Ch 1  ·  Q–
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Chapter 1 Miscellaneous Exercise Solutions

Relations and Functions

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 1 Miscellaneous Exercise

Class 12 Mathematics Miscellaneous Exercise NCERT Solutions Relations and Functions Class 12 Mathematics Chapter 1 CBSE Board Exam JEE Main CUET Functions Relations Equivalence Relations Injective Functions Surjective Functions Bijective Functions Inverse Functions
7 Questions
15–25 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks
Show that the function \(f:\mathbb{R}\rightarrow \{x\in\mathbb{R}:-1 < x < 1\}\) defined by \(f(x)=\frac{x}{1+|x|},\quad x\in\mathbb{R}\) is one-one and onto.
📘 Concept & Theory
Concept/Theory
›

To prove that a function is one-one and onto (bijective), two properties must be established.

  • One-One (Injective): Distinct elements of the domain must have distinct images. Equivalently,
    \[f(x_1)=f(x_2)\Rightarrow x_1=x_2\]
  • Onto (Surjective): Every element of the codomain must be the image of at least one element of the domain. That is, for every
    \[y\in(-1,1),\]
    there exists
    \[x\in\mathbb R\]
    such that
    \[f(x)=y.\]

Since the function contains the modulus sign, the analysis must be carried out separately for

  • \(x\ge 0\)
  • \(x<0\)

This is a standard technique used in CBSE Board examinations and competitive entrance examinations.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Express the function separately for positive and negative values of \(x\).

