Expert: Sherman D. - 12/13/2008

Question
1. Solve the inequality:

    10 + x² > x(x – 2)


2. The fifth term of an arithmetic series is 14 and sum of the first three terms of the series is -3.

a) Use algebra to show that the first term of the series is -6 and calculate the common difference of the series.
b) Given that nth term of the series is greater than 282, find the least possible value of n.


3. The fourth term of an arithmetic series is 3k, where k is a constant, and the sum of the first six terms of the series is 7k + 9.

a) Show that the first term of the series is 9 – 8k

b) Find an expression for the common difference of the series in terms of k.

Given that the seventh term of the series is 12, calculate:
c) The value of k
d) The sum of the first 20 terms of the series.


4.   

a) Show that eliminating y from the equation 2x + y = 8  and 3x² + xy = 1. Produces the equation:  
      
                        x² + 8x – 1 = 0

b) Hence solve the simultaneous equations  2x + y = 8   and  3x² + xy = 1. Giving your answer in the form a + b sqrt 17, where ‘a’ and ‘b’ are integers.


5. The diagram shows a sketch of the curve y = f(x). The point B (0, 0) lies on the curve and the point A (3, 4) is a maximum point. The line y = 2 is an asymptote.

Sketch the following and in each case give the co-ordinates of the new positions of A and B and state the equation of the asymptote:

a) f (2x)        b) ½ f(x)     c) f(x) -2     d)f(x+3)     e) f(x) + 1


6. The line l has equation 2x - y – 1 = 0. The line m passes through the point A (0, 4) and is perpendicular to the line l.

a) Find an equation of m and show that the lines l and m intersect at the point P (2, 3). The line n passes through the point B (3, 0) and is parallel to the line m.

b) Find an equation of n and hence find the co-ordinates of the point Q where the lines l and n intersect.


Answer
1.) Solve the inequality:
10 + x^2 > x(x – 2)

10 + x^2 > x^2 - 2x
-2x < 10
x > -5

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2. The fifth term of an arithmetic series is 14 and sum of the first three terms of the series is -3.

a) Use algebra to show that the first term of the series is -6 and calculate the common difference of the series.

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

a(5) = a1 + d(5 - 1)
14 = a1 + 4d

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

S(3) = (3/2)(a1 + an)
-3 = (3/2)(a1 + an)
-6 = 3(a1 + an)
-2 = a1 + an

a(3) = a1 + d(3 - 1)
a(3) = a1 + 2d

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

14 = -d - 1 + 4d
14 = 3d - 1
3d = 15
d = 5

a1 = -(5 + 1)
a1 = -6

ANS:
a1 = -6
d = 5

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b.) Given that nth term of the series is greater than 282, find the least possible value of n.

a(283) = -6 + 5(283 - 1)
a(283) = -6 + 5(282)
a(283) = 1404

ANS : 1404

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3. The fourth term of an arithmetic series is 3k, where k is a constant, and the sum of the first six terms of the series is 7k + 9.

a) Show that the first term of the series is 9 – 8k

a(4) = a1 + d(4 - 1)
3k = a1 + 3d
a1 = 3k - 3d

a(6) = a1 + d(6 - 1)
a(6) = a1 + 5d

S(6) = (6/2)(a1 + a1 + 5d)
7k + 9 = 3(2a1 + 5d)
7k + 9 = 3(2(3k - 3d) + 5d)
7k + 9 = 3(6k - 6d + 5d)
7k + 9 = 3(6k - d)
7k + 9 = 18k - 3d
-11k + 9 = -3d
(11/3)k - 3 = d

a1 = 3k - 3((11/3)k - 3)
a1 = 3k - 11k + 9
a1 = 9 - 8k

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b) Find an expression for the common difference of the series in terms of k.

d = (11/3)k - 3 or (11k - 3)/3

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Given that the seventh term of the series is 12, calculate:
c) The value of k

a(7) = 9 - 8k + ((11k - 3)/3)(7 - 1)
12 = 9 - 8k + 6((11k - 3)/3)
12 = 9 - 8k + 2(11k - 3)
12 = 9 - 8k + 22k - 6
12 = 3 + 14k
14k = 9
k = (9/14)

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d) The sum of the first 20 terms of the series.

a1 = 9 - 8(9/14)
a1 = 9 - (36/7)
a1 = (63 - 36)/7
a1 = (27/7)

a(20) = (27/7) + ((11(9/14) - 3)/3)(20 - 1)
a(20) = (27/7) + 19(((99/14) - 3)/3)
a(20) = (27/7) + 19((99 - 42)/14)/3)
a(20) = (27/7) + 19(3(57)/14)
a(20) = (54 + 3249)/14
a(20) = (3303/14)

thats what i got. to make sure, you can check it.

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a(2

4.

a) Show that eliminating y from the equation 2x + y = 8  and 3x² + xy = 1. Produces the equation:  
  
                       x² + 8x – 1 = 0

2x + y = 8
3x^2 + xy = 1

y = -2x + 8
3x^2 + (-2x + 8)x - 1 = 0
3x^2 - 2x^2 + 8x - 1 = 0
x^2 + 8x - 1 = 0

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b) Hence solve the simultaneous equations  2x + y = 8   and  3x² + xy = 1. Giving your answer in the form a + b sqrt 17, where ‘a’ and ‘b’ are integers.

using x^2 + 8x - 1 = 0 and the quadratic formula

x = (-b +/- sqrt(b^2 - 4ac)))/(2a)

x = (-8 +/- sqrt(8^2 - 4(1)(-1)))/(2(1))
x = (-8 +/- sqrt(64 + 4))/2
x = (-8 +/- sqrt(68))/2
x = (-8 +/- sqrt(4 * 17))/2
x = (-8 +/- 2sqrt(17))/2
x = -4 + sqrt(17)

a = -4 and b = 1

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5. The diagram shows a sketch of the curve y = f(x). The point B (0, 0) lies on the curve and the point A (3, 4) is a maximum point. The line y = 2 is an asymptote.

Sketch the following and in each case give the co-ordinates of the new positions of A and B and state the equation of the asymptote:

a) f (2x)        b) ½ f(x)     c) f(x) -2     d)f(x+3)     e) f(x) + 1

check with answers.yahoo.com

also for a graph, go to www.quickmath.com

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6. The line l has equation 2x - y – 1 = 0. The line m passes through the point A (0, 4) and is perpendicular to the line l.

a) Find an equation of m and show that the lines l and m intersect at the point P (2, 3). The line n passes through the point B (3, 0) and is parallel to the line m.

y = 2x - 1

(0,4), slope = (-1/2)
4 = (-1/2)(0) + b
b = 4

y = (-1/2)x + 4

2x - 1 = (-1/2)x + 4
4x - 2 = -x + 8
5x = 10
x = 2

y = 2(2) - 1
y = 4 - 1
y = 3

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b) Find an equation of n and hence find the co-ordinates of the point Q where the lines l and n intersect.

(3,0), m = (-1/2)
0 = (-1/2)(3) + b
0 = (-1/2) + b
b = (1/2)

y = (-1/2)x + (1/2) or y = (-x + 1)/2 or y = -(x - 1)/2 or (-1/2)(x - 1) or (1/2)(-x + 1)

just take your pick, they are all the same equation.