Consider a function `f:[0,pi/2]->R` given by `f(x)=sin x and g:[0,pi/2]->R` given by `g(x)=cos x.` Show that f and g are one-one, but `f+g` is not one-one.

Explanation

If `x_1​=x_2`​ and `x_1​,x_2`​`in[0,2pi​]`
then `sinx_1​=sinx_2`​ and `cosx_1​=cosx_2​`

`therefore` f is one-one and g is one-one.


But `(f+g)(x)=f(x)+g(x)=sinx+cosx.`

`therefore(f+g)(0)=(sin0+cos0)=1`
and

`(f+g)(pi/2​)=(sinpi/2​+cospi/2​)=1`

`therefore(f+g)` is not one-one.