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

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.

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.

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.

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)aT=(n+1)a

The function  f(x)=cosx is

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

#### SOLUTION

Solution : D

Try drawing cosx graph. It’s not periodic.

The inverse of f(x)=(5(x8)5)13 is

A.

5(x8)5

B.

8+(5x3)15

C.

8(5x3)15

D.

(5(x8)15)3

#### SOLUTION

Solution : B

y=f(x)=(5(x8)5)13
then y3=5(x8)5(x8)5=5y3
x=8+(5y3)15
Let, z=g(x)=8+(5x3)15
To check, f(g(x))=       [5(x8)5]13
=(5[(5x3)15]5)13=(55+x3)13=x
similarly , we can show that g(f(x))=x
Hence, g(x)=8+(5x3)15 is the inverse of f(x)

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

A. xR1 y|x|=|y|
B. xR2 yxy
C. xR3 yxy
D. xR4 yx<y

#### SOLUTION

Solution : A

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

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).

Given, f(x) = log1+x1x 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+3x23xx3=log (1+x)3(1x)3=3 f(x)

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.

Which one of the following function is not invertible?

A. f:RR,f(x)=3x+1
B. f:R[0,),f(x)=x2
C. f:R+R+,f(x)=1x3
D. None of the above

#### 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

The range of the function f(x)=2+x2x,x2 is

A. R
B. R{1}
C. R{1}
D. R{2}

#### SOLUTION

Solution : B

y=2+x2x2yyx=2+xx(y+1)=2y2x=2y2y+1f1(x)=2x2x+1
Range of f= Domain of f1=R{1}

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

A. [π3,π3]
B. [2π3,π3]
C. [2π3,π6]
D. [π6,π6]

#### SOLUTION

Solution : B

Put x=sin1x
2f(x)+f(1x2)=sin1x(1)
On Putting x=cos1x
2f(1x2)+f(x)=cos1x(2)
Eq.(1)×24f(x)+2f(1x2)=2sin1x(3)
On subtracting Eq. 2 from Eq. 3 we get  -
3f(x)=2sin1xcos1x
f(x)=23sin1x13(π2sin1x)
=sin1xπ6
fmax=π2π6=π3,fmin=π2π6=4π6=2π3
=[2π3,π3]

The range of f(x)=tan1(x2+x+a)  xϵ R is a subset of  [0,π2) then the range of a is -

A. [3,14]
B. (π2,π2)
C. [3,1]
D. [14,)

#### SOLUTION

Solution : D

tan1(x2+x+a)0x2+x+a0
D 014a0a14      ; D is discriminant of quadratic equation.
aϵ[14,)

The range of the function f(x)=x+3|x+3|,x3 is

A. {3,3}
B. R{3}
C. All positive integers
D. {1,1}

#### SOLUTION

Solution : D

f(x)=1 when x+3>0
f(x)=1 when x+3<0
Range ={1,1}