OP Malhotra Class 12 Maths Solutions Chapter 3 Binary Operations Exercise 3 (C)

 

Question 1. Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by a * b = min {a, b}. Write the operation table of the operation *.
Answer: The set is \( A = \{1, 2, 3, 4, 5\} \), and the operation \( a * b \) finds the smaller value between \( a \) and \( b \). The operation table shows the result for every combination of elements from the set. This table helps us see all possible outcomes of the operation clearly.

In simple words: The table shows the smaller number when you compare any two numbers from the set. For example, \( 3 * 4 \) is 3 because 3 is smaller than 4.

🎯 Exam Tip: When constructing an operation table for "min" or "max" operations, remember that the result of \( a * b \) will always be one of the elements \( a \) or \( b \) themselves.

 

Question 2. Construct the composition table for the operation * defined on A = {1, 2, 3} by a * b = a + b². Is * a binary operation?
Answer: The set given is \( A = \{1, 2, 3\} \), and the operation is defined as \( a * b = a + b^2 \). We will fill the table using this rule. A binary operation means that when you combine any two elements from the set using the operation, the result must also be an element of the same set. Let's see if all the results in our table belong to set A.

Clearly, not all elements in this table belong to the set \( A = \{1, 2, 3\} \). For example, 5, 10, 6, 11, 4, 7, 12 are all outside the set. Therefore, * is not a binary operation on set A.In simple words: We added the first number to the square of the second number. Some of the answers (like 5 or 10) are not in the original list {1, 2, 3}. Because of this, it is not a binary operation.

🎯 Exam Tip: For an operation to be binary on a set, *all* results must belong to that same set. Even one result outside the set means it's not a binary operation.

 

Question 3. Consider a binary operation * on the set {1, 2, 3, 4, 5} defined by a * b = HCF {a, b}. Write the operation table.
(i) Is it a binary operation
(ii) Is it commutative ?
(iii) Compute (2 * 3) * 4 and 2 * (3 * 4)
(iv) Compute (2 * 3) * (4 * 5)
Answer: The set is \( A = \{1, 2, 3, 4, 5\} \), and the operation \( a * b \) finds the Highest Common Factor (HCF) of \( a \) and \( b \). The HCF is the largest number that divides both numbers evenly.

(i) Since all entries in the table are elements of set A, * is a binary operation on A. This means the HCF of any two numbers in the set is also in the set.
(ii) The composition table is symmetrical around its main diagonal (from top-left to bottom-right). This shows that \( a * b = b * a \) for all elements. Thus, * is commutative on A.
(iii) First, we calculate (2 * 3). The HCF of 2 and 3 is 1.
\( \implies \) So, \( (2 * 3) * 4 = 1 * 4 \). The HCF of 1 and 4 is 1.
Next, we calculate (3 * 4). The HCF of 3 and 4 is 1.
\( \implies \) So, \( 2 * (3 * 4) = 2 * 1 \). The HCF of 2 and 1 is 1.
(iv) First, we calculate (2 * 3). The HCF of 2 and 3 is 1.
Next, we calculate (4 * 5). The HCF of 4 and 5 is 1.
\( \implies \) So, \( (2 * 3) * (4 * 5) = 1 * 1 \). The HCF of 1 and 1 is 1.In simple words: We made a table showing the biggest common factor for any two numbers. All answers stayed inside our original number set, and the table looked the same if you folded it across the middle, so it is a binary and commutative operation. We then calculated some combinations using the HCF rule.

🎯 Exam Tip: To check for commutativity from a table, visually inspect if the table is symmetric about its main diagonal. If an operation is defined as HCF or LCM, it will always be commutative.

 

Question 4. Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table
(i) Compute (a) (2 * 3) * 4 and (b) 2 * (3 * 4).
(ii) Is * commutative ?
(iii) Compute (2 * 3) * (4 * 5).
Answer: We are given the operation table for * on the set \( \{1, 2, 3, 4, 5\} \). We will use this table to find the results of the operations.

(i)
(a) We find \( (2 * 3) * 4 \). From the table, \( 2 * 3 = 1 \).
\( \implies \) So, \( (2 * 3) * 4 = 1 * 4 \). From the table, \( 1 * 4 = 1 \).
(b) We find \( 2 * (3 * 4) \). From the table, \( 3 * 4 = 1 \).
\( \implies \) So, \( 2 * (3 * 4) = 2 * 1 \). From the table, \( 2 * 1 = 1 \).
(ii) We check if the table is symmetrical about its main diagonal.
For example, \( 2 * 3 = 1 \) and \( 3 * 2 = 1 \). Also, \( 4 * 2 = 2 \) and \( 2 * 4 = 2 \).
Clearly, the composition table is symmetrical about the main diagonal. This means the order of numbers does not change the result. Hence, the operation * is commutative.
(iii) We compute \( (2 * 3) * (4 * 5) \).
From the table, \( 2 * 3 = 1 \).
From the table, \( 4 * 5 = 1 \).
\( \implies \) So, \( (2 * 3) * (4 * 5) = 1 * 1 \). From the table, \( 1 * 1 = 1 \).In simple words: We used the given table to find results. For part (i), we found that both ways of grouping numbers gave the same answer. For part (ii), we looked at the table and saw that if you swap the numbers, the answer stays the same, so the operation is commutative. For part (iii), we found the answers for two pairs and then combined them.

🎯 Exam Tip: For problems involving operation tables, always carefully trace the row and column intersection to find the correct value for each sub-operation. Remember that associativity (checking if \( (a*b)*c = a*(b*c) \)) requires specific calculations, while commutativity can often be observed by symmetry.

 

Question 5. Let * be defined on the set A ={ 1, 2, 3, 4, 5, 6} by a* b = max. {a, b}.
(i) Is * a binary operation ?
(ii) Is * a commutative ?
(iii) Find the identity element.
(iv) Find invertible elements with respect to * and find their inverses.
Answer: The set is \( A = \{1, 2, 3, 4, 5, 6\} \), and the operation \( a * b \) finds the maximum value between \( a \) and \( b \). The operation table is constructed by finding the larger of the two numbers.

(i) Since all entries in the composition table belong to set A (i.e., the maximum of any two numbers in the set is also in the set), * is a binary operation on A.
(ii) The composition table is symmetrical about its principal diagonal. This means that \( a * b = b * a \) for all elements. Therefore, * is commutative on A.
(iii) An identity element \( e \) is an element such that \( a * e = a \) and \( e * a = a \) for all \( a \) in the set.
For the maximum operation, the smallest element in the set, which is 1, acts as the identity element.
When we take \( max\{a, 1\} \), the result is \( a \).
From the table, the row headed by 1 is the same as the top-most row, and the column headed by 1 is the same as the left-most column. Thus, 1 is the identity element.
(iv) An element \( a \) is invertible if there exists an element \( b \) (its inverse) such that \( a * b = e \) (the identity element).
Here, the identity element is 1.
For an element \( a \) to have an inverse \( b \), we need \( max\{a, b\} = 1 \).
This condition can only be met if \( a = 1 \) and \( b = 1 \). If \( a \) is any number greater than 1, \( max\{a, b\} \) will be at least \( a \), and thus greater than 1.
So, only 1 is an invertible element, and its inverse is 1 (since \( 1 * 1 = max\{1, 1\} = 1 \)).
All other rows and columns (except the first row and column) do not contain the identity element 1. So, all other elements are not invertible.In simple words: We made a table where each box shows the bigger number. This is a binary operation because all answers are in the original list. It is also commutative because the table looks balanced. The number 1 is the special "identity" element because finding the maximum of any number and 1 just gives that number back. Only the number 1 has an inverse, which is 1 itself, because only \( max\{1,1\} \) equals 1.

🎯 Exam Tip: For the 'max' operation, the smallest element in the set is always the identity element, and only the identity element itself is invertible. For 'min' operation, the largest element is the identity element, and only the identity element is invertible.

 

Question 6. A binary operation * is defined on the set {0, 1, 2, 3, 4, 5, 6} as \( a * b = \left\{\begin{array}{cc} a+b & \text { if } a+b<7 \\ a+b-7 & \text { if } a+b \geq 7 \end{array}\right. \) Write the composition table of operation. Is it a binary operation? Is it commutative? Prove that zero is the identity for this operation and each element a* 0 of the set is invertible with 7 - a being inverse of a.
Answer: The set is \( A = \{0, 1, 2, 3, 4, 5, 6\} \). The operation is like addition modulo 7, but written slightly differently. It adds two numbers: if the sum is less than 7, that's the result; if the sum is 7 or more, it subtracts 7 from the sum. This keeps the result within the set \( \{0, 1, 2, 3, 4, 5, 6\} \). This type of operation is very common in abstract algebra.

(i) All elements in the binary operation table belong to set A. Thus, * is a binary operation on A.
(ii) The composition table is symmetrical about the principal diagonal. Thus, * is commutative on A.
(iii) The row headed by 0 is identical to the top-most row, and the column headed by 0 is identical to the left-most column. Thus, 0 is the identity element.
(iv) For every row and column, the identity element 0 is present. This means every element is invertible. Let \( b \) be the inverse element of \( a \).
Then \( a * b = 0 = b * a \).
We check for \( a * (7 - a) \):
\( a * (7 - a) = a + (7 - a) - 7 \) (since \( a + (7 - a) = 7 \ge 7 \))
\( = 7 - 7 = 0 \)
We check for \( (7 - a) * a \):
\( (7 - a) * a = (7 - a) + a - 7 \) (since \( (7 - a) + a = 7 \ge 7 \))
\( = 7 - 7 = 0 \)
Thus, \( (7 - a) \) is the inverse of \( a \).In simple words: We have an operation that adds numbers and then subtracts 7 if the sum is too big, to keep results between 0 and 6. All results are in the set, so it's a binary operation. The table is symmetrical, meaning the order of numbers doesn't matter (it's commutative). The number 0 is special because adding 0 doesn't change a number. Every number has an inverse; if you have 'a', its inverse is '7-a', because when you combine them, you get 0.

