First slide

Functions (XII)

Question

$If f (x) is continuous and increasing function such that domain of g (x) = \sqrt{f (x) - x} be R and h (x) = \frac{1}{1 - x}, then$
$the domain of ϕ (x) = \sqrt{f (f (f (x))) - h (h (h (x)))} is$

Moderate

A

$R$
B

${0, 1}$
C

$R - {0, 1}$
D

$R - (0, 1)$

Solution

$h (x) = \frac{1}{1 - x}, x \neq 1 \Rightarrow h (h (x)) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{x - 1}{x}, x \neq 0, 1$
$\Rightarrow h (h (h (x))) = \frac{1}{1 - (\frac{x - 1}{x})} = x, x \neq 0, 1$
$\Rightarrow g (x) = \sqrt{f (x) - x} \geq 0, f o r a l l x \in R \Rightarrow f (x) \geq x \Rightarrow f (f (x)) \geq f (x) \geq x (\sin c e f (x) i s i n c r e a \sin g f u n c t i o n)$
$\Rightarrow s i m i l a r l y f (f (f (x))) \geq x \Rightarrow f (f (f (x))) - x \geq 0, f o r a l l x \in R - \{0, 1\} \Rightarrow φ (x) = \sqrt{f (f ((f (x)))) - x} i s d e f i n e d f o r a l l x \in R - \{0, 1\}$

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