Let f(x) = begin{cases} -a & text{if } -a leq | vedclass

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(A) Given $f(x) = \begin{cases} -a & -a \leq x \leq 0 \ x+a & 0 < x \leq a \end{cases}$.First,we find $f(|x|)$:For $x \in [-a, 0]$,$|x| \in [0, a]$,s

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neither one-one nor onto.

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(A) Given $f(x) = \begin{cases} -a & -a \leq x \leq 0 \ x+a & 0 First,we find $f(|x|)$:For $x \in [-a, 0]$,$|x| \in [0, a]$,so $f(|x|) = |x| + a = -x + a$.For $x \in (0, a]$,$|x| \in (0, a]$,so $f(|x|) = |x| + a = x + a$.Thus,$f(|x|) = \begin{cases} -x+a & -a \leq x \leq 0 \ x+a & 0 Next,we find $|f(x)|$:For $x \in [-a, 0]$,$f(x) = -a$,so $|f(x)| = |-a| = a$.For $x \in (0, a]$,$f(x) = x+a$,so $|f(x)| = |x+a| = x+a$.Thus,$|f(x)| = \begin{cases} a & -a \leq x \leq 0 \ x+a & 0 Now,$g(x) = \frac{f(|x|) - |f(x)|}{2}$:For $x \in [-a, 0]$,$g(x) = \frac{(-x+a) - a}{2} = \frac{-x}{2}$.For $x \in (0, a]$,$g(x) = \frac{(x+a) - (x+a)}{2} = 0$.So,$g(x) = \begin{cases} -x/2 & -a \leq x \leq 0 \ 0 & 0 Analyzing $g(x)$:$1$. One-one: For $x \in (0, a]$,$g(x) = 0$. Since $g(0.1) = 0$ and $g(0.2) = 0$,it is not one-one.$2$. Onto: The range of $g(x)$ is $[0, a/2]$. Since the codomain is $[-a, a]$,the range is not equal to the codomain,so it is not onto.Therefore,$g(x)$ is neither one-one nor onto.

Column-$I$	Column-$II$
$A$. $f$ is one-one and onto,if	$1$. $A = R^{+}, B = R$
$B$. $f$ is one-one but not onto,if	$2$. $A = B = R$
$C$. $f$ is onto but not one-one,if	$3$. $A = R, B = R^{+}$
$D$. $f$ is neither one-one nor onto,if	$4$. $A = B = R^{+}$

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

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