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प्रश्न
Find all points of discontinuity of the function f(t) = `1/("t"^2 + "t" - 2)`, where t = `1/(x - 1)`
योग
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उत्तर
We have, f(t) = `1/("t"^2 + "t" - 2)`
Where t = `1/(x - 1)`
∴ f(t) = `1/((1/(x - 1))^2 + 1/(x - 1) - 2)`
= `(x - 1)^2/(1 + (x - 1) - 2(x - 1)^2)`
= `(x - 1)^2/(-(2x^2 - 5x + 2))`
= `(x - 1)^2/((2x - 1)(2 - x))`
So, f(t) is discontinuous at 2x – 1 = 0
⇒ x = `1/2` and 2 – x = 0
⇒ x = 2
Also f(t) is discontinuous at x = 1, where t = `1/(x - 1)` is discontinuous.
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