  2. Prove that equal function values imply equal inputs.

  3. Take an arbitrary element of the codomain.

  4. Construct its pre-image depending upon whether it is positive or negative.

  5. Verify that the obtained pre-image belongs to the domain.

  6. Hence conclude that the function is bijective.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  47 steps
  1. The function is
    \[f(x)=\frac{x}{1+|x|}\]
  2. Since
    \[|x|=\begin{cases}x,&x\ge0,\\ -x,&x<0,\end{cases}\]
  3. therefore,
    \[ f(x)= \begin{cases} \dfrac{x}{1+x},&x\ge0,\\[2ex] \dfrac{x}{1-x},&x<0. \end{cases} \]
  4. Proving that \(f\) is One-One
  5. Suppose
    \[f(x_1)=f(x_2)\]
  6. We shall consider different cases.
    Case I : Both \(x_1\ge0\) and \(x_2\ge0\)
  7. Then
    \[\frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}.\]
  8. Cross multiplying,
    \[x_1(1+x_2)=x_2(1+x_1)\]
  9. Expanding both sides,
    \[x_1+x_1x_2=x_2+x_1x_2\]
  10. Subtracting \(x_1x_2\) from both sides,
    \[x_1=x_2\]
  11. Case II : Both \(x_1<0\) and \(x_2<0\)
  12. Then
    \[\frac{x_1}{1-x_1}=\frac{x_2}{1-x_2}.\]
  13. Cross multiplying,
    \[x_1(1-x_2)=x_2(1-x_1)\]
  14. Expanding,
    \[x_1-x_1x_2=x_2-x_1x_2\]
  15. Subtracting \(-x_1x_2\) from both sides,
    \[x_1=x_2\]
  16. Case III : One number is non-negative and the other is negative
  17. Suppose
    \[x_1\ge0,\quad x_2 < 0\]
  18. Then
    \[f(x_1)=\frac{x_1}{1+x_1}\ge0\]
  19. because both numerator and denominator are non-negative.
    \[f(x_2)=\frac{x_2}{1-x_2} < 0\]
  20. because the numerator is negative while the denominator is positive.
  21. Hence
    \[f(x_1)\neq f(x_2)\]
    which contradicts the assumption that
    \[f(x_1)=f(x_2)\]
  22. Similarly, the case
    \[x_1 < 0,\quad x_2\ge0\]
    is impossible
  23. Therefore,
    \[f(x_1)=f(x_2)\Longrightarrow x_1=x_2\]
  24. Hence, the function is one-one.
  25. Proving that \(f\) is Onto
  26. Let
    \[y\in(-1,1)\]
  27. We must find \(x\in\mathbb R\) such that f(x)=y
  28. Case I : \(0\le y<1\)
  29. Choose
    \[x=\frac{y}{1-y}\]
  30. Since \(0\le y<1\) we have
    \[1-y>0\]
  31. therefore,
    \[x\ge0\]
  32. Now
    \[f(x)=\frac{x}{1+x}\]
  33. Substituting the value of \(x\),
    \[f\!\left(\frac{y}{1-y}\right)=\frac{\dfrac{y}{1-y}}{1+\dfrac{y}{1-y}}.\]
  34. Simplify the denominator.
    \[ \begin{aligned} 1+\frac{y}{1-y}&=\frac{1-y+y}{1-y}\\&=\frac1{1-y}. \end{aligned} \]
  35. Hence
    \[ \begin{aligned} f\!\left(\frac{y}{1-y}\right)&=\frac{\dfrac{y}{1-y}}{\dfrac1{1-y}}\\ &=y\end{aligned} \]
  36. Thus every
    \[y\in[0,1)\]
    has a pre-image.
  37. Case II : \(-1
  38. Choose
    \[x=\frac{y}{1+y}\]
  39. Since \(-1 < y < 0\) we have
    \[1+y>0\]
    and therefore
    \[x<0\]
  40. Hence
    \[f(x)=\frac{x}{1-x}\]
  41. Substitute the value of \(x\)
    \[f\!\left(\frac{y}{1+y}\right)=\frac{\dfrac{y}{1+y}}{1-\dfrac{y}{1+y}}\]
  42. Simplify the denominator.
    \[\begin{aligned}1-\frac{y}{1+y}&=\frac{1+y-y}{1+y}\\&=\frac1{1+y}\end{aligned}\]
  43. Therefore
    \[\begin{aligned}f\!\left(\frac{y}{1+y}\right)&=\frac{\dfrac{y}{1+y}}{\dfrac1{1+y}}\\&=y\end{aligned}\]
  44. Thus every \(y\in(-1,0)\) also has a pre-image.
  45. Case III : \(y=0\)
  46. Take \(x=0\) then
    \[f(0)=0\]
  47. Hence every element of the codomain has a pre-image.
  48. Therefore, \(f\) is onto.
  49. Conclusion
  50. Since the function is both one-one and onto, it is a bijection.
    \[\boxed{\text{Hence }f:\mathbb R\rightarrow(-1,1)\text{ is one-one and onto.}}\]
🎯 Exam Significance
Exam Significance
›
  • This problem demonstrates the complete procedure for proving a function is bijective.
  • It develops the skill of handling modulus functions by splitting them into suitable cases.
  • Questions involving inverse functions often require first proving that the given function is bijective.
  • The concepts of injective, surjective and bijective mappings are frequently tested in CBSE Board examinations, JEE Main, NDA, CUET and university entrance examinations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Whenever a function contains \(|x|\), split the analysis according to the sign of \(x\).

  2. For injectivity, prove that equal outputs imply equal inputs.

  3. For surjectivity, start with an arbitrary element of the codomain and construct its pre-image.

  4. Always verify that the obtained pre-image belongs to the domain.

  5. A function that is both one-one and onto is called a bijection and possesses an inverse function.
↑ Top
1 / 7  ·  14%
Q2 →
Q2
NUMERIC3 marks
Show that the function \(f:\mathbb{R}\rightarrow\mathbb{R}\) defined by \(f(x)=x^3\) is injective.
📘 Concept & Theory
Concept/Theory
›

A function is said to be injective (one-one) if different elements of the domain have different images.

Mathematically,

\[ f(x_1)=f(x_2)\Longrightarrow x_1=x_2. \]

To prove that a function is injective, we usually begin by assuming that two function values are equal and then show that the corresponding inputs must also be equal.

