Relations And Functions Test

Result

Relations And Functions Test - 3

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Determine the domain of the following relation.

^{_{R = {(a, b): a} _{a < 5, b = 4}}}

A
{1, 2, 3, 4}

B
(1, 2, 3, 4)

C
{1, 2, 3}

D
{1, 2, 3, 4, 5}

Solution

The domain of the relation is {1, 2, 3, 4}.
Question 2
1 / -0

Which of the following functions is an even function?

A
f(x) = x(x² – x⁴)

B
f(x) = sin(5x + x³)

C
f(x) = x² – 3x + 4

D
f(x) = x² +

Solution

For a function to be an even function, f(–x) = f(x) for all x.
Consider (1): f(x) = x(x² – x⁴) = x³ – x⁵
f(–x) = (–x)³ – (–x)⁵ = –x³ + x⁵ = –(x³ – x⁵), f(–x) ≠ f(x)
Thus, f(x) is not an even function. It is an odd function as f(–x) = –f(x).
Consider (2): f(x) = sin(5x + x³)
f(–x) = sin(–5x + (–x)³) = sin(–5x – x³) = sin(–(5x + x³)) = –sin(5x + x³)
f(–x) ≠ f(x)
This function is not an even function. It is an odd function.
Consider (3): f(x) = x² – 3x + 4; f(–x) = x² + 3x + 4, f(–x) ≠ f(x)
Also, f(–x) ≠ –f(x)
This function is neither an even function nor an odd function.
Consider (4): f(x) = x² + |x|
f(–x) = x² + |–x| = x² + |x| = f(x)
This function is an even function.
Question 3
1 / -0

If R is a relation from a non-empty set A to a non-empty set B, then

A
R = A B

B
R = A B

C
R = A B

D
R A B

Solution

A relation from a non-empty set A to a non-empty set B is defined as a subset of A B or a relation R from the non-empty set A to another non-empty set B is a subset of their Cartesian product A B, i.e. R A B
Question 4
1 / -0

The domain of the real function – is

A
(– 3, 1)

B
[– 3, 1]

C
(– 3, 2]

D
[– 3, 1)

Solution

_{Let f(x) =}_{^, then f(x) is defined for all x s.t. 2 - x 0

x 2}
And let g(x) = , then g(x) is defined for all x s.t. 9 - x² > 0
x²- 9 < 0

- 3 < x < 3

Domain of f(x) - g(x) = - 3 < x 2
Question 5
1 / -0

If f(x + y) = f(x) + f(y) and f(1) + f(2) + f(3) + ……. + f(10) = 1, find the value of f(1).

A
0

B

C
1

D

Solution

Given: f(x + y) = f(x) + f(y)
f(1 + 1) = f(2) = f(1) + f(1) = 2f(1)
And f(2 + 1) = f(3) = f(2) + f(1) = 3f(1)
Similarly, f(4) = 4f(1),
f(5) = 5f(1)
….......
f(10) = 10f(1)

Given expression:
f(1) + f(2) + f(3) … f(10) = 1
f(1) + 2f(1) + 3f(1) …. 10f(1) = 1
(1 + 2 + 3 + 4 + … + 10)f(1) = 1
f(1) =
Question 6
1 / -0

The range of the function f(x) = |x - 1| is

A
(- _∞, _∞)

B
(0, _∞)

C
[0, _∞)

D
(- _∞, 0)

Solution

Since |x| assumes every value in [0, ), so |x - 1| also takes every value in [0, ). Hence, range of f(x) = |x - 1| is [0, ).
Question 7
1 / -0

If f(x) = , then f(1994) is

A
log 1994

B
1994 log 3

C
3

D
None of these

Solution

Given: f(x) =
log f(x) = log (3^{1 + logx}) - log (x^{log 3})
log f(x) = (1 + log x) log 3 - log 3 log x = log 3
f(x) = 3, which is a constant
f(1994) = 3, which will also be a constant.
Question 8
1 / -0

The range of , where x 0, is

A
{-1, 1}

B
[-1, 1]

C
R - (-1, 1)

D
(-1, 1)

Solution

If x is -ve, then |x| = -x = -1.
If x is +ve, then |x| = x = 1.
Hence, the range is {-1, 1}.
Question 9
1 / -0

The range of the function f(x) = cot^-1 (e^x + e^-x) is

A

B

C

D

Solution

f(x) = cot^-1 (e^x + e^-x)
f '(x) = = -ve
f'(x) < 0 f(x) decreases
∵ e^x + e^-x ≥ 2
∴ cot^-1 (e^x + e^-x) ≤ cot^-1 2
y ≤ cot^-1
Also, cot^-1 x > 0
∴ The range of the function is .
Question 10
1 / -0

