If the interval in which $f(x) =\frac{x}{4}+\frac{4}{x};x≠0$ is strictly increasing is $(-∞,a) ∪ (b,∞)$, then

$a = -b$

The correct answer is Option (2) → $a = -b$

Given: $f(x)=\frac{x}{4}+\frac{4}{x},\; x\neq 0$

$f'(x)=\frac{1}{4}-\frac{4}{x^{2}}$

Require $f'(x)>0$:

$\frac{1}{4}-\frac{4}{x^{2}}>0 \Rightarrow \frac{1}{4}>\frac{4}{x^{2}} \Rightarrow x^{2}>16 \Rightarrow |x|>4$

Hence the function is strictly increasing on $\;(-\infty,-4)\cup(4,\infty)\;$, so $a=-4$ and $b=4$.