CBSE Class 12 Mathematics Relations And Functions Assignment Set 11

Chapter 1

Relations and Functions

Very Short Answer Type Questions (1 Mark)

Question 1. If A is the set of students of a school then write, which of following relations are. (Universal, Empty or neither of the two). R1 = {(a, b) : a, b are ages of students and |a – b|  0} R2 = {(a, b) : a, b are weights of students, and |a – b| < 0} R3 = {(a, b) : a, b are students studying in same class}
Answer:
\( R_1 \): Since the absolute difference between any two real numbers is always greater than or equal to zero, this relation includes all possible pairs of students. Thus, it is a universal relation.
\( R_2 \): Since the absolute value of any real number can never be negative, no pair of students can satisfy the condition. Thus, it is an empty relation.
\( R_3 \): Some students in a school study in the same class, while others do not. This means the relation contains some pairs but not all possible pairs, so it is neither universal nor empty.
In simple words: The first relation is universal because everyone's age difference is at least zero. The second is empty because a weight difference can never be less than zero. The third is neither because only some students share a class.

Exam Tip: Always justify your choice by verifying the mathematical limits of absolute values (e.g. absolute values are always non-negative).

Question 2. Is the relation R in the set A = {1, 2, 3, 4, 5} defined as R = {(a, b) : b = a + 1} reflexive?
Answer:
The relation is not reflexive. For a relation to be reflexive, every element \( a \) in set A must satisfy \( (a, a) \in R \). Here, the condition is \( b = a + 1 \). If we substitute \( b = a \), we get \( a = a + 1 \), which is mathematically impossible. For instance, for \( 1 \in A \), the pair \( (1, 1) \) does not belong to the relation since \( 1 \neq 1 + 1 \).
In simple words: A relation is reflexive if every number is paired with itself. Since no number can be equal to itself plus one, this relation is not reflexive.

Exam Tip: To prove a relation is not reflexive, providing a single counterexample (like \( (1, 1) \notin R \)) is sufficient to earn full marks.

Question 3. If R, is a relation in set N given by R = {(a, b) : a = b – 3, b > 5}, then does elements (5, 7)  R?
Answer:
No, the element \( (5, 7) \) is not in the relation R. The given condition is \( a = b - 3 \) with \( b > 5 \). If we take \( b = 7 \) (which satisfies \( b > 5 \)), then \( a \) must be \( 7 - 3 = 4 \). Since the given first element is 5 instead of 4, the pair \( (5, 7) \) does not satisfy the relation.
In simple words: For the second number 7, the first number should be 7 minus 3, which is 4. Since the pair given is \( (5, 7) \) instead of \( (4, 7) \), it is not in the relation.

Exam Tip: Always verify both parts of the relation's criteria (the equation and any inequality conditions on the variables) before deciding if an ordered pair belongs to it.

Question 4. If f : {1, 3}  {1, 2, 5} and g : {1, 2, 5}  {1, 2, 3, 4} be given by f = {(1, 2), (3, 5)}, g = {(1, 3), (2, 3), (5, 1)} Write down gof.
Answer:
The composite function is \( gof = \{(1, 3), (3, 1)\} \). This is obtained because the image of 1 under \( f \) is 2, and the image of 2 under \( g \) is 3; similarly, the image of 3 under \( f \) is 5, and the image of 5 under \( g \) is 1.
In simple words: Follow the path of each starting number. 1 goes to 2, and 2 goes to 3, so 1 ends at 3. 3 goes to 5, and 5 goes to 1, so 3 ends at 1.

Exam Tip: Write down intermediate steps like \( g(f(1)) = g(2) = 3 \) to show the logical flow of the composite function.

Question 5. Let g, f : R  R be defined by g(x) = x+2/3 , f(x) = 3x – 2. Write fog.
Answer:
The composition is \( (fog)(x) = x \) for every real number \( x \). We evaluate this as follows:
\( (fog)(x) = f(g(x)) = f\left(\frac{x+2}{3}\right) \)
\( (fog)(x) = 3\left(\frac{x+2}{3}\right) - 2 \)
\( (fog)(x) = x + 2 - 2 = x \)
Thus, the composite function \( (fog)(x) \) is the identity function on \( \mathbb{R} \).
In simple words: Put the formula for \( g(x) \) inside the formula for \( f(x) \). The 3 on the top and bottom cancel out, leaving you with just \( x \).

Exam Tip: Be sure to write the domain condition \( \forall x \in \mathbb{R} \) in your final answer to make it mathematically complete.

Question 6. If f : R  R defined by f(x) = 2x – 1/5 be an invertible function, write f–1(x).
Answer:
The inverse function is \( f^{-1}(x) = \frac{5x+1}{2} \). To find this, we let:
\( y = \frac{2x - 1}{5} \)
Multiplying by 5 gives:
\( 5y = 2x - 1 \)
Rearranging to solve for \( x \):
\( 2x = 5y + 1 \)

\(\implies x = \frac{5y + 1}{2} \)
Replacing \( y \) with \( x \), we get:
\( f^{-1}(x) = \frac{5x + 1}{2} \)
In simple words: Set the equation equal to \( y \), swap the variables, and solve for \( x \) to find the inverse formula.

Exam Tip: Expressing the final answer in terms of \( x \) (rather than \( y \)) is the standard notation required to receive full marks.

