Class 12 Relation and Functions:Miscellaneous Exercise with Solutions
1.Class 12 Relation and Functions:Miscellaneous Exercise with Solutions
In this article “Class 12 Relation and Functions”, we will solve specific questions of relation and functions.Also read historical brief note of relation and functions.
“Before diving into these solutions, make sure you understand the foundational concept of “Relations Class 12 Examples with Solution” and have reviewed the essential “Types of Functions in Class 12“. This will make following today’s step-by-step solutions much easier.”
2.Class 12 Relation and Functions
Illustration : 1. Show that the function f : R \to \{x \in R : -1 < x < 1\} defined by f(x) = \frac{x}{1 + |x|}, \ x \in R is one-one and onto function.
Solution : Here f : R \to \{x \in R : -1 < x < 1\} \\
\text{and } f(x) = \frac{x}{1 + |x|} \\
\text{Step 1 : Let, } x \ge 0 \Rightarrow |x| = x \\
\therefore f(x) = \frac{x}{1 + x} \\
x_1, x_2 \in R, \text{ then} \\
f(x_1) = f(x_2) \\
\Rightarrow \frac{x_1}{1 + x_1} = \frac{x_2}{1 + x_2} \\
\Rightarrow x_1(1 + x_2) = x_2(1 + x_1) \\
\Rightarrow x_1 + x_1 x_2 = x_2 + x_1 x_2 \\
\Rightarrow x_1 = x_2 \\
\text{When } x < 0 \Rightarrow |x| = -x \text{ then} \\
f(x) = \frac{x}{1 - x} \\
f(x_1) = f(x_2) \\
\Rightarrow \frac{x_1}{1 - x_1} = \frac{x_2}{1 - x_2} \\
\Rightarrow x_1(1 - x_2) = x_2(1 - x_1) \\
\Rightarrow x_1 - x_1 x_2 = x_2 - x_1 x_2 \\
\Rightarrow x_1 = x_2 \\
\therefore f \text{ is one-one.}\\
(b) \ \text{When } x \ge 0, \text{ Let } y = f(x) \\
\Rightarrow \frac{x}{1 + x} = y \\
\Rightarrow x = (1 + x)y \\
\Rightarrow x = y + xy \\
\Rightarrow x = \frac{y}{1 - y} \\
\text{When } x < 0, \ y = \frac{x}{1 - x} \\
\Rightarrow y(1 - x) = x \\
\therefore x(y + 1) = y \\
\Rightarrow x = \frac{y}{1 + y} \\
\text{In both conditions, each element} \text{of codomain is image of element} \text{of domain.} \\
\therefore f \text{ is onto.} \\
\therefore f \text{ is one-one onto function.} \\
\text{Illustration : 2. Show that the} \text{function } f : R \to R \text{ given by} f(x) = x^3 \text{ is injective.} \\
\text{Solution : Here } f : R \to R \\
f(x) = x^3 \\
x_1, x_2 \in R \text{ (domain), then} \\
f(x_1) = f(x_2) \\
\Rightarrow x_1^3 = x_2^3 \\
\Rightarrow x_1 = x_2 \\
\therefore f \text{ is injective.} \\
\text{Illustration : 3. Given a non} \text{empty set } X, \text{ consider } P(X) \\ \text{which is the set of all subsets} \text{of } X \text{. Define the relation } R \\ \text{ in} P(X) \text{ as follows :} \text{For subsets } A, B \text{ in } P(X), A R B \text{ if and only if } A \subset B \text{. Is} R \text{ an equivalence relation on} P(X) \text{? Justify your answer.} \\
\text{Solution :} (i) \text{ Here } A \subset A \Rightarrow R \text{ is} \\
\text{reflexive.} \\
(ii) \ A \subset B \not\Rightarrow B \subset A \Rightarrow R \text{ is not} \\
\text{symmetric.} \\
(iii) \ A \subset B, B \subset C \Rightarrow A \subset C \Rightarrow R \text{ is} \\
\text{transitive.} \\
\text{Hence } R \text{ is not an equivalence} \\
\text{relation.} \\
\text{Illustration : 4. Find the number} \text{of all onto functions from} \text{the set } \\ \{1, 2, 3, \dots, n\} \text{ to itself.} \\
\text{Solution :} \text{Let } A = \{1, 2, 3, \dots, n\} \\
\text{and } B = \{1, 2, 3, \dots, n\} \\
\text{Each element of } B \text{ is image} \\
\text{of the other element in} \text{ domain} A. \\
\text{Relation in elements of } A \text{ and } B \\
\text{like this} \\
n(n-1)(n-2) \dots 3 \cdot 2 \cdot 1 = n! \\
\text{can be.} \\
\therefore \text{ Number of binary operation will } n! .
By solving the above examples, one can understand the Class 12 Relation and Functions for Students.
