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| More Solved Examples |
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| The arrow diagram represents a relation from set A to set B. |
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| (i) Represent the relation in roster form. |
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| (ii) Is this relation a function? Give reason for your answer. |
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| (i) {(1, 2), (2, 4), (5, 2), (3, 10), (4, 10)} |
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| (ii) Yes it is a function. Every first element is mapped and it is many-one. The function is many-one into. |
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Let f be defined by f(x)  find |
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(i) f(2) (ii)  (iii)  (iv) x when  . |
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(i)  [f(2) means find the value of y when x = 2]. |
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(ii)  |
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(iii)  |
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(iv)  |
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 2x + 1 = 3x |
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 x = 1 |
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Ans: (i)  (ii)  (iii)  (iv) x = 1. |
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| Which of the following relations are functions: |
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(i)  |
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(ii) { for all values of y} |
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| (iii) {(x, y) : y = 4 for all values of x} |
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(iv) {(x, y) : y = 5x - 6  }. |
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| (i) Not a function: For one value of x there are many y values and for one value of y there are many x values (many-many). |
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| (ii) Not a function: For one value of x there are many values of y (one-many). |
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| (iii) Function: Many values of x equal to one value of y (many - one). |
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| (iv) Function: For one value of x there is one value of y (one-one). |
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| Given A = {0, 1, 2}, B = {5, 1, 6}, {(0, 5), (0, 1), (1, 6)} is not a function from A to B. State two reasons for the above statement. |
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| Reasons: |
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| (i) (0, 5), (0, 1) pairs make it one - many, one - many is not a function. |
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| (ii) Pre-image 2 does not have an image in B. |
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| It becomes easy to discuss a function with the help of an arrow diagram. |
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| Given A = {-2, -1, 0, 1, 2} and B = {-3, -1, 1, 5}. list the elements of S = {(x, y): y = 2x2 - 3,
x Î A and y Î B}. Is S a function? If so classify it. |
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y = 2x2 - 3  f(x) = 2x2 - 3 |
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| f(-2) = 2(-2)2 - 3 = 5 f(1) = 2(1)2 - 3 = -1 |
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| f(-1) = 2(-1)2 - 3 = -1 f(2) = 2(2)2 - 3 = 5 |
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| f(0) = 2(0)2 - 3 = -3 |
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 The roster form is S = {(-2, 5), (-1, -1), (0, -3), (1, -1), (2, 5)}. |
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| Yes, S is a function. It is many-one and into function. |
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| Domain = {-2, -1, 0, 1, 2} = A ; |
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| Range = {-3, -1, 5}; Co-Domain = {-3, -1, 1, 5} = B. |
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| (i) Find the domain and the range of the function. |
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| (ii) Draw the Cartesian graph of f. |
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| (iii) State whether f is one-one or many-one function. |
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| (i) Domain = {-3 < x < 2, x
Î R} |
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| To find the range, f(-3) = -(-3) - 2 = 1 (-3, 1) |
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| f(-2) = -(-2) - 2 = 0 (-2, 0) |
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| f(-1) = -(-1) - 2 = -1 (-1, -1) |
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| f(1) = +(1) - 2 = -1 (1, -1) |
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| f(2) = 2 - 2 = 0 (2, 0) |
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 Range = {-2 < y < 1,
y Î R} |
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| (ii) |
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| (iii) It is many-one function |
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