368 Answers to Exercises and Solutions to Problems
and
f(1) = -p-q 1 -(1/2)(&l) e-p 5 -q+(1/2)(Jz-1) 5 (&i-1)-1 < 0.
Now, write x = cos0, so that
f(cos ~9) = sin B - pcos t9 - q = JC&Fsin(0 + 4) - q,
where d-sin 4 = -p and J-OS 4 = 1. f assumes its maximum
value when B + 4 = 7r/2, i.e. when
cos e = sin 4 = -p/J-,
and we have that
f(-P/&7) = AT2 - q*
Let 0 5 u, v 5 1. Then If(u) - f(v)1 5 &- 1. In particular,
~~-l=f(-p/~~)-f(O)<JZ-l~~~IJZ
*p2_<l*p>--1
and
JS+p = f(-P/Jiq) - f(1) 5 h- 1
* &&(Jz-1)-P
+ l+pās(3-2fi)-2(fi-l)p+p2
=k 2(dLl)p_<2-2Jzap5 -1.
Putting these facts together yields p = -1. Now,
f (0) 1 -(W)(Jz - 1) * q I (W(Jz+ 1)
and
f(-P/Jl+p2) = f(l/Jz) = A-4 I (l/2)@-1) * q 2 (1/2)@+1>,
so that q = (l/2)(4+ 1).
So far, we have established what p and q must be if the inequality is valid
for each x; we must now show that this choice works, i.e. that
l&Z+ x - (l/2)(&+ 1)l _< (l/2)1&- 11 for 0 5 x 5 1.
Since 0 5 (1 - &x)~ = 1 - 24~ + 2z2, it follows that
1-x2<2-2~x+x2*~iG5+/5-x.