First slide

Functions (XII)

Question

The domain of the function $f (x) = \cos^{- 1} (\frac{2 - | x |}{4}) + [\log (3 - x)]^{- 1} is$

Moderate

A

[– 6, 3)\{2}
B

[– 6, 2) ∪ (2, 3]
C

[– 6, 3]
D

[– 6, 3)

Solution

$\cos^{- 1} (\frac{2 - | x |}{4}) is defined for - 1 \leq \frac{2 - | x |}{4} \leq 1 \begin{array}{l} \Rightarrow - 4 \leq 2 - | x | \leq 4 \Rightarrow - 6 \leq - | x | \leq 2 \\ \Rightarrow - 2 \leq | x | \leq 6 \Rightarrow | x | \leq 6 \Rightarrow - 6 \leq x \leq 6 . . . . (1) \end{array}$
$\frac{1}{\log (3 - x)}$ is defined if log (3 – x) ≠ 0 and 3 – x > 0
$\Rightarrow 3 - x \neq e^{0} = 1 and x < 3 \Rightarrow x \neq 2 and x < 3$
From Eq. (1) and (2), we get domain of f
= [– 6, 6] ∩ ((– ∞, 3) \ {2}) = [– 6, 3)\{2}.

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