Further mathematics eassy
F/maths Obj
1CBBDCAADCB
11DCBDCBCDCC
21ABDCDABBCB
31CDCADBCAAD
1a) 8m + 2³m
(2³)m + 2³m = 1/4
2³m + 2³m = 1/4
2(2³m) = 1/4
2³m = 1/8
2³m = 8 – ¹
2³m = 2-³
Since bases are the same, powers are same
3m = – 3
m = – 3/3
m = -1
1b) log15 base 4 = x
x = log15 base/log4 base 10 = 1.1761/0.6020
= 1.9536
therefore x approximately 1.954
2a)
(-2) = m(-2)² + n(-2)+2=0
4m-2n = – – – – – – – – (1)
F(1) = m(1)²+n(1)+2=3
m+n+2=3
m+n=3-2
m+n=1 – – – – – – – – – – – -(2)
X Equate (2) by 4
4m+4n=4 – – – – – – – – – – (3)
4m-2n= -2
-4m+4n=4/-6n = – 6
n=1
4)
Given y= 5/x² + 3
Y = 5(x² +3)-1
dy/dx = anxn-1
y = 5(x² + 3)-1
dy = 5(-1) (x2 +3)-1^1
dy/dx = 10x (x² + 3)-²
dy/dx = – 10/(x² +3)²
5a)
^nP5 ÷ ^nC4 = 24
n!/(n-5)! ÷ n!/(n-4)!4! =24
n!/(n-5)! * (n-4)!4!/n! =24
(n-4)!4!/(n-5)!=24
(n-4)(n-5)!4!/(n-5)! = 24
n-4=1
n=4+1
n=5
b)PR= 5C3 (1/6)³ (5/6)^5-3
5!/(5-3)!3! (1/6)³ (5/6)²
5!/2!3! (1/6)³ (5/6)²
10 (1/216) (25/36)
=0.03215
Pr = 1-0.03215
=0.9678
6) make a table containing
MARKS,TALLY,F & FX
UNDER F – 2,9,4,2,2,1
UNDER FX – 2,18,12,8,10,6
€F= 20
€FX = 56
6b) Mean = €fx / €f
= 56 / 20
= 2.8
7(a)
Given
h=15.4t – 4.9t²
Velocity v = dh/dt = 15.4 – 9.8t
At maximum height, V =0
15.4 – 9.8t = 0
9.8t = 15.4
t = 15.4/9.8
t = 1.6secs
Time to reach maximum height = 1.6secs
b) maximum height = 15.4 t – 4.9t²
= 15.4(1.6) – 4.9(1.6)²
=24.64 – 12.544
= 12.096
Maximum height = 12.1m
*********
9a)X+6/(x+1)^2 = A/x+1 + B/(x+1)² + C/(x+1)³
X+6/ (x+1)³ =A(x+1)² + B(x+1) + C / (x+1)³
X+6 =A (x+1)² + B(x+1) + C
Let X+1=0 , X=-1
-1+6= A (-1+1)² + B(-1+1) +C
5= C
Therefore:- C=5
X+6= A(X2+2x+1) + Bx + B + C
X+6= Ax²+2Ax+A+Bx+B+C
comparing the coefficient of X²
A=0
Comparing the coefficient of X
1=2A+B
1=2(0)+B
B=1
x+6/(x+3)² = o/x+1 + 1/(x+1)² + 5(x+1)³
1/(x+1)² + 5/(x+1)³
9b)
S²1 1/(x+1)² + 5/(x+1)³ DX
1n (x+1)³ + 5 1n (X+1)⁴ |²1
1n 3(3) + 5 1n(3)⁴ – 1n +2³ – 5 1n +2⁴
3.2958+21.9722-2.0794-13.8629
=9.3297
10a)
3x^2+x-2 <= 0
3x^2+3x-2x-2 <= 0
3x(x+1) -2 (x+1) <= 0
(3x-2)(x+1) <= 0
3x-2 <= 0 or 8+1 <= 0
3x <= 2 or x <= -1
X<= 2/3
Therefore range of value is
-1 >= x <= 2/3
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
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
——————
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