Question 7. If f(x) = x/x+1 \(\forall\) x  –1, Write fo f(x).
Answer:
The composite function is \( (fof)(x) = \frac{x}{2x+1} \) where \( x \ne -\frac{1}{2} \). We compute this by substituting the function into itself:
\( (fof)(x) = f(f(x)) = f\left(\frac{x}{x+1}\right) \)
\( (fof)(x) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} \)
Taking the LCM in the denominator:
\( (fof)(x) = \frac{\frac{x}{x+1}}{\frac{x + (x+1)}{x+1}} = \frac{x}{2x+1} \)
In simple words: Substitute the entire fraction into its own variable slot, then simplify the complex fraction by multiplying the top and bottom.

Exam Tip: Don't forget to include the new domain restriction \( x \ne -\frac{1}{2} \) that arises from simplifying the denominator.

Question 8. Let * is a Binary operation defined on R, then if (i) a * b = a + b + ab, write 3 * 2 (ii) a * b = (a+b)2/3 , Write (2 * 3) * 4.
Answer:
(i) For the first binary operation, we substitute \( a = 3 \) and \( b = 2 \) into the formula:
\( 3 * 2 = 3 + 2 + (3)(2) = 5 + 6 = 11 \)

(ii) For the second binary operation, we first compute \( 2 * 3 \):
\( 2 * 3 = \frac{(2+3)^2}{3} = \frac{25}{3} \)
Now, we compute \( \left( \frac{25}{3} \right) * 4 \):
\( \left( \frac{25}{3} \right) * 4 = \frac{\left( \frac{25}{3} + 4 \right)^2}{3} \)
Taking the LCM inside the parentheses:
\( = \frac{\left( \frac{37}{3} \right)^2}{3} = \frac{\frac{1369}{9}}{3} = \frac{1369}{27} \)
In simple words: (i) Add the numbers and add their product. (ii) First apply the formula to 2 and 3, then use that result with 4 in the same formula.

Exam Tip: For multi-step binary operations, strictly follow the order of operations by resolving the parentheses first.

Question 9. If n(A) = n(B) = 3, Then how many bijective functions from A to B can be formed?
Answer:
The number of bijective functions from set A to set B is given by \( n! \), where \( n \) is the number of elements in each set.
Since \( n = 3 \), the total number of bijective functions is:
\( 3! = 3 \cdot 2 \cdot 1 = 6 \)
In simple words: The first element has 3 choices, the second has 2, and the third has 1 choice. Multiplying these gives 6 total bijective mappings.

Exam Tip: Remember that bijective functions can only be formed if the domain and codomain have the exact same number of elements.

Question 10. If f (x) = x + 1, g(x) = x – 1, Then (gof) (3) = ?
Answer:
The value of \( (gof)(3) \) is equal to 3. We calculate this step-by-step:
First, find \( f(3) \):
\( f(3) = 3 + 1 = 4 \)
Now, substitute this result into \( g(x) \):
\( g(f(3)) = g(4) = 4 - 1 = 3 \)
In simple words: Put 3 into \( f(x) \) to get 4, then put 4 into \( g(x) \) to get back to 3.

Exam Tip: Evaluating the inner function first is the standard, safest way to solve composite function value questions.

Question 11. Is f : N  N given by f(x) = x2 is one-one? Give reason.
Answer:
Yes, the function \( f \) is one-one. Since the domain is the set of natural numbers \( \mathbb{N} \), if \( f(x_1) = f(x_2) \implies x_1^2 = x_2^2 \). Since natural numbers are strictly positive, taking the square root gives \( x_1 = x_2 \). Since distinct elements have distinct images, the function is one-one.
In simple words: Yes, because every positive number has a unique square, so no two different natural numbers can have the same squared value.

Exam Tip: Always mention the domain set (like \( \mathbb{N} \)) in your reason, as the same function \( f(x) = x^2 \) is not one-one if the domain is \( \mathbb{R} \) (since \( 2^2 = (-2)^2 \)).

Question 12. If f : R  A, given by f(x) = x2 – 2x + 2 is onto function, find set A.
Answer:
The set A is the interval \( [1, \infty) \). Since the function is onto, its codomain A must equal its range. Let us find the range by completing the square:
\( f(x) = x^2 - 2x + 2 = (x - 1)^2 + 1 \)
Since \( (x - 1)^2 \ge 0 \) for all real numbers \( x \):
\( (x - 1)^2 + 1 \ge 1 \)
Thus, the minimum value of \( f(x) \) is 1, meaning the range is \( [1, \infty) \). Therefore, \( A = [1, \infty) \).
In simple words: The minimum value of this quadratic expression is 1. Since the function is onto, the target set A must be everything from 1 upwards, i.e., \( [1, \infty) \).

Exam Tip: Completing the square is a very reliable algebraic technique to find the minimum value and range of a quadratic function.

Question 13. If f : A  B is bijective function such that n (A) = 10, then n (B) = ?
Answer:
The cardinality of set B is \( 10 \). For a function to be bijective (both one-one and onto), there must be a perfect one-to-one pairing between elements. This requires the domain and codomain to have the exact same number of elements, so \( n(B) = n(A) = 10 \).
In simple words: A bijective function couples every element in A with exactly one element in B, leaving none out. So both sets must have the same size.

Exam Tip: Remember this fundamental property: bijectivity between finite sets implies equal cardinality.

Question 14. If n(A) = 5, then write the number of one-one functions from A to A.
Answer:
The number of one-one (injective) functions from a set of size \( n \) to itself is \( n! \). Since \( n(A) = 5 \), the total number of such functions is:
\( 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \).
In simple words: The first element can map to any of the 5 elements, the second to any of the remaining 4, and so on, giving 120 total ways.

Exam Tip: For any finite set, any one-one function from the set to itself is also onto, making the number of one-one functions equal to the number of bijections.

