Let `f(1):R to R, f_(2):[0,oo) to R, f_(3):R to R and f_(4):R to [0,oo)` be a defined by
`f_(1)(x)={{:(,|x|,"if "x lt 0),(,e^(x),"if "x gt 0):}:f_(2)(x)=x^(2),f_(3)(x)={{:(,sin x,"if x"lt 0),(,x,"if "x ge 0):}` and `f_(4)(x)={{:(,f_(2)(f_(1)(x)),"if "x lt 0),(,f_(2)(f_(1)(f_(1)(x)))-1,"if "x ge 0):}` Then, `f_(4)` is

Options

A. onto but not one-one
B. neither continuous nor one-one
C. differentiable but not one-one
D. continuous and one-one

Answer

A

Explanation

For `x lt 0` we have
`f_(1)(x)=|x|-x and f_(2)(x)=x^(2)`
`therefore f_(4)(x)=f_(2)(f_(1)(x))=f_(2)(-x)=(-x)^(2)=(-x)^(2)=x^(2)`
For `x gt 0`, we have
`f_(1)(x)=e^(x) and f_(x)(x)=x^(2)`
`therefore f_(4)(x)=f_(2)(f_(1)(x))-1=f_(2)(e^(x))-1=(e^(x))^(2)-1=e^(2x)-1`
`"Thus", f_(4)(x)={{:(,x^(2),"if "x lt 0),(,e^(2x)-1,"if "x ge 0):}`
The graph, of `f_(4):R to [0,oo)` is shown in Fig. 15