Free Relations and Functions 02 Practice Test - 12th Grade - Commerce 

Question 1

Let X be a family of sets and R be a relation on X defined by 'A is disjoint from B'. Then R is

A.

Reflexive

B.

Symmetric 

C.

Anti-symmetric

D.

Transitive

SOLUTION

Solution : B

Clearly, the relation is symmetric but it is neither reflexive nor transitive.

Question 2

Let R be a relation on the set N of natural numbers defined by nRm

⇔ n is a factor of m (i.e. n(m). Then R is 

A.

Reflexive and symmetric

B.

Transitive and symmetric

C.

Equivalence

D.

Reflexive, transitive but not symmetric

SOLUTION

Solution : D

Since n | n for all n in N,

therefore R is reflexive.

Since 2 | 6 but 6 | 2, therefore R is not symmetric.

Let n R m and m R p ⇒ n|m and m|p ⇒ n|p ⇒ nRp

 So. R is transitive. 

Question 3

Inverse exists for a function which is

A. Injective
B. Surjective
C. Bijective
D. Many-one

SOLUTION

Solution : C

We have seen that inverse exists only when function is one-one and onto, i.e. Bijective.

Question 4

If  f (x) +f (x+a) + f(x+2a) +....+ f (x+na) = constant; ∀x ϵR  and  a>0 and f(x) is periodic,then period of f(x), is

A. (n+1) a
B. en+1a 
C. na
D. ena

SOLUTION

Solution : A

f(x)+f(x+a)+f(x+2a)+....+f(x+na)=k
replacing  x=x+a
f(x+a)+f(x+2a)+....+f(x+(n+1)a)=k
On subtracting second equation from first one -
f(x)−f(x+(n+1)a)=0f(x)=f(x+(n+1)a∴T=(n+1)a

Question 5

The function  f(x)=cos√x is 

A. Periodic with period 2√π
B. Periodic with period √π
C. Periodic with period  4π2 
D. Not a periodic function

SOLUTION

Solution : D

Try drawing cos√x graph. It’s not periodic. 

Question 6

The inverse of f(x)=(5−(x−8)5)13 is

A.

5−(x−8)5

B.

8+(5−x3)15

C.

8−(5−x3)15

D.

(5−(x−8)15)3

SOLUTION

Solution : B

y=f(x)=(5−(x−8)5)13
then y3=5−(x−8)5⇒(x−8)5=5−y3
⇒ x=8+(5−y3)15
Let, z=g(x)=8+(5−x3)15
To check, f(g(x))=       [5−(x−8)5]13
=(5−[(5−x3)15]5)13=(5−5+x3)13=x
similarly , we can show that g(f(x))=x
Hence, g(x)=8+(5−x3)15 is the inverse of f(x)

Question 7

Which of the following relations in R is an equivalence relatilon?

A. xR1 y⇔|x|=|y|
B. xR2 y⇔x≥y
C. xR3 y⇔xy
D. xR4 y⇔x<y

SOLUTION

Solution : A

It is simple to check that only, R1 is an equivalence relation.

Question 8

Let A be the non – empty set of children in a family.  The relation ‘x is a brother of y’ in A is

A. reflexive
B. symmetric
C. transitive
D. an equivalence relation

SOLUTION

Solution : C

Let R denotes the relation ‘is brother of ‘.  Let X ϵ A.  If x is a girl, then we cannot say that x is brother of x.
∴ (x,x) /ϵ R    ∴ R is not reflexive.
Let (x, y)    ϵ R                   ∴ x is a brother of y.
∴ y may or may not be a boy.
∴ we cannot say that (y, x) ϵ R.
∴ R is not symmetric.
Let (x, y) ϵ and (y, z)  ϵ R.
∴ x is a brother of y and y is a brother of z
⇒ is brother of z ⇒ (x, z)  ϵ R
∴ R is transitive.
∴ The correct answer is (c).

Question 9

Given, f(x) = log1+x1−x and g(x) =3x+x31+3x2 then (fog) (x) equals

A. -f(x)
B. 3f(x)
C. [f(x)]3
D. None of these

SOLUTION

Solution : B

(fog)(x)=f(g(x))=f(3x+x31+3x2)=log 1+3x2+3x+x31+3x2−3x−x3=log (1+x)3(1−x)3=3 f(x)

Question 10

Let f:[0,∞)→ [0,2] be defined by f(x)=2x1+x , then f  is

A. one – one but not onto
B. onto but not one – one
C. both one – one and onto
D. neither one – one nor onto

SOLUTION

Solution : A

We can draw the graph and see that the given function is one - one.
Since, Range ⊏ Codomain ⇒ f is into
∴ f is only injective.
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