🎯 Exam Tip: Operations like \( a * b = (a+b) \pmod{n} \) are fundamental examples of binary operations. To prove identity element \( e \), show \( a*e = a \) and \( e*a = a \). To prove inverse, show \( a*b = e \) and \( b*a = e \).

Examples

 

Question 1. If a relation in R in a set A is reflexive symmetric and transitive, then t is called an ............ relation.
Answer: If a relation R in a set A is reflexive, symmetric, and transitive, then it is called an equivalence relation. Equivalence relations are very important because they help group elements that are "equivalent" in some way.
In simple words: When a relation follows all three rules-reflexive, symmetric, and transitive-it is called an equivalence relation.

🎯 Exam Tip: Remember the three properties: reflexive (related to itself), symmetric (if A is related to B, B is related to A), and transitive (if A is related to B and B to C, then A is related to C). All three must hold for an equivalence relation.

 

Question 2. A is called transitive if (a, b) ∈ R and (b, c) ∈ R \( \implies \) ............ for all a, b, c
Answer: A relation R is called transitive if, for all elements a, b, and c in the set: if \( (a, b) \in R \) and \( (b, c) \in R \),
\( \implies \) then \( (a, c) \in R \). This means if the first element is related to the second, and the second is related to the third, then the first must also be related to the third.
In simple words: If A is linked to B, and B is linked to C, then A must also be linked to C for the relation to be transitive.

🎯 Exam Tip: Transitivity is crucial for many mathematical structures. When checking for transitivity, ensure you consider all possible combinations of three elements where the condition (a,b) and (b,c) exist.

 

Question 3. Let the relation R be defined is N by a R b, if 2a + 3b = 30, then R = ...................
Answer: The relation R is defined on the set of natural numbers N such that \( a R b \) if \( 2a + 3b = 30 \). We need to find all pairs \( (a, b) \) from natural numbers (1, 2, 3, ...) that satisfy this equation. We can rewrite the equation to solve for \( b \):
\( 3b = 30 - 2a \)
\( \implies b = \frac{30 - 2a}{3} \)
Now, we check values for \( a \) from the natural numbers, and for each \( a \), we find \( b \). Both \( a \) and \( b \) must be natural numbers.
When \( a = 1 \implies b = \frac{30 - 2(1)}{3} = \frac{28}{3} \notin N \).
When \( a = 2 \implies b = \frac{30 - 2(2)}{3} = \frac{26}{3} \notin N \).
When \( a = 3 \implies b = \frac{30 - 2(3)}{3} = \frac{24}{3} = 8 \in N \). So, \( (3, 8) \in R \).
When \( a = 4 \implies b = \frac{30 - 2(4)}{3} = \frac{22}{3} \notin N \).
When \( a = 5 \implies b = \frac{30 - 2(5)}{3} = \frac{20}{3} \notin N \).
When \( a = 6 \implies b = \frac{30 - 2(6)}{3} = \frac{18}{3} = 6 \in N \). So, \( (6, 6) \in R \).
When \( a = 7 \implies b = \frac{30 - 2(7)}{3} = \frac{16}{3} \notin N \).
When \( a = 8 \implies b = \frac{30 - 2(8)}{3} = \frac{14}{3} \notin N \).
When \( a = 9 \implies b = \frac{30 - 2(9)}{3} = \frac{12}{3} = 4 \in N \). So, \( (9, 4) \in R \).
When \( a = 10 \implies b = \frac{30 - 2(10)}{3} = \frac{10}{3} \notin N \).
When \( a = 11 \implies b = \frac{30 - 2(11)}{3} = \frac{8}{3} \notin N \).
When \( a = 12 \implies b = \frac{30 - 2(12)}{3} = \frac{6}{3} = 2 \in N \). So, \( (12, 2) \in R \).
When \( a = 13 \implies b = \frac{30 - 2(13)}{3} = \frac{4}{3} \notin N \).
When \( a = 14 \implies b = \frac{30 - 2(14)}{3} = \frac{2}{3} \notin N \).
For any other values of \( a \in N \), if \( a \) is larger than 14, then \( 30 - 2a \) will be negative, making \( b \) negative, which is not a natural number.
Thus, the relation \( R \) is: \( R = \{(3, 8), (6, 6), (9, 4), (12, 2)\} \).In simple words: We looked for pairs of natural numbers \( (a, b) \) that fit the rule \( 2a + 3b = 30 \). We listed all the pairs where both \( a \) and \( b \) are whole numbers greater than zero.

🎯 Exam Tip: When defining a relation on natural numbers, always ensure both elements in each pair \( (a, b) \) are positive integers. Carefully test values and remember to stop when \( b \) becomes zero or negative, as these are not natural numbers.

 

Question 4. Every function is ............ invertible. Only ............ functions are invertible
Answer: Not every function is invertible. Only bijective functions are invertible. A function is bijective if it is both one-to-one (injective, meaning each output comes from a unique input) and onto (surjective, meaning every possible output is achieved by at least one input).
In simple words: Not all functions can be reversed. Only special functions that are "one-to-one" (each input gives a different output) and "onto" (all possible outputs are used) can be reversed. These are called bijective functions.

🎯 Exam Tip: To be invertible, a function must pass both the horizontal line test (for one-to-one) and map its entire domain onto its codomain (for onto). This ensures a unique inverse function exists.

 

Question 5. If f : R → R is defined by f(x) = 2x + 3, then f¯¹ (x) is given by ....................
Answer: We are given the function \( f: R \to R \) defined by \( f(x) = 2x + 3 \). To find the inverse function \( f^{-1}(x) \), we follow these steps:
1. Let \( y = f(x) \), so \( y = 2x + 3 \).
2. Swap \( x \) and \( y \) to represent the inverse relationship: \( x = 2y + 3 \).
3. Solve this new equation for \( y \).
\( x = 2y + 3 \)
\( \implies x - 3 = 2y \)
\( \implies y = \frac{x - 3}{2} \)
4. This \( y \) is \( f^{-1}(x) \).
So, \( f^{-1}(x) = \frac{x - 3}{2} \). This type of linear function is always invertible.In simple words: We have a function that multiplies by 2 and adds 3. To find its opposite (inverse), we swap \( x \) and \( y \) and then solve for \( y \). The inverse function is then found by subtracting 3 and dividing by 2.

🎯 Exam Tip: The key steps to finding an inverse function are: (1) set \( y=f(x) \), (2) swap \( x \) and \( y \), and (3) solve for \( y \). Always check if the original function is one-to-one and onto to ensure an inverse exists.

 

Question 6. If f(x) = (x + 3), x ∈ R and g(x) = x - 3, x ∈ R, then fog (3) = ....................
Answer: We are given two functions: \( f(x) = x + 3 \) and \( g(x) = x - 3 \). We need to find the value of the composite function \( (fog)(3) \). The composition \( fog(x) \) means applying \( g \) first, then applying \( f \) to the result.
\( (fog)(x) = f(g(x)) \)
First, find \( g(3) \):
\( g(3) = 3 - 3 = 0 \)
Now, substitute this result into \( f(x) \):
\( (fog)(3) = f(g(3)) = f(0) \)
\( f(0) = 0 + 3 = 3 \)
So, \( (fog)(3) = 3 \). This shows how one function's output becomes the other's input.In simple words: We have two functions. First, we put 3 into the \( g \) function, which gives us 0. Then, we put that 0 into the \( f \) function, which gives us 3. So, the answer is 3.

🎯 Exam Tip: Remember that \( (fog)(x) = f(g(x)) \). Always evaluate the inner function first (g(x) in this case) and then use that result as the input for the outer function (f(x)).

 

Question 7. If f : A→B and g : B ← C be the bijective functions, then (gof)-¹ is ....................
Answer: We are given two bijective functions: \( f: A \to B \) and \( g: B \to C \). A bijective function is both one-to-one and onto, meaning it has an inverse.
The composite function \( gof \) maps from A to C: \( gof: A \to C \). Since both \( f \) and \( g \) are bijective, their composition \( gof \) is also bijective, and therefore \( (gof)^{-1} \) exists.
We know the property for the inverse of a composition of functions: \( (gof)^{-1} = f^{-1}og^{-1} \).
Let's prove this:
Let \( x \in A, y \in B, z \in C \) such that \( f(x) = y \) and \( g(y) = z \).
From \( f(x) = y \implies f^{-1}(y) = x \).
From \( g(y) = z \implies g^{-1}(z) = y \).
Then \( (gof)(x) = g(f(x)) = g(y) = z \).
So, \( (gof)^{-1}(z) = x \) ... (1)
Now consider \( (f^{-1}og^{-1})(z) = f^{-1}(g^{-1}(z)) \).
We know \( g^{-1}(z) = y \).
\( \implies (f^{-1}og^{-1})(z) = f^{-1}(y) \).
We know \( f^{-1}(y) = x \).
\( \implies (f^{-1}og^{-1})(z) = x \) ... (2)
From (1) and (2), we can see that \( (gof)^{-1}(z) = (f^{-1}og^{-1})(z) \) for all \( z \in C \).
Thus, \( (gof)^{-1} = f^{-1}og^{-1} \). This property is very useful for finding inverses of combined functions.In simple words: If you combine two functions and then want to reverse the whole thing, you reverse each function separately and then apply them in the opposite order. So, \( (gof)^{-1} \) is the same as \( f^{-1}og^{-1} \).

