Let `f:(-(pi)/2,(pi)/2)toR` be given by `f(x)=(log(secx+tanx))^(3)`. Then

Options

A. f(x) is an odd function
B. f(x) is a one-one function
C. f(x) is an onto function
D. f(x) is an even function

Answer

A, B, C

Explanation

We have ,
` f(x)={log(secx+tanx)}^(3), x in (-(pi)/(2),(pi)/(2))`
`:. f(-x)={-log(secx+tanx)}^(3)={log((1)/(secx+tanx))}^(3)={-log(secx+tanx)}^(3)=-f(x)`
So, f(x) is an odd function and hence option (a) is correct .
Let `g(x)=secx+tanx,x in (-(pi)/(2),(pi)/(2))`. THen,
`g'(x)=secx(secx+tanx)0` for all `x in (-(pi)/(2),(pi)/(2))`
`implies g(x)` is strictly increasing an `(-(pi)/(2),(pi))`.
Also, `phi(x)=log_(e)x` is increasing on R. Therefore ,
`(phi og)(x)=phi(g(x))=f(x)` is increasing on `(-(pi)/(2),(pi)/(2))`.
Hence , f(x) is increasing on `(-(pi)/(2),(pi)/(2))`. Consequently , f(x) is one-one on `(-(pi)/(2),(pi)/(2))` . So, option (b) is correct .
Clearly , `g(x) in (0,oo) ` for all ` x in (-(pi)/(2),(pi)/(2))`. So, `f(x)=log(g(x)) in (-oo,oo)`. Hence ,f(x) is an onto function . So, option (c ) is correct .