Maths Questions(20 points)

1) It is given that f(x) = 3x - 2
a) Verify that [f(2)]^2≠f(2^2)
f(2) = 6 - 2 = 4
[f(2)]^2 = 16
f(2^2) = f(4) = 3*4 - 2 = 10
[f(2)]^2≠f(2^2)
b) Is it true that f(√4)= √f(4)?
f(√4) = 3√4 - 2 = 4
√f(4) = √10
f(√4) does not equal √f(4)
2) It is given that g(x)=(x+k)(x-2)and g(-1)= -1.
a) Find the value of k
g(-1) = (-1+k)(-1-2) = -1
(k - 1)(-3) = -1
k - 1= 1/3
k = 4/3
b) Hence, find the value of 3g(4)+6g(-2)
3g(4) + 6g(-2)
= 3(4 + 4/3)(4 - 2) + 6(-2+4/3)(-2-2)
= (12 + 4)(2) + (-12 + 8)(-4)
= 32 + 16
= 48
3) It is given that f(x)=2(x+m),g(x)=x+m and f(8)=6g(4)
a) Find the value of m
f(8) = 6g(4) => 2(8 + m) = 6(4 + m)
16 + 2m = 24 + 6m
4m = -8
m = -2
b) Hence, find the value(s) of x such that f(x)‧g(x)=50
f(x)g(x) = 2(x - 2)(x - 2) = 50
(x - 2)^2 = 25
x - 2 = 5 or x - 2 = -5
x = 7 or -3
4) It is given that f(x)=x^2-kx
a) Find f(x+2) and f(x-2) in trems of k
f(x + 2) = (x + 2)^2 - k(x + 2)
= x^2 + 4k + 4 - kx - 2k
= x^2 + (4 - k)x + (4 - 2k)
f(x - 2) = (x - 2)^2 - k(x - 2)
= x^2 - 4x + 4 - kx + 2k
= x^2 - (4 + k)x + (4 + 2k)
b) If f(x+2)-f(x-2)=kx-32, find the value of k
f(x + 2) - f(x - 2) = kx - 32
x2 + (4 - k)x + (4 - 2k) - x^2 + (4 + k)x - (4 + 2k) = kx - 32
8x - 4k = kx - 32
k = 8
c) Hence, solve for x if f(x+2)=f(x-2)+40
f(x + 2) = f(x - 2) + 40
f(x + 2) - f(x - 2) = 40
8x - 32 = 40
8x = 72
x = 9
5) It is given that f(x/2)=3x/2+5
a) Find f(1)and f(2)
f(1) = f(2/2) = 3*2/2 + 5 = 8
f(2) = f(4/2) = 3*4/2 + 5 = 11
b) Find f(u)
Let u = x/2, then x = 2u
f(x/2) = 3x/2 + 5
f(2u/2) = 3(2u/2) + 5
f(u) = 3u + 5

1) It is given that f(x) = 3x - 2

a) Verify that [f(2)]^2≠f(2^2)

LHS = (f(2))^2=(3*2-2)^2=4^2=16
RHS = f(2^2)=f(4)=3*4-2=12-2=10≠LHS

b) Is it true that f(√4)= √f(4)?

LHS=3√4-2=3(2)-2=4
RHS=√(3*4-2)=√10≠LHS

2) It is given that g(x)=(x+k)(x-2)and g(-1)= -1.
a) Find the value of k

g(-1)=-1 => (-1+k)(-1-2)=-1 =>k-1=1/3=>k=4/3//

b) Hence, find the value of3g(4)+6g(-2)

3g(4)+6g(-2)
=3(4-4/3)(4-2)+6(-2-4/3)(-2-2)
=(12-3)(2)+(-12-8)(-4)
=(9)(2)-(20)(4)
=-62//

3) It is given that f(x)=2(x+m),g(x)=x+m and f(8)=6g(4)
a) Find the value of m
f(8)=6f(4)
=>2(8+m)=12(4+m)
=>16+2m=48+12m
=>8+m=24+6m
=>-16=5m
=>m=-16/5//

b) Hence, find the value(s) of x such that f(x)‧g(x)=50
f(x)*g(x)=50
=>2(x+m)^2=50
=>(x+m)^2=25
=>x+m=-5 or 5
=>x-16/5=-5 or 5
=>x=41/5 or -9/5//

4) It is given that f(x)=x^2-kx
a) FInd f(x+2) and f(x-2) in trems of k
=>f(x+2)=(x+2)^2-k(x+2)=x^2+(4-k)x+(4-2k)
and f(x-2)=(x-2)^2-k(x-2)=x^2-(k+4)x+(4+2k)

b) If F(x+2)-f(x-2)=kx-32, find the value of k
f(x+2)-f(x-2)=kx-32
=>(x^2+(4-k)x+(4-2k))-(x^2-(k+4)x+(4+2k))=-2kx-32
=>8x-4k=kx-32
By comparing coef , we have
k=8
c) Hence, solve for x if f(x+2)=f(x-2)+40
f(x+2)=f(x-2)+40
=>f(x+2)-f(x-2)=40
=>8x-32=40
=>x=9//

5) It is given that f(x/2)=3x/2+5
a) Find f(1)and f(2)
put x=2 , f(2/2)=f(1)=(3*2)/2+5=3+5=8
put x=4 , f(4/2)=f(2)=(3*4)/2+5=11//

b) Find f(u)
f(u)=f(2u/2)=(3*2u)/2+5=3u+5//