Another useful fact is that the cubic function

\[ y=x^3 \]

is strictly increasing on the entire set of real numbers. Hence, it cannot assign the same value to two different real numbers.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Assume that two images under the function are equal.

  2. Write the resulting algebraic equation.

  3. Factor the difference of cubes.

  4. Analyze all possible factors.

  5. Show that the only possible conclusion is \(x_1=x_2\).

  6. Conclude that the function is injective.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Let \(x_1,x_2\in\mathbb{R}\) and suppose
    \[f(x_1)=f(x_2)\]
  2. Since
    \[f(x)=x^3\]
  3. therefore,
    \[x_1^3=x_2^3\]
  4. Subtracting \(x_2^3\) from both sides, we obtain
    \[x_1^3-x_2^3=0\]
  5. Using the identity
    \[a^3-b^3=(a-b)(a^2+ab+b^2)\]
  6. we get
    \[(x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0\]
  7. Hence, either
    \[x_1-x_2=0\]
    or
    \[x_1^2+x_1x_2+x_2^2=0\]
  8. Now consider the second factor.
  9. Observe that
    \[x_1^2+x_1x_2+x_2^2=\frac{1}{2}(x_1-x_2)^2+\frac{3}{2}x_1^2+\frac{3}{2}x_2^2-\frac{1}{2}(x_1^2+x_2^2)\]
  10. or more conveniently,
    \[x_1^2+x_1x_2+x_2^2=\left(x_1+\frac{x_2}{2}\right)^2+\frac{3x_2^2}{4}\]
  11. Since each square is non-negative,
    \[\left(x_1+\frac{x_2}{2}\right)^2\ge0\]
    and
    \[\frac{3x_2^2}{4}\ge0\]
  12. Therefore,
    \[x_1^2+x_1x_2+x_2^2\ge0\]
  13. This expression can be equal to zero only when
    \[x_1=0\quad\text{and}\quad x_2=0\]
  14. which again gives
    \[x_1=x_2\]
  15. Hence, in every possible case,
    \[x_1=x_2\]
  16. Therefore,
    \[f(x_1)=f(x_2)\Longrightarrow x_1=x_2\]
  17. Hence, the function is injective (one-one).
    \[\boxed{\therefore\; f(x)=x^3\text{ is an injective function.}}\]
🎯 Exam Significance
Exam Significance
›
  • This question illustrates the standard method of proving that a function is one-one using algebraic identities.
  • It reinforces the factorization formula for the difference of cubes.
  • Such proofs are frequently asked in CBSE Board examinations.
  • The concept of injective functions forms the basis for studying inverse functions, which are important in JEE Main, CUET, NDA and university entrance examinations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. To prove injectivity, always begin with the assumption \(f(x_1)=f(x_2)\).

  2. Apply suitable algebraic identities wherever possible.

  3. The identity

    \[a^3-b^3=(a-b)(a^2+ab+b^2)\]
    is frequently used in such proofs.

  4. The cubic function is strictly increasing over the entire real line, making it one-one.

  5. An injective function assigns distinct outputs to distinct inputs.
← Q1
2 / 7  ·  29%
Q3 →
Q3
NUMERIC3 marks
Given a non-empty set \(x\) consider \(P(X)\) the set of all subsets of \(X\). Define a relation \(R\) on \(P(x)\) by \(A\,R\,B \iff A\subset B\) where \(A,B\in P(X)\). Is \(R\) an equivalence relation on \(P(X)\)? Justify your answer.
📘 Concept & Theory
Concept/Theory
›

A relation on a set is called an equivalence relation if it satisfies all the following three properties.

  • Reflexive: Every element must be related to itself.
    \[a\;R\;a,\qquad \forall a.\]
  • Symmetric: If one element is related to another, then the second must also be related to the first.
    \[aRb\Longrightarrow b\;R\;a.\]
  • Transitive: If one element is related to a second and the second is related to a third, then the first must also be related to the third.
    \[a\;R\;b\ \text{and}\ b\;R\;c\Longrightarrow a\;R\;c.\]

To determine whether a relation is an equivalence relation, each of these three properties must be verified separately.

