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WAEC 2017 Answers On Further Mathematics Now Available – May/June Expo
NIGERIAN ANSWERS
1a)
g(x)=y
y=x+6
x=y-6
g^- f(x-6)
=4-5(x-6)/2=4-5x+30/2
=34-5x/2
1b)
coodinate=(x1+x2/2 ,y1+y2/2)
=(7-2/2,7-5/2)=(5/2,2/2)
=(5/2,1)
FURTHER MATHS THEORY
(1)
(√3 x + 1 ) - (√x + 4 ) = 1
Squaring both sides
(√3 x + 1 )^ 2 - (√ x +4 )^ 2
= 1 ^ 2
3x + 1 - x + 4 = 1
3x - x + 1 + 4 = 1
2x + 5 = 1
2x = 1 - 5
x = - (4 / 2)
x = - 2
2)
x+1/3x^2-x-2
3x^2-x-2/-6(-3 2)
3x^2-3x+2x-2
3x(x-1)+2(x-1)
(3x+2)(x-1)
x+1/(3x+2)(x-1)=A/3x+2+B/x-1
x+1/(3x+2)(x-1)=A(x-1)+B(3x+2)/(3x+2)(x-1)
x+1=A(x-1)+B(3x+2)
3x+2=0
x=-2/3
8a )
60 , 56 , 70 , 63 , 50 , 72 , 65 , 60
mean =£ x /n = 60 + 56 + 70 + 63 + 50 +72 +65
+60 /8
mean =62
8b )
variance = £ (x - x ^ - )^ 2 / n
=( 62 - 60 )^ 2 + (62 - 56 )^ 2 + (62 - 70 )^ 2+
(62 - 63 )^ 2+ (62 - 50 )^ 2+ ( 62 - 72 )^ 2 +
(62 - 63 )^ 2+ (62 - 60 )^ 2/ 8
=362/ 8 =45 . 25
SD = sqr variane = sqr 45 . 25 = 6 . 73
10a)
(1+x)^7
7Co(1)^7(x)^0 + 7C1(1)^6(x) +
7C2(1)^5(x)^2 + 7C3(1)^4(x)^3 +
7C4(1)^3(x)^4 + 7C5(1)^2(x)^5 + 7C6(1)
(x)^6 + 7C7(1)^0(x)^7
= 1+7x + 21x^2+35x^3 + 35x^4+21x^5 +
7x^6+x^7
(10b)
35 21 7
a=35
d=T2-T1
=21-35
11a)
Kp2=72
K!/(k-2)!=72
K(k-1)(K-2)!/(K-2)!=72
K^2-K=72
K^2-K-72=0
K^2-9k+8k-72=0
K(K-9)+8(k-9)=0
(K+8)(K-9)=0
k=-8,K=9
We consider positive value of K=9
11b)
The equation 2cos^2tita-5costita=3
Let cos tita=x
2x^2-5x=3
using quadratic formular
a=2,b=-5,c=-3
5+_root(25+24)/4
=5+_root(49)/4
=(5+_7)/4
=(5+7)4=3 or (5-7)/4=-2/4=-1/2
since x cos tita
cos tita=-0.5
tita=cos^-1(-0.5)
tita=120degrees
——————
15a)
at max height
V=0m/s
g=10m/s^2
V^2=u^2-2gh
0^2=30^2-2 10H
0=900-20H
20H=900
H=900/20
H=45m
15b)
time taken to get to max height
V=u-gt
0=30-10t
10t=30
t=30/10
t=3secs
Timetaken to return=2t
=2 3=6secs
15c)
H=40m
H=ut-1/2gt^2=40
30t-1/210t^2=40
30t-5t^2=40
5t^2-30t+40=0
t^2-2t-4t+8=0
(t^2-2t)-(4t+8)=0
t(t-2)-4(t-2)=0
(t-2)(t-4)=0
t-2=0 or t-4=0
t=2secs or t=4secs
5a)
pr(age)=4/5
pr(fully)=3/4
pr(must)=2/3
pr(age not admitted)=1-4/5
=1/5
pr(fully not admitted)=1-3/4
=1/4
pr(must not admitted)=1-2/3
=1/3
Therefore pr(none admitted)=1/5*1/4*1/3
=1/60
5b)
pr(only age and fully gained
admission)=4/5*3/4*1/3
=1/5
4)
(x^2+5x+1)sqroot(2x^3+mx^2+nx+11)=(2x-5)
remainder:30x+16
(x^2+5x+1)(2x-5)
=2x^3+10x^2+2x-5x^2-25x-5
=2x^3+10x^2-5x^2-25x-5
=2x^3+5x^2-23x+30x+16-5
=2x^3+5x^2+7x+11
Therefore m=5, n=7
12a)
tabulate
Marks| 1-10, 11-20, 21-30, 31-40, 41-50,
51-60, 61-70, 71-80, 81-90, 91-100
F| 3, 17, 41, 85, 97, 115, 101, 64, 21, 6
C.B| 0.5-105, 10.5-205, 20.5-305, 30.5-405,
40.5-505, 50.5-605, 60.5-705, 70.5-805,
80.5-905, 90.5-1005
C.F| 0+3=3, 3+17=20, 20+41=61, 61+85=146,
146+77=243, 243+115=358, 358+101=459,
459+64=523, 523+21=544, 544+6=550
13ai)
M=2
P=5
C=3
total=10
If the books of the same subject are to stand
together
No of arrangements=2!*5!*3!*3!
=2*120*6*6
=8640arrangements
(13aii)
Only the physics textbook must stand together
No of arrangements=5!*6!
=120*720
=86400arrangements
(13b)
P=13/20
q=1-13/20=7/20
pr(atleast 3 speak E)=1-Pr(2 speak E)
=(1-8C1p^1q^7+8C2p^2q^6)
=1-(8*(13/20)*(7/20)^7+28(13/20)^2*(7/20)^6
=1-(0.003346+0.0217467)
=1-0.0251
=0.9749
=0.975(3s.f)
(9a)
1/1-cos tita + 1/1+cos tita
=1+cos tita + 1-cos tita//(1-cos tita) (1+cos tita)
= 2/1+cos tita - cos tita - cos^2 tita
= 2/1-cos^2 tita
Recall that :
Cos^2 tita + sin^2 tita = 1
.:. Cos^2 tita = 1-sin^2 tita
.:. 1/1-cos^2 tita + 1/1+cos tita
= 2/1-(1-sin^2 tita)
(9b)
At stationary points,
dy/dx=0.
y=x^0(x-3)
Let u=x^2,v=x-3.
du/dx=2x dv/dx=1.
dy/dx= Udv/dx + Vdu/dx
dy/dx=x^2(1)+(x-3)(2x)
.:. dy/dx=x^2+2x^2-6x
dy/dx=3x^2-6x
At stationary point,
dy/dx=0..
.:.3x^2-6x=0
Equation of line=> 3x^2-6x=0
10b)
X1 Y2
(3, 1)
r=sqr(x2-x1)^2+(y2-y1)^2
r=sqr(3+3)^2+(1-1)^2
r=sqr6^2+0=sqr36=6
the equatuon of a circle
(x-a)^2+(y-b)^2=r^2
(x-(-3))^2+(y-1)^2=6^2
(x+3)^2+(y-1)^2=36
x^2+6x*9+y^2-2y+1=36
x^2+y^2+6x-2y+9+1-36=0
x^2+y^2+6x-2y-26=0
