Let `f(x)=2x-sinx` and `g(x)`=`3^(sqrtx)` . Then

Options

A. range of gof is R
B. gof is one-one
C. both f and g are one-one
D. both f and g are onto

Answer

1,2,3,4

Explanation

f(X)=`2x -sin x or f(x)=2 - cos x gt 0 forall x` .Hence f(X) is strictly increasing .Hence the function is one in and onto
`g(x)=x^(1//3)or g(x)=(1)/(3)x^(-2//3)gt0 forallx`
Hence g(x) si strictly increasing and hence one one onto also gof is one-one
`gof(x)=(2x-sin x )^(1//3)` has range R as the range of 32x - sin x is R.