Relations and Functions Test

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Relations and Functions Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

Let f : R → R and f(x) = x²and g: R → R such that g(x) = sinx. Then, g o f = ?

A
(sin x)²

B
sin x²

C
sin x³

D
x² + sin x

Solution

(g o f)(x) = g(f(x)) = g(x²) = sin x².
Question 2
1 / -0

Let f and g be given by f = {(5,2) , (6 , 3)} and g = { (2,5), (3,6)}. Find gof.

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

B
{(5,5) , (6,6)}

C
{(2,5), (3,6)}.

D
{(5,5)(5,2) , (6,6)}

Solution

Domain of gof = Domain of f = {5,6}

Now, gof(5) = g {f(5)} = g (2) = 5 ; gof(6) = g {f(6)} = g (3) = 6

gof = { (5,5) , (6,6)}
Question 3
1 / -0

Given below is binary composition table a* b = LCM of a and b on S = { 1,2,3,4}. Then, from the table determine which one of these options is correct.

* 1 2 3 4

1 1 2 3 4

2 2 2 6 4

3 3 6 3 12

4 4 4 12 4

A
a* b   is a commutative binary operation

B
a* b   is a distributive binary operation

C
a* b   is not a binary operation

D
a* b   is an associative binary operation

Solution

All elements in the table do not belong  to the set  S.
Question 4
1 / -0

If a relation f:X→Y is a function , then for g:Y→X to be a function ,function f need to be

A
one-one and onto

B
many one

C
one-one

D
onto but not one-one

Solution

By definitiong:Y→X will be a function if every element of Y has a unique correspondence in X.This is possible only if for f the codomain = range , i.e it is an onto function .

Also because of requirement of uniqueness of correspondence , f needs to be a one-one function.
Question 5
1 / -0

f: R→R and g:R→R . f and g are defined as f(x) =2x - 3 and g(x) = x² + 3x + 1. Then, fog(x) is

A
2x²+6x -1

B
4x -9

C
x⁴+6x³+14x²+15x +5

D
4x²-6x -1

Solution

fog(x) = f[g(x)] =2(x² + 3x + 1) -3 = 2x²+6x +2 -3 = 2x²+6x -1
Question 6
1 / -0

If f: A→B and g:B→C are onto , then gof:A→C is:

A
an into function

B
a bijective function

C
an onto function

D
a many-one and onto function

Solution

Let c∈C

g:B→C is onto

⇒c∈Chas a pre image

⇒There exists a b ∈B such that g(b) = c

b ∈B and f: A→B is onto

⇒ b∈ B has a pre image

⇒There exists an a ∈A such that f(a) = b

c = g(b) = g[f(a)] = gof (a)

So, c∈Chas a pre image a∈A , such that gof(a) = c

→gof is an onto function

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Relations and Functions Test - 5

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Relations and Functions Test - 5

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Question 1

(sin x)2

sin x2

sin x3

x2 + sin x

Solution

Question 2

{(5,2) , (6 , 3)}

{(5,5) , (6,6)}

{(2,5), (3,6)}.

{(5,5)(5,2) , (6,6)}

Solution

Question 3

a* b is a commutative binary operation

a* b is a distributive binary operation

a* b is not a binary operation

a* b is an associative binary operation

Solution

Question 4

one-one and onto

many one

one-one

onto but not one-one

Solution

Question 5

2x2+6x -1

4x -9

x4+6x3+14x2+15x +5

4x2-6x -1

Solution

Question 6

an into function

a bijective function

an onto function

a many-one and onto function

Solution

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(sin x)²

sin x²

sin x³

x² + sin x

2x²+6x -1

x⁴+6x³+14x²+15x +5

4x²-6x -1