Let f : R → R be defined by f(x) = x² + 1. Is f one-one? Is it onto? Justify.
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One-one: f(1) = f(−1) = 2, so two distinct inputs give the same output ⇒ f is NOT one-one. Onto: f(x) = x² + 1 ≥ 1 for all real x, so negative reals (e.g. −1) have no preimage ⇒ f is NOT onto. f is neither one-one nor onto. ∎
Q2
CBSE Board Exam 20243 Marks
Let R be a relation on N defined by R = {(x, y) : x + 2y = 8, x, y ∈ N}. Write R in roster form and find its domain and range.
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x + 2y = 8 ⇒ x = 8 − 2y. For y = 1, 2, 3: x = 6, 4, 2 (all ∈ N); y = 4 gives x = 0 ∉ N. R = {(6,1), (4,2), (2,3)}. Domain = {6, 4, 2}, Range = {1, 2, 3}
📅 Year 2023
Q1
CBSE Board Exam 20233 Marks
Check whether the relation R on R defined by R = {(a, b) : a ≤ b³} is reflexive, symmetric or transitive.
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Reflexive: a ≤ a³ fails for a = 0.5 (0.5 ≤ 0.125 is false) ⇒ NOT reflexive. Symmetric: a ≤ b³ does not imply b ≤ a³ (e.g. a=1, b=2) ⇒ NOT symmetric. Transitive: a ≤ b³ and b ≤ c³ does not imply a ≤ c³ (e.g. a=2,b=1.5,c=1: 2 ≤ 3.375 and 1.5 ≤ 1 ok, but 2 ≤ 1 false) ⇒ NOT transitive. R is neither reflexive, nor symmetric, nor transitive.