Expert: Sherman D. - 1/19/2009

Question
1)Simplify:

a) 10/sqrt5 + sqrt 20
b) 1/sqrt2 (2sqrt2 – 1) + sqrt2 (1-sqrt8)



2) Find the greatest or the least values of these functions:

a) f(x) = x² - 3x + 5
b) g(x) = 3 – 2x - x²



3) The sequence U1, U2, U3… Where U1 is a real number, is defined by Un+1 = Un² - 1. If U1 = 3, find U2, U3 and U4



4) The sum of the first two terms of an A.P. is 181 and the sum of the first four terms is 52. Find the sum of the first 8 terms.



5) Solve the simultaneous equations:
  
  2x + 3y = 5
  x² + 3xy = 4



6) Solve the inequality (x + 1)(x – 2)(x – 3) > 0



7) Find the derived function of :

a) f(x) = 3x² + 2x
b) f(x) = 1/x²



8) Find the equation of the line passing through the points A (3, 4) and B (-1, 2). Also, find :

i) The distance AB
ii) The equation of the normal at A


Answer
1)Simplify:

a) 10/sqrt5 + sqrt 20

if by this you mean
10/(sqrt(5) + sqrt(20))
10/(sqrt(5) + sqrt(5 * 4))
10/(sqrt(5) + 2sqrt(5))
10/(3sqrt(5))

Multiply top and bottom by sqrt(5)

10sqrt(5)/15
2sqrt(5)/3

but if you meant

(10/sqrt(5)) + sqrt(20)
(10 + sqrt(5 * 20))/sqrt(5)
(10 + sqrt(100))/sqrt(5)
(10 + 10)/sqrt(5)
20sqrt(5)/5
4sqrt(5)

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b) 1/sqrt2 (2sqrt2 – 1) + sqrt2 (1-sqrt8)

if by this you meant

(1/sqrt(2))(2sqrt(2) - 1) + sqrt(2)(1 - sqrt(8))
((2sqrt(2) - 1)/sqrt(2)) + sqrt(2) - sqrt(16)
2 - (1/sqrt(2)) + sqrt(2) - 4
(-1/sqrt(2)) + sqrt(2) + (-2 or 6)
(-1 + 2 + (-2 or 6)sqrt(2))/sqrt(2)
(1 + (-2 or 6)sqrt(2))/sqrt(2)
(sqrt(2) + (-2 or 6)(2)/2
(sqrt(2) + (-4 or 12))/2
(1/2)sqrt(2) + (-2 or 6)

sorry i didn't know if they wanted to keep it as a -4, or use both sqrt values which would look like this -(4) and -(-4)

if you meant

(1/(sqrt(2)(2sqrt(2) - 1)) + sqrt(2)(1 - sqrt(8))
you should have wrote it like that, or at least a better way for me to understand it.

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2) Find the greatest or the least values of these functions:

a) f(x) = x² - 3x + 5

the easiest way to find the max or min vertex, is to use this formula

x = -b/(2a)

x = -(-3)/(2(1))
x = 3/2

f(3/2) = (3/2)^2 - 3(3/2) + 5
f(3/2) = (9/4) - (9/2) + 5
f(3/2) = (9 - 18 + 20)/4
f(3/2) = (-9 + 20)/4
f(3/2) = 11/4

ANS : (1.5,2.75)

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b) g(x) = 3 – 2x - x²
becomes
g(x) = -x^2 - 2x + 3

x = (-(-2))/(2(-1))
x = (2)/(-2)
x = -1

g(1) = -(-1)^2 - 2(-1) + 3
g(1) = -1 + 2 + 3
g(1) = 4

ANS : (1,4)

If your wondering where i got the x = -b/(2a), look up the vertex formula. It should be there.

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3) The sequence U1, U2, U3… Where U1 is a real number, is defined by Un+1 = Un² - 1. If U1 = 3, find U2, U3 and U4

check with answers.yahoo.com

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4) The sum of the first two terms of an A.P. is 181 and the sum of the first four terms is 52. Find the sum of the first 8 terms.

S(n) = (n/2)(a1 + an)

S(2) = (2/2)(a1 + an)
181 = a1 + an
a(n) = -a1 + 181

S(4) = (4/2)(a1 + an)
52 = 2(a1 + an)
26 = a1 + an
a(n) = -a1 + 26

a(n) = a1 + d(n - 1)

a(2) = a1 + d(2 - 1)
-a1 + 181 = a1 + d
d = -2a1 + 181

a(4) = a1 + d(4 - 1)
-a1 + 26 = a1 + 3d
3d = -2a1 + 26
d = (-2/3)a1 + (26/3)

-2a1 + 181 = (-2/3)a1 + (26/3)
-6a1 + 543 = -2a1 + 26
-4a1 = -517
a1 = 129.25

d = -77.5

a(n) = 129.25 - 77.5(n - 1)

Checking

a(2) = 129.25 - 77.5
a(2) = 51.75

a(4) = 129.25 - 77.5(3)
a(4) = -103.25

S(2) = 129.25 + 51.75 = 181
S(4) = 2(129.25 - 103.25) = 2(26) = 2

so

a(8) = 129.25 - 77.5(8 - 1)
a(8) = 129.25 - 77.5(7)
a(8) = -413.25

S(8) = (8/2)(129.25 - 413.25)
S(8) = 4(-284)
S(8) = -1136

ANS : -1136

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5) Solve the simultaneous equations:

2x + 3y = 5
x² + 3xy = 4

2x + 3y = 5
3y = -2x + 5
y = (-2/3)x + (5/3) or y = (-2x + 5)/3

x^2 + 3((-2x + 5)/3)x = 4
x^2 - 2x^2 + 5x = 4
-x^2 + 5x - 4 = 0
-(x^2 - 5x + 4) = 0
-((x - 4)(x - 1)) = 0
x = 4 or 1

2(4) + 3y = 5
8 + 3y = 5
3y = -3
y = -1
or
2(1) + 3y = 5
2 + 3y = 5
3y = 3
y = 1

ANS : (4,-1) or (1,1)

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6) Solve the inequality (x + 1)(x – 2)(x – 3) > 0

Using www.quickmath.com

-1 < x < 2
or
x > 3

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7) Find the derived function of :

a) f(x) = 3x² + 2x
b) f(x) = 1/x²

if by this you mean to integrate it, then

a.)
f(x)∫ = x^3 + x^2 + C

b.)
f(x)∫ = -1/x

but if you mean to find the derivative, then

a.)
f(x)' = 6x + 2

b.)
f(x)' = -2/x^3

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8) Find the equation of the line passing through the points A (3, 4) and B (-1, 2). Also, find :

(3,4) and (-1,2)
m = (2 - 4)/(-1 - 3)
m = -2/(-4)
m = 1/2

(3,4) and (1/2)
4 = (1/2)(3) + b
4 = (3/2) + b
8 = 3 + 2b
2b = 5
b = (5/2)

ANS : y = (1/2)x + (5/2)

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i) The distance AB

(3,4) and (-1,2)

D = sqrt((-1 - 3)^2 + (2 - 4)^2)
D = sqrt((-4)^2 + (-2)^2)
D = sqrt(16 + 4)
D = sqrt(20)
D = 2sqrt(5)

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ii) The equation of the normal at A
sorry i've never done "equation of the normal"