Let f ( x ) = { - a if - a ? x ? 0 x + a if 0 0 and g ( x ) = ( f ( | x | ) - | f ( x ) | ) 2 . Then the function g : [ - a , a ] ? [ - a , a ] is [2024]

Question

Let $f (x) = {\begin{matrix} - a if - a \leq x \leq 0 \\ x + a if 0 < x \leq a \end{matrix}$ where $a > 0$ and $g (x) = \frac{(f (| x |) - | f (x) |)}{2}$ . Then the function $g :$ $[- a, a] \to [- a, a]$ is [2024]

Smart Achievers · Accepted Answer

(4)

$y = f (x) = {\begin{matrix} - a, & - a \leq x \leq 0 \\ x + a, & 0 < x \leq a \end{matrix}$

$y = f | x | = {\begin{matrix} - a, & - a \leq | x | \leq 0 \\ | x | + a, & 0 < | x | \leq a \end{matrix}$

but $| x | < 0$ is not possible.

$f | x | = {\begin{matrix} - x + a, & - a \leq x < 0 \\ x + a, & 0 < x \leq a \end{matrix}$

$y = | f (x) | = {\begin{matrix} a, & - a \leq x \leq 0 \\ x + a, & 0 < x \leq a \end{matrix}$

$g (x) = \frac{f | x | - | f (x) |}{2} = {\begin{matrix} \frac{- x + a - a}{2} = - \frac{x}{2} if - a \leq x \leq 0 \\ \frac{x + a - x - a}{2} = 0 if 0 < x \leq a \end{matrix} \begin{matrix} \end{matrix}$

$g : [- a, a] \to [- a, a]$ is neither one-one nor onto as set $[0, a]$ has only one image i.e. 0.

Solution

Q.
Let $f (x) = {\begin{matrix} - a if - a \leq x \leq 0 \\ x + a if 0 < x \leq a \end{matrix}$ where $a > 0$ and $g (x) = \frac{(f (| x |) - | f (x) |)}{2}$ . Then the function $g :$ $[- a, a] \to [- a, a]$ is [2024]

1 onto.

2 both one-one and onto.

3 one-one.

4 neither one-one nor onto.

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