Question 15. R = {(a, b) : a, b  N, a  b and a divides b}. Is R reflexive? Give reason?
Answer:
No, the relation R is not reflexive. For a relation to be reflexive, every element \( a \in \mathbb{N} \) must satisfy \( (a, a) \in R \). However, the relation requires \( a \ne b \), which means no element can be paired with itself, so \( (a, a) \notin R \) for any natural number \( a \).
In simple words: No, because a number cannot be unequal to itself, which violates the \( a \ne b \) rule required by the relation.

Exam Tip: Clearly state that the condition \( a \ne b \) directly prevents the reflexive condition \( (a, a) \in R \) from being met.

Question 16. Is f : R  R, given by f(x) = |x – 1| is one-one? Give reason?
Answer:
The given function is not one-one (injective). For a function to be one-one, different inputs must always produce different outputs. However, if we evaluate the function at two distinct points, say \( x = 3 \) and \( x = -1 \):
\( f(3) = |3 - 1| = |2| = 2 \)
\( f(-1) = |-1 - 1| = |-2| = 2 \)
Since \( f(3) = f(-1) = 2 \) even though \( 3 \ne -1 \), different domain elements map to the same image, confirming that the function is not one-one.
In simple words: The absolute value function strips away negative signs, so two different starting numbers like 3 and -1 both end up giving the same result of 2. Because of this, it is not a one-to-one mapping.

Exam Tip: Showing that two different inputs produce the same output is the standard and most effective way to disprove that a function is one-one.

Question 17. f : R  B given by f(x) = sin x is onto function, then write set B.
Answer:
The set B must be the closed interval \( [-1, 1] \). For the function \( f(x) = \sin x \) to be onto (surjective), the codomain set B must be exactly equal to the range of the function. Since the value of the sine function always lies between \( -1 \) and \( 1 \) inclusive for all real inputs, the range is \( [-1, 1] \). Therefore, the codomain set is \( B = [-1, 1] \).
In simple words: Since a sine wave only goes as high as 1 and as low as -1, the output set B must be everything between -1 and 1 to make the function onto.

Exam Tip: For any function to be onto, the codomain must be defined as exactly equal to the function's mathematical range.

Question 18. If f(x) = log(1+x/1-x) , show that f(2x/1+x^2) = 2f(x).
Answer:
Let us substitute the input term \( \frac{2x}{1+x^2} \) into the given function \( f(x) \):
\( f\left(\frac{2x}{1+x^2}\right) = \log\left(\frac{1 + \frac{2x}{1+x^2}}{1 - \frac{2x}{1+x^2}}\right) \)
Simplifying the fractions inside the logarithm:
\( f\left(\frac{2x}{1+x^2}\right) = \log\left(\frac{\frac{1+x^2+2x}{1+x^2}}{\frac{1+x^2-2x}{1+x^2}}\right) \)
Canceling the common denominator \( 1+x^2 \):
\( = \log\left(\frac{x^2+2x+1}{x^2-2x+1}\right) \)
Recognizing the numerator and denominator as perfect squares:
\( = \log\left(\frac{(1+x)^2}{(1-x)^2}\right) = \log\left(\frac{1+x}{1-x}\right)^2 \)
Using the logarithm power rule \( \log(u^2) = 2\log(u) \):
\( = 2\log\left(\frac{1+x}{1-x}\right) = 2f(x) \).
Hence, it is shown that \( f\left(\frac{2x}{1+x^2}\right) = 2f(x) \).
In simple words: Substitute the complex fraction into the log formula, combine the terms using a common denominator, and recognize the perfect squares. The exponent of 2 can then be moved to the front of the log.

Exam Tip: Be comfortable with algebraic identities like \( a^2 \pm 2ab + b^2 = (a \pm b)^2 \) to simplify fractional terms in composite proofs.

Question 19. If ‘*’ is a binary operation on set Q of rational numbers given by a * b = ab/5 then write the identity element in Q.
Answer:
The identity element is \( e = 5 \). Let \( e \in \mathbb{Q} \) be the identity element for the binary operation \( * \). By the definition of the identity element, we must have:
\( a * e = a \quad \text{and} \quad e * a = a \quad \forall a \in \mathbb{Q} \)
Using the definition of the operation \( a * b = \frac{ab}{5} \):
\( \frac{a \cdot e}{5} = a \)
Since \( a \) is any rational number, we can divide both sides by \( a \) (for \( a \ne 0 \)):
\( \frac{e}{5} = 1 \)

\(\implies e = 5 \).
Since \( 5 \in \mathbb{Q} \), the identity element exists and is equal to 5.
In simple words: The identity number is the one that leaves other numbers unchanged. Solving the equation shows that multiplying by 5 and then dividing by 5 leaves any number exactly as it was.

Exam Tip: Always make sure to confirm that your calculated identity element belongs to the specified domain set (like \( 5 \in \mathbb{Q} \)).

Question 20. If * is Binary operation on N defined by a * b = a + ab  a, b  N. Write the identity element in N if it exists.
Answer:
The identity element does not exist. Let us assume an identity element \( e \in \mathbb{N} \) exists. Then, by definition:
\( a * e = e * a = a \quad \forall a \in \mathbb{N} \)
Using the definition of the operation \( a * e = a + ae \):
\( a + ae = a \implies ae = 0 \)
Since \( a \) is a natural number and cannot be zero, we must have \( e = 0 \). However, \( 0 \) is not a natural number (\( 0 \notin \mathbb{N} \)).
Furthermore, checking the other direction with \( e * a \):
\( e * a = e + ea \ne a \) in general.
Since the identity conditions are not met, the identity element does not exist in \( \mathbb{N} \).
In simple words: For a number \( e \) to be the identity, it must work in both directions. Solving the equation yields 0, which is not a natural number, meaning no valid identity exists in this set.

