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All questions of Relations and Functions for Commerce Exam

If f(x) = x2 and g(x) = cosx, which of the following is true?
  • a)
    f + g is even function
  • b)
    f – g is an odd function
  • c)
    f + g is not defined
  • d)
    f + g is an odd function
Correct answer is option 'A'. Can you explain this answer?

Raghav Bansal answered
if f(x) is an odd function
So, f(−x)=−f(x)
F(−x)=cos(f(−x))
=cos(−f(x))
=cos(f(x))
=F(x)
So cos(f(x)) is an even function
So, f(x) and g(x) is an even function
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If A = {1, 2, 3}, and B = {3, 6} then the number of relations from A to B is
  • a)
    32
  • b)
    23
  • c)
    23
  • d)
    26
Correct answer is option 'D'. Can you explain this answer?

Preeti Iyer answered
The number of relations between sets can be calculated using 2mn where m and n represent the number of members in each set.
So, number of relations from A to B is 26.

Let R be a relation defined as
R = {(x, y): y = 2x, x is natural number < 5} then Range of R is given as ,
  • a)
    {2, 4, 6, 8}
  • b)
    {2, 4, 6, 8, 10}
  • c)
    {1, 2, 3, 4}
  • d)
    N
Correct answer is option 'A'. Can you explain this answer?

Gaurav Kumar answered
X is a natural number and x < 5, the number is 1, 2, 3, 4
y = 2x
► x(1), y = 2 × 1 = 2
► x(2), y = 2 × 2 = 4
► x(3), y = 2 × 3 = 6
► x(4), y = 2 × 4 = 8
Range = {2, 4, 6, 8}

Which of the following is not an example of polynomial function ?
  • a)
    g(x) = 8x2 + 5x – 2
  • b)
    p(x) =5x3 + 3x2 + x – 2
  • c)
    h(x) =3 root of x minus 1 over x
  • d)
    f(x) = 3x2 + x – 2
Correct answer is option 'C'. Can you explain this answer?

Lavanya Menon answered
A polynomial function is a function which involves only non-negative integer powers or only positive integer exponents of a variable in an equation.
In option C, powers of x are negative and fractional.

 If X = {1, 2, 3, 4}, Y = {1, 2, 3,…., 20}, and f: X –> Y be the correspondence which assigns each element in X the value equal to its square, then the domain, co-domain and range of f is
  • a)
    Domain = {1, 2, …, 20}, Range = {1, 2, 3, 4}, Co domain = {1, 2, 3, 4}
  • b)
    Domain = {1, 4, 9, 16}, Range = {1, 2, 3, 4}, Co domain = {1, 2, 3, …, 16}
  • c)
    Domain = {1, 2, 3, 4}, Range = {1, 4, 9, 16}, Co domain = {1, 2, 3, …, 20}
  • d)
    Domain = {1, 2, 3, 4}, Range = {1, 4, 9, 16}, Co domain = {1, 2, 3, …, 16}
Correct answer is option 'B'. Can you explain this answer?

Raghav Bansal answered
The correct option is B
Since the function is defined from 
XY.So the domain will be the set X which is {1,2,3,4}. Codomain is  what may possibly come out of a function. So codomain is the whole set Y which is {1,2,3,...,20}. Range is The range is the set of possible output values which will be the values in the set of Y which have a pre-image. That is, the value of squares. So the range is {1,4,,9,16}. 

Which of the following relations is not a function?
  • a)
    R= {(1,2), (3,4),(2,1),(5,1)}
  • b)
    R={(1,2), (1,4)(3,1),(5,1)}
  • c)
    R= {(1,2), (3,4)(2,1),(5,2)}
  • d)
    R= {(2,1), (4,4)(3,1),(5,1)}
Correct answer is option 'B'. Can you explain this answer?

Gaurav Kumar answered
A function is a relation in which no two ordered pairs have the same first element. A function associates each element in its domain with one and only one element in its range. R={(1,2), (1,4)(3,1),(5,1)} is not  a function because all the first elements are not different.

The function f : R → R defined by y = f(x) = 5 for each x ∈ R is
  • a)
    Modulus function
  • b)
    Signum function
  • c)
    Identity function
  • d)
    Constant function
Correct answer is option 'D'. Can you explain this answer?

Pioneer Academy answered
Constant Function is defined as the real valued function.
f: R→R, y = f(x) = c for each x∈R and c is a constant.
So ,this function basically associate each real number to a constant value.
It is a linear function where f(x1) = f(x2) for all x1,x2 ∈R

If A = [a, b], B = [c,d], C = [d, e] then {(a, c), (a, d), (a,e), (b,c), (b, d), (b, e)} is equal to
  • a)
    A∪(B∩C)
  • b)
    A∩(B∪C)
  • c)
    A×(B∩C)
  • d)
    A×(B∪C)
Correct answer is option 'D'. Can you explain this answer?

