At the point x = 1 , then function
`f(x) = {{:(x^(3) - 1, 1lt x lt oo),(x - 1 , 1 - oolt x le 1):}` is

Options

A. continouous and differentiable
B. continous and not differentiable
C. discontinouss and diffentiable
D. discontinuous and not differentiable

Answer

B

Explanation

`LHL=underset(x to 1^(-))limf(x)=underset(x to 1)lim(x-1)=0`
`RHL=underset(x to 1)lim f(x)`
`underset(xto1)lim(x^(3)-1)=0`
Also,` f(1)=1-1=0`
`thereforef` is continuous at `x=1`
Now, `Lf'(1)=underset(hto0)lim(f(1-h)-f(1))/(-h)`
`=underset(hto 0)lim((1-h)-f(1))/(-h)`
`=underset(h to 0)lim (-h)/(-h)=1`
and `Rf'(1)=underset(hto 0)lim(f(1+h)-f(1))/(h)`
`=underset(hto0)lim((1+h)^(3)-1-0)/(h)`
`=underset(hto0)lim(1+h^(3)+3h+3h^(2)-1)/(h)`
`=underset(hto0)limh^(2)+3+3h=3`
Clearly, Lf'` (1) ne Rf'(1)`
`therefore f(x)` is not differentiable at `x=1`