f (x) = \( \dfrac{1}{\sqrt{x^2-1}}\)
We know the square of a real number is never negative.
f (x) takes real values only when x2 – 1 ≥ 0
x2 – 12 ≥ 0
(x + 1) (x – 1) ≥ 0
x ≤ –1 or x ≥ 1
x ∈ (–∞, –1] ∪ [1, ∞)
In addition, f (x) is also undefined when x2 – 1 = 0 because denominator will be zero and the result will be indeterminate.
x2 – 1 = 0 ⇒ x = ± 1
So, x ∈ (–∞, –1] ∪ [1, ∞) – {–1, 1}
x ∈ (–∞, –1) ∪ (1, ∞)
Domain (f) = (–∞, –1) ∪ (1, ∞)
Answered by Aaryan | 1 year agoLet R = {(a, b) : a, b, ϵ N and a < b}.Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.
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