Question

若函数f(x)=αsin2x+btanx+1,且f(－3)=5,则f(π+3)=?

Answer 1

f(π+3)=αsin2（π+3)+btan(π+3)+1

=αsin(2π+6)+btan(π+3)+1

=αsin6+btan3+1

=-{[αsin-6+btan-3+1]+1}+1

=-f(-3)+1

=-5+1

=-4