Here, the relation is defined using the subset relation

\[A\subset B.\]

Recall that every set is a subset of itself, and if one set is a subset of another, the reverse need not be true.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Check whether the relation is reflexive.

  2. Check whether it is symmetric.

  3. Check whether it is transitive.

  4. Conclude whether all three conditions of an equivalence relation are satisfied.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  24 steps
  1. The relation is defined by
    \[A\,R\,B\iff A\subset B\]
  2. Checking Reflexivity
  3. For every subset
    \[A\in P(X)\]
  4. we know that every set is a subset of itself.
    \[A\subset A\]
  5. Therefore,
    \[A\,R\,A\]
  6. Hence, the relation is reflexive.
  7. Checking Symmetry
  8. Suppose
    \[A\,R\,B\]
  9. Then
    \[A\subset B\]
  10. To satisfy symmetry, it must also be true that
    \[B\subset A\]
  11. This is not true in general.
  12. For example, let
    \[X=\{1,2\}\]
  13. and choose
    \[A=\{1\},\quad B=\{1,2\}\]
  14. Clearly,
    \[A\subset B,\]
  15. so
    \[A\,R\,B\]
  16. However,
    \[B\nsubseteq A\]
  17. Therefore,
    \[B\,R\,A\]
    does not hold.
  18. Hence, the relation is not symmetric.
  19. Checking Transitivity
  20. Suppose
    \[A\,R\,B\quad\text{and}\quad B\,R\,C\]
  21. Then
    \[A\subset B\]
    and
    \[B\subset C\]
  22. By the transitive property of subsets,
    \[A\subset C\]
  23. Therefore,
    \[A\,R\,C\]
  24. \[Hence, the relation is transitive.\]
  25. Conclusion
  26. The relation satisfies
    • Reflexive ✓
    • Symmetric ✗
    • Transitive ✓
  28. Since it is not symmetric, all the conditions of an equivalence relation are not satisfied.
    \[\boxed{\therefore\ R\ \text{is not an equivalence relation on}\ P(X).}\]
🎯 Exam Significance
Exam Significance
›
  • This question tests the understanding of equivalence relations using the subset relation.
  • Students should remember that checking only one or two properties is insufficient; all three properties must hold.
  • Counterexamples are frequently used to disprove symmetry or transitivity in Board examinations.
  • Questions on relations and their properties are commonly asked in CBSE Board examinations, JEE Main, CUET, NDA and other entrance examinations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Every set is a subset of itself, so the subset relation is reflexive.

  2. The subset relation is generally not symmetric.

  3. The subset relation is transitive.

  4. A relation must be reflexive, symmetric and transitive simultaneously to be an equivalence relation.

  5. Failure of even one property implies that the relation is not an equivalence relation.
← Q2
3 / 7  ·  43%
Q4 →
Q4
NUMERIC3 marks
Find the number of all onto functions from the set \(A=\{1,2,3,\ldots,n\}\) to itself.
📘 Concept & Theory
Concept/Theory
›

A function

\[f:A\rightarrow B\]

is called onto (surjective) if every element of the codomain has at least one pre-image in the domain.

In this question,

\[A=B=\{1,2,3,\ldots,n\},\]

so both the domain and the codomain contain exactly

\[n\]

elements.

An important theorem states:

  • If a function is from a finite set to itself and is onto, then it is also one-one.
  • Similarly, every one-one function from a finite set to itself is automatically onto.
  • Hence, every onto function from a finite set onto itself is a bijection.

The number of bijections from a set having

\[n\]

elements onto itself is equal to the number of permutations of its elements.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Observe that both the domain and codomain have the same number of elements.

