Maths MC

2009-01-08 5:32 am
The roots of 2x^2+4x-6=0 are the x-coordinates of the intersecting points of the graphs of y=2x^2+2x+3 and
A. y=2x+11
B. y=2x+5
C. y=-2x+9
D. y=-2x+3

Let f(x) and g(x) be two polynomials.When x+2 divides f(x) and g(x), the remainders are -4 and 4 respectively. Which of the followings is/are divisible by x+4
I. g[f(x+2)+2]-4
II. f[2-g(x/2)]+4
III. f(x/2)+g(x+2)
A. I and II only
B. I and III only
C. II and III only
D. I, II, III

Be quick~~!

回答 (3)

2009-01-08 5:49 am
✔ 最佳答案
want to find the root of 2x^2+4x-6=0
y=2x^2+2x+3
CONSIDER y=-2x+9, then the intersection point is given by the root of
-2x+9=2x^2+2x+3 or 2x^2+4x-6=0
So the answer is C. y=-2x+9
2
f(x) and g(x) be two polynomials.When x+2 divides f(x) and g(x), the remainders are -4 and 4 respectively
Let f(x)=(x+2)a(x)-4 and g(x)=(x+2)b(x)+4
For I. g[f(x+2)+2]-4
g[f(x+2)+2]-4
=(f(x+2)+4)b(f(x+2)+2)+4
=[(x+4)a(x+2)]b(f(x+2)+2)+4
which is divisible by x+4
II. f[2-g(x/2)]+4
f[2-g(x/2)]+4
=(4-g(x/2))a(2-g(x/2))
=[(x/2+2)b(x/2)]a(-(x/2+2)b(x/2)-2)
which is divisible by x+4
III. f(x/2)+g(x+2)
f(x/2)+g(x+2)
=(x/2+2)a(x/2)-4+(x+4)b(x+2)+4
=(x/2+2)a(x/2)+(x+4)b(x+2)
which is divisible by x+4
So the answer is D
2009-01-08 5:56 am
The roots of 2x^2+4x-6=0 are the x-coordinates of the intersecting points of the graphs of y=2x^2+2x+3 and ........

The key here is to organize the corresponsible simultaneous equation, and to merge them into one.

A:
y=2x^2+2x+3
y=2x+11
Thus, the merged equation: 2x^2+2x+3-2x-11=0
Merged equation: 2x^2-8=0 → Wrong choice here.

B:
y=2x^2+2x+3
y=2x+5
Thus, the merged equation: 2x^2+2x+3-2x-5=0
Merged equation: 2x^2-2=0 → Wrong choice here.

C:
y=2x^2+2x+3
y=-2x+9
Thus, the merged equation: 2x^2+2x+3+2x-9=0
Merged equation: 2x^2+4x+6=0 → Right choice!


D:
y=2x^2+2x+3
y=-2x+3
Thus, the merged equation: 2x^2+2x+3+2x-5=0
Merged equation: 2x^2+4x-2=0 → Wrong choice here.
--------------------
Let f(x) and g(x) be two polynomials.When x+2 divides f(x) and g(x), the remainders are -4 and 4 respectively. Which of the followings is/are divisible by x+4?

Thus, f(-2)=-4 and g(-2)=4 [According to the Remainder Theorem:if f(n)=z, the remainders will be z when x-n divides f(x).]
By the Remainder Theorem, the function f(x) is divisible by x+n provided that f(-n)=0.

Hence, if the function f(-4)=0, f(x) is then divisible by x+4.

I. g[f(x+2)+2]-4
= g[f(-4+2)+2]-4
= g[f(-2)+2]-4
= g(-4+2)-4
= g(-2)-4
= 4-4
= 0

II. f[2-g(x/2)]+4
= f[2-g(-4/2)]+4
= f[2-g(-2)]+4
= f(2-4)+4
= f(-2)+4
=-4+4
=0

III. f(x/2)+g(x+2)
= f(-4/2)+g(-4+2)
= f(-2)+g(-2)
=-4+4
=0

A. I and II only
B. I and III only
C. II and III only
D. I, II, III → Right choice!
參考: Evaluation by myself
2009-01-08 5:54 am
1.C
y=2x^2+2x+3
y=-2x+9
then 2x^2+2x+3=-2x+9
then2x^2+4x-6=0

2.D


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