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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+p”s(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.