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Sets Relations and Functions | Formulas – Exercises, Questions with Explanation

Sets and Functions

In competitive examinations, generally two-three questions are asked from Sets Relations and Functions Topic. On this page, we have provided Sets and Functions Formulas, Exercises that contains questions with explanation, so that candidates can easily understand the concept that is used for solving the questions and answers.

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Brief Definition of Set, Relation and Function:

Set: A collection of well-defined objects which are distinct from each other.

Relation: If M and N are two non-empty sets, then a relation R from M to N is a subset of M x N. If R ⊆ M X N and (m,n) ∈ R, it indicates that “m” is related to “n” by the relation R, and written as mRn.

Function: Function is a special class of relation. A function f takes an input x, and returns a single output, say f(x).

Sets and Functions

Sets Relations and Functions Formulas:

Sets in Maths:

A set is a collection of well-defined objects. The objects of a set are taken as distinct only on the ground of simplicity.

A set of sets is frequently called a family or collection of sets. For example, suppose we have a family of sets consisting A1, A2, A3,….. up to An, that is the family {A1, A2, A3,….., An } and could be denoted as

S = {Ai | i belongs to N and i ≤ i ≤ n}

Notation: A set is denoted by a capital letter and represented by listing all its elements between curly brackets such as { }.

Relations in Maths:

Relation is helpful to find the relationship between input and output of a function.

A relation R, from a non-empty set P to another non-empty set Q, is a subset of P X Q.

For example, Let P = {a, b, c} and Q = {3, 4} and

Let R = {(a, 3), (a, 4), (b, 3), (b, 4), (c, 3), (c, 4)}

Here R is a subset of A x B. Therefore R is a relation from P to Q.

Functions in Maths:

A function is simply used to represent the dependence of one quantity on the other andR easily defined with the help of the concept of mapping. In simple words, a function is a relation which derives one output for each input.

A function from set P to set Q is a rule that assigns to each element of set P, one and only one element of set Q.

Mathematically: If f: A -> B where y = f(x), x ∈ P and y ∈ Q. Here y is the image of x under f.

Get Here: Unitary Method Questions

Sets Relations and Functions Questions with Explanation

Question 1: In a college of 300 students, every student reads 5 newspapers and every newspaper is read by 60 students. The number of newspapers is ________.

Solution:

Let the number of newspapers be x.

If every student reads one newspaper, the number of students would be x (60) = 60x

Since every student reads 5 newspapers, the number of students = [x * 60] / [5] = 300

x = 25

Question 2: Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)} then R−1 o R is ________.

Solution:

First find R−1,

R−1 = {(5, 4) ; (4, 1) ; (6, 4) ; (6, 7) ; (7, 3)}.

Obtain the elements of R−1 o R.

Pick the element of R and then of R−1.

Since (4, 5) ∈ R and (5, 4) ∈ R−1, we have (4, 4) ∈ R−1 o R

Similarly, (1, 4) ∈ R, (4, 1) ∈ R−1 ⇒ (1, 1) ∈ R−1 o R

(4, 6) ∈ R, (6, 4) ∈ R−1 ⇒ (4, 4) ∈ R−1 o R,

(4, 6) ∈ R, (6, 7) ∈ R−1 ⇒ (4, 7) ∈ R−1 o R

(7, 6) ∈ R, (6, 4) ∈ R−1 ⇒ (7, 4) ∈ R−1 o R,

(7, 6) ∈ R, (6, 7) ∈ R−1 ⇒ (7, 7) ∈ R−1 o R

(3, 7) ∈ R, (7, 3) ∈ R−1 ⇒ (3, 3) ∈ R−1 o R,

Hence, R−1 o R = {(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)}.

Question 3: If f (x) = cos (log x), then find the value of f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)].

