2015_maths_alevel_nov_paper2

Extracted Content: 2015_maths_alevel_nov_paper2

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ZIMBABWE SCHOOL EXAMINATIONS COUNCIL

General Certificate of Education Advanced Level

CONFLOENTIAL

MARKING SCHEME

aa

NOVEMBER 2015

MATHEMATICS ||: 3 $164/2

de

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! a sty" ”~ Tay! , , dy ad ae dy MI + DAy - Ty 9 7(2) yz ~ dt Us 4 4: , borg f , Deady! = (Idzy- ry’) ' dx dy 2n- Ty’ AJ dy \Any~2Ay' 6 a Q-7 Mi a Pr on 14-2 -

5S OS Al (.
10 2
ey HHT (44 {|

XH.
2 Gy) Reosdcos%-R sin sin” = 4 cos O- 6 sine

Raine = G

Rose = 4

3
tan” ™ 2 ;
3
“a = wn >
bs;
= 56° ys, 4
Reon? « + Rin? ~ = 16436
2 = '
R 52 adh |
BR = 2/13 cod / ° a po or { wd otha j
; 5 i".
4cos0- 6sin § = 2/13 cos(0 +56") be i
cn aad
oe
(ii) 213 cos (0456) = 5 N
5
cos (9456) = pe ME
2/14 co
6+56° = 46°, 313°, 406°... Abie oo,
; wee)
A= 259°, 350°. fA, , ”l —

i (6] ca oo
ya NM GA, _, +e ( (.- /
‘ f js bes . \ /

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2(mx)- 3x +13 = 0

x,(2m-3) = -13

3y+2x413 = 0

3(mx)+2x+13 =

3m+2

1
5(4+%) = 0.

-13
2m-=3 3m+2

-26m'439-39m1=26 = 0”

65m = 13 . 3 @

4(2 - 1). 2°

4

0

= 0

Led Meas fee ‘

Stlyy CAR
Mens i haus, Mente a

-13 Al
2m-3

et

Ale
“
= «
a

Verify statement true for n= 2

2)

P M1 for both.

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LHS = RAS .. the statement is true for n = 2. Al {>.< Assuming its true for n= k and stating 2(2)+3(27)+4(27)+.... +k) = 4(k-1).2"? Bl To prove validity forn=k+ 1, ie. 2(2')+3(2?) + enconth(2)+(k41)2 = 4(K)(2") t 2(2')+3(27) + cree +k(2"") 4 (K+ 1)2" = 4(k~1)2"* +(k4+1)2" MI aseis ; bao = 4(27)(k-1)(2")+(k-+1)2" a per an

= (k-1)(2")+(k-+1)(2")

= 2*(k-1+k+1)

= 2* (2k)
i> (2k)(2)(2') gee
\> ee
Ne = kel .
4(k)(2 ) VO Veeck a
es {f true for n=k also true for n= KH. Since true for n = 2 also true for
: n= 3'and true for 4 and soon. ‘Therefore by mathematical induction its i
true for all natural numbers n 22. we Al Loney _
. N 7 eS
* Gus ee m 7” s
2 ii. ace
5 (ay ff (sin2x+sinx) de pe
= [-3208 2x-co8s| Mi ates an Ans
2 10 yen Vee

7 5 e0sz— cos +7-c0s0 +0080

VUE vy VU CU

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Extracted Text

We.

ey

My

x 2 | | |
" Jeentersn “ (i on Cou « ut a 6 Nn

= xf’ ( sin? 2+ 2sin2xsinx+ sin? x) de & ox i

7 af $(1~cos4s}+deosssin® s+ (1-cos22)dx bay

y+ ene \

( cert oe teh
= 4 [3-Lsinax 4 ime | inde 2 My UA , bo
2 3 2 2 0 dR. Sa, SN ae AL
= a(Z-0+842-0)-(0-0+0+0) M1 S5cxbeeeh thot ©
4 3.4 Coe € 7 SAY
Elixe (Lats \ %
= mz 4
»(2+4] «
= = (32 +8) af (Ag)
[5]
f

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i Cc |

tl

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Gi) Arg (Z) = wn'{2)

Cc

= -36.9° oy > cy Gy Bl

7 (a) Det = 6+5

- 1]
Inverse = Af 3 1 Bl
11\ -5 2
if 3 1) x Lf ax! 14x! ~22 wr
11, -5 2 Jl -14x'-22 J 11( -5x'~28x'-44
. | -t1'-22 |
1}{ 33x! 44 |
‘ 7 é
x j ( x'-2 | ay
y} \ -3x'-4 ] Lea
/ : oe A
¢ Ne.
x = XZ ssasnaes (60) M1 Wiad Le ; Z
oa
yer 3x! AL. (2) ne ad
from (1) Sal 3% See \
xis -x-2 7 os
e \
Substitute ir. (2)
y = -3(-x-2)-4
y = 3xt+6-4
y = 3x+2 Al oF
ce
° (Ey vo fel '
a)
¢ eh

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(i11}

af -l
@ = 180-cos'( =]
3V3
= 180 - 101.1

= 78.9

”
'
‘A
+
a
"
=

y+24A-64+10 = 1
y = 44-9
fo A
r= | -9 [Al 4 ,
5 3
O+2- 0 a
-9+4A| -|-6| = |[-3+44
543A -9 443A
I: 1):
-344a/+/4| = 0
443A 3

