First slide

Functions (XII)

Question

The domain of f(x)= $\frac{\sqrt{- \log_{0.3} {(x - 1)}_{}}}{\sqrt{x^{2} + 2 x + 8}}$ is

Moderate

A

(1,4)
B

(-2,4)
C

(2,4)
D

[2, $\infty$ )

Solution

We have f(x)= $\frac{\sqrt{- \log_{0.3} {(x - 1)}_{}}}{\sqrt{x^{2} + 2 x + 8}}$ ,for domain
$\frac{- \log_{0.3} (x - 1)}{x^{2} + 2 x + 8} \geq 0 \Rightarrow \frac{\log_{0.3} (x - 1)}{x^{2} + 2 x + 8} \leq 0, \sin c e x^{2} + 2 x + 8 i s$
always positive as x²+2x+8=(x+1)²+7
$∴$ log_0.3(x-1) $\leq$ 0
(x-1) $\geq$ (0.3)⁰
$∴$ x $\geq$ 2

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