Algebra Test

Result

Algebra Test - 13

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Let A = {4, 8, 16, 32, 64} and B = {1, 2, 3, 4, 5, 6, 7, 8} and R be the relation "is square of" from A to B. Which of the following represents relation R in roster form?

A
{(4, 2) (8, 4) (16, 8) (64, 8)}

B
{(4, 2), (16, 4), (64, 8)}

C
{(4, 8), (8, 8)}

D
{(4, 2), (8, 4), (16, 8), (32, 6), (64, 8)}

Solution

A = {4, 8, 16, 32, 64} B = {1, 2, 3, 4, 5, 6, 7, 8}
Roster form of R:
R = {(4, 2) (16, 4), (64, 8)}
Question 2
1 / -0

Consider A = {1, 2, 3, 4, 5, 6}, B = {7, 8, 9, 10, 11} and R = {(a, b): a A, b B and a + b is a prime number}. Find the domain of R.

A
{1, 2, 3}

B
{1, 2, 3, 4}

C
{1, 2, 3, 5}

D
{1, 2, 3, 4, 5, 6}

Solution

A = {1, 2, 3, 4, 5, 6} B = {7, 8, 9, 10, 11}
R = {(a, b): a A, b B and a + b is a prime number.}
(1 + 10), (2 + 9), (2 + 11), (3, + 8), (3 + 10), (4 + 7), (4 + 9), (5 + 8), (6 + 7) and (6 + 11) are prime numbers.
R = {(1, 10), (2, 9) (2, 11), (3, 8) (3, 10), (4, 7), (4, 9), (5, 8) (6, 7) (6, 11)}
Thus, domain of R = {1, 2, 3, 4, 5, 6}
Question 3
1 / -0

Consider P = {1, 3, 5, 7, 9}, Q = {10, 12, 15, 20, 25, 30} and R = {(a, b), a P and bQ and 5a = b}. Find the range of R.

A
{10, 15, 20, 25, 30}

B
{25, 30}

C
{15, 25}

D
{10, 15, 25, 30}

Solution

P = {1, 3, 5, 7, 9} and Q = {10, 12, 15, 18, 20}
R = {(3, 15), (5, 25)} [As, 3 5 = 15, 5 5 = 25]
Range of R = {15, 25}
Question 4
1 / -0

If M = {12, 13, 15, 17}, N = {6, 9, 11, 13, 15} and R is relation "divisible by" from P to Q, which of the following represents relation R.

A
{(2, 6) (12, 9) (13, 13) (15, 15)}

B
{(12, 6) (12, 9) (12, 15) (13, 13) (15, 15)}

C
{(12, 6) (13, 13) (15, 15)}

D
{(12, 9) (12, 15) (15, 15)}

Solution

M = {12, 13, 15, 17} and N = {6, 9, 11, 13, 15}
R = {(12, 6), (13, 13) (15, 15)} [As 12 is divisible by 6, 13 by 13 and 15 by 15]
Question 5
1 / -0

Let A = {2, 3, 5}, B = {4, 5, 6, 7} and R = {(a, b); aA, bB and a b is an even number.}. Find the range of R.

A
{4, 5}

B
{4, 6, 7}

C
{4, 5, 6, 7}

D
{5, 6}

Solution

A = {2, 3, 5} B = {4, 5, 6, 7}
R = {(2, 4) (2, 5) (2, 6) (2, 7), (3, 4) (3, 6) (5, 4) (5, 6)}
Thus, range of R = {4, 5, 6, 7}
Question 6
1 / -0

Let A = {a: a - 1 2, a positive integers} and B = {4, 5, 6, 7, 8}.
Which of the following represents R = {(a, b): aA, b B and 2a + 3 > b}?

A
{(1, 4), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5)}

B
{(1, 4), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8)}

C
{(1, 4), (2, 4), (2, 5), (3, 4), (3, 5), (3, 6)

D
{(1, 4), (2, 5), (2, 6), (3, 6), (3, 7), (3, 8)}

Solution

A = {a : a - 1 2, a positive integers}
= {1, 2, 3} [As a - 1 2, a 3]
B = {4, 5, 6, 7, 8}
R = {(a, b), 2a + 3 > b where a A, b A}
R = {(1, 4), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5) (3, 6), (3, 7), (3, 8)
Question 7
1 / -0

Which of the following is not the mapping from A to B?

A

B

C

D

Solution

Option (3) is not the mapping from A to B because member 7 of set A has two images in set B.
Question 8
1 / -0

Let f : x x³, where x P = {2, 4, 5}. Find the function f.

