If a function y=f(x) is defined as `y=(1)/(t^(2)-t-6)and t=(1)/(x-2), t in R`. Then f(x) is discontinuous at

Options

A. `2,(2)/(3), (7)/(3)`
B. `2, (3)/(2), (7)/(3)`
C. `2, (3)/(2), (3)/(7)`
D. None of these

Answer

B

Explanation

Clearly, the function `t=(1)/(x-2)` is discontinuous at
x = 2 and the function `y=(1)/(t^(2)-t-6)` is discontinuous at the points, where
`t^(2)-t-6=0 rArr (t+2)(t-3)=0`
`rArr t=-2,3`
When, `t =-2`
`rArr (1)/(x-2)=-2 rArr x =(3)/(2)`
and when `t=3`,
`(1)/(x-2)=3 rArr x=(7)/(3)`
Therefore, the value of x which make the function y discontinuous are `x=2, (3)/(2) and (7)/(3).`