Sets, Relations and Functions Test

Result

Sets, Relations and Functions Test - 9

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Weekly Quiz Competition

Question 1
2 / -0.83

The domain of the function is

A
{2 n π: n ∈ I}

B
{ }

C
R

D
none of these

Solution

For f to be defined term under square root has to be greater than zero.
⇒cosx −1 ≥0
⇒cosx ≥1
Only possible if cosx = 1
⇒x = 2n π∀n ∈Z
Question 2
2 / -0.83

The domain of the function

A
R

B
[0,∞)

C
(−∞,0]

D
none of these

Solution
Question 3
2 / -0.83

The domain of the function

A
{x=2n π:n ∈I}

B
[0,2 π]

C
R

D
none of these

Solution
Question 4
2 / -0.83

The domain of the function

A
{x ∈R:x=2n π+π/2 ,n ∈I}

B
−ϕ

C
{π/2}

D
{x ∈R:x=2n π±π/2,n ∈I}

Solution
Question 5
2 / -0.83

Let x be any real, then [x + y] = [x] + [y] holds for

A
y ∈ R, y ∈ Q.

B
y ∈ R

C
y ∈ Q

D
y ∈ I

Solution

Let y has some fractional part
x = 6.5
y = 0.9
[x] = 6, [y] = 0, [x]+[y] = 6
[x + y] = [7.4] = 7
[x + y] ≠[x]+[y]
Let x, y be integers
x = 6.5
y = 1
[x] = 6, [y] = 1, [x]+[y] = 7
[x + y] = [7.5] = 7
[x + y] = [x]+[y]
So, [x + y] = [x]+[y] holds for y ϵI
Question 6
2 / -0.83

The function f (x) = x + cos x

A
is defined for all reals

B
assumes only negative values

C
assumes only positive values

D
none of these

Solution

f(x) = x
g(x) = cosx
D(f) = R
D(g) = R
D(f+g) = D(f) ⋂D(g)
So , D(f+g) = R
Therefore, given function is defined for real numbers.
Question 7
2 / -0.83

A
1

B
2

C
-1

D
-2

Solution

lim(x →π) sinx/x-π
lim(h →0) sin(π-h)/π-h-π
lim(h →0) -sinh/h
⇒-1
Question 8
2 / -0.83

A function / from the set of natural numbers to integers defined by

A
one-one but not onto

B
onto but not one-one

C
neither one-one nor onto

D
one-one and onto both

Solution

Clearly/is both one - one and onto Because if n is odd, values are set of all non-negative integers and if n is an even, values are set of all negative integers.
Question 9
2 / -0.83

If f (x) = and g (x) = then domain of fog is

A
[- 1, 1]

B
(- 1, 1)

C

D

Solution
Question 10
2 / -0.83

If then domain of fof is

A
(−∞,0)

B
(0,∞)

C
(−∞,∞)

D
none of these

Solution