If `f(x)={[x]+sqrt({x}),x<1 1/([x]+{x}^2),xgeq1` , then [where [.] and {.] represent the greatest integer and fractional part functions respectively] `f(x)` is continuous at `x=1` `f(x)` is not continuous at `x=1` `f(x)` is differentiable at `x=1` `(lim)_(xvec1)f(x)` does not exist

Options

A. f(x) is continuous at x = 1 but not differentiable
B. f(x) is not continuous at x = 1
C. f(x) is differentiable at x = 1
D. `underset(xrarr1)(lim)f(x)` does not exist

Answer

A

Explanation

`f(x)={{:([x]+sqrt({x}),xlt1),((1)/([x]+{x}^(2)),xge1):}`
Consider the function f(x) in the interval (0,2).
`f(x)={{:(sqrtx,0ltxlt1),((1)/(1+(x-1)^(2)),1lexlt2):}`
`f(1)=1 and underset(xrarr1^(+))(lim)f(x)=1`
`thereforef(x)` is continuous at x = 1
`f'(1^(+))=underset(hrarr0^(+))(lim)(f(1+g)-f(1))/(h)=underset(hrarr0^(+))(lim)((1)/(1+h^(2))-1)/(h)`
`=underset(hrarr0^(+))(lim)(-h)/(1+h^(2))=0`
`f'(1^(-))=underset(hrarr0^(+))(lim)(f(1-g)-f(1))/(-h)=underset(hrarr0^(+))(lim)(sqrt(1-h)-1)/(-h)`
`=underset(hrarr0^(+))(lim)(1)/(sqrt(1-h)+1)=(1)/(2)`
`therefore "f(x) is not differentiable at x = 1"`