Samacheer Kalvi Class 11 Maths Solutions Chapter 1 Sets, Relations and Functions Exercise 1.5

Question 1. If A = \( \{(x,y): y = e^x, x \in R\} \) and B = \( \{(x, y): y = e^{-x}, x \in R\} \), then \( n(A \cap B) \) is
(a) Infinity
(b) 0
(c) 1
(d) 2
Answer: (c) 1
Consider the two curves: \( y = e^x \) and \( y = e^{-x} \). To find the intersection points, we set them equal: \( e^x = e^{-x} \). This can be written as \( e^x = \frac{1}{e^x} \). Multiplying both sides by \( e^x \) gives \( (e^x)^2 = 1 \). Taking the square root, \( e^x = \pm 1 \). Since \( e^x \) is always positive, we must have \( e^x = 1 \). This occurs only when \( x = 0 \). If \( x = 0 \), then \( y = e^0 = 1 \). So, the only point of intersection is \( (0, 1) \). The number of elements in the intersection set \( A \cap B \) is 1. Graphically, these exponential functions grow in opposite directions but always meet at one unique point.

In simple words: The two curves \( y = e^x \) and \( y = e^{-x} \) cross each other at only one point, which is \( (0, 1) \). So, there is just one element in their intersection.

🎯 Exam Tip: To find the intersection of two functions, set their expressions equal to each other and solve for the variable. Then find the corresponding y-value.

Question 2. If A = \( \{ ( x, y): y = \sin x, x \in R \} \) and B = \( \{ ( x, y): y = \cos x, x \in R\} \) then \( A \cap B \) contains
(a) no element
(b) infinitely many elements
(c) only one element
(d) cannot be determined
Answer: (b) infinitely many elements
To find the common points for sets A and B, we need to find where \( \sin x = \cos x \). This equation is true when \( \frac{\sin x}{\cos x} = 1 \), which simplifies to \( \tan x = 1 \). The general solution for \( \tan x = 1 \) is \( x = n\pi + \frac{\pi}{4} \), where \( n \) is any integer. For each value of \( n \), we get a different \( x \)-value, and thus a different point where \( \sin x = \cos x \). For example, when \( n=0 \), \( x = \frac{\pi}{4} \), then \( y = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \). When \( n=1 \), \( x = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \), then \( y = \sin(\frac{5\pi}{4}) = -\frac{1}{\sqrt{2}} \). Since there are infinitely many integer values for \( n \), there are infinitely many points where the sine and cosine graphs intersect. This means the intersection \( A \cap B \) has infinitely many elements.
In simple words: The graphs of \( y = \sin x \) and \( y = \cos x \) cross each other an endless number of times. Each time they cross, it means they share a common point. So, there are infinite common elements.

🎯 Exam Tip: Remember that trigonometric functions like sine and cosine are periodic, meaning their values repeat. This periodicity often leads to infinitely many solutions when setting them equal.

