Relations and Functions Test

Result

Relations and Functions Test - 17

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 f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Then gof is

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

B
{(1, 3), (3, 3), (4, 1)}

C
{(1, 3), (3, 1), (4, 1)}

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

Solution

(gof)(1)=g(f(1))=g(2)=3, (gof)(3)=g(f(3))=g(5)=1, (gof)(4)=g(f(4))=g(1)=3,
Question 2
1 / -0

Let f, g and h be functions from R to R, then

A
(f + g) oh = ho(f + g)

B
(f + g) oh = hof + goh

C
(f + g) oh = foh + hog

D
(f + g) oh = foh + goh

Solution

(f+g)oh: R→R And ( foh + goh): R → R for any x ∈ R ((f + g)oh) (x) = (f + g)(hx)) = f(h(x)) + g(h(x)) = foh(x) + goh(x) Threrefore, (f + g)oh = foh + goh
Question 3
1 / -0

If the set A contains 5 elements and the set B contains 6 elements, then the number of one – one and onto mappings from A to B is

A
120

B
0

C
30

D
720

Solution

Because the no. of elements in domain i.e. in A is less than the no. of elements in co-domain i.e. in B. Therefore, no bijection mapping is possible.
Question 4
1 / -0

Let A = {1, 2, 3,… n} and B = {a, b}. Then the number of surjections from A to B is

A
2ⁿ - 2

B
2ⁿ- 1

C
ⁿP₂

D
2ⁿ

Solution

Number of elements in A i.e in domain is n and the number of elements in B i.e in co-domain is 2. Therefore, number of onto functions defined from A to B is 2ⁿ- 2.
Question 5
1 / -0

Let f : A → B and g : B → C be the two bijective functions. Then (g o f)^-1 is

A
f⁻¹o g⁻¹

B
g^-1of^-1

C
g o f

D
f o g

Solution

Let f : A → B and g : B → C be the two bijective functions , then (go f) : A → C is invertible and (gof )⁻¹= f⁻¹o g⁻¹.
Question 6
1 / -0

Let f : [2, ∞) → R be the function defined by f(x) = x² - 4x + 5, then the range of f is

A
[1, ∞)

B
[4, ∞)

C
R

D
[5, ∞)

Solution

Range of f(x) = x² - 4x + 5 is [1,∞)[1,∞), because , if we take the minimum value of x from the domain i.e. x =2, we have f(2) =1.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 17

Result

Relations and Functions Test - 17

Score

-

Rank

-

SHARING IS CARING

Question 1

{(1, 3), (3, 1), (4, 3)}

{(1, 3), (3, 3), (4, 1)}

{(1, 3), (3, 1), (4, 1)}

{(1, 3), (3, 3), (4, 3)}

Solution

Question 2

(f + g) oh = ho(f + g)

(f + g) oh = hof + goh

(f + g) oh = foh + hog

(f + g) oh = foh + goh

Solution

Question 3

120

0

30

720

Solution

Question 4

2n - 2

2n - 1

nP2

2n

Solution

Question 5

f−1o g−1

g-1of-1

g o f

f o g

Solution

Question 6

[1, ∞)

[4, ∞)

R

[5, ∞)

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

2ⁿ - 2

2ⁿ- 1

ⁿP₂

2ⁿ

f⁻¹o g⁻¹

g^-1of^-1