Given,

To find: Points discontinuity of function f(t) where t =

As t is not defined at x = 1 as denominator becomes 0,at x = 1.

∴ x = 1 is a point of discontinuity

∵ f(t) =

⇒ f(t) =

Clearly f(t) is not going to be defined whenever denominator is 0 and thus will give a point of discontinuity.

∴ Solution of the following equation gives other points of discontinuities.

-2x² + 5x – 2 = 0

⇒ 2x² – 5x + 2 = 0

⇒ 2x² – 4x – x + 2 = 0

⇒ 2x(x – 2) – (x – 2) = 0

⇒ (2x – 1)(x – 2) = 0

∴ x = 2 or x = 1/2

Hence,

f(t) is discontinuous at x = 1, x = 2 and x = 1/2