The range of the function f(x) = -is equal to

A
[0, 2]

B
[- 2, 0]

C
[- 2, 2]

D
None of these

Solution

Note that D_f = [- 5, - 1]
(f (x) is real if - x² - 6x - 5 0 i.e. if (x + 5) (x + 1) 0)
For R_f, let y = f(x)
y = -
y² = - x² - 6x - 5, y ≤ 0
x² = 6x + (5 + y²) = 0, y ≤ 0
6² - 4.1 (5 + y²) ≥ 0
(x is real, disc ≥ 0)
y² ≤ 4
|y| ≤ 2 and y ≤ 0
R_f = [- 2, 0]
Question 11
1 / -0

The range of the function f(x) = is

A

B
R

C

D
None of these

Solution

The function f(x) = is defined for all x R.

1 ≤ 2 - cos 3x ≤ 3 for all x R

≤ ≤ 1 for all x R

≤ f(x) ≤ 1 for all x R

f(x) [, 1]
Range (f) = [, 1]
Question 12
1 / -0

Let A = {–2, –1, 0, 1, 2} and f : A Z be given by f(x) = x² – 2x – 3. Then the range of f is

A
{0, 5, –3, –4}

B
{0, –5, –3, –4}

C
(0, 5, 3, –4)

D
None of these

Solution

We have f(x) = x²− 2x − 3
∴ f(−2) = (−2)²− (2) × (−2) − 3 = 4 + 4 − 3 = 5
f(−1) = (−1)²− 2 × (−1) − 3 = 1 + 2 − 3 = 0
f(0) = (0)²− 2 × 0 − 3 = 0 + 0 − 3 = −3
f(1) = (1)²− 2 × 1 − 3 = 1 − 2 − 3 = −4
f(2) = (2)²− 2 × 2 − 3 = 4 − 4 − 3 = −3
So, range of f = {f(−2), f(−1), f(0), f(1), f(2)}
= {5, 0, −3, −4,−3}
= {0, 5, −3, −4}
Question 13
1 / -0

If a set A has n distinct elements, then the number of relations on A is

A

B
n²

C
2ⁿ

D
None of these

Solution

_{Since A contains n distinct elements, A × A contains n × n = n² distinct elements.
Since every subset of A × A is a relation on A, the number of relations on A is equal to the order of the power set of A × A, i.e..}
Question 14
1 / -0

The number of relations that can be defined on set A = {a, b, c, d} is

A
24

B
16

C
4⁴

D
2¹⁶

Solution

The number of relations on A is 2^{4 × 4} = 2¹⁶.
Question 15
1 / -0

Let A = {1, 2, 3}. The total number of distinct relations which can be defined over A is

A
6

B
8

C
2⁹

D
None of these

Solution

If a set A has n distinct elements, then the number of all relations is .
Question 16
1 / -0

The range of the function f(x) = |x| is

A
(, )

B
(0, )

C
[0, )

D
(, 0)

Solution

Given, function f(x) = ∣x∣
As we know, ∣x∣ ≥ 0
Hence, the range of f(x) = |x| is [0, ).
Question 17
1 / -0

2f(x) = |2x - 1| + |3x - 2| + |2 - x|

Find the range of the function given above.

A

B

C

D

Solution

2Plot the graph of the given equation.

To plot the graph, first decide the critical values for each part of the equation.
For |2x - 1|, the critical value, where the equation gives zero output, is: .
For |3x - 2|, the critical value is: .
For |2 - x|, the critical value is: x = 2^_.
Now, write the equation for each side of these critical points individually and plot it on the paper.

For ,

For ,

For ,

For ,

Thus, the graph will be as shown in the figure and the range will be: .
Question 18
1 / -0

Find the domain of the function .

A

B

C

D

Solution

Given:

Logarithmic function can take the value from .

Thus, the domain of the part is .

Therefore, the domain of the part is .

Therefore, the domain of the part is .

We know that .

So, we get

Therefore,

Hence, the domain of the given function is .
Question 19
1 / -0

What is the value of x which is not in the domain of the function ?

A

B

C

D

Solution

To define the given function, we have to eliminate the values which make the denominator 0.

Taking log of both sides, we get

Thus, this value must be excluded from the set. Hence, it is the correct answer.
Question 20
1 / -0

2A relation R is defined as: . Find the domain of the relation.

A

B

C

D

Solution

2Domain of the relation lies in .

Then,

We can see that this is an equation of ellipse. Therefore, x will be the major axis of the ellipse and y is the minor axis of the ellipse.

Therefore,