The function `f(x)=x-|x-x^(2)|` is

Options

A. continuous at x = 1
B. discontinuous at x = 1
C. not defined at x = 1
D. None of these

Answer

A

Explanation

Given, `f(x)=x=|x-x^(2)|`
At x = 1, `f(1)=1-|1-1|=1`
`lim_( x to 1^(-))f(x)=lim_(h to 0)[ (1-h)-|(1-h)-(1-h)^(2)|]`
`=lim_(h to 0) [(1-h)-|h-h^(2)|]=1`
`lim_(x to 1^(+)) f(x)=lim_(h to 0)[(1+h)-|(1+h)-(1+h)^(2)|]`
`=lim_(h to 0) [ 1+h-|-h^(2)-h|]l=1`
`because lim_(x to 1^(-)) f(x)=lim_(x to 1^(+)) =f(1)`
`therefore f(x) is continuous at x = 1.