  2. Use the theorem that an onto function between two finite sets of equal cardinality is one-one.

  3. Conclude that every onto function is a bijection.

  4. Count the number of bijections using permutations.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  20 steps
  1. The given function is \(f:A\rightarrow A\) where
    \[A=\{1,2,3,\ldots,n\}\]
  2. The set \(A\) contains exactly \(n\) elements.
  3. Suppose the function is onto.
  4. Since the domain and codomain have the same finite number of elements, every element of the codomain must receive exactly one image from the domain.
  5. Therefore, no two distinct elements of the domain can have the same image.
  6. Hence, the function is also one-one.
  7. Thus, every onto function from \(A\) to itself is a bijection.
  8. Now count the number of bijections.
  9. The first element of the domain can be mapped to any one of the \(n\) elements of the codomain.
  10. Therefore, the number of choices for the first element is \(n\)
  11. After one image has been chosen, the second element has \(n-1\) choices.
  12. The third element has \(n-2\) choices.
  13. Continuing in this way, the last element has only 1 choice
  14. Hence, the total number of onto functions is
    \[n(n-1)(n-2)\cdots2\cdot1\]
  15. Therefore,
    \[\boxed{\bbox[2pt]{n!}}\]
    is the required number of onto functions.
  16. is the required number of onto functions.
  17. An onto function from a finite set to itself must use every element of the codomain exactly once.
  18. Thus, each onto function is simply a rearrangement (permutation) of the elements of the set.
  19. The number of permutations of \(n\)
  20. distinct objects is \(n!\)
  21. Hence, the answer is again
    \[\boxed{\bbox[2pt]{n!}}\]
🎯 Exam Significance
Exam Significance
›
  • This question establishes the important relationship between injective, surjective and bijective functions for finite sets.
  • Students should remember that for finite sets having equal cardinality, onto ⇔ one-one ⇔ bijective.
  • The concept frequently appears in CBSE Board examinations and is also useful in JEE Main, CUET, NDA and other competitive examinations.
  • It also connects functions with permutations, an important topic in combinatorics.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. If two finite sets have the same number of elements, every onto function is automatically one-one.

  2. Every bijection corresponds to a unique permutation.

  3. The number of permutations of

    \[n\]

  4. distinct elements is

    \[n!\]

  5. Therefore, the number of onto functions from a finite set onto itself is

    \[\boxed{n!}.\]
← Q3
4 / 7  ·  57%
Q5 →
Q5
NUMERIC3 marks
Let \(A=\{-1,0,1,2\}\), \(B=\{-4,-2,0,2\}\) and \(f,g:A\rightarrow B\) be functions defined by \(f(x)=x^2-x,\qquad x\in A\) and \(g(x)=2\left|x-\frac12\right|-1,\qquad x\in A\). Are \(f\) and \(g\) equal? Justify your answer.
📘 Concept & Theory
Concept/Theory
›

Two functions are said to be equal if all the following conditions are satisfied.

  • They have the same domain.
  • They have the same codomain.
  • For every element of the domain, the corresponding function values are equal.

Mathematically,

\[f=g\]

if and only if

\[f(x)=g(x)\qquad\forall x\in A.\]

Thus, to determine whether two functions are equal, it is sufficient to compare their values at every element of the domain.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Compute the values of \(f(x)\) for every element of \(A\).

