13.51 The function $f(x)=a \sin x+b$ has a maximum value of 1 and a minimum value of -7. Then the maximum value of $b \sin ^{2} x-a \cos ^{2} x$ is (where $a, b$ are constants)
(A) 5 or 4.
(B) 4 or -5.
(C) 4 or -3.
(D) 5 or -3.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] From the given, if $a>0$, we have $b-a \leqslant a \sin x+b \leqslant a+b$, which implies $\left\{\begin{array}{l}b+a=1, \\ b-a=-7 .\end{array}\right.$ Solving this, we get $\quad\left\{\begin{array}{l}a=4, \\ b=-3 .\end{array}\right.$

If $a<0$, we have $\quad b+a \leqslant a \sin x+b \leqslant b-a$, which implies $\left\{\begin{array}{l}b-a=1, \\ b+a=-7 .\end{array}\right.$ Solving this, we get $\quad\left\{\begin{array}{l}a=-4, \\ b=-3 .\end{array}\right.$

Let $g(x)=b \sin ^{2} x-a \cos ^{2} x$. Then, when $a=4, b=-3$, $g(x)=-3 \sin ^{2} x-4 \cos ^{2} x=-3-\cos ^{2} x \leqslant-3$,

Thus,
$$
g(x)_{\max }=-3 \text {. }
$$

When $a=-4, \quad b=-3$,
$$
g(x)=-3 \sin ^{2} x+4 \cos ^{2} x=4-7 \sin ^{2} x \leqslant 4,
$$

Thus,
$$
g(x)_{\max }=4 \text {. }
$$

Therefore, the answer is $(\mathrm{C})$.