🎯 Exam Tip: Remember the "socks and shoes" rule for inverses of compositions: to undo putting on socks then shoes (gof), you must first undo the shoes, then undo the socks \( (f^{-1}og^{-1}) \). Always write the inverse functions in reverse order of their original composition.

 

Question 8. Let the functions f, g, h be defined from R to R such that f (x) = x² – 1, g(x) = \( \sqrt{x^2+1} \) and h(x) = \( \left\{\begin{array}{l} 0, \text { if } x<0 \\ x, \text { if } x>0 \end{array}\right. \) then ho (fog) (x) = ....................
Answer: We are given three functions defined from R to R:
\( f(x) = x^2 - 1 \)
\( g(x) = \sqrt{x^2+1} \)
\( h(x) = \left\{\begin{array}{l} 0, \text { if } x<0 \\ x, \text { if } x>0 \end{array}\right. \)
We need to find the composite function \( ho(fog)(x) \). This means we apply \( g \) first, then \( f \), then \( h \).
First, find \( (fog)(x) = f(g(x)) \):
\( f(g(x)) = f(\sqrt{x^2+1}) \)
\( = (\sqrt{x^2+1})^2 - 1 \)
\( = (x^2 + 1) - 1 \)
\( = x^2 \)
So, \( (fog)(x) = x^2 \).
Next, apply \( h \) to this result: \( ho(fog)(x) = h((fog)(x)) = h(x^2) \).
Now, we use the definition of \( h(x) \). Since \( x^2 \) is always greater than or equal to 0, for any real number \( x \):
If \( x \neq 0 \), then \( x^2 > 0 \), so \( h(x^2) = x^2 \).
If \( x = 0 \), then \( x^2 = 0 \). The function \( h(x) \) is defined for \( x<0 \) and \( x>0 \). If we consider \( h(0) \) to be part of the general pattern for \( x \geq 0 \), then \( h(0)=0 \).
So, \( h(x^2) = x^2 \) for all \( x \in R \) because \( x^2 \geq 0 \).
Therefore, \( ho(fog)(x) = x^2 \). This multi-step composition simplifies quite nicely.In simple words: We combined three functions. First, we put \( x \) into \( g \), then that answer into \( f \). This gave us \( x^2 \). Then, we put \( x^2 \) into \( h \). Since \( x^2 \) is never less than 0, the \( h \) function just gives us \( x^2 \) back. So, the final answer is \( x^2 \).

🎯 Exam Tip: When dealing with piecewise functions like \( h(x) \), pay careful attention to the conditions (e.g., \( x<0 \) vs. \( x>0 \)). The output of an inner function becomes the input for the next, and its value determines which part of the piecewise definition to use.

 

Question 9. Let * be a binary operation on N given by a * b = LCM (a, b) for all a, b ∈ N, then 5 * 7 = ....................
Answer: The binary operation * is defined on the set of natural numbers N, where \( a * b = LCM(a, b) \) (Least Common Multiple of \( a \) and \( b \)). We need to find the value of \( 5 * 7 \).
The numbers 5 and 7 are both prime numbers. The LCM of two distinct prime numbers is simply their product.
\( 5 * 7 = LCM(5, 7) \)
\( = 5 \times 7 \)
\( = 35 \)
So, \( 5 * 7 = 35 \). The LCM is an important concept for combining fractions and understanding number patterns.In simple words: The operation tells us to find the Least Common Multiple (LCM) of the two numbers. For 5 and 7, which are prime numbers, their LCM is just their multiplication, which is 35.

🎯 Exam Tip: Remember that for any two prime numbers \( p \) and \( q \), their LCM is always \( p \times q \). If numbers are not prime, use prime factorization to find their LCM efficiently.

 

Question 10. Let f be the greatest integer function defined as f (x) = [x] and g be the modulus function defined as g (x) = | x |, then the value of gof (-\frac { 4 }{ 3 }) is ....................
Answer: We are given two functions:
\( f(x) = [x] \) (the greatest integer function, which gives the largest integer less than or equal to \( x \))
\( g(x) = |x| \) (the modulus function, which gives the absolute value of \( x \))
We need to find the value of \( gof(-\frac{4}{3}) \). This means we apply \( f \) first, then apply \( g \) to the result.
First, find \( f(-\frac{4}{3}) \):
\( -\frac{4}{3} = -1.333... \)
The greatest integer less than or equal to \( -1.333... \) is \( -2 \).
So, \( f(-\frac{4}{3}) = [-1.333...] = -2 \).
Now, substitute this result into \( g(x) \):
\( gof(-\frac{4}{3}) = g(f(-\frac{4}{3})) = g(-2) \)
\( g(-2) = |-2| \) (the absolute value of -2)
\( = 2 \)
So, \( gof(-\frac{4}{3}) = 2 \). This shows how different types of number functions are evaluated in composition.In simple words: First, we find the greatest whole number that is less than or equal to \( -4/3 \) (which is -1.33). That number is -2. Then, we find the absolute value of -2, which is 2. So the final answer is 2.

🎯 Exam Tip: Be very careful with the greatest integer function \([x]\), especially for negative numbers. \([ -1.3 ]\) is \(-2\), not \(-1\). For \([x]\), always pick the integer to the left or at the point on the number line.

 

Question 11. If R= {(3, 3), (6, 6), (9, 9), (12, 12), (6,12), (3,9), (3,12), (3,6)}, is a relation on the set A = {3, 6, 9, 12}, then the function is
(a) an equivalence relation
(b) reflexive and symmetric
(c) reflexive and transitive
(d) only reflexive
Answer: (c) reflexive and transitive
Here's why:
Given set \( A = \{3, 6, 9, 12\} \).
Given relation \( R = \{(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)\} \).

1. **Reflexivity:** A relation is reflexive if \( (a, a) \in R \) for every \( a \in A \).
We have \( (3, 3), (6, 6), (9, 9), (12, 12) \in R \).
Thus, R is reflexive on A.

2. **Symmetry:** A relation is symmetric if for every \( (a, b) \in R \), we also have \( (b, a) \in R \).
We have \( (6, 12) \in R \), but \( (12, 6) \notin R \).
Therefore, R is not symmetric on A. This means option (a) and (b) are incorrect.

3. **Transitivity:** A relation is transitive if for every \( (a, b) \in R \) and \( (b, c) \in R \), we also have \( (a, c) \in R \).
Let's check the pairs:
- For (3, 6) and (6, 6) in R, we need (3, 6) in R (which is true).
- For (3, 6) and (6, 12) in R, we need (3, 12) in R (which is true).
- For (3, 9) and (9, 9) in R, we need (3, 9) in R (which is true).
- For (6, 6) and (6, 12) in R, we need (6, 12) in R (which is true).
By checking all possible combinations, it is clear that if \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \).
Thus, R is transitive on A.

Since R is reflexive and transitive but not symmetric, option (c) is the correct choice.
In simple words: This relation is reflexive because every number in the set is related to itself. It is transitive because if A is related to B and B to C, then A is also related to C. But it is not symmetric, because for example, 6 is related to 12, but 12 is not related to 6.

🎯 Exam Tip: When testing properties of a relation, systematically check each condition (reflexive, symmetric, transitive) for all elements or pairs. For symmetry, finding just one counterexample is enough to prove it's not symmetric.

 

Question 12. If R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} is a relation on the set A = {1, 2, 3, 4}, then the relation R is
(a) a function (ft) transitive
(c) not symmetric (d) reflexive Solution: Clearly (2, 4), (2, 3) ∈ R i.e. 2 have two images 4 and 3
Answer: (c) not symmetric
Here's why:
Given set \( A = \{1, 2, 3, 4\} \).
Given relation \( R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\} \).

1. **Is it a function?** A relation is a function if each element in the domain maps to exactly one element in the codomain.
From R, we have \( (2, 4) \in R \) and \( (2, 3) \in R \). This means element 2 maps to two different elements (4 and 3).
Therefore, R is not a function. This rules out option (a).

2. **Reflexivity:** A relation is reflexive if \( (a, a) \in R \) for every \( a \in A \).
Elements like \( (1, 1), (2, 2), (3, 3), (4, 4) \) are not present in R.
Therefore, R is not reflexive. This rules out option (d).

3. **Symmetry:** A relation is symmetric if for every \( (a, b) \in R \), we also have \( (b, a) \in R \).
We have \( (2, 3) \in R \), but \( (3, 2) \notin R \).
Therefore, R is not symmetric on A. This makes option (c) correct.

4. **Transitivity:** A relation is transitive if for every \( (a, b) \in R \) and \( (b, c) \in R \), we also have \( (a, c) \in R \).
We have \( (1, 3) \in R \) and \( (3, 1) \in R \), but \( (1, 1) \notin R \).
Therefore, R is not transitive. This means option (a) (transitive part) is also incorrect.

Based on the analysis, the only correct statement about R is that it is not symmetric.
In simple words: First, it's not a function because the number 2 points to two different numbers. It's not reflexive because numbers like (1,1) are missing. It's not symmetric because (2,3) is there but (3,2) is not. And it's not transitive because (1,3) and (3,1) are there, but (1,1) is missing. So, the only true choice is that it's not symmetric.

