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

Options

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

Answer

B

Explanation

For the continuity of f at x=2n+1, we must have `underset(x to (2n+1))lim f(x)=underset(x to (2n+1)^(+))lim f(x)=f(2n+1)`
`Rightarrow underset(x to (2n+1)^(-))lim a_(n)+sin pix=underset(x to (2n+1)^(+))lim b_(n+1)+cos pi x=a_(n)+sin x(2n+1)`
`Rightarrow a_(n)+sin(2n+1)pi=b_(n+1)+cos(2pi+1)pi =a_(n)+sin(2n+1)pi`
`Rightarrow a_(n)=b_(n+1)=1 Rightarrow a_(b)-b_(n+1)=-1`
Replacing n by n-1, we get
`a_(n-1)=b_(n)=-1`
For the continuity of f at x=2n, we must have
`underset(x to 2n^(-))lim f(x)=underset(x to 2n^(+))lim f(x)=f(2n)`
`Rightarrow underset(x to 2n^(-))lim b_(n)+cospi x=underset(x to 2n^(+))lim a_(n)+sin pix=a_(n)+sin 2npi`
`Rightarrow b_(n)+cos 2npi=a_(n)+sin2nxpi`
`Rightarrow b_(n)+1=a_(n) Rightarrow a_(n)-b_(n)=1`
So, option (c )is correct.
Hence, option (b) does not hold.