition
lz — (3 + i)l = lz ~(S + Sill.
lf 2 = x + iy, where x and y are real, ﬁnd and simplify the relation between x and y. Find also the values of x and y for which lzl is a minimum. (8 marks)
SECTION B (60 marks)
Answer any THREE questions from this section,
Each question carries 20 marks.
7. (a) Using De Moivre‘s theorem, or otherwise, show that
(i) cos40 = cos”) — 6cos’0 sin’9 + sin‘9
(ii) sin49 = 4cos’ll sinB — 4cos€ sinJH .
(5 made)
(b) Using (a), or otherwise, show that
”"49 = 4tan0 — 4tan30 .
I ~ 6tan’0 + tan‘G
(3 marks) (C) By putting x = tan!) and using the result of (b), show that the equation x‘+4x’—6x’—4x+l=0 ..................... (t)
can be transformed to
tan40 = l. ....................................... (u) Find the general solution of equation (it) in terms of 1r. Hence deduce the four roots of (t) leaving your answers in terms of 1r .
(12 marks) 33-CE-ADD MATHS ”El—2 8. Let
(a)
(b) (C) (b) rm = x’ +ax’ + bx — 72.
Given that x = 4 is a double root of :_f = 0I ﬁnd the values of a and b. (5 marks)
x I Show that f(x) can be expressed in the form (x + p)3 + q, and ﬁnd )2 and (1.
Hence ﬁnd the three roots of f(x) = 0 .
(9 marks) Represent the three roots XI, x1, x; of {(x) = 0 on an Argand diagram by the points A,
H, C, respectively, x, and Jr2 being complex conjugates and O < arg (x,) < n . By considering triangle ABC, or otherwise, determine arg (x2 — 4) . x, - 4
(6 marks)
Prove, by mathematical induction, that for all positive integers n ,
l><2+2X3 +3x4 +...+n(n +1) = 1311(n+l)(n+2).
(6 marks) On a battle ﬁeld, cannon-balls are stacked as shown in Figure 2. For a stack with n layers,
the balls in the bottom layer are arranged as shown in Figure 3 with n balls on each side.
For the second bottom layer, the arrangement is similar but each side consists of (n - 1) balls;
for the third bottom layer, each side has (n — 2) balls, and so on. The top layer consists of
only one ball. ‘ Figure 3 (i) Find the number of balls in the r-th layer counting from the top. (ii) Using the result of (a), or otherwise, ﬁnd the total number of cannonvballs in a stack
consisting of n layers. (iii) If the time required to deliver and ﬁre a cannon-ball taken from the nth layer is
2
7
layer. minutes, ﬁnd the time required to deliver and ﬁre all the cannon-balls in the r-th Hence ﬁnd the total time needed to use up all the cannon-balls in a stack of 10 layers.
(14 marks) Bﬂ-CE—ADD MATHS llEl—J ‘~— g... 10. In Figure 4, PQR is an isosceles triangle with P i is ﬁ' E m
we QR = 2" N ‘5 the mld‘mm of QR' HONG KONG EXAMINATIONS AUTHORITY L and M are variable points on PQ and PR ,
respectively, such that LM I QR . Let _‘ ‘
#J‘LA:$§7€:I’P$Q%
HONG KONG CERTIFICATE OF EDUCATION EXAMINATION I983 LM = x.
[if-t 1m 51 2% ADDITIONAL MATHEMATICS
1 gig: PAPER II 2'
1
l
1
3 (a) Find x such that the area of ALMN
is a maximum. i
1
s
.
. (8 marks) i (b) If the ﬁgure is revolved about PN, ﬁnd 1
. I so that the volume of the cone Q N R l :4‘5‘1’24.
generated by ALMN is a maximum. In Two hours
(6 mks) ﬂsL‘L‘E L’Fi‘mﬂﬁ't‘i’i}EET’r-~Ew’%-I—IL‘/} 11.15 a.m.—l.15 p.m.