Exam Tip: Remember that an identity element must satisfy both \( a * e = a \) and \( e * a = a \). If even one condition fails, the identity does not exist.

Short Answer Type Questions (4 Marks)

Question 21. Check the following functions for one-one and onto. (a) f : R  R, f(x) = 2x-3/7 (b) f : R  R, f(x) = |x + 1| (c) f : R – {2}  R, f(x) = 3x-1/x-2
Answer:
(a) The function \( f(x) = \frac{2x-3}{7} \) is bijective (both one-one and onto):
- **One-one:** Let \( f(x_1) = f(x_2) \implies \frac{2x_1-3}{7} = \frac{2x_2-3}{7} \implies 2x_1 - 3 = 2x_2 - 3 \implies 2x_1 = 2x_2 \implies x_1 = x_2 \). Thus, it is one-one.
- **Onto:** Let \( y \in \mathbb{R} \). We set \( y = \frac{2x-3}{7} \implies 7y = 2x - 3 \implies x = \frac{7y+3}{2} \). Since \( x \in \mathbb{R} \) exists for every real \( y \), the function is onto.

(b) The function \( f(x) = |x + 1| \) is neither one-one nor onto:
- **Not one-one:** Let \( x = 1 \) and \( x = -3 \). Then \( f(1) = |1+1| = 2 \) and \( f(-3) = |-3+1| = 2 \). Since \( f(1) = f(-3) \), it is not one-one.
- **Not onto:** The absolute value function can never output negative values, so the range is \( [0, \infty) \). Since the range is not equal to the codomain \( \mathbb{R} \), the function is not onto.

(c) The function \( f(x) = \frac{3x-1}{x-2} \) is one-one but not onto:
- **One-one:** Let \( f(x_1) = f(x_2) \implies \frac{3x_1-1}{x_1-2} = \frac{3x_2-1}{x_2-2} \implies (3x_1-1)(x_2-2) = (3x_2-1)(x_1-2) \implies 3x_1x_2 - 6x_1 - x_2 + 2 = 3x_1x_2 - 6x_2 - x_1 + 2 \implies -5x_1 = -5x_2 \implies x_1 = x_2 \). Thus, it is one-one.
- **Not onto:** Let \( y = \frac{3x-1}{x-2} \implies y(x-2) = 3x-1 \implies xy - 2y = 3x-1 \implies x(y-3) = 2y-1 \implies x = \frac{2y-1}{y-3} \). This expression is undefined for \( y = 3 \). Since \( y = 3 \) in the codomain \( \mathbb{R} \) has no pre-image, the function is not onto.
In simple words: (a) This linear function is perfectly bijective. (b) Absolute values lose signs, so it is neither. (c) The fraction is one-one, but since the output can never be exactly 3, it is not onto.

Exam Tip: When checking if a rational function is onto, express \( x \) in terms of \( y \) and find the value of \( y \) that makes the denominator zero.

Question 22. Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by a* b = H.C.F. of a and b. Write the operation table for the operation *.
Answer:
The operation table is constructed by finding the Highest Common Factor (H.C.F.) for each pair of numbers in the set \( \{1, 2, 3, 4, 5\} \):

In simple words: Fill in the grid by calculating the greatest common divisor for each row and column intersection. For example, the H.C.F. of 4 and 2 is 2, while the H.C.F. of 4 and 3 is 1.

Exam Tip: Be neat when drawing binary operation tables, and make sure that the entries on the main diagonal correspond to the H.C.F. of each element with itself.

Question 23. Let f : R - {-4/3}  R - {4/3} be a function given by f(x) = 4x/3x + 4 . Show that f is invertible with f–1(x) = 4x/4 – 3x .
Answer:
To prove that \( f \) is invertible, we must show that it is both one-one and onto:
- **One-one:** Let \( f(x_1) = f(x_2) \):
\( \frac{4x_1}{3x_1 + 4} = \frac{4x_2}{3x_2 + 4} \)
Cross-multiplying:
\( 4x_1(3x_2 + 4) = 4x_2(3x_1 + 4) \)
\( 12x_1x_2 + 16x_1 = 12x_1x_2 + 16x_2 \)
\( 16x_1 = 16x_2 \implies x_1 = x_2 \).
Thus, the function is one-one.
- **Onto:** Let \( y \in \mathbb{R} - \{4/3\} \). We set \( y = f(x) \):
\( y = \frac{4x}{3x + 4} \)
\( y(3x + 4) = 4x \implies 3xy + 4y = 4x \)
\( 4x - 3xy = 4y \implies x(4 - 3y) = 4y \)
\( x = \frac{4y}{4 - 3y} \).
Since \( y \ne 4/3 \), the denominator is never zero, and \( x \) is a well-defined real number. Hence, the function is onto.
Since \( f \) is bijective, it is invertible, and its inverse is:
\( f^{-1}(x) = \frac{4x}{4 - 3x} \).
In simple words: First show the function is one-one by proving that equal outputs require equal inputs. Then, solve the equation for \( x \) to find the onto inverse formula.

Exam Tip: Clearly state the separate proofs for one-one and onto, as both are required to establish invertibility in a 4-mark question.

