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Relations and Functions

Short Answer Type

241.

Let R be the relation in the set {1. 2, 3, 4} given by R = {(1,2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}.

Choose the correct answer.

(A)    R is reflexive and symmetric but not transitive.
(B)    R is reflexive and transitive but not symmetric.
(C)    R is symmetric and transitive but not reflexive.
(D)    R is an equivalence relation.

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242. 23. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is (A) 1 (B) 2 (C) 3 (D) 4

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243. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f { (1, 4). (2, 5), (3. 6)} be a function from A to B. Show that f is one-one.

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244.

L.et A be the set of all 50 students of class X in a school. Let f : A → N be function defined by f (x) = roll number of student x. Show that f is one-one but not onto.

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245. Show that the function f : N → N given by f(x) = 2x, is one-one but not onto.

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246. State whether the function f : N → N given by f(x) = 5 x is injective, surjective or both.

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247. Prove that the function f : R → R , given by f (x) = 2x, is one-one and onto.

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Long Answer Type

248.
Check the injectivity and surjectivity of the following functions :

(i) f : N → N given by f (x) = x²(ii) f : Z → Z given by f (x) = x²(iii) f : R → R given by f (x) = x² (iv) f : N → N given by f (x) = x³(v) f : Z → Z given by f (x) = x³

(i) f : N → N is given by f (x) = x²Let x₁ x_{2 ∈ N be such that f(x₁) ⁼ f(x₂)}∴ x₁² = x ₂² ⇒ x₂ ² x₁ ² = 0
⇒ (x₂ –x₁) (x₂ + x₁) = 0
⇒    x₂ – x₁ = 0    [x₁ + x₂ ≠ 0 as x₁, x₂ ∈ N]
⇒    x₂ = x₁ ⇒ x₁ = x₂∴ f is one-one, i.e.. f is injective.
Since range of f = { 1² , 2², 3²............}
= {1.4.9......} ≠ N.
∴ f is not surjective.
(ii) f : Z → Z is given by f (x) = x²Let x₁, x₂ ∈ Z be such that f(x₁)= f (x₂)
∴ x₂² =x₂² ⇒ x₂² = 0
⇒ (x₂ – x₁) (x₂ + x₁) = 0
⇒    x₂ = x₁    or x₂ = – x₁∴ f (x₁) = f(–x₁) ∀ x₁ ∈ Z
∴ f is not one-one, i.e. f is not injective.
Also range of f = { 0², 1², 2²,.....}
= {0, 1,4, 9,.........}
≠     Z
∴ f is not onto i.e.. f is not surjective (iii) f : R → R is given by f (x₁) = x² Let x₁, x₂ ∈ R be such that f (x₁) = f (x₂)
⇒    x₁² = x₂² ⇒ (x₂ – x₂) (x₂ + x₁) = 0
⇒    x₂ = x₁ or x₂ = – x₁⇒    f(x₁) = f (–x₁) ∀ x₁ ∈ R
∴ f is not one-one, i.e., f is not injective.
As range of f does not contain any negative real, therefore, range of ≠ R.
Hence. f is not onto, i.e., f is not surjective.
(iv) f : N → N is given by f (x) = x³ Let x₁ ,. x₂ ∈ N be such that f (x₁) = f(x₂)
⇒    x₁³ = x₂³ ⇒ x₁ = x₂∴ f is one-one, i.e., injective.
Also range of f = {1³, 2³, 3³,.........}
= {1,8,27,.....}
≠ N
∴ f is not onto, i.e.,f is not surjective.
(v) f : Z → Z is given by f (x) = x³ Let x₁, x₂ ∈ Z be such that f (x₁) = f (x₂)
⇒    x₁³ = x₂³ ⇒ x₁ = x₂Also range of f = {0³ ± 1³, ± 2³, ± 3³,....}
= {0, ± 1, ± 8, ± 27.............}
≠ Z ∴ f is not onto, i.e., f is not surfective.

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Short Answer Type

249. Show that the function f : R →R , defined as f (x) = x² , is neither one-to-one nor onto.

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250. Show that the modulus function f : R → R, given by f (x) = | x |, is neither one-one nor onto, where |x| is x, if x is positive or 0 and | x | is —x, if. x negative.

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