中四附加數學

(a)
f(1/2) < 0
5(1/2)^2+b(1/2)+c < 0
5+2b+4c<0
2b<-5-4c
b<(-5-4c)/2
So b is a negative number
So b^2>1/4(25+40c+16c^2)
b^2>4c^2+10c+25/4
Now
b^2-4ac
=b^2-20c
>4c^2+10c+25/4-20c
=4c^2-10c+25/4
=1/4(16c^2-40c+25)
=1/4(4c-5)^2
>0
So f(x) = 0 has two distinct real roots
(b)
(i)
f(x)=(x-A)(x-B)=x^2-(A+B)x+AB
So A+B=-b/5
AB=c/5
Since c>0 and b is negative both A and B should be larger than 0
Because f(0)>0, f(1/2)<0
So there is one root between 0 and 1/2, also since f(1/2)<0 and consider the graph of quadratic equation there should exist one value B>0 such that f(B)=0
So 0 < A < 1/2 < B
(ii)
∣ A - 1/2 ∣ = ∣ B - 1/2 ∣
Case (i)
A-1/2=B-1/2
A=B
contradict the result of b(i)
Case (ii)
A-1/2=-B+1/2
A=1-B
So 0<1-B<1/2
B>1/2
That is 1/2<B<1
A+B<3/2
So -b/5<3/2
b>-15/2
Since b<(-5-4c)/2
2b+5<-4c
c>-(2b+5)/4
sub the minimum value of b=-15/2
0<c<2.5 (because c>0 by definition)
上面個位做錯了

2008-01-02 22:49:18 補充：
there should exist one value B&gt;1/2 such that f(B)=0

(a) f(1/2) &lt; 0
5/4+b/2+c &lt; 0
5+2b+4c &lt; 0
∴4c &lt; -5-2b
20c &lt; -25-4b
-20c &gt; 25-4b---------(1)
Discriminant=b^2-4(5)(c)
=b^2-20c
&gt; b^2-4b+25
=(b-2)^2+21 (By (1))
&gt; 0 for all value of b
∴f(x) has two distinct real roots

b(i)∵A and B are roots of f(x)=0
∴f(A)=f(B)=0
∴f(x)=(x-A)(x-B)=0
Product of root=c/5, which is &gt; 0 ∵c &gt; 0
∴We get the result that A and B are either both -ve or +ve
∵f(1/2) &lt; 0
∴5+2b+4c &lt; 0
∴b &lt; (-4c-5)/2---------(2)
∴We can conclude that b is -ve
∵Sum of root=-b/5, which is +ve when b is -ve
∴Combining with the above results, both A and B are positive and so 0 &lt; A &lt; B(∵A &lt; B)
By considering the graph of f(x), which is a parabola opening upwares, a -ve value of x states that it is in between the two roots
∴1/2 is somewhere between A and B
∴0 &lt; A &lt; 1/2 &lt; B

b(ii)∣ A - 1/2 ∣ = ∣ B - 1/2 ∣
A-1/2=B-1/2 or 1/2-B
A=B(rej.) or A+B=1
∴Sum of root =-b/5=1
b=-5
From (2),
(-4c-5)/2 &gt; b
∴-4c-5 &gt; -10
∴c &lt; 5/4
∵c &lt; 0
∴ 0 &lt; c &lt; 5/4

2008-01-03 19:24:52 補充：
上面果位才是做錯了