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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

Here 1 is related to 2.
there are two possible cases :
Case I : When 1 is not related to 3, then the relation R₁ = {(1, 1), (1,2), (2, 1), (2, 2), (3, 3)} is only equivalence relation containing (1,2).
Case II : When 1 is related to 3, then
A x A = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)} is the only equivalence relation containing (1, 2). there are two equivalence relations on A containing (1, 2). (B) is correct answer.

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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

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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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