Let `f_1:R→R,f_2:[0,∞)→R, f_3:R→R` and `f_4:R→[0,∞)` be defined by `f_1(x)={ ∣x∣` if `x<0`; `e^x` if `x≥0`;`f_2(x)=x^2`; `f_3(x)={ sin x` if `x<0` ; `x` if `x≥0`; `f_4(x)={ f_2(f_1(x))` if `x<0` `f_2(f_1(x))` if `x≥0` ​then `f_4` is

Options

A. `{:(A,B,C,D),(r,p,s,q):}`
B. `{:(A,B,C,D),(p,r,s,q):}`
C. `{:(A,B,C,D),(r,p,q,s):}`
D. `{:(A,B,C,D),(p,r,q,s):}`

Answer

D

Explanation

PLAN
(i) For such questions, we need ot property define the functions and then we draw their graphs.
(ii) From the graphs, we can examine the function for continuity. defferentiability, one-one and onto.
`f_(1)(x)={(-x",",x lt 0),(e^(x)",",x ge 0):}`
`f_(2)(x)=x^(2),x ge 0`
`f_(3)(x)={(sin x",",x lt 0),(x",",x ge 0):}`
`f_(4)(x)={(f_(2)(f_(1)(x))",",x lt 0),(f_(2)(f_(1)(x))-1",",x ge 0):}`
Now, `f_(2)(f_(1)(x))={(x^(2)",",x lt 0),(e^(2x)",",x ge 0):}`
`rArr f_(4)(x)={(x^(2)",",x lt 0),(e^(2x)-1",",x ge 0):}`
As `f_(4)(x) " continuous", f'_(4)(x)={(2x",",x lt 0),(2e^(2x)",",x gt 0):}`

`f_(4)(0)` is not defined. Its range is `[0, oo)`.
Thus, range`="codomain"=[0,oo),"thus "f_(4)` is onto.
Also, horizontal line (drawn parallel ot X-axis) meets hte curve more than once, thus function is not one-one.