# Free Objective Test 02 Practice Test - 11th and 12th

### 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

Reflexive

Symmetric

Anti-symmetric

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

Reflexive and symmetric

Transitive and symmetric

Equivalence

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

#### 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

#### 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

#### 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

5−(x−8)5

8+(5−x3)15

8−(5−x3)15

(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?

#### 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

#### 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

#### 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

#### 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.

### Question 11

Which one of the following function is not invertible?

#### SOLUTION

Solution :B

The function f(x) = x2, x ϵ R is not one – one because

f(-4)=f (4) = 16

∴ It is not invertible

### Question 12

The range of the function f(x)=2+x2−x,x≠2 is

#### SOLUTION

Solution :B

y=2+x2−x⇒2y−yx=2+x⇒x(y+1)=2y−2⇒x=2y−2y+1⇒f−1(x)=2x−2x+1

∴ Range of f= Domain of f−1=R−{−1}

### Question 13

If 2f(sin x)+f(cos x)=x ∀ x ϵ R then range of f(x) is

#### SOLUTION

Solution :B

Put x=sin−1x

2f(x)+f(√1−x2)=sin−1x→(1)

On Putting x=cos−1x

⇒2f(√1−x2)+f(x)=cos−1x→(2)

Eq.(1)×2⇒4f(x)+2f(√1−x2)=2sin−1x→(3)

On subtracting Eq. 2 from Eq. 3 we get -

3f(x)=2sin−1x−cos−1x

f(x)=23sin−1x−13(π2−sin−1x)

=sin−1x−π6

fmax=π2−π6=π3,fmin=−π2−π6=−4π6=−2π3

=[−2π3,π3]

### Question 14

The range of f(x)=tan−1(x2+x+a) ∀ xϵ R is a subset of [0,π2) then the range of a is -

#### SOLUTION

Solution :D

tan−1(x2+x+a)≥0⇒x2+x+a≥0

⇒D ≤0⇒1−4a≤0⇒a≥14 ; D is discriminant of quadratic equation.

⇒aϵ[14,∞)

### Question 15

The range of the function f(x)=x+3|x+3|,x≠−3 is

#### SOLUTION

Solution :D

f(x)=1 when x+3>0

f(x)=−1 when x+3<0

Range ={−1,1}