20. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
(iii) {(1,3), (1,5), (2,5)}.
20. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
(iii) {(1,3), (1,5), (2,5)}.
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1 Answer
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20. (i) Domain of the given relation = {2,5,8,11,14,17}
Since every element of the domain has one and only one image, the given relation is a fxn.
So, domain = {2,5,8,11,14,17}
range = {1}
(ii) Domain of the given relation = {2,4,6,8,10,12,14}
Since every element of the domain has one and only one image, the given relation is a fxn.
So, domain = {2, 4, 6, 8, 10, 12, 14}
range = {1,2,3,4,5,6,7}
(iii) Domain of the given relation = {1,2}
As element 1 has more than one image i.e., 3 and 5, the given relation is not a fxn.
Similar Questions for you
Total number of possible relation =
Favourable relations =
Probability =
Circle S? : x² + y² - 10x - 10y + 41 = 0.
Center C? = (5, 5). Radius r? = √ (5² + 5² - 41) = √ (25 + 25 - 41) = √9 = 3.
Circle S? : x² + y² - 16x - 10y + 80 = 0.
Center C? = (8, 5). Radius r? = √ (8² + 5² - 80) = √ (64 + 25 - 80) = √9 = 3.
The solution checks if the center of one circle lies on the other.
Put C? (8, 5) into S? : 8² + 5² - 10 (8) - 10 (5) + 41 = 64 + 25 - 80 - 50 + 41 = 130 - 130 = 0. So C? lies on S?
Put C? (5, 5) into S? : 5² + 5² - 16 (5) - 10 (5) + 80 = 25 + 25 - 80 - 50 + 80 = 130 - 130 = 0. So C? lies on S?
This means bo
Kindly consider then following figure
36. Given, A={9,10,11,12,13}.
f(x)=the highest prime factor of n.
and f: A → N.
Then, f(9)=3 [? prime factor of 9=3]
f (10)=5 [? prime factor of 10=2,5]
f(11)=11 [? prime factor of 11 = 11]
f(12)=3 [? prime factor of 12 = 2, 3]
f(13)=13 [? prime factor of 13 = 13]
?Range of f=set of all image of f(x) = {3,5,11,13}.
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