Question 1 :
If the ordered pairs (x2 − 3x, y2 + 4y) and (-2,5) are equal, then find x and y.
Question 2 :
The Cartesian product A×A has 9 elements among which (–1, 0) and (0,1) are found. Find the set A and the remaining elements of A×A.
Question 3 :
Given that f(x) =
Question 4 :
Let A= {9, 10, 11, 12, 13, 14, 15, 16, 17} and let f : A-> N be defined by f (n) = the highest prime factor of n ∈ A. Write f as a set of ordered pairs and find the range of f.
Question 1 :
If the ordered pairs (x2 − 3x, y2 + 4y) and (-2,5) are equal, then find x and y.
Answer :
x2 - 3x = -2 x2 - 3x + 2 = 0 (x - 1)(x - 2) = 0 x = 1 and x = 2 |
y2 + 4y = 5 y2 + 4y - 5 = 0 (y + 5)(y - 1) = 0 y = -5 and y = 1 |
Question 2 :
The Cartesian product A×A has 9 elements among which (–1, 0) and (0,1) are found. Find the set A and the remaining elements of A×A.
Answer :
So, in this question since A X A has 9 elements, so the number of elements in the set A must obviously be 3.
Since the given elements of the cross product have -1, 0 and 1 as part of the entries, clearly these only must be the elements of the set A.
So A = {-1, 0, 1}
A X A
= {(-1, -1),(-1, 0),(-1, 1),(0, -1),(0, 0),(0, 1),(1, -1),(1, 0),(1, 1)}
Question 3 :
Given that f(x) =
(i) f (0) (ii) f (3) (iii) f (a+1) in terms of a.(Given that a ≥ 0)
Answer :
(i) f(0)
Instead of x, we have 0, it is less than 1
f(0) = 4
(ii) f(3)
Instead of x, we have 3, it is greater than 1
f(3) = √(3 - 1)
f(3) = √2
(iii) f (a+1)
Instead of x, we have a+1, it is greater than 1
f(a + 1) = √(a + 1 - 1)
f(a + 1) = √a
Question 4 :
Let A= {9, 10, 11, 12, 13, 14, 15, 16, 17} and let f : A-> N be defined by f (n) = the highest prime factor of n ∈ A. Write f as a set of ordered pairs and find the range of f.
Answer :
Elements of A 9 10 11 12 13 14 15 16 17 |
Highest prime factor 3 5 11 3 13 7 5 2 17 |
Set ordered pairs :
= {(9,3) (10, 5)(11, 11)(12, 3)(13, 13) (14, 7) (15, 5) (16,2) (17, 17)}
Range = {2, 3, 5, 7, 11, 13, 17}
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