Question 24. Let R be the relation on set A = {x : x  Z, 0  x  10} given by R = {(a, b) : (a – b) is multiple of 4}, is an equivalence relation. Also, write all elements related to 4.
Answer:
To show that \( R \) is an equivalence relation, we must prove it is reflexive, symmetric, and transitive:
1. **Reflexive:** For any \( a \in A \), \( a - a = 0 \), which is a multiple of 4. Thus, \( (a, a) \in R \), so \( R \) is reflexive.
2. **Symmetric:** Let \( (a, b) \in R \implies a - b = 4k \) for some integer \( k \). Then, \( b - a = -4k = 4(-k) \), which is also a multiple of 4. Thus, \( (b, a) \in R \), so \( R \) is symmetric.
3. **Transitive:** Let \( (a, b) \in R \) and \( (b, c) \in R \). Then \( a - b = 4k_1 \) and \( b - c = 4k_2 \) for some integers \( k_1, k_2 \). Adding these equations:
\( (a - b) + (b - c) = 4k_1 + 4k_2 \implies a - c = 4(k_1 + k_2) \), which is a multiple of 4. Thus, \( (a, c) \in R \), so \( R \) is transitive.
Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.
Now, let us find the elements related to 4, i.e., those \( x \in A \) such that \( (x, 4) \in R \):
\( x - 4 = 4k \implies x = 4k + 4 \)
Since \( 0 \le x \le 10 \), the possible values of \( x \) are \( 0, 4, 8 \).
Thus, the elements related to 4 are \( \{0, 4, 8\} \).
In simple words: Prove the relation is reflexive (difference is 0), symmetric (reversing doesn't change divisibility), and transitive (summing differences). The numbers between 0 and 10 whose difference with 4 is a multiple of 4 are 0, 4, and 8.

Exam Tip: Be sure to write the final equivalence class elements as a set \( \{0, 4, 8\} \) to show a complete and rigorous solution.

Question 25. Show that function f : A  B defined as f(x) = 3x+4/5x-7 where A = R - {7/5}, B = R - {3/5} is invertible and hence find f–1.
Answer:
To show that \( f \) is invertible, we prove it is one-one and onto:
- **One-one:** Let \( f(x_1) = f(x_2) \):
\( \frac{3x_1 + 4}{5x_1 - 7} = \frac{3x_2 + 4}{5x_2 - 7} \)
\( (3x_1 + 4)(5x_2 - 7) = (3x_2 + 4)(5x_1 - 7) \)
\( 15x_1x_2 - 21x_1 + 20x_2 - 28 = 15x_1x_2 - 21x_2 + 20x_1 - 28 \)
\( -21x_1 + 20x_2 = 20x_1 - 21x_2 \implies 41x_2 = 41x_1 \implies x_1 = x_2 \).
Thus, the function is one-one.
- **Onto:** Let \( y \in \mathbb{R} - \{3/5\} \). We set \( y = f(x) \):
\( y = \frac{3x + 4}{5x - 7} \)
\( y(5x - 7) = 3x + 4 \implies 5xy - 7y = 3x + 4 \)
\( 5xy - 3x = 7y + 4 \implies x(5y - 3) = 7y + 4 \)
\( x = \frac{7y + 4}{5y - 3} \).
Since \( y \ne 3/5 \), the denominator is never zero. Thus, every \( y \) in the codomain has a pre-image in the domain, making the function onto.
Since it is bijective, \( f \) is invertible, and its inverse is:
\( f^{-1}(x) = \frac{7x + 4}{5x - 3} \).
In simple words: Prove that the rational function is one-one by cross-multiplying. Then, rearrange the variables to express \( x \) in terms of \( y \) to find the inverse equation \( \frac{7x+4}{5x-3} \).

Exam Tip: Double-check your algebraic signs when cross-multiplying and moving terms across the equals sign to avoid simple arithmetic slips.

Question 26. Let * be a binary operation on Q. Such that a * b = a + b – ab. (i) Prove that * is commutative and associative. (ii) Find identify element of * in Q (if it exists).
Answer:
(i) Proof of Commutativity and Associativity:
- **Commutativity:** For any two rational numbers \( a, b \in \mathbb{Q} \):
\( a * b = a + b - ab \)
Since addition and multiplication of rational numbers are commutative:
\( b * a = b + a - ba = a + b - ab = a * b \)
Thus, the binary operation \( * \) is commutative.
- **Associativity:** For any three rational numbers \( a, b, c \in \mathbb{Q} \):
\( a * (b * c) = a * (b + c - bc) = a + (b + c - bc) - a(b + c - bc) = a + b + c - bc - ab - ac + abc \)
\( (a * b) * c = (a + b - ab) * c = (a + b - ab) + c - (a + b - ab)c = a + b - ab + c - ac - bc + abc \)
Since both expressions are identical, \( a * (b * c) = (a * b) * c \), which proves that the operation is associative.

(ii) Identity Element:
Let \( e \in \mathbb{Q} \) be the identity element. By definition, we must have:
\( a * e = a \implies a + e - ae = a \)
\( e(1 - a) = 0 \)
Since this must hold true for all rational numbers \( a \), we have:
\( e = 0 \)
Since \( 0 \in \mathbb{Q} \), the identity element of \( * \) on \( \mathbb{Q} \) exists and is equal to 0.
In simple words: Reversing the order of numbers doesn't change the outcome, proving the operation is commutative. Grouping them differently also gives the same result, proving it is associative. The identity element is 0 because performing the operation with 0 leaves the original number unchanged.

Exam Tip: Always show both steps of the associativity expansion clearly to demonstrate that the two paths yield the exact same polynomial expression.