A
{(2, 8), (4, 5), (5, 25)}

B
{(2, 4), (4, 16), (5, 25)}

C
{(2, 8), (4, 64), (5, 125)}

D
{(2, 6), (4, 12), (5, 15)}

Solution

f : x x³
Here, f(x) = x³ ... (1)
P = {2, 4, 5}
Substituting x = 2, 4, 5 in (1), we get
f(2) = (2)³ = 8, f(4) = (4)³ = 64, f(5) = (5)³ = 125
So, the images of the elements 2, 4 and 5 of P are 8, 64 and 125 respectively.
f = {(2, 8), (4, 64), (5, 125)}
Question 9
1 / -0

Which of the following is a function?

A
{(1, 1), (2, 3), (3, 3), (4, 5)}

B
{(1, 2), (1, 3), (2, 3)}

C
{(- 2, 4), (- 3, 5), (- 3, 2)}

D
{(1, 3), (2, 2), (3, 2), (4, 5), (1, 5)}

Solution

Option (1) is the mapping (function) because first elements are different and belong to unique elements.
Question 10
1 / -0

If f : p 4p² - 1, where p P = {- 1, - 2, 0, 1, 2}, and Q is the set of images of the elements of P, find the range of f.

A
{3, 15, -1}

B
{3, 5, 1}

C
{- 1, - 2, 0, 1, 2}

D
{- 3, - 2, 0}

Solution

Here, f(p) = 4p² - 1 ...(1)
P = {- 1, - 2, 0, 1, 2}
Substituting p = - 1, - 2, 0, 1, 2 respectively in (1), we get
f(- p) = 4(- 1)² - 1 = 4 - 1 = 3, f(- 2) = 4(- 2)² - 1 = 16 - 1 = 15
f(0) = 4(0)² - 1 = - 1, f(1) = 4(1)² - 1 = 3, f(2) = 4(2)² - 1 = 15
Images of elements - 1, - 2, 0, 1 and 2 are 3, 15, - 1, 3 and 15 respectively.
Thus, Q = {3, 15, - 1}
Also, f = {(-1, 3), (- 2, 15), (0, - 1), (1, 3), (2, 15)}
Range of f = {3, 15, - 1}
Question 11
1 / -0

If A = {0, 1, 2, 3, 4}, B = {6, 7, 8, 9} and R = {(x, y): x A, y B and 5x – y 10}, which of the following is correct arrow diagram for the relation R?

A

B

C

D

Solution

A = {0, 1, 2, 3, 4} and B = {6, 7, 8, 9}
R = {(x, y), xA, yB and 5x – y 10}
Then, R = {(4, 6), (4, 7), (4, 8) (4, 9)}
Question 12
1 / -0

Find the domain and range from the following arrow diagram.

A
Domain - {0, - 3, - 2, - 1}, Range - {0, 4, 16, 32, 36}

B
Domain - {0, - 1, - 2, - 3, 1, 2, 3}, Range - {0, 4, 16, 36}

C
Domain - {0, - 3, - 2, - 1, 1, 2, 3}, Range - {0, 4, 16, 32, 36, 64}

D
Domain - {0, 1, 2, 3}, Range {0, 4, 16, 32, 36}

Solution

Domain - {- 1, - 2, - 3, 0, 1, 2, 3}
Range - {0, 4, 16, 36}
Question 13
1 / -0

If f : x 6x² + 5, where xP = {- 1, 0, 2} and Q is the set of images of the elements of P, find the range of f.

A
{1, 5, 19}

B
{11, 5, 24}

C
{5, 11, 29}

D
{1, 11, 29}

Solution

f : x 6x² + 5
f(x) = 6x² + 5 ... (1) and P = {- 1, 0, 2}
Substituting - 1, 0 and 2 in (1), we get
f(- 1) = 6(- 1)² + 5
= 6 + 5
= 11
f(0) = 6(0)² + 5
= 5
f(2) = 6(2)² + 5
= 24 + 5
= 29
Thus, the images of elements - 1, 0 and 2 are 11, 5 and 29 respectively.
Q = {5, 11, 29}

f = {(- 1, 11), (0, 5) (2, 29)}
The range of f = {5, 11, 29}
Question 14
1 / -0

Consider A = {10, 12, 15} and B = {2, 3, 4, 5, 6}. Represent the relation "a multiple of" from the set A to the set B in roaster form and in an arrow diagram.

A
{(10, 2), (10, 5) (12, 2) (12, 3) (12, 4) (12, 6) (15, 3) (15, 5)}

B
{(10, 2) (10, 5), (12, 3) (12, 4) (15, 5)}

C
{(10, 5), (12, 4), (15, 5)}

D
{(10, 4) (10, 2) (10, 5) (12, 6) (12, 3), (15, 3) (15, 5)}

Solution

A = {10, 12, 15} B = {2, 3, 4, 5, 6]
Relation (R) {(10, 2) (10, 5) (12, 2) (12, 3) (12, 4) (12, 6) (15, 3) (15, 5)}
Question 15
1 / -0

Which of the following is a mapping?

A

B

C

D

Solution

C D is a mapping as each element of set C belongs to a unique element in set D.