🎯 Exam Tip: To identify if a relation is *not* a function, look for an element in the domain that has more than one image. For properties like symmetric or transitive, a single counterexample is sufficient to disprove the property.

 

Question 13. The maximum number of equivalence relations on the set A = {1, 2, 3} is
(a) 1
(b) 2
(c) 3
(d) 5
Answer: (d) 5
Here's why:
Given set \( A = \{1, 2, 3\} \). We need to find the number of equivalence relations possible on this set. An equivalence relation must be reflexive, symmetric, and transitive. The number of equivalence relations on a set with \( n \) elements is given by the \( n^{th} \) Bell number, denoted \( B_n \).
For \( n=3 \), the number of equivalence relations is \( B_3 = 5 \).
Let's list them:
1. \( R_1 = \{(1, 1), (2, 2), (3, 3)\} \) (This is the smallest equivalence relation, where each element is related only to itself). This partitions A into {{1}, {2}, {3}}.
2. \( R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\} \) (Relates 1 and 2). This partitions A into {{1, 2}, {3}}.
3. \( R_3 = \{(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)\} \) (Relates 2 and 3). This partitions A into {{1}, {2, 3}}.
4. \( R_4 = \{(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)\} \) (Relates 1 and 3). This partitions A into {{1, 3}, {2}}.
5. \( R_5 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\} \) (This is the universal relation, where all elements are related to each other). This partitions A into {{1, 2, 3}}.

Let's verify these are indeed equivalence relations:
**Reflexivity:** All five relations contain \( (1, 1), (2, 2), (3, 3) \). So, all are reflexive.
**Symmetry:**
- In \( R_1 \), all pairs \( (a, a) \) are symmetric.
- In \( R_2 \), \( (1, 2) \in R_2 \implies (2, 1) \in R_2 \). Symmetric.
- In \( R_3 \), \( (2, 3) \in R_3 \implies (3, 2) \in R_3 \). Symmetric.
- In \( R_4 \), \( (1, 3) \in R_4 \implies (3, 1) \in R_4 \). Symmetric.
- In \( R_5 \), all pairs have their symmetric counterparts. Symmetric.
So, all five relations are symmetric.
**Transitivity:** This property needs to be checked carefully for pairs like \( (a, b) \) and \( (b, c) \).
- \( R_1 \) is transitive.
- In \( R_2 \), if \( (1, 2) \) and \( (2, 1) \) are present, \( (1, 1) \) must be present (it is). If \( (2, 1) \) and \( (1, 2) \) are present, \( (2, 2) \) must be present (it is). Transitive.
- Similarly for \( R_3 \) and \( R_4 \).
- \( R_5 \) (the universal relation) is always transitive because if \( a, b, c \) are all related, then \( a \) is related to \( c \).
So, all five relations are transitive.

Since all five relations satisfy reflexivity, symmetry, and transitivity, there are 5 equivalence relations on set A.
In simple words: An equivalence relation must be reflexive, symmetric, and transitive. For a set with three numbers like {1, 2, 3}, there are 5 possible ways to create relations that follow all these three rules. These different ways are like grouping the numbers together.

🎯 Exam Tip: The number of equivalence relations on a set with \( n \) elements is given by the Bell numbers. For small sets, it's possible to list them by considering partitions of the set. For a set of 3 elements, the partitions are: {{1},{2},{3}}, {{1,2},{3}}, {{1,3},{2}}, {{2,3},{1}}, {{1,2,3}}.

 

Question 14. The relation R to R is defined as a R ft if a > b is
(a) an equivalence relation
(ft) reflexive, transitive but not symmetric
(c) symmetric, transitive but not reflexive
(d) neither reflexive nor transitive but symmetric
Answer: (ft) reflexive, transitive but not symmetric
Here's why:
Given a relation R defined on R (real numbers) such that \( a R b \) if \( a \ge b \).

1. **Reflexivity:** A relation is reflexive if \( a R a \) for all \( a \in R \).
Is \( a \ge a \) true for all real numbers \( a \)? Yes, it is always true.
Thus, R is reflexive.

2. **Symmetry:** A relation is symmetric if for every \( a R b \), we also have \( b R a \).
If \( a \ge b \), does this mean \( b \ge a \)?
For example, if \( a = 5 \) and \( b = 3 \), then \( 5 \ge 3 \) is true. But \( 3 \ge 5 \) is false.
Thus, R is not symmetric.

3. **Transitivity:** A relation is transitive if for every \( a R b \) and \( b R c \), we also have \( a R c \).
If \( a \ge b \) and \( b \ge c \), does this mean \( a \ge c \)? Yes, if \( a \) is greater than or equal to \( b \), and \( b \) is greater than or equal to \( c \), then \( a \) must be greater than or equal to \( c \). This is a fundamental property of inequalities.
Thus, R is transitive.

The relation R is reflexive and transitive but not symmetric. Therefore, option (ft) is the correct choice.
In simple words: The relation "greater than or equal to" is reflexive because any number is greater than or equal to itself. It is transitive because if \( A \ge B \) and \( B \ge C \), then \( A \ge C \). But it is not symmetric because if \( A \ge B \), it does not mean \( B \ge A \) (e.g., \( 5 \ge 3 \) but \( 3 \not\ge 5 \)).

🎯 Exam Tip: The 'greater than or equal to' ( \( \ge \) ) and 'less than or equal to' ( \( \le \) ) relations are classic examples of relations that are reflexive and transitive but not symmetric. Understand why these properties hold and fail.

 

Question 15. Let S denote the set of all straight lines in a plane. Let a relation R be defined by m R n \( \iff \) m \( \perp \) n, m, n ∈ S. Then, R is
(a) reflexive only
(b) symmetric only
(c) transitive only
(d) none of these
Answer: (b) symmetric only
Here's why:
Given set S is the set of all straight lines in a plane.
Relation R is defined by \( m R n \iff m \perp n \) (line \( m \) is perpendicular to line \( n \)).

1. **Reflexivity:** Is \( m \perp m \) true for all lines \( m \in S \)?
A line cannot be perpendicular to itself. If a line is perpendicular to itself, it implies it makes a 90-degree angle with itself, which is not possible.
Therefore, R is not reflexive.

2. **Symmetry:** Is it true that if \( m \perp n \), then \( n \perp m \)?
Yes, if line \( m \) is perpendicular to line \( n \), then line \( n \) is also perpendicular to line \( m \). The relationship is mutual.
Thus, R is symmetric.

3. **Transitivity:** Is it true that if \( m \perp n \) and \( n \perp p \), then \( m \perp p \)?
If \( m \perp n \), then \( m \) and \( n \) form a 90-degree angle.
If \( n \perp p \), then \( n \) and \( p \) form a 90-degree angle.
This implies that \( m \) and \( p \) are both perpendicular to \( n \). When two lines are perpendicular to the same line, they must be parallel to each other (or coincident).
So, if \( m \perp n \) and \( n \perp p \), then \( m \parallel p \) (m is parallel to p), not \( m \perp p \).
For example, let \( m \) be the x-axis, \( n \) be the y-axis, and \( p \) be another line parallel to the x-axis. \( m \perp n \) and \( n \perp p \), but \( m \parallel p \).
Therefore, R is not transitive.

The relation R is symmetric only.
In simple words: This relation means "is perpendicular to". A line cannot be perpendicular to itself, so it's not reflexive. If line A is perpendicular to line B, then line B is also perpendicular to line A, so it IS symmetric. But if line A is perpendicular to line B, and line B is perpendicular to line C, then line A is parallel to line C (not perpendicular), so it's NOT transitive. Therefore, it is only symmetric.

🎯 Exam Tip: Geometric relations like 'perpendicular' or 'parallel' are common examples. Always draw a mental picture or a quick sketch to test reflexivity, symmetry, and transitivity. Remember that if \( m \perp n \) and \( n \perp p \), then \( m \parallel p \).

 

Question 16. If f(x) = 8x³ and g (x) = x1/3, then find (gof) (x).
(a) x
(b) x²
(c) 2x
(d) None of these
Answer: (c) 2x
Here's why:
Given functions:
\( f(x) = 8x^3 \)
\( g(x) = x^{1/3} \) (which is the cube root of x, \( \sqrt[3]{x} \))
We need to find the composite function \( (gof)(x) \), which means \( g(f(x)) \). We substitute \( f(x) \) into \( g(x) \).
\( (gof)(x) = g(f(x)) \)
\( = g(8x^3) \)
Now, apply the definition of \( g(x) \), which is taking the cube root of its input:
\( g(8x^3) = (8x^3)^{1/3} \)
Using the exponent rule \( (ab)^n = a^n b^n \):
\( = 8^{1/3} \cdot (x^3)^{1/3} \)
\( = \sqrt[3]{8} \cdot x^{(3 \cdot 1/3)} \)
\( = 2 \cdot x^1 \)
\( = 2x \)
Therefore, \( (gof)(x) = 2x \). This composition simplifies nicely due to the cube and cube root operations.
In simple words: We put the first function \( (8x^3) \) into the second function \( (x^{1/3}) \). This means we take the cube root of \( 8x^3 \). The cube root of 8 is 2, and the cube root of \( x^3 \) is \( x \). So, the answer is \( 2x \).

🎯 Exam Tip: Remember that \( x^{1/3} \) means the cube root. When dealing with composite functions involving powers and roots, correctly apply exponent rules like \( (ab)^n = a^n b^n \) and \( (x^m)^n = x^{mn} \).

