First slide

Functions (XII)

Question

$Let f (x) be defined on [- 2, 2] and is given by f (x) = \{\begin{array}{cc} - 1, & - 2 \leq x \leq 0 \\ x - 1, & 0 \leq x \leq 2 \end{array}, then f (| x |) is defined as$

Moderate

A

$f (| x |) = \{\begin{array}{cc} 1 & - 2 \leq x \leq 0 \\ 1 - x, & 0 < x \leq 2 \end{array}$
B

$f (| x |) = x - 1 \forall x \in R$
C

$f (| x |) = \{\begin{array}{cc} - x - 1, & - 2 \leq x \leq 0 \\ x - 1, & 0 < x \leq 2 \end{array}$
D

$none of these$

Solution

$\begin{aligned} We have f (x) = \{\begin{array}{lc} - 1, & - 2 \leq x \leq 0 \\ x - 1, & 0 \leq x \leq 2 \end{array} \\ \begin{aligned} f (⌊ x) & = \{\begin{array}{cc} - 1, & - 2 \leq | x | \leq 0 \\ | x | - 1, & 0 \leq | x | \leq 2 \end{array} \\ \Rightarrow f (⌊ x ∣) & = | x | - 1, 0 \leq | x | \leq 2 \end{aligned} \\ (as - 2 \leq | x | \leq 0 is not possible) \\ \Rightarrow f (| x |) = \{\begin{array}{cc} - x - 1, & - 2 \leq x \leq 0 \\ x - 1, & 0 < x \leq 2 \end{array} \end{aligned}$

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