Question 3. The relation R defined on a set A = \( \{0, -1, 1, 2\} \) by \( x R y \) if \( |x^2 + y^2| \leq 2 \), then which one of the following is true.
(a) R = \( \{(0,0), (0,-1), (0,1), (-1,0), (-1,1), (1,2), (1,0)\} \)
(b) \( R^{-1} = \{(0, 0), (0, -1), (0, 1), (-1, 0), (1, 0)\} \)
(c) Domain of R is \( \{0,- 1, 1, 2\} \)
(d) Range of R is \( \{0, -1, 1\} \)
Answer: (d) Range of R is \( \{0, -1, 1\} \)
The set A is \( \{0, -1, 1, 2\} \). The relation is defined as \( x R y \) if \( |x^2 + y^2| \leq 2 \), where \( x, y \in A \). We need to find the range of R, which consists of all possible \( y \)-values from the set A that satisfy the relation. Let's check which values of \( y \) from A can be paired with any \( x \) from A such that \( |x^2 + y^2| \leq 2 \).
If \( y = 0 \): We can have \( (0,0) \) since \( |0^2 + 0^2| = 0 \leq 2 \). We can also have \( (-1,0) \) since \( |(-1)^2 + 0^2| = 1 \leq 2 \). Similarly \( (1,0) \).
If \( y = -1 \): We can have \( (0,-1) \) since \( |0^2 + (-1)^2| = 1 \leq 2 \). We can also have \( (-1,-1) \) since \( |(-1)^2 + (-1)^2| = 2 \leq 2 \). Similarly \( (1,-1) \).
If \( y = 1 \): We can have \( (0,1) \) since \( |0^2 + 1^2| = 1 \leq 2 \). We can also have \( (-1,1) \) since \( |(-1)^2 + 1^2| = 2 \leq 2 \). Similarly \( (1,1) \).
If \( y = 2 \): We need \( |x^2 + 2^2| \leq 2 \), so \( |x^2 + 4| \leq 2 \). This implies \( x^2 + 4 \leq 2 \), which means \( x^2 \leq -2 \). There is no real number \( x \) for which \( x^2 \) is negative, let alone less than or equal to -2. So, \( y=2 \) is not in the range. The possible values for \( y \) that satisfy the condition are \( \{0, -1, 1\} \). These y-values lead to valid pairs within the relation, forming the range. This process shows how to systematically check each potential element.
In simple words: The relation means that when you square \( x \) and \( y \) from the set A, add them, and take the absolute value, the result must be 2 or less. By checking each number in A for \( y \), we find that only 0, -1, and 1 work as outputs, making them the range.

🎯 Exam Tip: To find the range of a relation, test each element in the codomain (or the given set, if the codomain isn't specified) to see if it can be an 'output' (y-value) for at least one 'input' (x-value) that satisfies the relation's rule.

Question 4. If \( f(x) = |x - 2 | + |x + 2| \), \( x \in R \), then \( f(x) \) is
(a) \( f(x) = \begin{cases} -2x & \text{if } x \in (-\infty, -2] \\ 4 & \text{if } x \in (-2, 2] \\ 2x & \text{if } x \in (2, \infty) \end{cases} \)
(b) \( f(x) = \begin{cases} 2x & \text{if } x \in (-\infty, -2] \\ 4x & \text{if } x \in (-2, 2] \\ -2x & \text{if } x \in (2, \infty) \end{cases} \)
(c) \( f(x) = \begin{cases} -2x & \text{if } x \in (-\infty, -2] \\ -4x & \text{if } x \in (-2, 2] \\ 2x & \text{if } x \in (2, \infty) \end{cases} \)
(d) \( f(x) = \begin{cases} -2x & \text{if } x \in (-\infty, -2] \\ 2x & \text{if } x \in (-2, 2] \\ 2x & \text{if } x \in (2, \infty) \end{cases} \)
Answer: (a) \( f(x) = \begin{cases} -2x & \text{if } x \in (-\infty, -2] \\ 4 & \text{if } x \in (-2, 2] \\ 2x & \text{if } x \in (2, \infty) \end{cases} \)
To define \( f(x) = |x - 2| + |x + 2| \) without absolute values, we need to consider the critical points where the expressions inside the absolute values become zero. These points are \( x - 2 = 0 \implies x = 2 \) and \( x + 2 = 0 \implies x = -2 \). These points divide the real number line into three intervals: \( (-\infty, -2] \), \( (-2, 2] \), and \( (2, \infty) \).

**Case 1: \( x \in (-\infty, -2] \)**
In this interval, \( x \) is less than or equal to -2. So, \( x - 2 \) is negative (e.g., if \( x = -3 \), \( x - 2 = -5 \)) and \( x + 2 \) is also negative (e.g., if \( x = -3 \), \( x + 2 = -1 \)).
Therefore, \( |x - 2| = -(x - 2) = -x + 2 \), and \( |x + 2| = -(x + 2) = -x - 2 \).
Then \( f(x) = (-x + 2) + (-x - 2) = -x + 2 - x - 2 = -2x \).