Question 27. If * is a binary operation defined on R – {0} defined by a * b = 2a/b^2, then check * for commutativity and associativity.
Answer:
Let us test the given binary operation on the set of non-zero real numbers for both properties:
- **Commutativity:** Let us compute the operation in both orders:
\( a * b = \frac{2a}{b^2} \)
\( b * a = \frac{2b}{a^2} \)
In general, \( \frac{2a}{b^2} \ne \frac{2b}{a^2} \). For example, if we take \( a = 1 \) and \( b = 2 \):
\( 1 * 2 = \frac{2(1)}{2^2} = \frac{1}{2} \)
\( 2 * 1 = \frac{2(2)}{1^2} = 4 \)
Since \( \frac{1}{2} \ne 4 \), the binary operation is not commutative.
- **Associativity:** Let us check the grouping of three elements:
\( a * (b * c) = a * \left( \frac{2b}{c^2} \right) = \frac{2a}{\left( \frac{2b}{c^2} \right)^2} = \frac{2a \cdot c^4}{4b^2} = \frac{a c^4}{2b^2} \)
\( (a * b) * c = \left( \frac{2a}{b^2} \right) * c = \frac{2 \left( \frac{2a}{b^2} \right)}{c^2} = \frac{4a}{b^2 c^2} \)
Comparing the two results, we see they are unequal. Thus, the operation is not associative.
In simple words: The operation is neither commutative nor associative. Changing the order of the inputs or grouping them in a different sequence changes the final fraction.

Exam Tip: A single numerical counterexample is the most elegant and effective way to disprove both commutativity and associativity in exams.

Question 28. If A = N × N and binary operation * is defined on A as (a, b) * (c, d) = (ac, bd). (i) Check * for commutativity and associativity. (ii) Find the identity element for * in A (If it exists).
Answer:
(i) Commutativity and Associativity:
- **Commutativity:** Let \( (a, b), (c, d) \in \mathbb{N} \times \mathbb{N} \):
\( (a, b) * (c, d) = (ac, bd) \)
Since multiplication of natural numbers is commutative:
\( (c, d) * (a, b) = (ca, db) = (ac, bd) = (a, b) * (c, d) \)
Thus, the operation is commutative.
- **Associativity:** Let \( (a, b), (c, d), (e, f) \in \mathbb{N} \times \mathbb{N} \):
\( (a, b) * [ (c, d) * (e, f) ] = (a, b) * (ce, df) = (a(ce), b(df)) = (ace, bdf) \)
\( [ (a, b) * (c, d) ] * (e, f) = (ac, bd) * (e, f) = ((ac)e, (bd)f) = (ace, bdf) \)
Since both expressions are identical, the operation is associative.

(ii) Identity Element:
Let \( (x, y) \in \mathbb{N} \times \mathbb{N} \) be the identity element. By definition, we must have:
\( (a, b) * (x, y) = (a, b) \)
\( (ax, by) = (a, b) \)
Comparing components:
\( ax = a \implies x = 1 \)
\( by = b \implies y = 1 \)
Since \( (1, 1) \in \mathbb{N} \times \mathbb{N} \), the identity element exists and is \( (1, 1) \).
In simple words: Since coordinate-wise multiplication is commutative and associative, the entire operation inherits these properties. The identity element is \( (1, 1) \) because multiplying by 1 leaves the coordinates unchanged.

Exam Tip: When working with Cartesian product sets like \( \mathbb{N} \times \mathbb{N} \), remember that the identity element must be written as an ordered pair \( (e_1, e_2) \).

Question 29. Show that the relation R defined by (a, b) R(c, d)  a + d = b + c on the set N × N is an equivalence relation.
Answer:
To prove that \( R \) is an equivalence relation, we must demonstrate that it is reflexive, symmetric, and transitive:
1. **Reflexive:** For any \( (a, b) \in \mathbb{N} \times \mathbb{N} \), we have \( a + b = b + a \) because addition of natural numbers is commutative. This satisfies the condition, so \( (a, b) R (a, b) \). Hence, \( R \) is reflexive.
2. **Symmetric:** Let \( (a, b) R (c, d) \). By definition, this means:
\( a + d = b + c \)
Rearranging this equality:
\( c + b = d + a \)
This satisfies the definition for \( (c, d) R (a, b) \), proving that \( R \) is symmetric.
3. **Transitive:** Let \( (a, b) R (c, d) \) and \( (c, d) R (e, f) \). This implies:
\( a + d = b + c \quad \text{--- (1)} \)
\( c + f = d + e \quad \text{--- (2)} \)
Adding equations (1) and (2):
\( (a + d) + (c + f) = (b + c) + (d + e) \)
Canceling \( c \) and \( d \) from both sides:
\( a + f = b + e \)
This satisfies the relation, so \( (a, b) R (e, f) \), proving that \( R \) is transitive.
Since the relation \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: The relation is reflexive since any pair matches with itself, symmetric because reversing the pairs preserves the equation, and transitive because adding the equations cancels out the middle terms. This proves it is an equivalence relation.

Exam Tip: Adding the two coordinate equations is the most robust and standard algebraic way to prove transitivity for this highly repeated board exam question.

Question 30. Let * be a binary operation on set Q defined by a * b = ab/4, show that (i) 4 is the identity element of * on Q. (ii) Every non zero element of Q is invertible with a–1 = 16/a , a  Q – {0}.
Answer:
(i) Identity Element Proof:
Let \( e \in \mathbb{Q} \) be the identity element. By definition, we must have:
\( a * e = a \)
Using the operation formula:
\( \frac{ae}{4} = a \)
Since this must hold for any rational number \( a \) (for \( a \ne 0 \)):
\( \frac{e}{4} = 1 \implies e = 4 \)
Since \( 4 \in \mathbb{Q} \), 4 is the identity element of \( * \) on \( \mathbb{Q} \).