\text{Illustration : 5. Let } A = \{-1, 0, 1, 2\}, B = \{-4, -2, 0, 2\} \text{ and } f, g : A \to B \\ \text{be functions defined by} f(x) = x^2 - x, \ x \in A \text{ and}\\ g(x) = 2\left|x - \frac{1}{2}\right| - 1, \ x \in A \text{ . Are} \\ f \text{ and } g \text{ equal? Justify your answer.} \\ \text{Solution : Here } A = \{-1, 0, 1, 2\} \\ B = \{-4, -2, 0, 2\} \\ f, g : A \to B \\ \text{and } f(x) = x^2 - x, \ \forall x \in A \\ g(x) = 2\left|x - \frac{1}{2}\right| - 1, \ x \in A \\ \text{When } x = -1, \text{ then} \\ f(-1) = (-1)^2 - (-1) = 1 + 1 = 2 \\ g(-1) = 2\left|-1 - \frac{1}{2}\right| - 1 = 2\left|-\frac{3}{2}\right| - 1 \\ = 2 \times \frac{3}{2} - 1 = 3 - 1 = 2 \\ \text{When } x = 0, \ f(0) = 0 - 0 = 0 \\ g(0) = 2\left|0 - \frac{1}{2}\right| - 1 = 2 \times \frac{1}{2} - 1 = 0 \\ \text{When } x = 1, \ f(1) = 1^2 - 1 = 0 \\ g(1) = 2\left|1 - \frac{1}{2}\right| - 1 = 0 \\ \text{When } x = 2, \ f(2) = 2^2 - 2 = 2 \\ g(2) = 2\left|2 - \frac{1}{2}\right| - 1 = 3 - 1 = 2 \\ \therefore \forall a \in A \\ f(a) = g(a) \\ \therefore \text{function } f \text{ and } g \text{ are equal.} \\ \text{Illustration : 6. Let } A = \{1, 2, 3\} \text{.} \text{Then number of relations} \\ \text{containing } (1, 2) \text{ and } (1, 3) \text{which are reflexive and} \\ \text{symmetric but not transitive} \text{is} \\ (A) 1 \quad (B) 2 \quad (C) 3 \quad (D) 4 \\ \text{Solution : Here } A = \{1, 2, 3\} \\ \text{That relation which has} \\ (1, 2) \text{ and } (1, 3) \text{ and relation} \\ \text{is reflexive and symmetric} \\ \text{but not transitive.} \\ \{(1, 2), (1, 3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)\}\\ \dots \text{such only one relation.} \\ \therefore \text{Alternate } (A) \text{ is true.} \\ \text{Illustration : 7. Let } A = \{1, 2, 3\} \text{.} \text{Then number of equivalence} \\ \text{relations containing } (1, 2) \text{ is} \\ (A) 1 \quad (B) 2 \quad (C) 3 \quad (D) 4Solution: The given set is
A = {1, 2, 3} .
The smallest equivalence relation containing (1, 2) is given by;
R_1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Now, we are left with only four pairs
i.e., (2, 3), (3, 2), (1, 3), and (3, 1).
If we add anyone pair [say (2, 3)] to R_1 , then for symmetry, we must add (3, 2).
Also, for transitivity we are required to add (1, 3) and (3, 1).
Hence,
the only equivalence relation (bigger than R_1) is the universal relation.
This shows that the total number of equivalence relations containing (1, 2) is two.
The correct answer is B
By solving the above examples, one can understand the Class 12 Relation and Functions for Students.
3.Practice Problems Class 12 Relation and Functions for Students
(1.)What is range of function f(x)=2+x-[x-3],if [x] represents greatest integer function not greater than x
(2.)What is the range of the function g(\theta) = \frac{1}{3 - \cos 4\theta}
Answers:(1.)[5,6] (2.) \left[\frac{1}{4}, \frac{1}{2}\right]
By solving the above questions,one can understand the Class 12 Relation and Functions well because the concept is well understood when you solve the questions practically.
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4.Key Points of Class 12 Relation and Functions
(1.)R. Descartes (1637):Used the term ‘function’ in “Geometrie” to signify positive integral powers (x^n) of a variable x while studying curves (hyperbola,parabola,ellipse).
(2.)James Gregory (1667):Defined a function in “Vera Circuli et Hyperbolae Quadratura” as a quantity obtained from other quantities through algebraic or other mathematical operations.
(3.)G. W. Leibnitz (1673 & 1714):Initially used ‘function’ for quantities varying along a curve (coordinates,slopes,tangents,normals).
(4.)Later used it for quantities depending on a variable and was the first to coin the phrase ‘function of x’.
(5.)John Bernoulli (1718):First introduced a formal notation for a function of x using \phi(x).
(6.)Leonhard Euler (1734):Standardized the widely used symbols f,F, \phi ,\psi for functions in “Analysis Infinitorium”.
(6.)Joseph Louis Lagrange (1793):Introduced analytic functions and formal notations like f(x),F(x) in “Théorie des fonctions analytiques”.
(7.)Lejeune Dirichlet & Georg Cantor:Dirichlet gave a precise definition of function,which was later generalized and formalized using Set Theory by Georg Cantor into the modern set-theoretic definition used today.
Also Read This Article:- Function (mathematics)
5.Frequently Asked Questions Related to Class 12 Relation and Functions
Q:1.Define the one-one and onto function
Ans:A function is said to be one-one and onto (or bijective),if f is both one-one and onto.
Q:2.Who was the first to given the concept of function?
Ans:R.Descartes used the word function.
Q:3.What is the property of a finite set?
Ans:Given a finite set X,a function is one-one (respectively onto) if and only if onto respectively.This is the characteristic property of a finite set.
By answering the above questions,you can know about the primary terms of Class 12 Relation and Functions.
6.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):
🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
*”यदि x = 7 - 4\sqrt{3} है,तो \left(x + \frac{1}{x}\right) का मान बताइए।
(If x = 7 - 4\sqrt{3} ,then state the value of \left(x + \frac{1}{x}\right) .)”*
*पिछली प्रश्नोत्तरी का उत्तर*
शेष पारी= 5 - 1\frac{1}{4} \\
= \frac{5}{1} - \frac{5}{4} = \frac{20 - 5}{4} \\
= \frac{15}{4}
*Previous Quiz Solution*
Remaining innings= 5 - 1\frac{1}{4} \\
= \frac{5}{1} - \frac{5}{4} = \frac{20 - 5}{4} \\
= \frac{15}{4}
This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*
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