Jatin Sen answered
Understanding the Sets
To analyze the sets A, B, and C:
- A = [a, b]
- B = [c, d]
- C = [d, e]
Defining Operations
1. Union (U): Combines elements from both sets.
2. Intersection (∩): Contains elements common to both sets.
3. Cartesian Product (×): Forms pairs from elements of two sets.
Calculating B ∩ C
- B = [c, d]
- C = [d, e]
Common Element:
- The only common element is d.
- Therefore, B ∩ C = [d].
Calculating B U C
- B = [c, d]
- C = [d, e]
Combined Elements:
- B U C = [c, d, e].
Forming the Cartesian Product
- Now, we will find A × (B U C):
A = [a, b] and B U C = [c, d, e]
Pairs:
- From A and B U C, we get:
- (a, c)
- (a, d)
- (a, e)
- (b, c)
- (b, d)
- (b, e)
Final Cartesian Product:
- A × (B U C) = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e)}.
Conclusion
The set {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e)} is indeed equal to A × (B U C), confirming that option 'D' is the correct answer.

Which of the following is incorrect?
  • a)
    Modulus function, Domain: R; Range: [0, infinity)
  • b)
    Signum function, Domain: R; Range: {-1, 0,1}
  • c)
    Constant function, Domain: R; Range: R
  • d)
    Identity function, Domain: R; Range: R
Correct answer is option 'C'. Can you explain this answer?

Preeti Khanna answered
Constant Function is defined as the real valued function.
f : R→R, y = f(x) = c for each x∈R and c is a constant.
So, this function basically associate each real number to a constant value.
It is a linear function where f(x1) = f(x2) for all x1,x2 ∈ R
For f : R→R, y = f(x) = c for each x ∈ R
Domain = R
Range = {c}
The value of c can be any real number.

Number of subsets of a set of order three is
  • a)
    2
  • b)
    4
  • c)
    6
  • d)
    8
Correct answer is option 'D'. Can you explain this answer?

Raj Ghoshal answered
Number of Subsets of a Set of Order Three

To find the number of subsets of a set of order three, we need to consider all possible combinations of elements in the set.

Let's say we have a set S = {a, b, c}. We can represent each element of the set as a binary digit, where 0 represents the absence of the element and 1 represents the presence of the element.

- Empty Set: The empty set is a subset of every set, so it is included.
- Single Element Subsets: There are three single element subsets - {a}, {b}, and {c}.
- Two Element Subsets: There are three possible two-element subsets - {a, b}, {a, c}, and {b, c}.
- Three Element Subset: There is only one three-element subset - {a, b, c}.

Therefore, the total number of subsets of the set S = {a, b, c} is 1 (empty set) + 3 (single element subsets) + 3 (two-element subsets) + 1 (three-element subset) = 8.

Hence, the correct answer is option 'D' - 8.

In summary, to find the number of subsets of a set of order three, we consider all possible combinations of elements, including the empty set and subsets of different sizes. By applying this approach to the given set S = {a, b, c}, we find that there are 8 subsets in total.

The domain of the function f(x)=   is all real x such that
  • a)
    x<2
  • b)
    -2
  • c)
    -2 ≤ x ≤ 2
  • d)
    x>-2
Correct answer is option 'C'. Can you explain this answer?

Geetika Shah answered
4 - x2 = 0
=> (2+x)(2-x) = 0
x = +-2
Now select a test point, let it be x = 0. Then y = (4 - (0)2)1/2 = 2, so the function is defined on [-2,2]
Thus, the graph of y = (4 - x2)1/2 is a semicircle with radius 2 and domain is [-2,2]

If f (x)  equals
  • a)
    1/x
  • b)
    0
  • c)
    1
  • d)
    x
Correct answer is option 'A'. Can you explain this answer?

Pooja Shah answered
f(x) = (x−1)/(x+1)
1/f(x) = (x+1)/(x-1)
f(1/f(x)) = [(x+1)/(x-1) - 1]/[(x+1)/(x-1) + 1]
⇒ (x+1-x+1)/(x+1+x-1)
= 2/2x 
= 1/x

f(x) = x is called
  • a)
    an identity function
  • b)
    a constant function
  • c)
    inverse function
  • d)
    step function
Correct answer is option 'A'. Can you explain this answer?

Mira Sharma answered
In mathematics, an identity function, also called an identity relation or identity map or identity transformation, is a function that always returns the same value that was used as its argument. In equations, the function is given by f(x) = x.

Which of the following is not true?
  • a)
    Every relation from the set A to the set B is a function from the set A to the set B.
  • b)
    If f is a function from A to B and (a, b) ∈ f, then f(a) = b, where a is called the pre-image of b under f.
  • c)
    Every function from the set A to the set B is a relation from the set A to the set B
  • d)
    If f is a function from A to B and (a, b) ∈ f, then f(a) = b, where b is called the image of a under f.
Correct answer is option 'A'. Can you explain this answer?

