$y = \frac{1}{t^{2} + t - 2}$ where $t = \frac{1}{x - 1}$ then the number of points of discontinuities of $y = f (x), x ϵ R$ is

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Solution

The correct option is C 3
$t = \frac{1}{x - 1}$ is discontinuous at x=1
$y = \frac{1}{t^{2} + t - 2}$ is discontinuous when
$t^{2} + t - 2 = 0$
Or $(t + 2) (t - 1) = 0$
Or $t = - 2 t = 1$
$x = \frac{1}{t} + 1$
When $t = - 2$
$y = \frac{- 1}{2} + 1 = \frac{1}{2}$
When $t = 1$
$x = 1 + 1 = 2$
Hence function is discontinuous at 3 points

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Similar questions

y = \frac{1}{t^{2} + t - 2}

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