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

Options

A. continuous and differentiable
B. continuous and not differnetiable
C. discontinuous and differentiable
D. discontinuous and not differntiable

Answer

B

Explanation

` LHL =underset(x to 1)lim .f(x) = underset(x to 0)lim f(1 - h)`
`=underset(x to 0)lim (1 - h-1)=underset(x to 0)lim (- h)= 0 `
and RHL `underset(h to 1^(+)) lim + f(x)`
`underset(h to 0) lim f(1 + h) = underset(h to 0) (lim)(1 +h)^(3) - 1`
Also, ` f(x) = 1 - 1 = 0 `
` therefore ` f is continuous at x = 1
Now , `Lf'(1) = underset(h to 0)lim(f(1-h)-f(1))/(-h) `
` underset(h to0)lim((1 - h)-1-0)/(-h) = underset(hto0)lim(-h)/(-h) = 1`
and `Rf'(1) = underset(h to 0)lim(f(1+h)-f(1))/(h) `
` underset(h to0)lim((1 + h)^(3)-1-0)/(h) `
`= underset(hto0)lim(1 + h^(2) + 3h + 2+h^(2)-1)/(h) `
` = underset(hto 0) limh^(2) + 3 + 3h = 3`
Clearly , ` Lf'(1) ne Rf'(1)`
` therefore f (x) ` is not differentiable at x = 1