Pooja Shah answered
A relation f from a set A to a set B is said to be function if every element of set A has one and only one image in set B. In other words, a function f is a relation such that no two pairs in the relation has the same first element. The notation f : X →Y means that f is a function from X to Y.

  • a)
    1
  • b)
    2
  • c)
    -1
  • d)
    -2
Correct answer is option 'C'. Can you explain this answer?

Mohit Rajpoot answered
 lim(x → π) sinx/x-π
lim(h → 0) sin(π-h)/π-h-π
lim(h → 0) -sinh/h
⇒ -1

 If f(x) = x2 – x + 1; g(x) = 7x – 3, be two real functions then (f + g)(3) is
  • a)
    25
  • b)
    3
  • c)
    7
  • d)
    18
Correct answer is option 'A'. Can you explain this answer?

Riya Banerjee answered
 f(x) = x2 – x + 1; g(x) = 7x – 3
(f+g)(x) = (x2 - x + 1 + 7x - 3)
=(x2 - x + 7x + 1 - 3)
= x2 + 6x - 2
(f+g)(3) = x2 + 6x - 2
= (3)2 + 6(3) - 2
= 9 + 18 - 2
= 25

The interval in which the values of f (x) lie where f (x) = 3 sin 
  • a)
    [0, ]
  • b)
    [-3, 3]
  • c)
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Om Desai answered
f(x) = 3.sin(√π2/16 − x2)
since the quantity within the square root can not be negative.
therefore
π2/16 − x2 ≥ 0
x2 ≤ π2/16
−π/4 ≤ x ≤ π/4
the minimum value of the function is 3.sin0=0
and the maximum value of the function is 3sinπ/4 = 3/√2
therefore range is [0, 3/√2]

A condition for a function y = f (x) to have an inverse is that it should be
  • a)
    an even function
  • b)
    defined for all x
  • c)
    strictly monotone and continuous in the domain
  • d)
    continuous every where
Correct answer is option 'C'. Can you explain this answer?

Krishna Iyer answered
For a function to have its inverse in a given domain, it should be continuous in that domain and should be a one-one function in that domain.
If the function is one-one in the domain, then it has to be strictly monotonic.
For example y=sin(x) has its domain in xϵ[−π/2,π/2] since it is strictly monotonic and continuous in that domain.

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
Correct answer is option 'D'. Can you explain this answer?

Neha Joshi answered
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 ϵ I

Find the domain of the function f(x) = 
  • a)
    R – {-4}
  • b)
    R
  • c)
    R – {4}
  • d)
    R+
Correct answer is option 'C'. Can you explain this answer?

Lavanya Menon answered
 f(x) = 3x/2x-8
denominator can't be zero so
2x - 8 not equal to 0
x should not equal to 4
we have domain R - {4}

On the set Z of all integers define f ; Z → Z as follows : f (x) = {x/2 if x is even, and f (x) = 0 if x is odd , then f is
  • a)
    onto but not one-one
  • b)
    into
  • c)
    one-one but not onto
  • d)
    one-one and onto
Correct answer is option 'A'. Can you explain this answer?

Sneha Sarkar answered
To define a function f on the set Z of all integers, we need to specify how the function maps each integer to another value.

For example, we can define a function f as follows:
- f(n) = n + 1, where n is any integer in Z.

This function takes an integer n and returns the next integer in the sequence. For example, f(1) = 2, f(0) = 1, f(-1) = 0, and so on.

Another example of a function on Z could be:
- f(n) = n^2, where n is any integer in Z.

This function takes an integer n and returns its square. For example, f(2) = 4, f(-3) = 9, and so on.

There are infinite possible functions that can be defined on the set Z, depending on how we want to map the integers to other values. The specific definition of the function would depend on the desired outcome or purpose of the function.

The domain of the function  is
  • a)
    {2 n π: n ∈ I}
  • b)
    { }
  • c)
    R
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Geetika Shah answered
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 = 2nπ ∀ n∈Z

If f(x) =  is equal to
  • a)
    f (x)
  • b)
    2 f (x)
  • c)
    4 f(x)
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Om Desai answered
f(x) = log[(1+x)/(1-x)]    ……….(1)
Substitute 2x/1+x2 in place of x in equation(1)
f(2x1+x2)=log⁡(1+2x1+x2)(1−2x1+x2)
⟹f(2x/(1+x2))=log[1+(2x/(1+x2))/(1-(2x/1+x2))]
⟹f(2x/(1+x2))=log⁡((1+x2+2x)/(1+x2−2x))
⟹f(2x/(1+x2))=log((1+x)2 / (1−x))2
⟹f(2x/ (1+x2))=log⁡(1+x / 1−x)2
We know that log⁡mn=nlog⁡m
⟹f(2x/(1+x2))=2log⁡(1+x / 1−x)
From equation (1):
f(2x/(1+x2))=2f(x)

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