Let $f(\mathrm{x})=\left\{\begin{array}{cl}-\mathrm{a} & \text { if }-\mathrm{a} \leq \mathrm{x} \leq 0 \\ \mathrm{x}+\mathrm{a} & \text { if } 0 0$ and $\mathrm{g}(\mathrm{x})=((f|\mathrm{x}|)-|f(\mathrm{x})|) / 2$. Then the function $\mathrm{g}:[-\mathrm{a}, \mathrm{a}] \rightarrow [-\mathrm{a}, \mathrm{a}]$ is

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Sets Relations and Functions

Let

f (x) = {- a x + a if - a \leq x \leq 0 if 0 < x \leq a

where

a > 0

and

g (x) = ((f ∣ x ∣) - ∣ f (x) ∣) /2

.
Then the function

g : [- a, a] \to [- a, a]

Answer: (A)

f

is odd-symmetric-like on

[- a, a]

f (∣ x ∣)

mirrors

f

's positive-side branch, and

∣ f (x) ∣

reflects the negative-side branch (constant

- a

) up to

+ a

.
Sketching

y = f (∣ x ∣)

and

y = ∣ f (x) ∣

and forming

g (x) = \frac{f ( ∣ x ∣ ) - ∣ f ( x ) ∣}{2}

shows

g

0

[0, a]

and equals

- a

on part of

[- a, 0)

then rises — the resulting piecewise function is neither injective (repeats values) nor surjective onto all of

[- a, a]

.
So

g

is neither one-one nor onto.

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option