 

Question 18. Let f : N → R be the function defined by \( f(x) = \frac{2 x-1}{2} \) and g : Q → R be another function defined by \( g(x) = x + 2 \). Then \( (\text{gof}) \frac { 3 }{ 2 } \) is
(a) 1
(b) 2
(c) \( \frac { 7 }{ 2 } \)
(d) None of the options
Answer: (d) None of the options
In simple words: We need to calculate the value of the composite function g(f(x)) when x is 3/2. First, find f(3/2) by putting x=3/2 into the f(x) rule. Then, use that result as the input for the g(x) rule to get the final answer.

🎯 Exam Tip: Remember that \( (\text{gof})(x) \) means \( g(f(x)) \). Always evaluate the inner function first, then substitute its result into the outer function.

 

Question 19. If \( f(x) = x^2 - 1 \) and \( g(x) = (x + 1)^2 \), then \( (\text{gof}) x \) is equal to
(a) \( \frac { x-1 }{ x+1 } \)
(b) \( \frac { x+1 }{ x-1 } \)
(c) \( x^4 \)
(d) \( (x + 1)^4 \)
Answer: (c) \( x^4 \)
In simple words: To find the composite function \( (\text{gof})(x) \), we put the entire function \( f(x) \) inside \( g(x) \). This means wherever you see 'x' in \( g(x) \), you replace it with \( f(x) \). Then you simplify the expression.

🎯 Exam Tip: When simplifying composite functions, always expand and combine like terms carefully. Remember the order of operations, especially with exponents and parentheses.

 

Question 20. If \( f : R \rightarrow R \) and \( g : R \rightarrow R \) are defined by \( f(x) = x - 3 \) and \( g(x) = x^2 + 1 \), then the values of x for which \( g(f(x)) = 10 \) are
(a) 0, -6
(b) 2, -2
(c) 1, -1
(d) 0, 6
Answer: (d) 0, 6
In simple words: We are given two functions and an equation involving their combination. First, figure out what the combined function \( g(f(x)) \) looks like. Then, set this expression equal to 10 and solve for x to find the values that make the equation true.

🎯 Exam Tip: When solving equations involving composite functions, first determine the composite function's algebraic expression, then substitute it into the given equation. Don't forget to consider both positive and negative roots when dealing with squared terms.

 

Question 21. If R is the set of real numbers and the functions \( f : R \rightarrow R \) and \( g : R \rightarrow R \) be defined by \( f(x) = x^2 + 2x - 3 \) and \( g(x) = x + 1 \), then the value of x for which \( f(g(x)) = g(f(x)) \) is
(a) -1
(b) 0
(c) 1
(d) 2
Answer: (a) -1
In simple words: We have two functions, f and g. We need to find the value of x where applying g first and then f to x gives the same result as applying f first and then g to x. This involves setting two composite functions equal to each other and solving for x.

🎯 Exam Tip: Carefully compute each composite function, \( f(g(x)) \) and \( g(f(x)) \), before setting them equal. Algebraic mistakes during expansion or simplification are common, so double-check each step.

 

Question 22. Let \( f : R \rightarrow R \) be given by \( f(x) = \tan x \). Then, \( f^{-1}(1) \) is
(a) \( \frac { \pi }{ 4 } \)
(b) \( \left\{n \pi+\frac{\pi}{4}: n \in Z\right\} \)
(c) does not exist
(d) None of the options
Answer: (b) \( \left\{n \pi+\frac{\pi}{4}: n \in Z\right\} \)
In simple words: To find the inverse function value, we ask "for what x is \( \tan x \) equal to 1?". The tangent function repeats its values. So, there isn't just one answer, but a whole series of answers that differ by multiples of pi. We represent this set of answers using 'n' which stands for any integer.

🎯 Exam Tip: For trigonometric inverse functions, remember to provide the general solution unless a specific interval is given. The periodicity of trigonometric functions means there are often infinitely many solutions.

 

Question 23. The binary operation \( * : R \times R \rightarrow R \), is defined as \( a * b = 2a + b \), the \( (2 * 3) * 4 \) is
(a) 24
(b) 9
(c) 56
(d) 18
Answer: (d) 18
In simple words: This question defines a new way to combine two numbers using the star symbol. To solve \( (2 * 3) * 4 \), we first work out \( 2 * 3 \) using the rule, then take that result and combine it with 4 using the same rule.

🎯 Exam Tip: For binary operations with parentheses, always perform the operation inside the parentheses first. Be careful to apply the given rule correctly in each step.

 

Question 24. If \( a * b \) denotes the larger of 'a' and 'b' and if \( a \circ b = (a * b) + 3 \), then \( (5) \circ (10) \) is
(a) 18
(b) 53
(c) 13
(d) None of the options
Answer: (c) 13
In simple words: We have two operations here. First, find the larger number between 5 and 10 using the star operation. Then, take that larger number and add 3 to it according to the circle operation to get the final answer.

🎯 Exam Tip: Pay close attention to the definitions of compound operations. Break down the problem into smaller steps and solve each part according to its specific rule.

 

Question 25. Let \( * \) be a binary operation on set Q of rational number defined \( a \times b = \frac { ab }{ 5 } \). The identity element for \( * \) if any is
(a) 0
(b) 1
(c) 5
(d) None of the options
Answer: (c) 5
In simple words: The identity element 'e' is a special number that, when combined with any other number 'a' using the operation, leaves 'a' unchanged. We use the rule given for the operation to find this 'e'.

🎯 Exam Tip: To find the identity element 'e' for an operation, solve the equation \( a * e = a \) (or \( e * a = a \)) for 'e'. Ensure that the identity element you find actually belongs to the given set.

 

Question 26. If \( * \) binary operation defined on R by \( a * b = 1 + ab, \forall a, b \in R \), then the operation \( '*' \) is
(a) commutative but not associative
(b) associative but not commutative
(c) neither commutative nor associative
(d) both commutative and associative
Answer: (a) commutative but not associative
In simple words: A commutative operation means the order of numbers doesn't matter (a * b is the same as b * a). An associative operation means how you group three numbers doesn't matter ((a * b) * c is the same as a * (b * c)). We test our given rule for both properties.

🎯 Exam Tip: To check for commutativity, compare \( a * b \) with \( b * a \). To check for associativity, compare \( (a * b) * c \) with \( a * (b * c) \). If even one counterexample can be found for a property, then the operation does not have that property.

 

Question 27. Let A be the non-void set of children in a family and the relation 'a is a brother of b' on A is
(a) reflexive
(b) symmetric
(c) transitive
(d) None of the options
Answer: (c) transitive
In simple words: We check if the relation "is a brother of" fits three rules: can someone be their own brother (reflexive)? If A is B's brother, is B also A's brother (symmetric)? If A is B's brother and B is C's brother, is A also C's brother (transitive)? Only the last one holds true.

🎯 Exam Tip: For relation questions, mentally test simple examples for reflexivity, symmetry, and transitivity. A relation involving gender or specific roles often fails symmetry or reflexivity.

 

Question 28. If R is a relation on the set N, defined by \( \{(x, y) : 2x - y = 10\} \), then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) None of the options
Answer: (d) None of the options
In simple words: We test the relation \( y = 2x - 10 \) for the three properties: reflexive, symmetric, and transitive. Since we find that it does not satisfy any of them, we conclude that none of these properties apply to this specific relation. For example, (1,1) is not in R, so it's not reflexive. (6,2) is in R, but (2,6) is not, so it's not symmetric.

🎯 Exam Tip: When checking relation properties, always try to find a counterexample if a property doesn't hold. If the set is infinite (like N for natural numbers), a few examples are usually enough to reveal patterns or non-patterns.

 

Question 29. The number of equivalence relations on the set \( \{1, 2, 3\} \) containing \( (1, 2) \) and \( (2, 1) \) is
(a) 3
(b) 1
(c) 2
(d) None of the options
Answer: (c) 2
In simple words: An equivalence relation must be reflexive, symmetric, and transitive. Given the pairs (1,2) and (2,1), we need to add other pairs to satisfy all three rules. We find that there are two possible ways to do this to create complete equivalence relations.

🎯 Exam Tip: To construct an equivalence relation, start with the given pairs. Then, ensure reflexivity by adding all \( (a, a) \) pairs, ensure symmetry by adding \( (b, a) \) for every \( (a, b) \), and ensure transitivity by adding all necessary \( (a, c) \) pairs if \( (a, b) \) and \( (b, c) \) exist.

 

Question 30. On set \( A = \{1, 2, 3\} \), relations R and S are given by \( R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\} \) and \( S = \{(1, 1), (2, 2), (3, 3), (3, 1)\} \), then
(a) R U S is an equivalence relation.
(b) R U S is reflexive and symmetric but not transitive.
(c) R U S is symmetric and transitive but not reflexive.
(d) R U S is reflexive but neither symmetric nor transitive.
Answer: (d) R U S is reflexive but neither symmetric nor transitive.
In simple words: First, we combine all pairs from relations R and S into a new relation R U S. Then, we check if this new combined relation follows the rules for being reflexive, symmetric, and transitive. We find it is reflexive, but it fails the tests for being symmetric and transitive.

🎯 Exam Tip: When combining relations with union, create the new set of ordered pairs first. Then, systematically check each property (reflexive, symmetric, transitive) for the combined relation, using specific examples if a property fails.