  2. Compute the values of \(g(x)\) for every element of \(A\).

  3. Compare the corresponding values.

  4. Conclude whether the two functions are equal.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  18 steps
  1. The domain is
    \[A=\{-1,0,1,2\}\]
  2. Find the values of \(f(x)\)
  3. Since
    \[f(x)=x^2-x\]
    evaluate the function for each element of \(A\).
  4. For \(x=-1\)
    \[f(-1)=(-1)^2-(-1)=1+1=2.\]
  5. For \(x=0\)
    \[f(0)=0^2-0=0\]
  6. For \(x=1\)
    \[f(1)=1^2-1=1-1=0\]
  7. For \(x=2\)
    \[f(2)=2^2-2=4-2=2\]
  8. Hence,
    \(x\) \(-1\) \(0\) \(1\) \(2\)
    \(f(x)\) \(2\) \(0\) \(0\) \(2\)
  9. Find the values of \(g(x)\)
  10. Given —
    \[g(x)=2\left|x-\frac12\right|-1\]
  11. Evaluate the function for each element of \(A\)
  12. For \(x=-1\)
    \[ \begin{aligned} g(-1) &=2\left|-1-\frac12\right|-1\\ &=2\left|\frac{-3}{2}\right|-1\\ &=2\left(\frac32\right)-1\\ &=3-1\\ &=2. \end{aligned} \]
  13. For \(x=0\)
    \[ \begin{aligned} g(0) &=2\left|-\frac12\right|-1\\ &=2\left(\frac12\right)-1\\ &=1-1\\ &=0. \end{aligned} \]
  14. For \(x=1\)
    \[ \begin{aligned} g(1) &=2\left|\frac12\right|-1\\ &=2\left(\frac12\right)-1\\ &=1-1\\ &=0. \end{aligned} \]
  15. For \(x=2\)
    \[ \begin{aligned} g(2) &=2\left|2-\frac12\right|-1\\ &=2\left(\frac32\right)-1\\ &=3-1\\ &=2. \end{aligned} \]
  16. Thus,
    \(x\) \(-1\) \(0\) \(1\) \(2\)
    \(g(x)\) \(2\) \(0\) \(0\) \(2\)
  17. Compare the Function Values
  18. From the above tables,
    \[ \begin{aligned} f(-1)&=g(-1)=2,\\ f(0)&=g(0)=0,\\ f(1)&=g(1)=0,\\ f(2)&=g(2)=2. \end{aligned} \]
  19. Thus,
    \[f(x)=g(x)\quad\forall x\in A\]
  20. Since the domain and codomain are also the same, the two functions are equal.
  21. Conclusion
  22. \[\boxed{\bbox[2pt]{f=g}}\]
  23. Therefore, the functions \(f\) and \(g\) are equal.
🎯 Exam Significance
Exam Significance
›
  • This question emphasizes the definition of equality of functions.
  • Students learn that two functions may have different algebraic expressions but still represent the same function if they produce identical outputs for every element of the domain.
  • Such conceptual questions are frequently asked in CBSE Board examinations.
  • The idea is also useful in JEE Main, CUET, NDA and other entrance examinations where function properties are tested.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  4 points

  1. Two functions are equal only if they have the same domain, the same codomain and identical function values for every input.

  2. Different formulas may define the same function.

  3. Checking equality requires evaluating the function at every element of the domain when the domain is finite.

  4. Never conclude that two functions are different merely because their formulas look different.
← Q4
5 / 7  ·  71%
Q6 →
Q6
NUMERIC3 marks
Let \(A=\{1,2,3\}\) Then the number of relations containing \((1,2)\ \text{and}\ (1,3)\) which are reflexive and symmetric but not transitive is

(A) 1    (B) 2    (C) 3    (D) 4

📘 Concept & Theory
Concept/Theory
›

A relation on a set is

  • Reflexive if
    \[(a,a)\in R\]
    for every element \(a\) of the set.
  • Symmetric if
    \[(a,b)\in R\Longrightarrow (b,a)\in R.\]
  • Transitive if
    \[(a,b)\in R,\ (b,c)\in R \Longrightarrow (a,c)\in R\]

Whenever a relation is required to be reflexive and symmetric, those ordered pairs become compulsory. The remaining optional pairs are then chosen so that the relation is not transitive.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write all compulsory ordered pairs using reflexivity.