(ii) Inverse Element Proof:
Let \( b \in \mathbb{Q} \) be the inverse of any non-zero rational number \( a \). By definition:
\( a * b = e \)
Using \( e = 4 \) from part (i):
\( \frac{ab}{4} = 4 \)
\( ab = 16 \)

\(\implies b = \frac{16}{a} \)
Since \( a \ne 0 \), this division is well-defined and \( \frac{16}{a} \in \mathbb{Q} \).
Therefore, every non-zero element \( a \) is invertible with \( a^{-1} = \frac{16}{a} \).
In simple words: (i) Multiplying a number by 4 and then dividing by 4 leaves it unchanged, so 4 is the identity. (ii) For any non-zero number, multiplying it by \( 16/a \) and dividing by 4 gives back the identity 4, so the inverse is \( 16/a \).

Exam Tip: Remember to explicitly state that the inverse \( a^{-1} = \frac{16}{a} \) is valid because \( a \ne 0 \), as division by zero is undefined.

Question 31. Show that f : R+  R+ defined by f(x) = 1/2x is bijective where R+ is the set of all non-zero positive real numbers.
Answer:
To prove that the function \( f: \mathbb{R}_+ \rightarrow \mathbb{R}_+ \) defined by \( f(x) = \frac{1}{2x} \) is bijective, we must show that it is both one-one and onto:
- **One-one:** Let \( f(x_1) = f(x_2) \) for any \( x_1, x_2 \in \mathbb{R}_+ \):
\( \frac{1}{2x_1} = \frac{1}{2x_2} \)
Multiplying both sides by 2:
\( \frac{1}{x_1} = \frac{1}{x_2} \)
Taking the reciprocal of both sides:
\( x_1 = x_2 \)
Therefore, the function is one-one.
- **Onto:** Let \( y \in \mathbb{R}_+ \) (the codomain). We set \( y = f(x) \):
\( y = \frac{1}{2x} \)
Solving for \( x \):
\( 2x = \frac{1}{y} \)

\(\implies x = \frac{1}{2y} \)
Since \( y > 0 \) (as \( y \in \mathbb{R}_+ \)), \( \frac{1}{2y} > 0 \), which means \( x \in \mathbb{R}_+ \) (the domain). Thus, every element in the codomain has a pre-image in the domain, proving the function is onto.
Since \( f(x) \) is both one-one and onto, it is a bijective function.
In simple words: To prove bijectivity, first show it is one-one by proving that equal outputs require equal inputs. Next, show it is onto by proving that any positive output \( y \) has a corresponding positive starting value \( x = 1/2y \).

Exam Tip: Be sure to write down the domain constraints explicitly, as showing \( x > 0 \) for any \( y > 0 \) is what mathematically completes the "onto" proof.

Question 32. Consider f : R+  [–5, ) given by f(x) = 9x2 + 6x – 5 show that f is invertible with f–1 = \(\frac{\sqrt{x+6} - 1}{3}\) .
Answer:
To prove that the function is invertible, we express the equation \( y = f(x) \) in terms of \( x \):
\( y = 9x^2 + 6x - 5 \)
Let us complete the square on the right-hand side:
\( y = (3x + 1)^2 - 1 - 5 \)
\( y = (3x + 1)^2 - 6 \)
Adding 6 to both sides:
\( y + 6 = (3x + 1)^2 \)
Since the domain is \( \mathbb{R}_+ \) (positive real numbers), \( 3x + 1 > 0 \). Taking the positive square root of both sides:
\( \sqrt{y + 6} = 3x + 1 \)
\( 3x = \sqrt{y + 6} - 1 \)

\(\implies x = \frac{\sqrt{y + 6} - 1}{3} \)
Since \( x \ge 0 \implies \sqrt{y + 6} \ge 1 \implies y + 6 \ge 1 \implies y \ge -5 \), which matches the codomain interval \( [-5, \infty) \).
Thus, the function is invertible, and its inverse is:
\( f^{-1}(x) = \frac{\sqrt{x + 6} - 1}{3} \).
In simple words: Write the quadratic formula as a perfect square, rearrange the terms to solve for \( x \) using the positive square root, and swap the variables to find the inverse function.

Exam Tip: Completing the square is the most robust algebraic method to find the inverse of a quadratic function.

Question 33. If ‘*’ is a binary operation on R defined by a * b = a + b + ab. Prove that * is commutative and associative. Find the identify element. Also show that every element of R is invertible except –1.
Answer:
1. Commutativity:
For any \( a, b \in \mathbb{R} \):
\( a * b = a + b + ab \)
Since addition and multiplication of real numbers are commutative:
\( b * a = b + a + ba = a + b + ab = a * b \)
Thus, the operation \( * \) is commutative.

2. Associativity:
For any \( a, b, c \in \mathbb{R} \):
\( a * (b * c) = a * (b + c + bc) = a + (b + c + bc) + a(b + c + bc) = a + b + c + bc + ab + ac + abc \)
\( (a * b) * c = (a + b + ab) * c = (a + b + ab) + c + (a + b + ab)c = a + b + ab + c + ac + bc + abc \)
Since both expressions are identical, the operation \( * \) is associative.

