f is a continous function in [a, b] and g is a continuous functin in [ b, c]. A function h(x) is defined as
`h(x)={{:(f(x)",",x in [a,b)),(g(x)",", x in (a,b]):}`
Now, if `f(b)=g(b)`, then

Options

A. `h(b^(+))=g(b^(-))` and `h(b^(-))=f(b^(+))`
B. `h^(b^(-))=g(b^(+))` and `h(b^(+))=f(b^(-))`
C. `h(x)` can be made continuous at `x=b` by by defining it at the point `x=b`
D. `f(a)=g(c)`

Answer

B, C

Explanation

`h(x)` is continous in `[a,b)cup(b,c]`
`becauseh(b^(-))=f(b^(+))` are undefined
`h(b^(-))=f(b^(-))=f(b)=g(b)=g(b^(+))`
and `h(b^(+))=g(b^(+))=g(b)=f(b)=f(b^(-))`
Hence `h(b^(-))=h(b^(+))=f(b)=g(b)`
Thus, `h(x)` has removable discontinuity at `x=b`