Relations and Functions Test

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Relations and Functions Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

A function f : X → Y is said to be one – one and onto if

A
f is onto

B
f is one – one

C
f is both one – one and onto

D
f is either one – one or onto

Solution

A function f : X → Y is defined to be one – one (or injective),if fx₁≠ x₂in X ⇒ f (x₁) ≠ f(x₂) in Y. and R_f= Y.
Question 2
1 / -0

Let f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by gof, is defined as

A
g(f(x + 2)), ∀∀ x ∈∈ A

B
f(g(x)), ∀∀ x ∈∈ A

C
g(f(x)), ∀∀ x ∈∈ A

D
g(f(x²)), ∀∀ x ∈∈ A

Solution

Let f : A → B and g : B → C be two functions. Then the composition of f and g is defined as (gof)(x) = g(f(x)) for all x belongs to A.
Question 3
1 / -0

A function f : X → Y is defined to be invertible, if

A
there exists a function g : Y → X such that gof = I_x and fog= I_Y.

B
there exists a function g : Y → X fog = I_Y.

C
there exists a function g : Y → X such that gof = I_Xand fog ≠ I_Y

D
there exists a function g : Y → X such that gof = I_X

Solution

for any two functios f : X → Y ,g:Y→X such that that (gof)(x) = g(f(x)) = g(y) = x. (fog)(y) = f(g(y)) = f(x) = y.
Question 4
1 / -0

Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f (2) = 3, f(3) = 4, f (4) = f (5) = 5 and g (3) = g (4) = 7 and g (5) = g (9) = 11. Then gof is

A
{7, 7, 11, 11}

B
{7, 7, 15, 11}

C
{7, 7, 5, 11}

D
{15, 15, 11, 11}

Solution

gof(2) = g(f(2)) = g(3) = 7; gof(3) = g(f(3)) = g(4) =7; gof(4) = g(f(4)) =g(5) = 11; gof(5) = g(f(5)) = g(5) = 11 ;
Question 5
1 / -0

if f : R → R and g : R → R are given by f (x) = cos x and g(x) = 3x². Then

A
f is odd

B
gof = fog

C
gof ≠ fog

D
g is odd

Solution

(gof)(x)=g(f(x)) =g(cos x) = 3cos²x ; (fog) = f(g(x)) = f(3x²) = cos3x².
Question 6
1 / -0

Let f : X → Y and g : Y → Z be two invertible functions. Then gof is

A
I_y

B
invertible with (gof)^-1= f^-1og^-1

C
Non invertible

D
I_N

Solution

If f : X → Y and g : Y → Z be two bijective functions which are invertible then (gof)^-1= f^-1og^-1

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 16

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Relations and Functions Test - 16

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Question 1

f is onto

f is one – one

f is both one – one and onto

f is either one – one or onto

Solution

Question 2

g(f(x + 2)), ∀∀ x ∈∈ A

f(g(x)), ∀∀ x ∈∈ A

g(f(x)), ∀∀ x ∈∈ A

g(f(x2)), ∀∀ x ∈∈ A

Solution

Question 3

there exists a function g : Y → X such that gof = Ix and fog= IY.

there exists a function g : Y → X fog = IY.

there exists a function g : Y → X such that gof = IX and fog ≠ IY

there exists a function g : Y → X such that gof = IX

Solution

Question 4

{7, 7, 11, 11}

{7, 7, 15, 11}

{7, 7, 5, 11}

{15, 15, 11, 11}

Solution

Question 5

f is odd

gof = fog

gof ≠ fog

g is odd

Solution

Question 6

Iy

invertible with (gof)-1 = f-1og-1

Non invertible

IN

Solution

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g(f(x²)), ∀∀ x ∈∈ A

there exists a function g : Y → X such that gof = I_x and fog= I_Y.

there exists a function g : Y → X fog = I_Y.

there exists a function g : Y → X such that gof = I_Xand fog ≠ I_Y

there exists a function g : Y → X such that gof = I_X

I_y

invertible with (gof)^-1= f^-1og^-1

I_N