**Case 2: \( x \in (-2, 2] \)**
In this interval, \( x \) is greater than -2 and less than or equal to 2. So, \( x - 2 \) is negative (e.g., if \( x = 0 \), \( x - 2 = -2 \)) and \( x + 2 \) is positive (e.g., if \( x = 0 \), \( x + 2 = 2 \)).
Therefore, \( |x - 2| = -(x - 2) = -x + 2 \), and \( |x + 2| = x + 2 \).
Then \( f(x) = (-x + 2) + (x + 2) = -x + 2 + x + 2 = 4 \).

**Case 3: \( x \in (2, \infty) \)**
In this interval, \( x \) is greater than 2. So, \( x - 2 \) is positive (e.g., if \( x = 3 \), \( x - 2 = 1 \)) and \( x + 2 \) is also positive (e.g., if \( x = 3 \), \( x + 2 = 5 \)).
Therefore, \( |x - 2| = x - 2 \), and \( |x + 2| = x + 2 \).
Then \( f(x) = (x - 2) + (x + 2) = x - 2 + x + 2 = 2x \).

Combining these cases, the piecewise definition of \( f(x) \) is: \[ f(x) = \begin{cases} -2x & \text{if } x \in (-\infty, -2] \\ 4 & \text{if } x \in (-2, 2] \\ 2x & \text{if } x \in (2, \infty) \end{cases} \] This method of breaking down the function by intervals where absolute value expressions change sign is fundamental in calculus. The key is to correctly determine the sign of the expressions inside the absolute values for each interval.
In simple words: To remove the absolute value signs, we look at where \( x \) is on the number line. We split the number line into three parts based on where \( x-2 \) and \( x+2 \) become zero. Then, in each part, we simplify the expression. This gives us three different rules for \( f(x) \) depending on the value of \( x \).

🎯 Exam Tip: When dealing with functions involving multiple absolute values, always identify the critical points where each absolute value expression becomes zero. These points divide the real line into intervals, and you must analyze the function's behavior in each interval separately.

Question 5. Let R be the set of all real numbers. Let T = \( \{(x, y): x - y \text{ is an integer}\} \). Then which of the following is true?
(a) T is an equivalence relation but S is not an equivalence relation
(b) Neither S nor T is an equivalence relation
(c) Both S and T are equivalence relation
(d) S is an equivalence relation but T is not an equivalence relation.
Answer: (a) T is an equivalence relation but S is not an equivalence relation
Let's analyze the relation T defined by \( T = \{(x, y): x - y \text{ is an integer}\} \). We need to check if T is an equivalence relation, which means checking for reflexivity, symmetry, and transitivity.

1. **Reflexivity:** For any \( x \in R \), \( x - x = 0 \), which is an integer. So, \( (x, x) \in T \). Thus, T is reflexive.
2. **Symmetry:** If \( (x, y) \in T \), then \( x - y \) is an integer. Let \( x - y = k \) for some integer \( k \). Then \( y - x = -(x - y) = -k \). Since \( k \) is an integer, \( -k \) is also an integer. So, \( (y, x) \in T \). Thus, T is symmetric.
3. **Transitivity:** If \( (x, y) \in T \) and \( (y, z) \in T \), then \( x - y \) is an integer and \( y - z \) is an integer. Let \( x - y = k_1 \) and \( y - z = k_2 \) for some integers \( k_1, k_2 \). Then, \( (x - y) + (y - z) = k_1 + k_2 \), which simplifies to \( x - z = k_1 + k_2 \). Since \( k_1 \) and \( k_2 \) are integers, their sum \( k_1 + k_2 \) is also an integer. So, \( (x, z) \in T \). Thus, T is transitive.