 

Question 31. If \( A = \{1, 2, 3, 4\} \), then which one of the following is reflexive?
(a) \( \{(1, 1), (2, 3), (3, 3)\} \)
(b) \( \{(1, 1), (2, 2), (3, 3), (4, 4)\} \)
(c) \( \{(1, 2), (2, 1), (3, 2), (2, 3)\} \)
(d) \( \{(1, 2), (1, 3), (1, 4)\} \)
Answer: (b) \( \{(1, 1), (2, 2), (3, 3), (4, 4)\} \)
In simple words: A relation is called reflexive if every element in the set is related to itself. This means for a set with elements 1, 2, 3, 4, the relation must contain the pairs (1,1), (2,2), (3,3), and (4,4). Only option (b) has all these self-related pairs.

🎯 Exam Tip: To quickly identify a reflexive relation on a set A, simply look for all pairs of the form \( (a, a) \) where \( a \in A \). If any such pair is missing, the relation is not reflexive.

 

Question 32. Let f be the greatest integer function defined as \( f(x) = [x] \) and g be the modulus function defined as \( g(x) = |x| \), then the value of \( (\text{gof}) (-\frac { 5 }{ 4 }) \) is
(a) 2
(b) -2
(c) \( \frac { 5 }{ 4 } \)
(d) 1
Answer: (a) 2
In simple words: First, we find the greatest integer less than or equal to \( -\frac{5}{4} \). This result is then passed into the modulus function. The modulus function gives the absolute value of a number.

🎯 Exam Tip: When working with the greatest integer function \( [x] \), remember it gives the largest integer that is less than or equal to x. For negative numbers, this means rounding down to the next smaller integer (e.g., \( [-1.25] = -2 \)).

 

Question 33. If \( f : R \rightarrow R \) defined as \( f(x) = \frac{3x+5}{2} \) is an invertible function, write \( f^{-1} \).
(a) \( \frac{2y-3}{2} \)
(b) \( \frac{2y-2}{5} \)
(c) \( \frac{2y+5}{2} \)
(d) \( \frac{2y-5}{3} \)
Answer: (d) \( \frac{2y-5}{3} \)
In simple words: To find the inverse of a function, we swap x and y, and then solve for y. This new y is the inverse function. The inverse function basically "undoes" what the original function did.

🎯 Exam Tip: To find \( f^{-1}(x) \), set \( y = f(x) \), then swap x and y to get \( x = f(y) \). Finally, solve this new equation for y, which will be your \( f^{-1}(x) \).

 

Question 34. If \( f(x) = \frac{x+1}{x-1} \), then the value of \( f(f(x)) \) is equal to
(a) x
(b) 0
(c) -x
(d) 1
Answer: (a) x
In simple words: We need to apply the function \( f \) to its own output. This means we replace 'x' in \( f(x) \) with the entire expression for \( f(x) \) itself. After performing the substitution and simplifying, we find that the original 'x' value is returned.

🎯 Exam Tip: When evaluating composite functions like \( f(f(x)) \), perform the substitution carefully. Simplify the complex fraction by finding a common denominator in the numerator and denominator.

 

Question 35. Let \( X = \{1, 2, 3, 4\} \) and \( Y = \{a, b, c\} \). Then the mapping \( f: X \rightarrow Y \) defined as \( f(1) = a, f(2) = b, f(3) = a, f(4) = b \) is
(a) one-one into
(b) one-one onto
(c) many-one onto
(d) None of the options
Answer: (d) None of the options
In simple words: We check two things: if different inputs always give different outputs (one-one or many-one), and if every output in the target set is used (onto or into). Here, since inputs 1 and 3 both map to 'a', it's many-one. Since 'c' in Y is not an output, it's not onto (it's into). So the function is many-one into.

🎯 Exam Tip: A function is "one-one" (injective) if each element in the codomain is mapped to by at most one element in the domain. A function is "onto" (surjective) if every element in the codomain is mapped to by at least one element in the domain. If it's not onto, it's "into."

 

Question 36. If \( f : R \rightarrow R \) be given by \( f(x) = \tan x \), then \( f^{-1}(1) \) is
(a) \( \frac { \pi }{ 4 } \)
(b) \( n\pi + \frac { \pi }{ 4 }, n \in Z \)
(c) does not exist
(d) None of the options
Answer: (b) \( n\pi + \frac { \pi }{ 4 }, n \in Z \)
In simple words: To find the inverse function value, we ask "for what x is \( \tan x \) equal to 1?". The tangent function repeats its values. So, there isn't just one answer, but a whole series of answers that differ by multiples of pi. We represent this set of answers using 'n' which stands for any integer.

🎯 Exam Tip: For trigonometric inverse functions, remember to provide the general solution unless a specific interval is given. The periodicity of trigonometric functions means there are often infinitely many solutions.

 

Question 37. If \( f(x) = \frac{2x-3}{3x+4} \), then \( f^{-1}(-\frac { 4 }{ 3 }) = \)
(a) 0
(b) \( \frac { 3 }{ 4 } \)
(c) \( -\frac { 2 }{ 3 } \)
(d) None of the options
Answer: (d) None of the options
In simple words: First, we need to find the inverse function \( f^{-1}(y) \) by setting \( y = f(x) \) and solving for x in terms of y. Once we have the inverse function, we substitute \( y = -\frac{4}{3} \) into it to find the specific value.

🎯 Exam Tip: When finding \( f^{-1}(y) \), always swap x and y in the original function and then rearrange the equation to express y as a function of x. Be careful with algebraic manipulation, especially when dealing with fractions.

 

Question 38. Domain of the function \( \frac{1}{\sqrt{x+2}} \) is
(a) \( (-2,\infty) \)
(b) \( (0, \infty) \)
(c) \( (2,\infty) \)
(d) R
Answer: (a) \( (-2,\infty) \)
In simple words: For a function with a square root in the denominator, two rules apply: the part under the square root must be greater than or equal to zero, and the denominator cannot be zero. Combining these, the part under the square root must be strictly greater than zero.

🎯 Exam Tip: When finding the domain of functions involving square roots or fractions, remember two key restrictions: the argument of a square root must be non-negative, and the denominator of a fraction cannot be zero. For \( \frac{1}{\sqrt{g(x)}} \), \( g(x) \) must be strictly greater than zero.

 

Question 39. The domain of the function \( \sin^{-1} 3x \) is
(a) \( \left[-\frac{2}{3}, \frac{2}{3}\right] \)
(b) \( \left[-\frac{1}{3}, \frac{1}{3}\right] \)
(c) \( [-1, 1] \)
(d) \( [-3, 2] \)
Answer: (b) \( \left[-\frac{1}{3}, \frac{1}{3}\right] \)
In simple words: The inverse sine function can only work with values between -1 and 1. So, whatever is inside the \( \sin^{-1} \) must be in that range. We set up an inequality to find the possible values for x.

🎯 Exam Tip: The domain of \( \sin^{-1} u \) is \( [-1, 1] \). Therefore, for \( \sin^{-1} (g(x)) \), you must set \( -1 \leq g(x) \leq 1 \) and solve for x.

 

Question 40. The domain and range of the function \( f(x) = \frac{x}{1-x} \) are respectively.
(a) \( R - \{1\}, R \)
(b) \( R - \{1\}, R - \{-1\} \)
(c) \( R - \{1\}, R - \{1\} \)
(d) \( R, R - \{-1\} \)
Answer: (b) \( R - \{1\}, R - \{-1\} \)
In simple words: For the domain, we just need to make sure the bottom part of the fraction is not zero. For the range, we pretend the function equals 'y' and then swap things around to find out what values 'y' cannot be. This process helps us find all possible output values.

🎯 Exam Tip: To find the domain, exclude any values of x that make the denominator zero or lead to undefined operations like square roots of negative numbers. To find the range, set \( y = f(x) \), solve for x in terms of y, and then identify any values of y that would make x undefined.

 

Question 41. The range of the function \( f(x) = \cos(\frac { x }{ 3 }) \) is
(a) \( [-1, 1] \)
(b) \( \left[-\frac{1}{3}, \frac{1}{3}\right] \)
(c) \( [1,\infty) \)
(d) None of the options
Answer: (a) \( [-1, 1] \)
In simple words: The cosine function, no matter what number you put inside it, will always give an answer between -1 and 1. So, even though x is divided by 3, the range of the function remains the same.

🎯 Exam Tip: The range of basic trigonometric functions like \( \sin(ax+b) \) and \( \cos(ax+b) \) is always \( [-1, 1] \), as long as there are no external multipliers or additions to the function itself.

 

Question 42. Let \( f : R \rightarrow R \) be defined by \( f(x) = \frac { 1 }{ x } \forall x \in R \). Then f is
(a) onto
(b) not defined
(c) one-one
(d) bijective
Answer: (b) not defined
In simple words: The function \( f(x) = \frac{1}{x} \) cannot work if x is zero, because you can't divide by zero. Since the problem says the function is defined for all real numbers (R), but it doesn't work for x=0, the function cannot exist as stated.

🎯 Exam Tip: Always check the domain and codomain of a function definition. If a function is stated to map from R to R but has points where it's undefined (like \( 1/x \) at \( x=0 \)), then the function itself is not well-defined for that entire domain.

 

Question 43. If \( A = \{1, 2, 3, 4\} \), \( B = \{1, 2, 3, 4, 5, 6\} \) are two sets, and function \( f : A \rightarrow B \) is defined by \( f(x) = x + 2 \forall x \in A \), then the function f is
(a) bijective
(b) onto
(c) one-one
(d) many-one
Answer: (c) one-one
In simple words: We check if each input from set A gives a unique output in set B. If so, it's one-one. We also check if every element in set B is an output of some input from set A. If not, it's not onto. This function is one-one because different inputs always produce different outputs, but it's not onto because elements 1 and 2 in set B are not outputs.

