Question

The range of $f\left(x\right)=\frac{1}{4\cos x-3}ϵ(-\infty ,-\frac{1}{a}]\cup [b,\infty ]$ Find $a+b$

Answer 1

$-1\le \cos x\le 1$
or $-4\le 4\cos x\le 4$
or $-7\le 4\cos x-3\le 1$
or $-7\le 4\cos x-3<0$ or $0<4\cos x-3\le 1$ $\because 4\cos x-3\ne 0$
or $-\frac{1}{7}\ge \frac{1}{4\cos x-3}>-\infty$ or $\infty >\frac{1}{4\cos x-3}\ge 1$
or $\frac{1}{4\cos x-3}ϵ(-\infty ,-\frac{1}{7}]\cup [1,\infty ]$
Therefore, $a+b=7+1=8$
Ans: 8