Question

The domain of the function $f\left(x\right)\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}$ is equal to
(a) (−∞, −1) ∪ (1, 4)
(b) (−∞, −1] ∪ (1, 4]
(c) (−∞, −1) ∪ [1, 4]
(d) (−∞, −1) ∪ [1, 4)

Answer 1

$f\left(x\right)=\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}$
for $\sqrt{4-x}$
4 − x ≥ 0
i.e 4 ≥ x
i.e x ≤ 4
and $\frac{1}{\sqrt{x^2-1}}$
Since x² − 1 > 0 $\left(\because x^2-1=0\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}\frac{1}{\sqrt{x^2-1}}\right)$
i.e x² > 1
i.e x < − 1 or x > 1

for $\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}},$ Domain is (–∞, –1) ∪ (1, 4].