A-12+i16A+12+9A = 0

542 = 9

2 =0
0
TF = |-3
4

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MI Ad!
BEN
fy

Met SS
(r- Oe, !
Al

fr.

ae |

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10

it Nit ada h |
5
Al 5]
. Nyy |
ay 1,
l Sad Oa
jlalOr} = [-O4add MI caged sol
(04S: 543d \
\\.
.4Rr
5 -S+3(3+ sr) M1 for attempt to solve
St AR a vs
Sy —54+9497
r= 4
i = -l ‘ Al
| Point of intersection
(0,-9,-5) Al
[4]

ijk

3.10 5

1 4 3

lio 5| |3 5 3 10 AY
t - k a

43 J} Shea? q | =

10i— 47 + 2k Al

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3 | Ml

r |-2 13 Al

(4]

9 Resolve honzont

solve vertically

J2 . BL either one correct
l, 4 1 = lg
' )

Solve simultane

~ \ . ,
I, 4 . ml, = 2R "
2 v2
de
j le V3 f
1436283684 N i |
rf
3; 4B
V2\14V3/
17.59081076 N Al
. ne v3 4
10 i) parucle P: 7 "8-8 = 2a Bie Yu
particle Q: 3g -T 3a By i i

ve simultaneously

HN

Al

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Extracted Text

12

3(10-v3
(by) TT = 3g- ( 5 }s Miv _ substitute his (a)

(ss4)
— (8

7 25
= 19.69N Al [2]
lo
ie 200 Bls.oi “x pS soko fh
9 ; .
Ao nd &
yee EAS SPS
0 = (= j-2o( S $) use of V=u? -2as Ml
a= 8.000 (= 10.973) ms~ Al
729 . 7
a [3]
(Db) distance before braking
£20x1000 1 304 Sgt OF oy
——— — = 2 : ” Tae PON ie : ¢ 4
sox6o 2 3 PPS eb Gay
‘ .
2 é
100\" 8.000) - - 7 .
0 =|—| -2|=, IS: . ML ee oe Aas .
() (Oe er AM Be ea G
5 = Om _ a Al
3,0:
stops in
50 , 405 _ 1615 ,,(- 67.292 m) et yp [4]
3 8 a
12 horizontal motion
1§ =(30cos&) ¢ or equiv. Bl
vertical motion
, 2
5 = (30sin " =] a (sca) or equiv. Mi bogs
2 cos © 2° \2cos* . oa
\ a

pos bE SK(L LA {Gen # } ERY

i
iy ees

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13 9. ~ 15 tana—2*" (1+ tan’ a) 8 simplifies to 1.22625 tan’ a - 1 Stana +6.22625 =0 Al ° Solving quadratic et Al D9LIELIE DIAIE) ana =< 15+ ,/15* - 4(1.22625)(6.22625) MI 2 (1.22625) = 0.430214 , or 1.802202 Al a=23.278°, 85.157° footer’ JE i L 13 fa) Modal age group eesevess 8 - EC Bi id) 25-30years = SeeT Sx“, bs = 24aseters “O * 8 epee v0 TE a Bl (c Mothers i in the survey xe ~OXEL y SKXC LC r Ox oe exe = 5xOf+ tage pie 5x564 540524 + FBI 16 Mil 1

4 Ss Al
ee - [4]

14 (a) flardr+ f'a(2-xJ'dx = 1

ax? 4x - 4°
——} + — — =]
2 2 3
lo I
Coy ee

a,4_,
2 3
3a _4
6

6 No
a= z Ai [+

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14

' 2
fx Sears fr 8(2- x) ae
pans

GO x op doe 2 4b x |
x + = = +-
E ail 3|| 2 3 4]

14 (b) «Var (X) =
: Wy

2 2
1 5 2 6 2 9
[oxiadet [Page a-(F

af}

6 x

—x
5

81
6(1.%2 62 -S41-5)- :

~ 3473 5.3.5) 100
,
- 47 _ 81
50 100
-
100
= 0.13;
a flee )
15 Xx n(s(2) oar Nis -_
3) oe
() 1 PX&>6) = piz> =>
3
= P(Z>1,5)
= 1-9(1,5)
= 1~0.9332
= 0.0668

YJLGINICU vy VAINIICAGIINICI

6(42 4x4 x , ( }
4+—|——--— +>] -
5\ 3 4 5/), \10

MI Ap.
aft

L iy opp

Al

Al

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Extracted Text

15

4n-5
plat<x<5| = | 2 ames MI
2 2 2 a

Nw

3 3

= P(-0.75<Z <0)

#(0)-(1-9(0-75))

= 0.5-(1-0.7734)

= 0.2734
= 27.34 % Al
FP ¢
(ai) xX~ Bin (6,02734) Bl-/
M1

p(X=4) = 6 C,x0.2734* x 0.7266"

0.0442 a \

16 (a)

|
| : 0.3 hit target

windy Bl ny sess
0.2 0.7 misses target a i a
0.8 0.8 hit target
not windy Bl cqaet © eked!
02 misses target “

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