Since T is reflexive, symmetric, and transitive, T is an equivalence relation. The options mention a relation S, which is not fully defined in the question. However, based on the provided partial explanation, pairs like (0, 1) and (1, 2) are mentioned as not leading to an equivalence relation, implying S is not an equivalence relation. An equivalence relation partitions a set into disjoint equivalence classes. Thus, T is an equivalence relation. The question often implies S is a simple relation like \( \{(x,y): y = x+1 \} \), which is not an equivalence relation.
In simple words: The relation T checks if the difference between two real numbers is a whole number. This relation is 'reflexive' (a number relates to itself), 'symmetric' (if A relates to B, B relates to A), and 'transitive' (if A relates to B and B to C, then A relates to C). Because it has all three properties, T is an equivalence relation. The other relation S mentioned is stated not to be an equivalence relation.

🎯 Exam Tip: To prove a relation is an equivalence relation, you must rigorously check all three properties: reflexivity, symmetry, and transitivity. Provide clear examples or general arguments for each property.

Question 6. Let A and B be subsets of the universal set N, the set of natural numbers. Then \( A' \cup [(A \cap B) \cup B'] \) is
(a) A
(b) A'
(c) B
(d) N
Answer: (d) N
We need to simplify the expression \( A' \cup [(A \cap B) \cup B'] \). We can use properties of set operations to simplify this step-by-step. Let's remember De Morgan's laws and the distributive property.
First, let's simplify the inner part of the bracket: \( (A \cap B) \cup B' \). We can use the distributive property here or think about it visually. This is equivalent to \( (A \cup B') \cap (B \cup B') \). Since \( B \cup B' = N \) (the universal set), this simplifies to \( (A \cup B') \cap N \), which is just \( A \cup B' \). Alternatively, we know that \( A \cap B \) is the part where A and B overlap. \( (A \cap B) \cup B' \) includes all elements that are not in B, plus the overlap between A and B. This covers everything except elements that are in B but not in A. So this is \( A \cup B' \).
Now, substitute this back into the original expression: \( A' \cup (A \cup B') \).
Using the associative property of union, this becomes \( (A' \cup A) \cup B' \).
We know that \( A' \cup A = N \) (the universal set of natural numbers).
So, the expression further simplifies to \( N \cup B' \).
Since \( B' \) is a subset of \( N \), the union of \( N \) with any of its subsets is simply \( N \). Thus, \( N \cup B' = N \). This demonstrates how logical simplification of set expressions leads to the universal set.
In simple words: We are trying to simplify a set expression. By carefully using set rules like how 'not A' combines with 'A' (which equals the whole set), and how 'union' works, the entire complex expression simplifies down to the universal set, N.

🎯 Exam Tip: When simplifying complex set expressions, start with the innermost parentheses and apply De Morgan's laws, distributive laws, and basic identities like \( A \cup A' = N \) and \( A \cap N = A \).

Question 7. The number of students who take both the subjects Mathematics and Chemistry is 70. This represents 10% of the enrollment in Mathematics and 14% of the enrollment in Chemistry. The number of students takes at least one of these two subjects, is
(a) 1120
(b) 1130
(c) 1100
(d) Insufficient data
Answer: (b) 1130
Let \( n(M) \) be the number of students in Mathematics and \( n(C) \) be the number of students in Chemistry. We are given the following information:

1. Number of students taking both subjects, \( n(M \cap C) = 70 \).
2. This 70 represents 10% of the enrollment in Mathematics. So, \( 0.10 \times n(M) = 70 \).
From this, we can find \( n(M) = \frac{70}{0.10} = 700 \).
3. This 70 also represents 14% of the enrollment in Chemistry. So, \( 0.14 \times n(C) = 70 \).
From this, we can find \( n(C) = \frac{70}{0.14} = 500 \).

We need to find the number of students who take at least one of these two subjects, which is \( n(M \cup C) \). We use the formula for the union of two sets:
\( n(M \cup C) = n(M) + n(C) - n(M \cap C) \)
Substitute the values we found:
\( n(M \cup C) = 700 + 500 - 70 \)
\( n(M \cup C) = 1200 - 70 \)
\( n(M \cup C) = 1130 \)
The calculation shows that 1130 students take at least one of the subjects, Mathematics or Chemistry. This kind of problem often appears in real-world surveys to avoid double-counting.
In simple words: We know that 70 students take both subjects. This 70 helps us figure out the total number of students taking Math (700) and the total taking Chemistry (500). To find how many students take at least one subject, we add the Math and Chemistry totals and then subtract the students who were counted twice (the 70 students taking both). This gives us 1130.

