# Free Objective Test 01 Practice Test - 11th and 12th

Let P = {(x,y) x2+y2=1,x,yR}. Then P is.

A.

Reflexive

B.

Symmetric

C.

Transitive

D.

Anti-symmetric

#### SOLUTION

Solution : B

Here we can see that the relation is neither reflexive nor transitive but it is symmetric,

because x2+y2=1y2+x2=1

R is relation over the set of integers and it is given by (x, y) ϵ R R  |x - y| 1.  Then, R is

A. Reflexive and transitive
B. reflexive and symmetric
C. Symmetric and transitive
D. an equivalence relation

#### SOLUTION

Solution : B

As (x,x) ϵ R   |xx| 1
0 1 (True),
Thus, reflexive.
As (x,y) ϵ R     |xy| 1
|yx||1  (y,x) ϵ R,
Thus, symmetric.
Again, (x, y) ϵ R and (y, z) ϵ R
|xy| 1   and |yz|1/|xz| 1
Not transitive

Let R be a relation over the set N×n and it is defined by (a, b) R (c, d)   a+ d = b + c.  Then, R is

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

#### SOLUTION

Solution : D

(a, b) R (a, b) because a + b = b + a.  So, r is reflexive.
(a, b)R (c, d) a+d = b+c c+b = d+a
(c,d) R (a,b)
So, R is symmetric.
(a, b) R (c, d) and (c, d) R (e, f)
a + d = b + c, c + f = d + e
Adding,         a + d + c + f = b + c + d +e
a + f = b + e
(a, b) R (e, f).
R is transitive.

The period of the function |sinx|+|cosx| is

A. π2
B. 2π
C. 4π
D. π

#### SOLUTION

Solution : A

The smallest of π2,2π,4π,π is π2
Let f(x)=|sin x|+|cos x|.
f(x+π2)=sin(x+π2)+cos(x+π2)
=|cos x|+|sin x|
=|cos x|+|sin x|
=f (x)
The period of given function is π2

Which of the following functions are periodic?

A. f(x) = log x, x > 0
B. f(x) = ex, x ϵ R
C. f(x) = x - [x], x ϵ R
D. f(x) = x + [x], x ϵ R

#### SOLUTION

Solution : C

f(x) = log x, is not periodic.
f(x) = ex, is not periodic.
f(x) = x - [x] = {x}, has period 1
f(x) = x + [x], is not periodic

Let f:RR,g:RR, be two functions,  such that f(x) =2x – 3, g (x) = x3 + 5.
The function (fog)1 (x) is equal to

A. (x+72)13
B. (x72)13
C. (x27)13
D. (x72)13

#### SOLUTION

Solution : D

We have, f : R R, g: R R defined by f(x) = 2x - 3 and g (x) = x3 + 5
It can be checked that f(x) and g(x) are bijective functions
fo g is also bijective and (fog) = f(g(x)) = f (x3+5)=2(x3+5)3=2x3+7
(fog)(x)=y2x3+7=yx=(y72)13
(fog)1(x)=(x72)13,x ϵ R

If f: R   R ,g : R R and h: R   R are such that f(x)=x2 , g(x)=tanx and h(x)=logx, then the value of (h(g(f(x))))
if x=π4 will be

A. 0
B. 1
C. -1
D. π

#### SOLUTION

Solution : A

(ho(gof))(x)=h{g{f(x)}}
=h{tan x2}=log{tan x2}
At x=π4(ho(gof))(x)=log tan π4
= log 1 = 0

If the function f:[1,)[1,) is defined by f(x)=2x(x1), then  f1(x) is

A. (12)x(x1)
B. 12(1+1+4 log2 x)
C. (12)(11+4log2 x)
D. None of these

#### SOLUTION

Solution : B

f(x)=y2x(x1)=yx(x1) log2 2=log2y
x(x1)=log2 yx2xlog2 y=0
x=1±1+4 log2 y2
x=1+1+4 log2 y2
f1(x)=12(1+1+4 log2 x)

f:R×RR such that f(x + iy) =  x2+y2. Then, f is

A. many – one and into
B. one-one and onto
C. many – one and onto
D. one – one and into

#### SOLUTION

Solution : A

Since, f (x + iy) = f (x - iy) f is many – one
Also, Range =R+ and codomain = R.
Range codomain f is into.
Hence, f is many – one into.

With reference to a universal set, the inclusion of a subset in another, is relation, which is

A. Symmetric only
B. Equivalence relation
C. Reflexive only
D. None of these

#### SOLUTION

Solution : D

Since AA.
Relation is reflexive.
Since AB,BCAC
Relation is transitive.
But If AB, Doesn't imply  BA,
Relation is not symmetric

If f:RR and g:RR are given by f(x) = |x| and g(x) = [x], then g(f(x))f(g(x) is true for -

A. Z(,0)
B. (,0)
C. Z
D. R

#### SOLUTION

Solution : D

g(f(x))f(g(x))g(|x|)f[x][|x|]|[x]|
This is true for xϵR..

f(x)=x23x+4x2+3x+4 the range of f(x) is

A. [0,17]
B. (,17)(7,)
C. (,7)
D. [17,7]

#### SOLUTION

Solution : D

y=x23x+4x2+3x+4
yx2+3xy+4y=x23x+4
x2(y1)+3x(y+1)+4(y1)=0
D 09(y+1)24.4(y1)20
(3(y+1)4(y1)) (3(y+1)+4(y1))0
(y+7)(7y1)0
(y7)(y17)0
17y7

The range of the function f(x)=cos2x4+sinx4,x ϵ R is

A. [0,54]
B. [1,54]
C. (1,54)
D. [1,54]

#### SOLUTION

Solution : D

f(x)=1sin2x4+sinx4={sin2x4sinx4}+1={(sinx412)214}+1
=54(sinx412)2
Maximun f(x)=54
Minimum f(x)=54(112)2=5494=1
Range of f(x)=[1,54]

Let f(x)=x[x]1+x[x],x ϵ R, where [ x] denotes the greatest integer function. Then, the range of f is

A. (0,1)
B. [0,12)
C. [0,1]
D. [0,12]

#### SOLUTION

Solution : B

The graph of y=x[x] is as shown below When x is an integer, x[x]=0
Hence, f(x) = 0 when x is an integer
x[x] as x tends to an integer.
Let X =  x[x]
So, f(x)=X1+X,X ϵ [0,1)
As X1,X1+X12
Hence, the range of f(x) is [0,12)  .

Range of f(x)=tan(π[x2x])1+sin(cos x) is (where [x] denotes the greatest integer function)

A. (,)[0,tan 1]
B. (,)[tan 2,0]
C. [tan 2,tan 1]
D. {0}

#### SOLUTION

Solution : D

f(x)=tan(π[x2x])1+sin(cos x)={0} because of [x2x] is integer