Solution:

f (x) = cos (log x)

Now let y = f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)]

y = cos (log x) * cos (log 4) − [1 / 2] * [cos log (x / 4) + cos (log 4x)]

y = cos (log x) cos (log 4) − [1 / 2] * [cos (log x −log 4) + cos (log x + log 4)]

y = cos (log x) cos (log 4) − [1 / 2] * [2 cos (log x) cos (log 4)]

y = 0

Question 4: If f (x) = cos [π2] x + cos[−π2] x, then find the function of the angle.

Solution:

f (x) = cos [π2] x + cos[−π2] x

f (x) = cos (9x) + cos (−10x)

= cos (9x) + cos (10x)

= 2 cos (19x / 2) cos (x / 2)

f (π / 2) = 2 cos (19π / 4) cos (π / 4);

f (π / 2) = 2 * −1 / √2 * 1/ √2

= −1

Check Out: Quadratic Equation | Formulas, Tricks

Question 5: The function f : R → R defined by f (x) = ex is ________.

Solution:

Function f: R → R is defined by f (x) = ex.

Let x1, x2 ∈ R and f (x1) = f (x2) or ex1 = ex2 or x= x2.

Therefore, f is one-one.

Let f (x) = ex = y.

Taking log on both sides, we get x = logy.

We know that negative real numbers have no pre-image or the function is not onto and zero is not the image of any real number.

Therefore, function f is into.

Question 6: The function f: R → R is defined by f (x) = cosx + sin4x for x ∈ R, then what is f (R)?

Solution:

f (x) = cosx + sin4x

y = f (x) = cosx + sin2x (1 − cos2x)

y = cosx + sin2x − sin2x cos2x

y = 1 − sin2x cos2x

y = 1 − [1 / 4] * [sin22x]

3 / 4 ≤ f (x) ≤ 1, (Because 0 ≤ sin22x ≤ 1)

f (R) ∈ [3/4, 1]

Question 7: If A = [(x, y) : x2 + y2 = 25] and B = [(x, y) : x2 + 9y2 = 144], then A ∩ B contains _______ points.

Solution:

A = Set of all values (x, y) : x2 + y2 = 25 = 52

B = [x2 / 144] + [y2 / 16] = 1

i.e., [x/ (12)2] + [y2 / (4)2] = 1.

Clearly, A ∩ B consists of four points.

Question 8: Let R be the relation on the set R of all real numbers defined by a R b if and only if |a − b| ≤ 1. Then R is __________.

Solution:

|a − a| = 0 < 1

Therefore, a R a ∀ a ∈ R

Therefore, R is reflexive.

Again a R b, |a − b| ≤ 1 ⇒ |b − a| ≤ 1 ⇒ b R a

Therefore, R is symmetric.

Again 1 R [½] and [½] R1 but [½] ≠ 1

Therefore, R is not anti-symmetric.

Further, 1 R 2 and 2 R 3 but [1 / R3], [Because, |1 − 3| = 2 > 1]

Hence, R is not transitive.

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Question 9: Let f : R → R be defined by f (x) = 2x + |x|, then f (2x) + f (−x) − f (x) = _______.

Solution:

f (2x) = 2 (2x) + |2x| = 4x + 2 |x|,

y = x2 + 1,

f (x) = 2x + |x|

f (2x) + f (−x) − f (x) = 4x + 2 |x| + |x| −2x − 2x − |x|

= 2|x|

Question 10: If f (x) = a cos (bx + c) + d, then what is the range of f (x)?

Solution:

f (x) = a cos (bx + c) + d ..(i)

For minimum cos (bx + c) = −1

From (i), f (x) = −a + d = (d − a)

For maximum cos (bx + c) = 1

From (i), f (x) = a + d = (d + a)

Range of f (x) = [d − a, d + a]

Read Out: Square Root and Cube Root – Concepts

All the details mentioned above about Sets and Functions Exercises, Sets Relations and Functions Questions and Answers, Sets and Functions Formulas etc is well written by the team members of recruitmentresult.com Individuals must go through it for better preparation of the examination.

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