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

### Question 1

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

Reflexive

Symmetric

Transitive

Anti-symmetric

#### SOLUTION

Solution :B

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

because x2+y2=1⇒y2+x2=1

### Question 2

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

#### SOLUTION

Solution :B

As (x,x) ϵ R ⇒ |x−x|≤ 1

⇒ 0 ≤ 1 (True),

Thus, reflexive.

As (x,y) ϵ R ⇒ |x−y|≤ 1

⇒ |y−x||≤1 ⇒ (y,x) ϵ R,

Thus, symmetric.

Again, (x, y) ϵ R and (y, z) ϵ R

⇒|x−y|≤ 1 and |y−z|1/⇒|x−z|≤ 1

∴ Not transitive

### Question 3

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

#### 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.

### Question 4

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

#### 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

### Question 5

Which of the following functions are periodic?

#### 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

### Question 6

Let f:R→R,g:R→R, be two functions, such that f(x) =2x – 3, g (x) = x3 + 5.

The function (fog)−1 (x) is equal to

#### 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)=y⇒2x3+7=y⇒x=(y−72)13

∴ (fog)−1(x)=(x−72)13,x ϵ R

∴ The correct answer is (d).

### Question 7

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

#### 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

### Question 8

If the function f:[1,∞)→[1,∞) is defined by f(x)=2x(x−1), then f−1(x) is

#### SOLUTION

Solution :B

f(x)=y⇒2x(x−1)=y⇒x(x−1) log2 2=log2y

⇒x(x−1)=log2 y⇒x2−x−log2 y=0

⇒x=1±√1+4 log2 y2

∴x=1+√1+4 log2 y2

∴f−1(x)=12(1+√1+4 log2 x)

∴ The correct answer is (b).

### Question 9

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

#### 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.

### Question 10

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

#### SOLUTION

Solution :D

Since A⊆A.

∴ Relation ′⊆′ is reflexive.

Since A⊆B,B⊆C⇒A⊆C

∴ Relation ′⊆′ is transitive.

But If A⊆B, Doesn't imply B⊆A,

∴ Relation is not symmetric

### Question 11

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

#### SOLUTION

Solution :D

g(f(x))≤f(g(x))⇒g(|x|)≤f[x]⇒[|x|]≤|[x]|

This is true for xϵR..

### Question 12

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

#### SOLUTION

Solution :D

y=x2−3x+4x2+3x+4

yx2+3xy+4y=x2−3x+4

x2(y−1)+3x(y+1)+4(y−1)=0

D ≥0⇒9(y+1)2−4.4(y−1)2≥0

(3(y+1)−4(y−1)) (3(y+1)+4(y−1))≥0

(−y+7)(7y−1)≥0

(y−7)(y−17)≤0

17≤y≤7

### Question 13

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

#### SOLUTION

Solution :D

f(x)=1−sin2x4+sinx4=−{sin2x4−sinx4}+1=−{(sinx4−12)2−14}+1

=54−(sinx4−12)2

Maximun f(x)=54

Minimum f(x)=54−(−1−12)2=54−94=−1

Range of f(x)=[−1,54]

### Question 14

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

#### 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 X→1,X1+X→12

Hence, the range of f(x) is [0,12) .

### Question 15

Range of f(x)=tan(π[x2−x])1+sin(cos x) is (where [x] denotes the greatest integer function)

#### SOLUTION

Solution :D

f(x)=tan(π[x2−x])1+sin(cos x)={0} because of [x2−x] is integer