Question 71

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 $$g(x) = (f(x) - |f(x)|)/2$$. Then the function $$g : [-a, a] \rightarrow [-a, a]$$ is :

Name: 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) = (f(x) - |f(x)|)/2. Then the function g : [-a, a] \\rightarrow [-a, a] is :
Uploaded: 2026-05-29T18:19:16.347725+05:30
Duration: 3 min 28 s
Description: 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) = (f(x) - |f(x)|)/2. Then the function g : [-a, a] \\rightarrow [-a, a] is :

Solution

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

$$If(f(x)<0\Rightarrow g(x)=f(x)=-a)$$

$$If(f(x)>0\Rightarrow g(x)=0)$$

So range = ({-a, 0})

→ Not one-one, not onto

Neither one-one nor onto

Get AI Help

Video Solution

Create a FREE account and get:

Free JEE Mains Previous Papers PDF
Take JEE Mains paper tests

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 $$g(x) = (f(x) - |f(x)|)/2$$. Then the function $$g : [-a, a] \rightarrow [-a, a]$$ is :

Solution

Get AI Help

Video Solution

Download resources