WAEC 2017 Further Maths Answers

NIGERIA FURTHER MATHS ANSWERS*

_Pls check ur Country Answer, *Mr. Kunlessi Carez*_
F/maths Obj
1CBBDCAADCB
11DCBDCBCDCC
21ABDCDABBCB
31CDCADBCAAD

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)
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10a)
i) (x^2-1) (x+2)=0
(x-1) (x+1) (x+2)
x=1, or -1 or -2
ii) 2x-3/(x-1)(x+1)(+2)
=A/x-1+B/x+1+C/x+2
2x-3=A(x+1)(x+2)+B(x-1)(x+2)
+C(x-1)(x+1)
let x+1=0,x=-1
2(-1)-3=B(-1-1)(-1+2)
-5/2=-2B/-2 B=5/2
let x-1 =0 x=1
2(1)-3=A(1+1)(1+2)
-1=CA, A=-1/6
Let x+2=0 x=-2
2(-2)-3=C(-2-1)(-2+1)
-7=3C, C=-7/3
=======================
11a)
Given:
f(x)={(4x-x^2)dx
f(x)=2x^2 – x^3/3 + K
f(3)=2(3)^2 – (3)^2/3 + K =21
18 – 9 + K=11
9+K=21
K=21-9
K=12
Therefore
f(x)= -x^3 + 2x^2 + 12
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14ai)
SKETCH THE DIAGRAM
14aii)
Using lami's theory
T1/sin60=T2/sin30
48N/sin60=T2/sin30
48N/0.8660=T2/0.5
0.5(48)/0.8660=T2(0.8660)/0.8660
T2=24/0.8660
T2=27.7N
14b)
Using the equation of motion
H=U^2/2g
H=(20)^2/2*10
=20*20/20
H=20m
Timetaken to reach the maximum height
S=Ut+1/2at^2
20=0+1/2(100)t^2
20/5=5t^2/5
t^2=4
t=sqroot4
t=2S
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*GHANA WAEC ANSWERS*
OBJ……

1 A. 21 B.
2 B 22 C
3 D. 23 D
4 C 24 D
5 B. 25 D
6 C. 26 C
7 D. 27 A
8 A. 28 D
9 C. 29 D
10 C. 30 A
11 D. 31 D
12 C. 32 A
13 A. 33 C
14 A. 34 D
15 A. 35 B
16 D. 36 A
17 B. 37 D
18 A. 38 A
19 D. 39 A
20 D. 40 D

2)
(5,2)(-4,k)(2,1)
(y3-y2)/(x3-x2)=(y2-y1)/(x2-x1)
(1-k)/2-(-4)=(k-2)/(-4-5)
(1-k)/(2+4)=(k-2)/-9
(1-k)/6=(k-2)/-9
-9(1-k)=6(k-2)
-9+9k=6k-12
9k-6k=-12+9
3k=-3
k=-1

==============================

3a)
If f(x+2)=6x^2+5x-8)
To find f(5)
Therefore f(x+2)=f(5)
where x+2=5
x=5-2
x=3
therefore f(5)=6(3)^2+5(3)-8
=6(9)+15-8
=54+7
=61
3b)
(7root2+3root3)/(4root2-2roo3)*(4root2+2root3)/(4root2+2root3)
(24*2+14root6+12root6+6*3)/(16*2+8root6-8root6-4*3)
(48+26root6+18)/(32-12)
=(66+26root6)/20
=66/20+(26root6/20)
=33/10+(13root6/10)
=3.3+1.3root6
==============================
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
==============================
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
=============================
6a)
1+4+k+k+4+11/5=k+1
20+2k/5=k+1 cros multiply
20+2k=5(k+1)
20+2k=5k+5
20-5=5k-2k
15/3 = 3k/3
k=15/3 =5
the numbers are 1,4,5,9,11 the mean X=6
6b)
tabulate
x| 1, 4, 5, 9, 11
x-x| -5, -2, -1, 3, 5
(x-x)^2| 25, 4, 1, 9, 25
total| 64
standard deviation
=sqr £(x-x)^2/n
=sqr64/5
=sqr12.8
=3.58
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7)
m1=3
u1=8m/s
m2=?
u2=5m/s
v=6m/s
m1u1+m1u2=(m1+m2)v
3*8+m2*5=(3+m2)6
24+5m2=18+6m2
24-18=6m2-5m2
m2=6
(7b)
m2u2-m1u1=V(m1+m2)
6*5-3*8=V(3+6)
30-24=9v
9v=6
v=6/9
v=0.67m/s
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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
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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