The correct option is C ex
f(x)=sinx
f′(x)=cosx
Now
f′(x)<0
Hence
cos(x)<0
xϵ(π2,3π2).
Similarly for f(x)=cosx.
Now
f(x)=ex
f′(x)=ex
And
ex>0 for all real x.
Hence
f′(x)>0 for all real x.
Hence
f(x) is an increasing function for all real values of x.
Consider
f(x)=2−x
f′(x)
=−2−x
Now
f′(x)<0 for all real values of x.
Hence it is a decreasing function for all real values of x.