Sets, Relations and Functions Test

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Sets, Relations and Functions Test - 10

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Weekly Quiz Competition

Question 1
2 / -0.83

If f(x) = ax + b and g(x) = cx + d, then f[g(x)] –g[f(x)] is equivalent to

A
f(c) + g(a)

B
f(a) –g(c)

C
f(d) –g(b)

D
f(b) –g(b)

Solution

We have f(x) = ax+b, g(x) = cx+d
Therefore, f{g(x)} = g{f(x)}
⇔f(cx+d) = g(ax+b)
⇔a(cx+d)+b = c(ax+b)+d
⇔ad+b = cb+d
⇔f(d) - g(b)
Question 2
2 / -0.83

If , then inverse of f is:

A

B

C

D

Solution

let y = (4x + 3)/(6x - 4)
⇒ y * (6x - 4) = 4x + 3
⇒ 6xy - 4y = 4x + 3
⇒ 6xy - 4y - 4x = 3
⇒ 6xy - 4x = 3 + 4y
⇒ x(6y - 4) = 4y + 3
⇒ x = (4y + 3)/(6y - 4)
So, f^-1 (x) = (4x + 3)/(6x - 4)
Question 3
2 / -0.83

If f(x) = (3 –x² )^1/2 , then fof is:

A
½x

B
x

C
-x^1/2

D
x^1/2
Question 4
2 / -0.83

If f: R →R and g: R →R be defined as f(x) = x + 1 and g(x) = x –1. Then for all x ∈R

A
f o g = g o f

B
g o f does not exist

C
f o g does not exist

D
f o g = g o g

Solution

Given f:R →R and g: R →R
=>fog : R →R and gof : R →R
We know that I(R) : R →R
So the domains of gof, fog I(R) are the same.
fog(x) = f(g(x)) = f(x-1) = x-1+1 = I(R)x …..(1)
gof(x) = g(f(x)) = g(x-1) = x-1+1 = I(R)x ……(2)
From (1) and (2), we get
fog(x) = gof(x) = I(R) for all x belongs to R
Hence fog = gof
Question 5
2 / -0.83

If f and g two functions such that they are one-one then g o f is

A
A one-one function

B
A bijective function

C
A many-one function

D
A many-one and onto function

Solution

f:A →B and g:B →C are both one-to-one functions.
Suppose a₁ ,a₂ ∈A such that (gof)(a₁ )=(gof)(a₂ )
⇒g(f(a₁ )) = g(f(a₂ )) (definition of composition)
Since gg is one-to-one, therefore,
f(a₁ ) = f(a₂ )
And since ff is one-to-one, therefore,
a₁ = a₂
Thus, we have shown that if (gof)(a₁ )=(gof)(a₂ then a₁ =a₂
Hence, gof is one-to-one function.
Question 6
2 / -0.83

Let A = {a,b,c} and B = {1,2,3} and f: A →B is defined by f={(a,2), (b,1), (c,3)}. Find f^-1

A
{(2,a),(1,b), (3, c)}

B
Can not be inverted as it is not one-one

C
Can not be inverted as it is not onto

D
{(a,2), b,1), (c,3)}.
Question 7
2 / -0.83

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

A
A many-one and onto function

B
A bijective function

C
An into function

D
An onto function

Solution

since g : B →C is onto
Suppose z implies C, there exist a pre image in B
Let the pre image be y
Hence y implies B such that g(y) = z
Similarly f : A →B is onto
Suppose y implies B, there exist a pre image in A
Let the pre image be x
Hence y implies A such that f(x) = y
Now, gof : A →C gof = g(f(x))
= g(y)
= z
So, for every x in A, there is an image z in C
Thus, gof is onto.
Question 8
2 / -0.83

A function in which second element of order pair are distinct is called

A
Onto function

B
One-one function

C
Identity function

D
Inverse function
Question 9
2 / -0.83

If f be a mapping defined by f(x) = x² + 5, then f^-1 (x) is:

A
(x - 5)^1/2

B
(x² + 5)^1/2

C
(x + 5)^1/2

D

Solution

f(x) = x² + 5 = y(let)
⇒y = x² +5
⇒x² = y-5
⇒x = (y-5)^1/2
⇒f-1(x) = (x-5)^1/2
Question 10
2 / -0.83

Let A = {2 , 3 , 6}. Which of the following relations on A are reflexive?

A
R₁ = {(2, 2), (3, 3), (6, 6)}

B
R₃ = {(2, 2), (3, 6), (2, 6)}

C
R₂ = {(2, 2), (3, 3), (3, 6), (6, 3)}

D
None of these

Solution

R₁ is a reflexive on A , because (a,a) ∈R₁ for each a ∈ A