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  • Isthegivenrelation afunction? Givereasons foranswer.(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) | npositive integer}(v) t = {(x, 3) | x is a real number.

Is the given relation a function? Give reasons for your answer.

(i) h = \left \{(4, 6), (3, 9), (- 11, 6), (3, 11)\right \}

(ii) f = \left \{(x, x) | x \: \: is \: \: a \: \: real\: \: number\right \}

(iii) g = n, (1/n) |n \: \: is\: \: a\: \: positive\: \: integer

(iv)s = \left \{(n, n2) | n \: \: is\: \: a\: \: positive\: \: integer\right \}

(v) t = \left \{(x, 3) | x \: \: is \: \: a \: \: real\: \: number\right \}.

Answers (1)

(i) Given data: h = \left \{(4,6),(3,9),(-11,6),(3,11)\right \}

‘h’ is not a function since there are two images- 9 & 11 for the relation 3.

(ii) f = \left \{(x,x)/ x \: \: is\: \: a\: \: real\: \: no.\right \}

Here, f is a function because every element of the domain has a unique image.

(iii) g = \left \{(n, 1/n)/ n \: \: is\: \: a\: \: positive\: \: integer\right \}

‘g’ is a function because there is a unique image, ‘1/n’ for every element in the domain.

(iv) 

S = \left \{(n, n2)/ n \: \: is\: \: a\: \: positive\: \: integer.\right \}

‘S’ is a function since square of any integer is a unique no. & thus, for every element in the domain there is a unique image.

(V) T = \left \{(x,3)/ x \: \: is\: \: a\: \: real\: \: number\right \}

Here ‘t’ is a constant function, since we can observe that there is a constant no. 3 for every real element in the domain.

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