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If f(x)={m x+1,xlt=pi/2sinx+n ,x >pi/...

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

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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.`
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