S5 Math Ex.

[匿名]

2013-06-19 2:24 am

Please help me to solve the following questions:
1.Let f(x) be a quadratic function. If the coordinates of the vertex of the graph
of y = f(x) are (3,2), which of the following must be true?

a. The roots of the equation f(x) - 2 = 0 are rational numbers
b. The roots of the equation f(x) = 0 are integers
c. There are real roots in the equation f(x) - 3 = 0
d. There are no real roots in the equation f(x) + 1 = 0

2. If x^2 + y^2 - 2x + 8y - k = 0 represents a real circle, then

3. If P is a moving point in the rectangular coordinate plane such that P is
equidistant from the points ( -10 , 5 ) and ( 12 , -6 ). then the locus of P is a

a. square b. circle c. parabola d. straight line

4. If the solution of ax - b > 0 is x <2, the solution of bx - a > 0 is

5. A collection of discs consists of 10 different pop music discs and 6 different
classical music discs. How many ways are there to select 5 discs including
at least 1 classical music disc?

6. A box contains six ballos numbered 1, 1, 2, 2, 3 and 3 respectively. Mary
repeats drawing two balls at a time randomly from the box with replacement
until the numbers drawn are the same. Find the probability that she needs
exactly two times.

回答 (1)

Tony

2013-06-19 4:13 pm

✔ 最佳答案

1. vertex of the graph of y = f(x) are (3,2), and f(x) is a quadratic function
f(x) = a(x-3)^2+2
f(x) - 2 = a(x-3)^2
1(a) true(=must true),
x=3(rational number) is double roots of the equation f(x) - 2 = 0,
1(b) false(≠must false),
let a=1,
f(x) = (x-3)^2+2 = x^2-6x+11 has no real roots, so roots are not integer
1(c) false
f(x)-3 = a(x-3)^2-1 = ax^2-6ax+(9a-1)
discrimination=(-6a)^2 - 4(a)(9a-1)=36a^2-36a^2+4a=4a
if a<0,then discrimination<0,no real roots
1(d) false
f(x)+1 = a(x-3)^2+3 = ax^2-6ax+(9a+3)
discrimination=(-6a)^2 - 4(a)(9a+3)=36a^2-36a^2-12a= -12a
if a<0,then discrimination>0,it has real roots

2. what do you want to ask? do you want to find range of k?
x^2 + y^2 - 2x + 8y - k = 0
(x^2 - 2x + 1 - 1) + (y^2 +8y +16 - 16) - k = 0
[(x - 1)^2 - 1] + [(y + 4)^2 -16] - k = 0
(x - 1)^2 + (y + 4)^2 = [ √(k +17) ] ^ 2
if it represents a real circle, k + 17>0 => k>-17
the center of the circle is (1,-4)

3. the answer is D
let P be (x,y)
√ { [x - (-10)]^2 + (y - 5)^2 } = √ { (x - 12)^2 + [y - (-6)]^2 }
{ [x + 10)]^2 + (y - 5)^2 } = { (x - 12)^2 + [y + 6)]^2 }
x^2 + 20x +100 + y^2 - 10y +25 = x^2 - 24x +144 +y^2 +12y +36
(20 + 24)x + (-10 - 12)y + (100 + 25 - 144 - 36) = 0
44x - 22y -55 = 0
4x - 2y -5 = 0

4. do you have typing error on x<2? should it be x>2>
the answer is x > 1/2
the solution of ax - b > 0 is x > 2
ax > b
x > b/a=2
bx - a > 0
x > a/b=1/2

5. (6C1)(10C4)=1260

6. 3/15
the possible result: 1 1, 2 2 , 3 3
the required probability = (2C2) / (6C2) + (2C2) / (6C2) + (2C2) / (6C2)
=3(2C2) / (6C2) = 3/15

參考: myself

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