  2. Use symmetry to determine additional compulsory pairs.

  3. Identify the only optional ordered pair.

  4. Check transitivity in each possible case.

  5. Count the valid relations.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  19 steps
  1. The set is
    \[A=\{1,2,3\}\]
  2. Since the relation is reflexive, the following ordered pairs must belong to the relation:
    \[(1,1),\ (2,2),\ (3,3)\]
  3. The relation is also given to contain
    \[(1,2)\quad\text{and}\quad(1,3)\]
  4. Since it is symmetric, it must also contain
    \[(2,1)\quad\text{and}\quad(3,1)\]
  5. Thus the compulsory ordered pairs are
    \[ \begin{aligned} R=\{&(1,1),(2,2),(3,3),\\ &(1,2),(2,1),(1,3),(3,1)\}. \end{aligned} \]
  6. The only remaining ordered pairs are
    \[(2,3)\quad\text{and}\quad(3,2)\]
  7. Because the relation must remain symmetric, these two pairs must either
    • both be absent, or
    • both be present.
  8. Case 1. Both \((2,3)\) and \((3,2)\) are absent
  9. The relation becomes
    \[ \begin{aligned} R=\{&(1,1),(2,2),(3,3),\\ &(1,2),(2,1),(1,3),(3,1)\}. \end{aligned} \]
  10. Now observe that
    \[(2,1)\in R\]
    and
    \[(1,3)\in R\]
  11. By transitivity, this would require
    \[(2,3)\in R\]
    which is absent.
  12. Hence the relation is not transitive.
  13. Therefore, this case satisfies all the conditions.
  14. Case 2. Both \((2,3)\) and \((3,2)\) are present
  15. The relation becomes
    \[ \begin{aligned} R=\{&(1,1),(2,2),(3,3),\\ &(1,2),(2,1),(1,3),(3,1),\\ &(2,3),(3,2)\}. \end{aligned} \]
  16. This is the universal relation on the set except that no ordered pair is missing.
  17. Now every required transitive implication is satisfied.
  18. Hence this relation is transitive.
  19. Therefore, this case does not satisfy the question.
  20. Conclusion
  21. Only one relation is reflexive, symmetric and not transitive.
    \[\boxed{\text{Number of such relations}=1}\]
  22. Hence, the correct option is
    \[\boxed{\textbf{(A) }1}\]
🎯 Exam Significance
Exam Significance
›
  • This problem combines the three important properties of relations in a single question.
  • It demonstrates how compulsory ordered pairs are determined using reflexivity and symmetry.
  • Such objective questions are frequently asked in CBSE Board examinations, JEE Main, CUET, NDA and university entrance examinations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  4 points

  1. Reflexivity immediately forces all diagonal ordered pairs.

  2. Symmetry forces the reverse of every non-diagonal ordered pair.

  3. Transitivity must be checked systematically using all possible chains.

  4. In counting problems, first identify compulsory pairs and then analyze the remaining optional pairs.
← Q5
6 / 7  ·  86%
Q7 →
Q7
NUMERIC3 marks
Let \(A=\{1,2,3\}\). Then the number of equivalence relations containing \((1,2)\) is

(A) 1    (B) 2    (C) 3    (D) 4

📘 Concept & Theory
Concept/Theory
›

An equivalence relation on a set must satisfy the following three properties.

  • Reflexive:
    \[(a,a)\in R\]
    for every element of the set.
  • Symmetric:
    \[(a,b)\in R\Longrightarrow (b,a)\in R.\]
  • Transitive:
    \[(a,b)\in R,\ (b,c)\in R\Longrightarrow (a,c)\in R.\]

An important fact is that every equivalence relation partitions a set into disjoint equivalence classes, and every partition determines a unique equivalence relation.

Therefore, instead of listing all ordered pairs, it is often easier to determine the possible partitions.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Use the fact that \((1,2)\in R\).