🎯 Exam Tip: For finite sets, test one-one by seeing if any two different domain elements map to the same codomain element. Test onto by listing all image elements and checking if they cover the entire codomain. If any element in the codomain is missed, the function is not onto.

 

Question 44. The function \( f : R \rightarrow R \) given by \( f(x) = x^3 - 1 \) is
(a) a one-one function
(b) an onto function
(c) a bijection
(d) neither one-one nor onto
Answer: (c) a bijection
In simple words: A function is "one-one" if every different input gives a different output. It is "onto" if every possible output value is actually reached by some input. This function satisfies both conditions, meaning it's a bijection, a perfect one-to-one correspondence between the domain and the codomain.

🎯 Exam Tip: To check if \( f(x) = x^3 - 1 \) is one-one, assume \( f(a) = f(b) \) and show \( a = b \). To check if it's onto, for any \( y \) in the codomain, find an \( x \) in the domain such that \( f(x) = y \). A function that is both one-one and onto is called a bijection.

 

Question 45. If the binary operation \( * \) on the set of integers Z, is defined by \( a * b = a + 3b^2 \), then find the value \( 2 * 4 \).
(a) 8
(b) 6
(c) 16
(d) 50
Answer: (d) 50
In simple words: This question gives us a specific rule for combining two numbers using the star symbol. We just need to replace 'a' with the first number (2) and 'b' with the second number (4) in the given rule and then calculate the result.

🎯 Exam Tip: Binary operations are simply new rules for combining elements. Always substitute the given values into the definition of the operation precisely and follow the standard order of mathematical operations (PEMDAS/BODMAS).

 

Question 46. State the reason for the relation R on the set \( \{1, 2, 3\} \) given by \( R = \{(1, 2), (2, 1)\} \) not to be transitive.
Answer: A relation R is transitive if, whenever \( (a, b) \in R \) and \( (b, c) \in R \), it implies that \( (a, c) \in R \) for all \( a, b, c \) in the set. For the given relation \( R = \{(1, 2), (2, 1)\} \) on the set \( \{1, 2, 3\} \):
We have \( (1, 2) \in R \) and \( (2, 1) \in R \). According to the transitivity rule, for these two pairs, the pair \( (1, 1) \) must also be in R.
However, \( (1, 1) \notin R \).
Since the condition for transitivity is not met, the relation R is not transitive.
In simple words: Transitivity means if A is related to B, and B is related to C, then A must be related to C. In our relation, 1 is related to 2, and 2 is related back to 1. For it to be transitive, 1 should also be related to 1. Since (1,1) is missing from the relation, it is not transitive.

🎯 Exam Tip: To prove a relation is not transitive, you only need to find one counterexample. Look for pairs \( (a,b) \) and \( (b,c) \) in the relation where the corresponding \( (a,c) \) pair is missing.

 

Question 47. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 5)} be a function from A to B. State whether f is one-one or not.
Answer: The function is given as \( f = \{(1, 4), (2, 5), (3, 5)\} \). For a function to be one-one, every distinct element in set A must map to a distinct element in set B. Here, we can see that \( f(2) = 5 \) and \( f(3) = 5 \). Since two different input values (2 and 3) lead to the same output value (5), the function is not one-one. Functions that are not one-one are sometimes called many-one functions because multiple inputs map to a single output.
In simple words: A function is one-one if each different input gives a different output. Here, both '2' and '3' give the same answer '5', so this function is not one-one.

🎯 Exam Tip: To check if a function is one-one, look for any two distinct inputs that produce the same output. If you find such a pair, the function is not one-one.

 

Question 48. What is the range of the function \( f(x) = \frac{|x-1|}{x-1} \)?
Answer: The function \( f(x) = \frac{|x-1|}{x-1} \) can be understood by looking at two main cases, keeping in mind that \( x \ne 1 \) because of the denominator. First, if \( x > 1 \), then \( (x-1) \) is positive, so \( |x-1| = x-1 \). In this case, \( f(x) = \frac{x-1}{x-1} = 1 \). Second, if \( x < 1 \), then \( (x-1) \) is negative, so \( |x-1| = -(x-1) \). In this case, \( f(x) = \frac{-(x-1)}{x-1} = -1 \). This type of function is an example of a piecewise function, where the rule changes based on the input value. Therefore, the function can only output values of 1 or -1. The range of the function is \( \{-1, 1\} \).
In simple words: This function gives either 1 (when \( x \) is bigger than 1) or -1 (when \( x \) is smaller than 1). So, the only possible answers it can give are -1 and 1.

🎯 Exam Tip: For functions involving absolute values, always consider cases where the expression inside the absolute value is positive, negative, or zero to correctly determine the function's behavior.

 

Question 49. Find the identity element in the set 'A' of all positive rational numbers for the operation defined by \( a * b = \frac{3ab}{2} \) for all \( a, b \in Q-\{0\} \).
Answer: An identity element, let's call it \( e \), for a binary operation is a special value that, when combined with any other element \( a \) from the set, leaves \( a \) unchanged. So, for the operation \( a * b = \frac{3ab}{2} \), we need to find \( e \) such that \( e * a = a \) and \( a * e = a \). Let's use \( e * a = a \):
\( \frac{3ea}{2} = a \)
Now, we multiply both sides by 2 to clear the fraction:
\( 3ea = 2a \)
Since \( a \) is a positive rational number, it's not zero, so we can divide both sides by \( a \):
\( 3e = 2 \)
Finally, divide by 3 to find \( e \):
\( e = \frac{2}{3} \)
The identity element is unique for a given binary operation and set.
In simple words: The identity element is like a "neutral" number for the operation. When you combine it with any number, that number stays the same. For this operation, the identity element is \( \frac{2}{3} \).

🎯 Exam Tip: Remember to solve for 'e' by setting \( e * a = a \) (or \( a * e = a \)). Ensure you mention that \( a \ne 0 \) if you divide by \( a \).

 

Question 50. If \( f : R \rightarrow R \) be given by \( f (x) = (3 – x^3)^{1/3} \), then find \( fof (x) \).
Answer: To find the composition \( fof(x) \), we need to apply the function \( f \) to the result of \( f(x) \). This means we substitute \( f(x) \) into itself. The function is \( f(x) = (3 - x^3)^{1/3} \).
So, \( fof(x) = f(f(x)) \)
Substitute \( f(x) \) into the expression for \( f(x) \):
\( fof(x) = (3 - (f(x))^3)^{1/3} \)
Now, replace \( f(x) \) with its definition \( (3 - x^3)^{1/3} \):
\( fof(x) = (3 - ( (3 - x^3)^{1/3} )^3 )^{1/3} \)
The cube and cube root cancel each other out: \( ( (3 - x^3)^{1/3} )^3 = 3 - x^3 \)
So, the expression becomes:
\( fof(x) = (3 - (3 - x^3))^{1/3} \)
Remove the inner parentheses:
\( fof(x) = (3 - 3 + x^3)^{1/3} \)
Simplify the terms inside:
\( fof(x) = (x^3)^{1/3} \)
Again, the cube and cube root cancel:
\( fof(x) = x \)
When \( fof(x) = x \), it means the function \( f \) is its own inverse.
In simple words: We put the whole function inside itself. The cube root and the power of 3 cancel each other out, leaving just \( x \). This means the function undoes itself.

🎯 Exam Tip: When evaluating composite functions like \( fof(x) \), substitute the entire expression for the inner function into the outer function's variable carefully.

 

Question 51. Let * be an operation defined as \( * = R \times R \rightarrow R \) such that \( a* b = 2a + b \), \( a, b \in R \). Check if * is a binary operation. Find if it is associative also.
Answer: We need to check two properties for the operation \( a * b = 2a + b \): whether it is a binary operation and whether it is associative.
**1. Is it a binary operation?**
A binary operation takes two elements from a set and produces a single element within the same set. Here, the set is R (real numbers). If \( a \) and \( b \) are any real numbers, then \( 2a \) is a real number, and \( 2a + b \) will also always be a real number. Since the result \( (2a + b) \) is always in R, the operation \( * \) is a binary operation on R.
**2. Is it associative?**
An operation is associative if for any elements \( a, b, c \) in the set, \( (a * b) * c = a * (b * c) \). Let's calculate both sides:
**Left side:** \( (a * b) * c \)
First, calculate \( a * b = 2a + b \).
Now, substitute this result into \( (a * b) * c \), treating \( (2a+b) \) as the first element:
\( (2a + b) * c = 2(2a + b) + c \)
\( = 4a + 2b + c \)
**Right side:** \( a * (b * c) \)
First, calculate \( b * c = 2b + c \).
Now, substitute this result into \( a * (b * c) \), treating \( (2b+c) \) as the second element:
\( a * (2b + c) = 2a + (2b + c) \)
\( = 2a + 2b + c \)
Since \( 4a + 2b + c \ne 2a + 2b + c \) (unless \( a=0 \)), the left side is not equal to the right side. For example, let \( a = 1, b = 2, c = 3 \):
\( (1 * 2) * 3 = (2(1) + 2) * 3 = (2 + 2) * 3 = 4 * 3 = 2(4) + 3 = 8 + 3 = 11 \)
\( 1 * (2 * 3) = 1 * (2(2) + 3) = 1 * (4 + 3) = 1 * 7 = 2(1) + 7 = 2 + 7 = 9 \)
Since \( 11 \ne 9 \), the operation \( * \) is not associative. Associativity is an important property that allows us to group operations differently without changing the result.
In simple words: This operation always gives a real number, so it is a binary operation. But it is not associative because the order in which we do the operations changes the final answer.