🎯 Exam Tip: For problems involving the number of elements in sets, always use the Principle of Inclusion-Exclusion formula: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \). Make sure to correctly calculate individual set sizes from percentages or ratios.

Question 8. If \( n[(A \times B) \cap (A \times C)] = 8 \) and \( n(B \cap C) = 2 \), then \( n(A) \) is
(a) 2
(b) 4
(c) 8
(d) 16
Answer: (b) 4
We are given two pieces of information: 1. \( n[(A \times B) \cap (A \times C)] = 8 \) 2. \( n(B \cap C) = 2 \) We need to find \( n(A) \).

First, let's use the property of Cartesian products intersection: \( (A \times B) \cap (A \times C) = A \times (B \cap C) \) This means that the intersection of two Cartesian products with a common first set is the Cartesian product of that common set with the intersection of the second sets. It's a useful rule for simplifying set expressions.

Now, let's find the number of elements in this simplified set: \( n[(A \times B) \cap (A \times C)] = n[A \times (B \cap C)] \) The number of elements in a Cartesian product \( X \times Y \) is \( n(X) \times n(Y) \). So, \( n[A \times (B \cap C)] = n(A) \times n(B \cap C) \)

We can now substitute the given values into this equation: \( 8 = n(A) \times 2 \)

To find \( n(A) \), divide both sides by 2: \( n(A) = \frac{8}{2} \)
\( n(A) = 4 \)
So, the number of elements in set A is 4. This property allows for efficiently determining set sizes without listing all elements.
In simple words: We have a rule that \( A \times B \) intersecting with \( A \times C \) is the same as \( A \) multiplied by the intersection of \( B \) and \( C \). We are told the count for the first part is 8, and the count for \( B \cap C \) is 2. So, \( n(A) \) multiplied by 2 equals 8, which means \( n(A) \) must be 4.

🎯 Exam Tip: Remember the property \( (A \times B) \cap (A \times C) = A \times (B \cap C) \). This identity is crucial for solving problems involving the cardinality of intersecting Cartesian products.

Question 9. If \( n(A) = 2 \) and \( n(B \cup C) = 3 \), then \( n[(A \times B) \cup (A \times C)] \) is
(a) 23
(b) 3<
(c) 6
(d) 5
Answer: (c) 6
We are given: 1. \( n(A) = 2 \) 2. \( n(B \cup C) = 3 \) We need to find \( n[(A \times B) \cup (A \times C)] \).

First, let's use the property of Cartesian products union: \( (A \times B) \cup (A \times C) = A \times (B \cup C) \) This property states that the union of two Cartesian products with a common first set is the Cartesian product of that common set with the union of the second sets. This is a powerful tool for simplifying such expressions.

Now, we can find the number of elements in this simplified set: \( n[(A \times B) \cup (A \times C)] = n[A \times (B \cup C)] \) The number of elements in a Cartesian product \( X \times Y \) is \( n(X) \times n(Y) \). So, \( n[A \times (B \cup C)] = n(A) \times n(B \cup C) \)

Substitute the given values into this equation: \( n[(A \times B) \cup (A \times C)] = 2 \times 3 \)
\( n[(A \times B) \cup (A \times C)] = 6 \)
So, the number of elements in the union of these Cartesian products is 6. This demonstrates the efficiency of using set properties to solve cardinality problems.
In simple words: We have a rule that says \( A \times B \) union \( A \times C \) is the same as \( A \) multiplied by the union of \( B \) and \( C \). Since we know the count for \( A \) is 2 and the count for \( B \cup C \) is 3, we simply multiply these numbers together: \( 2 \times 3 = 6 \).