*aemamauteas This paper must be answered in English (c) Show that the volume of the cone generated by revolving the ALMN speciﬁed in (a) about
PN is only -§—; of the volume generated in (b). (6 marks) I]. Figure 5 shows a rail POQ with
LPOQ = 120°. A rod AB of length
Wm is free to slide on the rail with its
end A on 01’ and end B on Ca. Let
0A = x metres and OH = y metres. Answer ALL questions in Section A and any THREE
questions from Section B. All working must be clearly shown. (a) (i) Find a relation between x
and y and hence ﬁnd the
value of y when x = 2. (11) Find dx . Figure 5 Given that x and y are functions of time t (in seconds), show that SECTION A (40 marks)
Answer ALL questions in this section. u:-2x+gsi_x_
d1 .r+2ydt' (10 marks)
1. A triangle has vertices P(k, ~l), Qt7, 1]) and R(1, 3). Given that the area of the triangle is (b) The end A is pushed towards 0 with a uniform speed of %m/s. When A is at a distance 20 Units. ﬁnd the two Values of k. of 2 metres from 0. ﬁnd the speed of the end B. (5 marks)
(4 marks)
2. Use the substitution :1 = x2 to ﬁnd the indeﬁnite integral
(6) Suppose the perpendicular distance from 0 to the rod is p metres. Show that
[x sin1(x1)dx .
= L! .1 _ (5 marks)
p 2 J;
Hence ﬁnd ﬁg when X = 2. 3. Use the substitution :1 = l + 3x’ to evaluate
(6 marks) 1
J x3 1 + 3x1 dx .
o (5 marks) END OF PAPER 93 CE ADD MATHS ”5,4 BS-CE-Add Maths IIIE)—1 Figure I shows the curve y = x1 - 4x.
A straight line L intersects the curve at
the points P0, '3) and Q(S, 5). Find (a) the equation of L . and (b) the area of the shaded regions Figure l (6 marks) ' — = 0 and
Find the eqUations of the two lines which are both parallel to the line 3:: 2y tangent to the ellipse 4x2 + y2 = “3- (6 marks) A circle C passes through the point P0, 2) and the points of intersection of the circles C.:x’+y’-3X+2y-2=0
and C1zx2+y’+x+3y—10=0. Find the equations of (a) the circle C. t C at l’.
and (b) the ”“3““ ° (6 marks) Show that sin’nﬂ -- sin’mt} = sin (n + m)65in(r: ~ mm. Hence, or otherwise, solve the equation sin239 - sinlm - sinO = 0 for 0 < 9 < 1!. (7 marks) 83-CE-Add Maths "(El—2 sccrron n (60 marks) Answer any THREE questions from this section.
Each question carries 20 marks. 8. Figure 2 shows a tent consisting of two
inclined square planes ABCD and EFCD I)
standing on the horizontal ground ABFE.
The length of each side of the inclined
planes is a. N is a point on CF such
that AN J. CF. Let NF = )c(ae 0),
LCFB = 6 and M be a point on BF
such that NM 1 BF. (a) By considering AABM, express AM
in terms of a, x and 0. (4 marks) A (h) By considering AANF, express AN
in terms of a, x and 0. Figure 2
(5 marks) (c) Using the results of (a) and (b), or otherwise, show that x = 2acos’9 . (5 marks) (d) Given that x = % , ﬁnd (correct to the nearest degree) the inclination of AN to the horizontal .
(6 marks) 9. A(1, -2) and 3(4, 4) are two points on the parabola y2 = 4):. P is a point on the line AB such that AP : PB = l : k. A line L, through A is perpendicular to the tangent at A. Another line L; through E is perpendicular to the tangent at B. L. and L, intersect at N.
Let 0 be the origin. (a) Find the coordinates of the point N and the slope of 0N , (8 marks) (b) (i) Express the slope of 01’ in terms of k. (ii) Express tan LPON in terms of k when
(l) LPON is acute,
(2) LPON is obtuse. (7 marks)
(c) Find the value of k in each of the following cases :
(i) when LPON = 45°;
ii when OPN is a strai t line.
