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Find all point of discountinuity of t...

Find all point of discountinuity of the function `f(t) = (1)/(t^(2)+t-2)`, where `t = (1)/(x-1)`.

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The correct Answer is:
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We have , `f(t) = (1)/(t^(3)+t-2)` and `t = (1)/(x-1)`
` :. f(t) = (1)/(((1)/(x^(2)+1-2x))+((1)/(x-1))-2/1)`
`= (1)/(((1+x-1+[-2(x-1)^(2)])/((x^(2)+1-2x))))`
`= (x^(2)+1-2x)/(x-2x^(2)-2+4x)`
` = (x^(2)+1-2x)/(-2x^(2)+5x-2)`
`= ((x-1)^(2))/(-(2x^(2)-5x+2))`
`= ((x-1)^(2))/((2x-1)(2-x))`
So, f(t) is discountinuous at `2x - 1 = 0 rArr x = 1//2`
and `2 -x = 0 rArr x = 2`.
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