99 Given the function $f(x)=a \sin x+b \cos x$ satisfies $f\left(\frac{\pi}{3}\right)=1$. The range of the minimum value of the function $f(x)$ is
A. $(-\infty, 0)$
B. $(-\infty, 0]$
C. $(-\infty,-1)$
D. $(-\infty,-1]$
99 D. From the condition, $\sqrt{3} a+b=2$. And
$$
f(x)_{\min }=-\sqrt{a^{2}+b^{2}}=-2 \sqrt{a^{2}-\sqrt{3} a+1},
$$

From this, the maximum value of $f(x)_{\min }$ can be found to be -1, so the answer is D.