If `f(x)={m x+1,xlt=pi/2sinx+n ,x >pi/2` is continuous at `x=pi/2,` then (A) m=1,n=0 (B)` m=(npi)/2+1` (C) `n=(mpi)/2` (D)`m= n=pi/2`

Explanation

As given `f(x)={m x+1,xlt=pi/2sinx+n ,x >pi/2`
is continuous at `x=pi/2,`
then
Since, f(x) is continuous at `x=pi/2`

​Therefore `lim_(x->(pi/2)^-)(mx+1)= lim_(x->(pi/2)^-)(sinx+n)`

`impliesm(pi/2)+1=sin(pi/2)​+n`

`implies(mpi)/2=n.`

Hence correct option is (c)