Relations and Functions Test

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Relations and Functions Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

The domain of definition of the function y = f(x) = √−x is :

A
(−∞,10]

B
[0,∞)

C
(−∞,0]

D
(1,∞)

Solution

y is defined if −x ⩾ 0 ,i.e.if x ⩽ 0 , i.e. x ∈(−∞,0].
Question 2
1 / -0

The range of the function f(x) = [sin x] is

A
[1, 1]

B
{–1, 1}

C
(–1, 1)

D
{ – 1, 0, 1}

Solution

The only possible integral values of sin x are { -1 ,0, 1 }. As − 1 ≤ sin θ ≤ 1
Question 3
1 / -0

The range of the function f(x) = x – [x] is

A
[0, 1

B
[0, 1]

C
[0,2]

D
(0, 1)

Solution

Since [x] ≤ x, therefore , x – [x] ≥ 0. Also, x – [x] <1,∴ 0 ⩽ x − [x] < 1. Therefore ,Range of f = [0,1).
Question 4
1 / -0

The range of function f(x) = [x] is :

A
Q (set of rational numbers)

B
Z

C
R

D
R – Z

Solution

R_f= Z, ∵ [x] = integral part of x.
Question 5
1 / -0

If f(x) = x – x², where f : R → R, then f( a + 1 ) – f( a – 1), ∀a ∈ R is :

A
10a - 8

B
2a + 4

C
4 – 2a

D
2 – 4a

Solution

f(a + 1) − f (a − 1) = (a + 1) − (a + 1)²− {(a − 1) − (a − 1)²} = 2 − 4a
Question 6
1 / -0

The domain of the function f/g is

A
{x ∈R : g (x) ≠ 0 } = R – { x ∈R : g (x) = 0 }.

B
D_f∩ D_g

C
D_f∪ D_g

D
D_f- D_g

Solution

domain of the function f/g is equals to {x ∈R : g (x) ≠ 0 } = R – { x ∈R : g (x) = 0 }.

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Relations and Functions Test - 11

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Relations and Functions Test - 11

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Question 1

(−∞,10]

[0,∞)

(−∞,0]

(1,∞)

Solution

Question 2

[1, 1]

{–1, 1}

(–1, 1)

{ – 1, 0, 1}

Solution

Question 3

[0, 1

[0, 1]

[0,2]

(0, 1)

Solution

Question 4

Q (set of rational numbers)

Z

R

R – Z

Solution

Question 5

10a - 8

2a + 4

4 – 2a

2 – 4a

Solution

Question 6

{x ∈R : g (x) ≠ 0 } = R – { x ∈R : g (x) = 0 }.

Df ∩ Dg

Df ∪ Dg

Df - Dg

Solution

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D_f∩ D_g

D_f∪ D_g

D_f- D_g