3. Identity Element:
Let \( e \in \mathbb{R} \) be the identity element. By definition:
\( a * e = a \implies a + e + ae = a \)
\( e(1 + a) = 0 \)
Since this must hold for all real numbers \( a \), we must have:
\( e = 0 \)
Since \( 0 \in \mathbb{R} \), the identity element is 0.

4. Inverse Element:
Let \( b \) be the inverse of any element \( a \). By definition:
\( a * b = e \implies a + b + ab = 0 \)
\( b(1 + a) = -a \)

\(\implies b = \frac{-a}{1 + a} \)
This division is defined for all real numbers except when the denominator is zero:
\( 1 + a = 0 \implies a = -1 \)
Therefore, every element of \( \mathbb{R} \) is invertible except \( -1 \), and the inverse of \( a \) is \( \frac{-a}{1+a} \).
In simple words: The operation is commutative and associative because real number addition and multiplication are commutative and associative. The identity element is 0. Any number \( a \) has an inverse except \( -1 \) because dividing by \( 1+a \) is undefined when \( a = -1 \).

Exam Tip: Be sure to write the restriction \( a \ne -1 \) clearly when stating the inverse formula, as division by zero is mathematically undefined.

Question 34. If f, g : R  R defined by f(x) = x2 – x and g(x) = x + 1 find (fog) (x) and (gof) (x). Are they equal?
Answer:
We find both composite functions step-by-step:
- **Find \( (fog)(x) \):**
\( (fog)(x) = f(g(x)) = f(x+1) \)
\( (fog)(x) = (x+1)^2 - (x+1) \)
\( (fog)(x) = x^2 + 2x + 1 - x - 1 = x^2 + x \)
- **Find \( (gof)(x) \):**
\( (gof)(x) = g(f(x)) = g(x^2 - x) \)
\( (gof)(x) = (x^2 - x) + 1 = x^2 - x + 1 \)
Comparing the two results:
\( x^2 + x \ne x^2 - x + 1 \)
Therefore, \( (fog)(x) \ne (gof)(x) \), so the composite functions are not equal.
In simple words: Substitute \( g(x) \) into the \( x \) slots of \( f(x) \) to get \( x^2+x \). Then substitute \( f(x) \) into \( g(x) \) to get \( x^2-x+1 \). Since the formulas are different, the compositions are not equal.

Exam Tip: Expand the polynomial terms carefully to ensure that you do not make simple arithmetic errors when simplifying the final expressions.

Question 35. f : [1, \(\infty\))  [2, \(\infty\)) is given by f(x) = x + 1/x , find f–1(x).
Answer:
Let us set the equation equal to \( y \) and solve for \( x \):
\( y = x + \frac{1}{x} \)
Multiplying the entire equation by \( x \):
\( yx = x^2 + 1 \)
\( x^2 - yx + 1 = 0 \)
This is a quadratic equation in \( x \). Using the quadratic formula:
\( x = \frac{-(-y) \pm \sqrt{(-y)^2 - 4(1)(1)}}{2(1)} \)
\( x = \frac{y \pm \sqrt{y^2 - 4}}{2} \)
Since the domain of \( f \) is \( [1, \infty) \), we must have \( x \ge 1 \). Choosing the positive sign satisfies this condition:
\( x = \frac{y + \sqrt{y^2 - 4}}{2} \)
Replacing \( y \) with \( x \), we get the inverse function:
\( f^{-1}(x) = \frac{x + \sqrt{x^2 - 4}}{2} \).
In simple words: Multiply by \( x \) to write the equation as a standard quadratic equation. Solve for \( x \) using the quadratic formula, and choose the plus sign to keep the input within the domain interval.

Exam Tip: Be sure to mathematically justify why the minus sign in the quadratic formula is discarded (it would yield values less than 1, violating the domain constraint).

Question 36. f : R  R, g : R  R given by f(x) = [x], g(x) = |x| then find (fog)(-2/3) and (gof)(-2/3) .
Answer:
We evaluate both composite function values step-by-step:
- **Evaluate \( (fog)\left(-\frac{2}{3}\right) \):**
\( (fog)\left(-\frac{2}{3}\right) = f\left(g\left(-\frac{2}{3}\right)\right) \)
Using the absolute value function \( g(x) = |x| \):
\( g\left(-\frac{2}{3}\right) = \left|-\frac{2}{3}\right| = \frac{2}{3} \)
Now, apply the greatest integer function \( f(x) = [x] \):
\( f\left(\frac{2}{3}\right) = \left[\frac{2}{3}\right] = 0 \quad \left(\text{since } 0 \le \frac{2}{3} < 1\right) \)
- **Evaluate \( (gof)\left(-\frac{2}{3}\right) \):**
\( (gof)\left(-\frac{2}{3}\right) = g\left(f\left(-\frac{2}{3}\right)\right) \)
Using the greatest integer function \( f(x) = [x] \):
\( f\left(-\frac{2}{3}\right) = \left[-\frac{2}{3}\right] = -1 \quad \left(\text{since } -1 \le -\frac{2}{3} < 0\right) \)
Now, apply the absolute value function \( g(x) = |x| \):
\( g(-1) = |-1| = 1 \)
Therefore, \( (fog)\left(-\frac{2}{3}\right) = 0 \) and \( (gof)\left(-\frac{2}{3}\right) = 1 \).
In simple words: For \( fog \), first take the absolute value of -2/3 to get 2/3, then apply the greatest integer function to get 0. For \( gof \), first find the greatest integer of -2/3 which is -1, and then take the absolute value to get 1.

Exam Tip: Be very careful with negative inputs in the greatest integer function; remember that \( [-0.67] = -1 \) (the largest integer less than or equal to the number).

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