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JEE Main 20248 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Functions

Let f(x)=\ array ccc -a & if & -a ≤ x ≤ 0 x+a & if & 0 0 and...

Let f(x)=\left\{−a if −a≤x≤0x+a if 0<x≤a\begin{array}{ccc}-\mathrm{a} & \text{ if } & -\mathrm{a} \leq x \leq 0 \\ x+\mathrm{a} & \text{ if } & 0 < x \leq \mathrm{a}\end{array}\right. where a>0\mathrm{a}>0 and g(x)=(f(x∣)−∣f(x)∣)/2\mathrm{g}(x)=(f(x \mid)-|f(x)|) / 2. Then the function g:[−a,a]→[−a,a]g:[-a, a] \rightarrow[-a, a] isLet f(x)=\ array ccc -a & if & -a ≤ x ≤ 0 x+a & if & 0 0 and g(x)=(f(x )-|f(x)|) / 2 . Then the function g:[-a, a]Let f(x)=\ array ccc -a & if & -a ≤ x ≤ 0 x+a & if & 0 0 and g(x)=(f(x )-|f(x)|) / 2 . Then the function g:[-a, a]Let f(x)=\ array ccc -a & if & -a ≤ x ≤ 0 x+a & if & 0 0 and g(x)=(f(x )-|f(x)|) / 2 . Then the function g:[-a, a]Let f(x)=\ array ccc -a & if & -a ≤ x ≤ 0 x+a & if & 0 0 and g(x)=(f(x )-|f(x)|) / 2 . Then the function g:[-a, a]

Options

  1. Aneither one-one nor onto.
  2. Bonto.
  3. Cboth one-one and onto.
  4. Done-one.

Correct answer

A. neither one-one nor onto.

Step-by-step solution

y=f(x)y=f(x) Let f(x)=\ array ccc -a & if & -a ≤ x ≤ 0 x+a & if & 0 0 and g(x)=(f(x )-|f(x)|) / 2 . Then the function g:[-a, a]
y=f∣x∣y=f|x|Let f(x)=\ array ccc -a & if & -a ≤ x ≤ 0 x+a & if & 0 0 and g(x)=(f(x )-|f(x)|) / 2 . Then the function g:[-a, a]
y=∣f(x)∣y=|f(x)|Let f(x)=\ array ccc -a & if & -a ≤ x ≤ 0 x+a & if & 0 0 and g(x)=(f(x )-|f(x)|) / 2 . Then the function g:[-a, a]
g(x)=f(∣x∣)−∣f(x)∣2g(x)=\frac{f(|x|)-|f(x)|}{2}Let f(x)=\ array ccc -a & if & -a ≤ x ≤ 0 x+a & if & 0 0 and g(x)=(f(x )-|f(x)|) / 2 . Then the function g:[-a, a]
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