( ) Eh (5 marks) aa-cE-Add Maths "(El-3 \crozwam«wuwmn m... Mum ....,..._ __._... —s M m. . ID. A straight line through the point R(_1, —l) has a variable slope m. It intersects the circle
x1 + y’ = l
at A and B. Let P be the mid-point of AB. (21) Find the coordinates of P in terms of m. (9 marks)
(b) The locus of I' is a part of a curve C. Find the equation of C and name it (6 marks)
(c) Sketch the locus of I’. (5 marks)
11‘ (a) Show that M = 2c0529 + I.
stnﬂ
By putting 9 = i— + t» in the above identity, show that
cosBQ - sinSQ = 1, Zsin2¢.
cost» + sin¢
(7 marks)
(b) Using the substitution at = 3% -> u, show that
5
1 c0539 d¢= a sin3u du.
o cosdh + sin¢ 5 cost: 1‘ sinu
2
Hence, or otherwise, show that
I a
1 cosSQ M, = L 1cosSQ—sin39ddj'
0 coszﬁ + sino 2 a cosd) + sind)
(8 marks)
(0) Using the results in (a) and (b), evaluate
n
5 cos} 9 dd? .
a cosnb + sind:
(5 marks)
12. let {(x) be a function of x and let k and s be constants.
(a) By using the substitution y = x + ks, show that
r (kn): {(x +ks)dx = J I(x)dx.
o ks
Hence show that, for any positive integer n,
‘- HS
[[Kx) + I(x +5) + + {(x + (n ~ Ilstx = [ f(x)dx.
o o
(10 marks)
1
- 2 dx . . . = .
(h) Evaluate -—=—.—_—. by ustng the substitution x smﬁ.
o \/l_ _ X, Using this result together with (a), evaluate 1
1" 1 + 1 + _-_l.__—— + + ‘ dx
#5 T"f#7 —’JI _ 7 .
I (1“ x/l—-(x+il;) \/l“(x+—Z—) \ﬂ‘(x+"2"l) 0 2n (10 marks) END OF PAPER 83-CE-Add Maths II(EI~4 HONG KONG EXAMINATIONS AUTHORITY HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 1984 Iﬁﬂﬂ¥i$ Efﬁg—
ADDITIONAL MATHEMATICS. PAPER | 8.30 Elm—10.30 am (2 hours) ’ This paper must be answered in English Answer ALL questions in Section A and any THREE
questions from Section B. All working must be clearly shown. 84-CE-ADD MATHS I——1
9 1983 NUMERICAL ANSWERS Addilional Malhe’malirs I 1. 3. 10. ll. -2<)\<—1 —42
x351 and 1—ﬁ<x<l+\/7 X+2)'—10=0 v=2.y=4
(c) a=—(——)—4”121",n=0,il.t2.
1r 5n 91r 131r
_ _‘ __ t ._._
t'c1n16,tan16,anl6,anl6 (a) a=—12,b=48
(b) p=-4.q=-8 x=6 or sin/3 (c) an; (X1 ::)= 120° 11 (b) (i) gun) (ii) é—n(n+1)(n+2)
(iii) (r + 1) minutes
65 minutes
(a) x = I
(b) x = %r
(a) (n x1 +1;1 +xy =
Whenx=2,y=l. (ii) in = _2x + 2
dx x +2y (b) i m/S 8
(c) a; 1983 1. 2. 10. ll. 12. k: Additional Murhenmlits ll 30! —7 1 sin 2):1 + c 8 (b) 13
y = g—x i 5
(a) x2+y1+5x+4y—18=0
(b) 7x+8y—23=0
= _1r_ 1 21
8 0,10,2,10011r
(3) AM = Vai + (2n - x)’ 00529
(b) AN = V112 + 4a,j 00529 ~Jri
(d) 19°
(8) N = (5 . 2)
Slope 01' ON =%
. = 4 - 2k
(1)) (I) Slope of GP 4 + k
.. _ 12 -12k
[’0 — 1r _.___
(n) tanL N 28+k
according as LPON is acute
or obtuse
. ___ 39
(c) (1) M
(ii) k = 1
a = m — m2 m -1
() (1+m7'l+m1)
(b) The circle 1:2 + y1 + x + y = 0
(C) ‘ 1
(b) 0‘1: N: N: 101 ...

View
Full Document

- Spring '17
- Marketing, triangle, LPON