  2. Since the relation is symmetric, conclude that \((2,1)\in R\).

  3. Hence, 1 and 2 must belong to the same equivalence class.

  4. List all possible partitions satisfying this condition.

  5. Count the corresponding equivalence relations.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. The set is
    \[A=\{1,2,3\}\]
  2. Since
    \[(1,2)\in R\]
    and \(R\) is an equivalence relation, symmetry gives
    \[(2,1)\in R\]
  3. Therefore, 1 and 2 belong to the same equivalence class.
  4. Now examine the position of the remaining element 3.
  5. Case 1. Element 3 forms a separate equivalence class
  6. The partition is
    \[\{\{1,2\},\{3\}\}\]
  7. The corresponding equivalence relation is
    \[ \{ (1,1),(2,2),(3,3), (1,2),(2,1) \} \]
  8. This is a valid equivalence relation.
  9. Case 2. Element 3 belongs to the same equivalence class as 1 and 2
  10. The partition is
    \[\{\{1,2,3\}\}\]
  11. The corresponding equivalence relation contains every ordered pair of
    \[A\times A\]
  12. Thus, the relation is
    \[ \{ (1,1),(1,2),(1,3), (2,1),(2,2),(2,3), (3,1),(3,2),(3,3) \} \]
  13. This is also an equivalence relation.
  14. Are there any other possibilities? → No
  15. Since 1 and 2 must remain in the same equivalence class, the only choice is whether 3 joins this class or remains separate.
  16. Hence, exactly two equivalence relations satisfy the given condition.
  17. Conclusion
  18. \[\boxed{\text{Number of equivalence relations}=2}\]
  19. Therefore, the correct option is
    \[\boxed{\textbf{(B) }2}\]
🎯 Exam Significance
Exam Significance
›
  • This question illustrates the close relationship between equivalence relations and partitions of a set.
  • Students should remember that every partition corresponds to exactly one equivalence relation.
  • This approach is much faster than checking reflexivity, symmetry and transitivity for every possible relation.
  • Such objective questions are frequently asked in CBSE Board examinations, JEE Main, CUET, NDA and university entrance examinations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  4 points

  1. An equivalence relation partitions a set into equivalence classes.

  2. If \((a,b)\) belongs to an equivalence relation, then \(a\) and \(b\) lie in the same equivalence class.

  3. Once elements are forced into one equivalence class, only the remaining elements need to be arranged.

  4. For this problem, only two partitions are possible:

    \[\{\{1,2\},\{3\}\}\]
    and
    \[\{\{1,2,3\}\}.\]
← Q6
7 / 7  ·  100%
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NCERT Class 12 Maths Miscellaneous Exercise Solutions
NCERT Class 12 Maths Miscellaneous Exercise Solutions — Complete Notes & Solutions · academia-aeternum.com
The NCERT Class 12 Mathematics Chapter 1 – Relations and Functions Miscellaneous Exercise Solutions provide a complete revision of the most important concepts from the chapter in a single place. This exercise combines questions on relations, equivalence relations, injective (one-one), surjective (onto), bijective functions, inverse functions, and counting of functions, helping students strengthen both conceptual understanding and problem-solving skills. Each solution is presented with the…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Frequently Asked Questions

    The Miscellaneous Exercise covers important concepts such as relations, equivalence relations, injective (one-one), surjective (onto), bijective functions, inverse functions, and counting of functions, integrating all topics from the chapter.

    It serves as a comprehensive revision of the entire chapter and includes conceptual and proof-based questions that are frequently asked in the CBSE Board Examination.

    Yes. Every solution is explained in a detailed, step-by-step manner with complete mathematical reasoning, ensuring that no intermediate step is omitted.

    These solutions strengthen conceptual understanding and logical reasoning, making them useful for competitive examinations such as JEE Main, CUET, NDA, BITSAT, and university entrance tests.

    An equivalence relation is a relation that satisfies three properties simultaneously: reflexivity, symmetry, and transitivity.

    A function is one-one if distinct inputs have distinct outputs, and it is onto if every element of the codomain has at least one pre-image in the domain. The solutions demonstrate systematic methods to prove both properties.

    Proofs develop logical thinking and mathematical reasoning, which are essential for solving higher-order Board examination questions and competitive entrance examination problems.

    Yes. Two functions are equal if they have the same domain, the same codomain, and identical outputs for every input, even if their algebraic expressions are different.

    For a finite set containing n elements, the number of onto functions from the set to itself is n!, since every onto function is also a bijection.

    Yes. The solutions are prepared according to the latest NCERT textbook and CBSE Class 12 Mathematics syllabus, making them suitable for current academic and examination requirements.

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