Q.

Let f(x)={-a if -ax0x+a if 0<xa where a>0 and g(x)=(f(|x|)-|f(x)|)2. Then the function g: [-a,a][-a,a] is           [2024]

1 onto.  
2 both one-one and onto.  
3 one-one.  
4 neither one-one nor onto.  

Ans.

(4)

   y=f(x)={-a,-ax0x+a,0<xa

   y=f|x|={-a,-a|x|0|x|+a,0<|x|a

   but |x|<0 is not possible.

   f|x|={-x+a,-ax<0x+a,0<xa

   y=|f(x)|={a,-ax0x+a,0<xa

   g(x)=f|x|-|f(x)|2={-x+a-a2=-x2 if -ax0x+a-x-a2=0 if 0<xa

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