🎯 Exam Tip: To prove an operation is not associative, a single counterexample with specific numbers is sufficient and often easier than general algebraic proof.

 

Question 52. Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by \( a * b = \min \{a, b\} \). Write the operation table of operation for the operation *.
Answer: The operation \( a * b = \min \{a, b\} \) means that when you combine two numbers, the result is the smaller of the two numbers. We need to create a table showing the result for every possible pair of numbers from the set \( A = \{1, 2, 3, 4, 5\} \). This table will have rows and columns labeled by the elements of the set. This type of operation is often used in discrete mathematics and logic, representing "AND" like behavior.
The operation table is given as under:

In simple words: This table shows the smaller number for each pair. For example, if you combine 3 and 5, the answer is 3 because 3 is smaller than 5.

🎯 Exam Tip: When constructing an operation table, ensure that each cell represents the result of the operation on its corresponding row and column elements exactly as defined.

 

Question 53. Let * be a binary operation on N given by \( a * b = HCF (a, b) \), \( \forall a, b \in N \). Write the value of 22 * 4.
Answer: The operation \( a * b \) is defined as finding the Highest Common Factor (HCF) of \( a \) and \( b \). We need to find the value of \( 22 * 4 \), which means we need to find the HCF of 22 and 4. To do this, we list the factors of each number:
Factors of 22: 1, 2, 11, 22
Factors of 4: 1, 2, 4
The common factors are 1 and 2. The highest of these common factors is 2. So, HCF(22, 4) = 2. HCF is useful for simplifying fractions and finding common multiples in number theory.
Therefore, \( 22 * 4 = 2 \).
In simple words: The HCF of 22 and 4 is the largest number that divides both 22 and 4. That number is 2.

🎯 Exam Tip: Always list all factors for both numbers to accurately identify the Highest Common Factor, especially for smaller numbers.

 

Question 54. Let {a, b, c} and the relation R be defined as R = {{a, a), (b, c), (a, b)}. Write the minimum number of ordered pairs which should be added to R to make R reflexive and transitive.
Answer: We are given the set \( A = \{a, b, c\} \) and the relation \( R = \{(a, a), (b, c), (a, b)\} \). We need to make R both reflexive and transitive by adding the fewest possible ordered pairs. Understanding these properties is fundamental for classifying different types of relations, like equivalence relations.
**1. For R to be reflexive:**
A relation R on set A is reflexive if every element in A is related to itself. This means for every \( x \in A \), the pair \( (x, x) \) must be in R. Our set A has elements \( a, b, c \).
Currently, R contains \( (a, a) \).
It is missing \( (b, b) \) and \( (c, c) \).
So, we must add \( (b, b) \) and \( (c, c) \) to R.
Now, \( R = \{(a, a), (b, c), (a, b), (b, b), (c, c)\} \).
**2. For R to be transitive:**
A relation R is transitive if whenever \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \) must also be in R.
Let's check the current R: \( \{(a, a), (b, c), (a, b), (b, b), (c, c)\} \).
- Consider \( (a, b) \in R \) and \( (b, c) \in R \). For transitivity, \( (a, c) \) must be in R. It is currently missing. So, we must add \( (a, c) \).
- If we consider \( (a, a) \) and \( (a, b) \), we need \( (a, b) \) in R, which is already there.
- If we consider \( (b, b) \) and \( (b, c) \), we need \( (b, c) \) in R, which is already there.
- If we consider \( (c, c) \) and no pair starting with \( c \) (other than \( (c,c) \)), then no further transitivity check is needed for \( (c,c) \).
- The pair \( (a, b) \) and \( (b, b) \) would imply \( (a, b) \) which is present.
After adding \( (a, c) \), the relation becomes \( R = \{(a, a), (b, c), (a, b), (b, b), (c, c), (a, c)\} \). All conditions for transitivity are now met.
The minimum number of ordered pairs to add is 3: \( (b, b), (c, c), \) and \( (a, c) \).
In simple words: We need to add pairs so that every element is related to itself (reflexive) and also if A is related to B and B to C, then A must be related to C (transitive). We add \( (b, b) \), \( (c, c) \) for reflexivity, and then \( (a, c) \) for transitivity.

🎯 Exam Tip: When checking for transitivity, systematically list all pairs \( (x,y) \) and \( (y,z) \) in the relation, and then verify if \( (x,z) \) is also present or needs to be added.

 

Question 55. Let \( A = \{a, b, c, d\} \) and \( f : A \rightarrow A \) be given by \( f = \{(a, b), (b, d), (c, a), (d, c)\} \), write \( f^{-1} \).
Answer: We are given the set \( A = \{a, b, c, d\} \) and the function \( f: A \rightarrow A \) defined by \( f = \{(a, b), (b, d), (c, a), (d, c)\} \). This means that \( f(a) = b, f(b) = d, f(c) = a, \) and \( f(d) = c \). To find the inverse function, \( f^{-1} \), we simply reverse the mapping for each pair. If \( f(x) = y \), then \( f^{-1}(y) = x \). An inverse function only exists if the original function is bijective (both one-to-one and onto). In this case, each element in A maps to a unique element in A, and every element in A has a pre-image, so \( f \) is bijective and its inverse exists.
We just swap the elements in each ordered pair to find \( f^{-1} \):
Original pairs for \( f \):
\( (a, b) \)
\( (b, d) \)
\( (c, a) \)
\( (d, c) \)
Inverse pairs for \( f^{-1} \):
\( (b, a) \)
\( (d, b) \)
\( (a, c) \)
\( (c, d) \)
So, \( f^{-1} = \{(b, a), (d, b), (a, c), (c, d)\} \).
In simple words: To find the inverse function, you just swap the first and second items in each pair. For example, if \( f \) maps \( a \) to \( b \), then \( f^{-1} \) maps \( b \) back to \( a \).

🎯 Exam Tip: To find the inverse of a function given as a set of ordered pairs, simply reverse the order of elements in each pair.

 

Question 56. If \( f : R - \{\frac{3}{5}\} \rightarrow R \) be defined as \( f(x) = \frac{3x-5}{5x-3} \) then show that \( f^{-1} (x) = f(x) \).
Answer: To show that \( f^{-1}(x) = f(x) \), we first need to find the inverse function \( f^{-1}(x) \). We do this by setting \( y = f(x) \), and then solving for \( x \) in terms of \( y \). The process of finding an inverse function essentially "undoes" the original function's operations in reverse order.
Let \( y = f(x) \), so \( y = \frac{3x-5}{5x-3} \)
Multiply both sides by \( (5x-3) \):
\( y(5x-3) = 3x-5 \)
Distribute \( y \) on the left side:
\( 5xy - 3y = 3x - 5 \)
Now, we want to isolate \( x \) terms on one side and \( y \) terms on the other. Move all terms with \( x \) to the left side and constant \( y \) terms to the right:
\( 5xy - 3x = 3y - 5 \)
Factor out \( x \) from the terms on the left side:
\( x(5y - 3) = 3y - 5 \)
Divide by \( (5y - 3) \) to solve for \( x \):
\( x = \frac{3y-5}{5y-3} \)
Since \( x = f^{-1}(y) \), we have:
\( f^{-1}(y) = \frac{3y-5}{5y-3} \)
Finally, to express \( f^{-1}(x) \), we replace \( y \) with \( x \):
\( f^{-1}(x) = \frac{3x-5}{5x-3} \)
By comparing this result with the original function \( f(x) = \frac{3x-5}{5x-3} \), we can see that \( f^{-1}(x) \) is indeed equal to \( f(x) \).
In simple words: To find the inverse, we swap \( x \) and \( y \) and then solve for \( y \). After doing all the steps, we find that the inverse function is exactly the same as the original function.

🎯 Exam Tip: When proving \( f^{-1}(x) = f(x) \), clearly show all algebraic steps involved in finding the inverse function. This property indicates a special symmetry for the function.

 

Question 57. If \( f(x) = 4 – (x – 7)^3 \), write \( f^{-1} (x) \).
Answer: To find the inverse function \( f^{-1}(x) \), we follow these steps: first, set \( y = f(x) \), then rearrange the equation to solve for \( x \) in terms of \( y \). Finally, replace \( y \) with \( x \) to get the inverse function. Functions that are their own inverse are special and have a symmetrical property, often seen in transformations like reflections.
Let \( y = f(x) \):
\( y = 4 - (x - 7)^3 \)
Now, we want to isolate \( (x-7)^3 \). Subtract 4 from both sides:
\( y - 4 = - (x - 7)^3 \)
Multiply both sides by -1 to remove the negative sign:
\( 4 - y = (x - 7)^3 \)
To get rid of the cube, take the cube root of both sides:
\( (4 - y)^{1/3} = x - 7 \)
Finally, add 7 to both sides to solve for \( x \):
\( x = 7 + (4 - y)^{1/3} \)
Since \( x \) is the inverse function in terms of \( y \), we can write \( f^{-1}(y) = 7 + (4 - y)^{1/3} \).
To express the inverse function in terms of \( x \), we replace \( y \) with \( x \):
\( f^{-1}(x) = 7 + (4 - x)^{1/3} \)
In simple words: We change \( f(x) \) to \( y \), then move things around to find \( x \). We subtract 4, change the sign, then take the cube root, and finally add 7. This gives us the inverse function.

🎯 Exam Tip: When finding inverse functions involving powers, remember to use the corresponding root (e.g., cube root for a cube) to isolate the variable correctly.