1. $|b| \leqslant 1$ is the ( ) for the inequality $a \cos x + b \cos 3 x > 1$ to have no solution with respect to $x$.
(A) necessary but not sufficient condition
(B) sufficient but not necessary condition
(C) necessary and sufficient condition
(D) neither sufficient nor necessary condition
,- 1 (A). $+b \alpha \cos 3 x \leqslant 1$ holds for all real numbers $x$.
By substituting $x=0, x$ into $\left\{\begin{array}{l}a+b \leqslant 1, \\ -a-b \leqslant 1\end{array}\right.$, we get $-1 \leqslant a+b \leqslant 1$.
By substituting $x=\frac{\pi}{3}, \frac{2 \pi}{3}$ into $\left\{\begin{array}{l}\frac{a}{2}-b \leqslant 1, \\ -\frac{a}{2}+b \leqslant 1\end{array}\right.$, we get
$$
\sim 1 \leqslant \frac{a}{2}-b \leqslant 1, -2 \leqslant -a+2 b \leqslant 2 \text{. }
$$
(1) + (2) gives $-3 \leqslant 3 b \leqslant 3, |b| \leqslant 1$.

The condition is not sufficient. When $|b| \leqslant 1$, the original inequality can have solutions (for example, when $b=\frac{1}{2}, a=3$, $3 \cos x+\frac{1}{2} \cos 3 x>1$ has a solution $x=0$).