`f(x)={((5^(cosx)-1)/((pi)/(2)-x)",", x ne (pi)/(2)), (log 5"," , x =(pi)/(2)):}` at `x =(pi)/(2)` is

Options

A. discontinuous
B. imaginary
C. continuous
D. not defined

Answer

C

Explanation

Given ,` f(x)={((5^(cosx)-1)/((pi)/(2)-x)",",x ne (pi)/(2)), (log 5",", x =(pi)/(2)):}`
Now, `LHL=lim_(x to (pi^(-))/(2)) f(x)=lim_(h to 0) (5^(cos((pi)/(2)-h))-1)/((pi)/(2)-((pi)/(2)-h))`
`=lim_(h to 0) (5^(sin h)-1)/(h)=lim_(h to 0) (5^(sin h) log 5 xx cos h)/(1) =log 5`
`RHL= lim_(x to (pi^(+))/(2)) f(x) =lim_(h to 0) (5^(cos((pi)/(h)+h)-1))/((pi)/(2)-((pi)/(2)+h))`
`=lim_(h to 0) (5^(-sinh) -1)/(-h) =lim_(h to 0) (5^(-sin h)log 5(-cos h))/(-1) =log 5`
`and f((pi)/(2)) =log 5 `
`therefore LHL=RHL=f((pi)/(2))`
Hence, f(x) is continuous at `x =(pi)/(2).`