F 4數學功課

1.Given f(t)=1/9t2次+1,find the values of
a. f(1/3)

f(t) = 1/9 * t^2 + 1
f(1/3) = 1/9 * (1/3)^2 + 1
=1/9 * 1/9 + 1
= 1 1/81

b. f(1)

f(t) = 1/9 * t^2 + 1
f(1) = 1/9 * (1)^2 + 1
= 1/9 + 1
= 1 1/9

2.It is given that f(x)=3x-2
a.Find the values of f(2)and f(4)

f(x)=3x-2
f(2) = 3 * 2 - 2
= 4
f(4) = 3 * 4 - 2
= 10

b.Verify that [f(2)]2次/=f(2 2次)

f(2)^2 = 4^2
= 16

f(2^2) = f(4)
= 10

f(2)^2 > f(2^2)

f(x^2) = 3x^2 - 2

f(x)^2 = (3x -2)^2
= 9x^2 - 12x + 4
= 3x^2 - 2 + (6x^2 - 12x + 6)
= 3x^2 - 2 + 6(x^2 - 2x + 1)
= 3x^2 - 2 + 6(x - 1)^2
= f(x^2) + 6(x - 1)^2

since for any real number x, 6(x - 1)^2 is bigger or equal to 0
i.e. 6(x - 1)^2 will never smaller than 0 for any real number x
so f(x)^2 is bigger or equal to f(x^2)
ｆ(ｘ)^２　>=　ｆ(ｘ^２)

3.If f(x)=x(x+1),find the values of
a. 2f(-4)

2 * f(-4) = 2 * (-4) (-4 + 1)
= 2 * (-4)(-3)
= 2 * 12
= 24

b. f(2)+[f(-3)]2次

f(2) + [f(-3)]^2 = 2(2+1) + ((-3)(-3+1))^2
= 2(3) + ((-3)(-2))^2
= 6 + (6)^2
= 42

4.If h(x)=3x2次+m and h(3)=1,find the value of m

h(3) = 3 * (3)^2 + m = 1
m = 1 - 3 * (3)^2
m = 1 - 27
m = -26

5.If g(x)=(x+k)(x-2)and g(-1)=-1,find thevalue of k

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

6.If f(x)=3(2 3x次),find the values of
a. f(0)

f(x)=3(2^3x), <--------------------------- is this right ???
f(0) = 3 (2^0)
= 3*1
= 3

b. f(- 2/3)

f(- 2/3) = 3(2^(3 * (-2/3)))
= 3(2^(-2)))
= 3(1/(2^2))
= 3(1/4)
= 3/4

6.If H(x)=4x2次-8x+1,find the values of
a. H(0)

H(x)=4x^2-8x+1
H(0) = 1

b. H(p-2)

H(p-2) = 4(p-2)^2 - 8(p-2) + 1
= 4(p^2-2p-2p+4) - 8(p-2) + 1
= (4p^2-8p-8p+16) - (8p-16) + 1
= 4p^2-16p+16 - 8p +16 + 1
= 4p^2 - 24p + 33

7.It is given that f(x)=(x+2)(x-2)+ax+b. If f(2)=2 and f(-2)=4,find the values of a and b

f(2) = (2+2)(2-2) + a * 2 + b = 2
2a + b = 2
b = 2 - 2a --------------------- (1)
f(-2) = (-2+2)(-2-2) + a * (-2) + b = 4
-2a + b = 4
b = 4 + 2a ----------------------(2)
from eq(1) & eq (2), we get
2 - 2a = 4 + 2a
4a = -2
a = - 1/2
b = 2 - 2(-1/2)
= 2 + 1
= 3

8.It is given that f(x)=(x+3)(x+2)-k2次 and f(k)=2k
a. Find the value of k

f(x)=(x+3)(x+2)-k^2
f(k) = (k+3)(k+2) - k^2 = 2k
k^2 +3k + 2k + 6 - k^2 = 2k
5k + 6 = 2k
3k = -6
k = -2

b. Hence,solve for x if 2x-f(x)=0

2x-f(x)=0
f(x) = 2x
(x+3)(x+2)-(-2)^2 = 2x
x^2 + 3x + 2x +6 - 4 = 2x
x^2 + 3x + 2 = 0
(x+1)(x+2) = 0
x+1=0 or x+2=0
x = -1 or -2

9. In the figure,the height of a trapezium is x cm and the lengths of AD and BC are (2x+1) cm and (3x-2) cm respectively
a. Write a function S(x) to express the area (in cm2次) of the trapezium

Area of trapezium = 1/2 * height * (AD + BC)
S(x) = 1/2 * x * ((2x + 1) + (3x - 2))
S(x) = 1/2 * x * (5x - 1)
S(x) = (5x^2 - x)/2

b. Find the lengths of AD and BC if the area of the trapezium is 60cm2次

S(x) = (5x^2 - x)/2 = 60
(5x^2 - x) = 120
5x^2 - x - 120 = 0
(x-5)(5x+24) = 0
x - 5 = 0 or 5x + 24 = 0
x = 5 or - 24/5
x = 5 or -4.8 (rejected) (since x will never be -ve number or length of BC will be -ve, which is impossible)

AD and BC are (2x+1) cm and (3x-2) cm respectively
AD = 2(5) + 1 cm
= 11 cm
BC = 3(5) - 2 cm
= 13 cm

I hope this can help your understanding. =)