🎯 Exam Tip: Recognize the distributive property for Cartesian products over set union: \( (A \times B) \cup (A \times C) = A \times (B \cup C) \). This helps simplify the expression before calculating cardinality.

Question 10. If two sets A and B have 17 elements in common, then the number of elements common to the set \( A \times B \) and \( B \times A \) is
(a) \( 2^{17} \)
(b) \( 17^2 \)
(c) 34
(d) Insufficient data
Answer: (b) \( 17^2 \)
We are given that two sets A and B have 17 elements in common. This means \( n(A \cap B) = 17 \). We need to find the number of elements common to the set \( A \times B \) and \( B \times A \). This is asking for \( n[(A \times B) \cap (B \times A)] \).

We use the property that for any two sets P and Q, the intersection of their Cartesian products is given by: \( (P \times Q) \cap (R \times S) = (P \cap R) \times (Q \cap S) \) In our case, \( P=A \), \( Q=B \), \( R=B \), and \( S=A \). So, \( (A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A) \)

We know that \( n(A \cap B) = 17 \). Also, the intersection of B and A is the same as the intersection of A and B, so \( n(B \cap A) = n(A \cap B) = 17 \).

Now, substitute these values into the expression for the number of elements in the intersection: \( n[(A \times B) \cap (B \times A)] = n[(A \cap B) \times (B \cap A)] \) \( n[(A \times B) \cap (B \times A)] = n(A \cap B) \times n(B \cap A) \) \( n[(A \times B) \cap (B \times A)] = 17 \times 17 \)
\( n[(A \times B) \cap (B \times A)] = 17^2 = 289 \)
This property is derived from the definition of ordered pairs: an element \( (x,y) \) is in \( (A \times B) \cap (B \times A) \) if and only if \( (x,y) \in (A \times B) \) and \( (x,y) \in (B \times A) \). This means \( x \in A \) and \( y \in B \), AND \( x \in B \) and \( y \in A \). Combining these, \( x \in A \cap B \) and \( y \in B \cap A \). This directly leads to the formula used. This is a common result when dealing with Cartesian products and set intersections.
In simple words: If two sets A and B share 17 elements, then the number of matching pairs when you combine A with B and also B with A is found by multiplying 17 by 17. This is because any common pair \( (x,y) \) means \( x \) is in both A and B, and \( y \) is also in both A and B. So we get \( 17^2 \).

🎯 Exam Tip: Remember the general identity for the cardinality of the intersection of Cartesian products: \( n[(A \times B) \cap (C \times D)] = n[(A \cap C) \times (B \cap D)] = n(A \cap C) \times n(B \cap D) \). This simplifies many problems.

Question 11. For non-empty sets A and B, if \( A \subset B \) then \( (A \times B) \cap (B \times A) \) is equal to
(a) \( A \cap B \)
(b) \( A \times A \)
(c) \( B \times B \)
(d) None of these
Answer: (b) \( A \times A \)
We are given that A and B are non-empty sets, and \( A \subset B \) (A is a subset of B). We need to find what \( (A \times B) \cap (B \times A) \) is equal to.

Using the property of Cartesian products intersection from the previous question: \( (A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A) \)

Now, let's consider the given condition: \( A \subset B \).
If A is a subset of B, then: 1. The intersection of A and B is A itself: \( A \cap B = A \). (If all elements of A are in B, then the elements common to both are just the elements of A.) 2. The intersection of B and A is also A itself: \( B \cap A = A \). (This is the same as \( A \cap B \)).

Substitute these results back into the simplified intersection of Cartesian products: \( (A \times B) \cap (B \times A) = A \times A \)
Thus, when \( A \subset B \), the expression simplifies to \( A \times A \). This shows how fundamental set relationships simplify more complex expressions.
In simple words: If set A is completely inside set B, then what A and B share in common is just A. Similarly, what B and A share is also just A. When we look at the common parts of \( A \times B \) and \( B \times A \), it ends up being \( A \times A \), because every \( x \) and \( y \) in the common part must belong to A.

