If `f(x)=cos^(-1)(sgn((2[x])/(3x-[x])))`, where sgn ( ) denotes the signum function and [.] dentoes the greatest integer functions, then which of the following is/are correct?

Options

A. a. `underset(xto1)(lim)f(x)=0`
B. b. `underset(xto1^(-))(lim)f(x)=0`
C. c. `f(x)` is derivable at `x=-1`
D. d. `f(x)` is derivable at `x=1`

Answer

A, C

Explanation

`f(1)=0`
`=underset(hto0)(lim)cos^(-1)(sgn((2)/(3+3h-1)))`
`=underset(hto0)(lim)cos^(-1)(sgn((2)/(2+3h)))=0`
`L.H.L=underset(hto0)(lim)f(1-h)=underset(hto0)(lim)`
`cos^(-1)(sgn((0)/(3-3h)))=cos^(-1)(0)=(pi)/(2)`
`becausef(x)` is discontinuous hence non-derivable at `x=1`
`becausef^(')(-1^(+))=underset(hto0)(lim)(f(-1+h)-f(-1))/(h)=0`
and `f^(')(-1^(-))=underset(hto0)(lim)(f(-1-h)-f(-1))/(h)=0`
`impliesf^(')(-1^(+))=f^(')(-1^(-))=0`
`becausef(x)` is derivable at `x=-1`