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Is the given relation a function? Give reasons for your answer.(i) h = {(4, 6), (3, 9), (– 11, 6), (3, 11)}(ii) f = {(x, x) | x is a real number}(iii) g = n, (1/n) |n is a positive integer(iv) s = {(n, n2) | n is a positive integer}(v) t = {(x, 3) | x is a real number. |
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Answer» (i) According to the question, h = {(4, 6), (3, 9), (– 11, 6), (3, 11)} Therefore, element 3 has two images, namely, 9 and 11. A relation is said to be function if every element of one set has one and only one image in other set. Hence, h is not a function. (ii) According to the question, f = {(x, x) | x is a real number} This means the relation f has elements which are real number. Therefore, for every x ∈ R there will be unique image. A relation is said to be function if every element of one set has one and only one image in other set. Hence, f is a function. (iii) According to the question, g = n, (1/n) |n is a positive integer Therefore, the element n is a positive integer and the corresponding 1/n will be a unique and distinct number. Therefore, every element in the domain has unique image. A relation is said to be function if every element of one set has one and only one image in other set. Hence, g is a function. (iv) According to the question, s = {(n, n2) | n is a positive integer} Therefore, element n is a positive integer and the corresponding n2 will be a unique and distinct number, as square of any positive integer is unique. Therefore, every element in the domain has unique image. A relation is said to be function if every element of one set has one and only one image in other set. Hence, s is a function. (v) According to the question, t = {(x, 3) | x is a real number. Therefore, the domain element x is a real number. Also, range has one number i.e., 3 in it. Therefore, for every element in the domain has the image 3, it is a constant function. A relation is said to be function if every element of one set has one and only one image in other set. Hence, t is a function. |
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