Q. For every integer n, let an and bn be real numbers. Let function `f:R->R` be given by a `f(x)={a_n+sin pix, for x in [2n,2n+1]`, `bb_n+cos pix, for x in (2n+1,2n)` for all integers n.

Options

A. `a_(n-1)-b_(n-1)=0`
B. `a_(n)-b_(n)=1`
C. `a_(n)-b_(n+1)=1`
D. `a_(n-1)-b_(n)=-1`

Answer

B, D

Explanation

`{:(f(2n)=a_(n)),(f(2n^(+))a_(n)),(f(2n^(-))=b_(n)+1):}}{:(a_(n)=b_(n)+1),(a_(n)-b_(n)=1),("So B is correct"):}`
`{:(f(2n+1)=a_(n)),(f((2n+1)^(-))a_(n)),(f((2n+1)^(+))=b_(n+1)-1):}}{:(a_(n)=b_(n+1)-1),(a_(n)-b_(n+1)=-1),(a_(n-1)-b_(n)=-1):}`
So D is correct