One F.4 Amaths question

First assume the contrary is true, i.e. the two graphs intersect with each other.
Therefore, mx^2 + 4 = mx + 6
==> mx^2 - mx - 2 = 0
==> the two graphs intersect at x = [m (+-) sq root (m^2 + 8m) ] / 2m

Now if the determinant of the above equation is less than 0, i.e, (m^2 + 8m) < 0, then there is no real x values at which the two graphs intersect.
Therefore, if (m^2 + 8m) < 0, then the two graphs do not intersect.
Now consider the values of m as follows.
If m is +ve, (m^2 + 8m) can never be less than 0. Therefore m is never +ve.
If m is -ve, (m^2 + 8m) can only be less than 0 if -8 < m < 0.
The case of m = 0 should be considered separately. If m = 0, the two graphs also do not intersect (because the first graph becomes y = 4 and the second becomes y = 6. They become two vertical parallel lines.)

The answer is therefore, -8 < m <= 0.

Consider mx+6 = -mx^2+4 or mx^2 +mx+2 =0 .
To have no intersection, the discriminant must be less than zero.
This means m^2 -8m <0 or (m -8) m <0
This implies that 0< m <8

y=-mx^2+4 ...(1)
y=mx+6 ...(2)
combine (1) and (2),
-mx^2+4=mx+6
mx^2+mx+2=0
x= [-m +/- ( m^2-4(m)(2) )^-0.5]/2m
they don't interscet each other when there are no real roots for x
therefore,
m^2-8m<0
0<8
p.s. hope u can read this, 'cause its a bit difficult to type it in computer screen

2006-12-31 20:03:49 補充：
0

2006-12-31 20:04:27 補充：
0

2006-12-31 20:04:52 補充：
0 < m < 8