+279 1. Given f(x)=3ax^2+2bx+c with a≠0, when 0≤x≤1 |f(x)|≤1, please determine the maximum value of a.
2. Given
- m is the greater root of the equation (1999x)^2-1998x2000x-1=0
- n is the lesser root of the equation x^2+1998x-1999=0,
please determine the value of m-n!
+118703 1. Given f(x)=3ax^2+2bx+x with a≠0, when 0≤x≤1 |f(x)|≤1, please determine the maximum value of a.
f(0)=0
f(1)=3a+2b+1
−1≤3a+2b+1and3a+2b+1≤+1−2≤3a+2band3a+2b≤0−2−2b≤3aand3a≤−2b−2−2b3≤aanda≤−2b3−2b3−23≤aanda≤−2b3−2b3−23≤a≤−2b3
I want the biggest 'a' so it follows that
a=−2b3b=−3a2
It also follows that a is positive and b is negative
so the question becomes
f(x)=3ax2+2bx+xf(x)=3ax2+2∗−3a2x+xf(x)=3ax2−3ax+xorf(x)=x(3ax−3a+1)
with a≠0, when 0≤x≤1 |f(x)|≤1, please determine the maximum value of a.
and f(1)= 3a+2b+1 = 3a-3a+1 = 1
So f(x) is a concave up parabola passing through (0,0) and (1,1)
f′(x)=6ax−3a+1statptiswhenf′(x)=06ax−3a+1=06ax=3a−1x=3a−16aorx=12−16a
Now a is positive and as big as possible so I am going to assume that an 'a' exists such that
the stationary point will lie between x=0 and x=0.5
This means that the stationary point (minimim) lies in the designated region of the graph.
So the minimum value must be greater than or equal to -1 (we already know it is less than 1)
The minimum value of f(x) is
f(3a−16a)=3a−16a[(3a∗3a−16a)−3a+1]≥−13a−16a[(3a−1−6a+22)]≥−1rememberingthata>0(3a−1)(3a−1−6a+2)≥−12a(3a−1)(−3a+1)≥−12a(3a−1)(3a−1)≤12a9a2−6a+1≤12a9a2−18a+1≤0If this was set to y it would be a concave up parabola so the function will be less than zero between the roots. Find the roots.a=18±√324−3618a=1±2√23So the biggest value of a isa=1+2√23a≈1.9428
Here is the graph
