If ` a .lim_(xrarr 1) x^(1//1-x)+b=e^(-1)(a ge 1,bge 0)` , then

Options

A. `a=1,b=e^-1`
B. `a=2,b=e^-1`
C. `a=-1,b=e^-1`
D. `a=1,b=0`

Answer

D

Explanation

We have
`a. lim_(xto1) x^(1//1-x+b=e^(-1))`
` rArr a. lim_(xto1)[1+(x-1)]^(1//1-x)+b=e^(-1)`
` rArr ae ^(lim_(xto1)(x-1)/(1-x))+b=e^(-1)`
` rArr a.e ^-1+b=e^-1rArr a=1and b=0`.