🎯 Exam Tip: When given a subset relationship (e.g., \( A \subset B \)), immediately recall its implications for set operations like intersection and union (e.g., \( A \cap B = A \), \( A \cup B = B \)). This simplifies expressions significantly.

Question 12. The number of relations on a set containing 3 elements is
(a) 9
(b) 81
(c) 512
(d) 1024
Answer: (c) 512
Let a set S contain \( n \) elements. The number of relations on this set S is given by the formula \( 2^{n^2} \). A relation on a set S is essentially a subset of the Cartesian product \( S \times S \).

In this question, the set contains 3 elements. So, \( n = 3 \).

Substitute \( n = 3 \) into the formula:
Number of relations \( = 2^{3^2} \)
\( = 2^9 \)
To calculate \( 2^9 \):
\( 2^1 = 2 \)
\( 2^2 = 4 \)
\( 2^3 = 8 \)
\( 2^4 = 16 \)
\( 2^5 = 32 \)
\( 2^6 = 64 \)
\( 2^7 = 128 \)
\( 2^8 = 256 \)
\( 2^9 = 512 \)
So, there are 512 possible relations on a set containing 3 elements. Each ordered pair in the Cartesian product \( S \times S \) can either be in the relation or not, leading to the power of 2 in the formula.
In simple words: If a set has 3 elements, then the total number of possible pairs we can make from those elements is \( 3 \times 3 = 9 \). For each of these 9 pairs, we can either include it in a relation or not include it. Since there are 2 choices for each of the 9 pairs, the total number of different relations is \( 2 \) multiplied by itself 9 times, which is 512.

🎯 Exam Tip: Memorize the formula for the total number of relations on a set with \( n \) elements, which is \( 2^{n^2} \). Also, be quick at calculating powers of 2 for small exponents.

Question 13. Let R be the universal relation on a set X with more than one element. Then R is
(a) not reflexive
(b) not symmetric
(c) transitive
(d) none of the above
Answer: (c) transitive
A universal relation R on a set X is a relation where every element of X is related to every other element of X (including itself). In other words, R is the entire Cartesian product \( X \times X \). Since X has more than one element, let \( X = \{a, b, c, ...\} \).

Let's check the properties of the universal relation:
1. **Reflexivity:** For every \( x \in X \), is \( (x, x) \in R \)? Yes, because in a universal relation, every element is related to every other element, including itself. So, R is reflexive.
2. **Symmetry:** For every \( x, y \in X \), if \( (x, y) \in R \), is \( (y, x) \in R \)? Yes, because if \( (x, y) \) is an ordered pair in \( X \times X \), then \( (y, x) \) is also an ordered pair in \( X \times X \). So, R is symmetric.
3. **Transitivity:** For every \( x, y, z \in X \), if \( (x, y) \in R \) and \( (y, z) \in R \), is \( (x, z) \in R \)? Yes, because by definition of the universal relation, every pair of elements from X is in R. So, if \( x, y, z \) are elements in X, then \( (x, y) \), \( (y, z) \), and \( (x, z) \) are all automatically in R. Thus, R is transitive.

Since the universal relation is reflexive, symmetric, and transitive, it is an equivalence relation. Therefore, the statement "R is transitive" is true. This property holds for any universal relation, regardless of the number of elements in the set (as long as it's non-empty).
In simple words: A universal relation means that every element in the set is related to every other element, including itself. Because all possible connections exist, it naturally satisfies being 'reflexive' (self-connected), 'symmetric' (two-way connection), and 'transitive' (if A connects to B and B to C, then A connects to C). Since it satisfies all three, it is transitive.

🎯 Exam Tip: The universal relation is a classic example of an equivalence relation. Understanding why it satisfies reflexivity, symmetry, and transitivity helps solidify the definitions of these properties.

Question 14. Let X = \( \{1,2,3,4\} \) and R = \( \{(1, 1), (1, 2), (1, 3), (2, 2), (3, 3), (2, 1), (3, 1), (1, 4), (4, 1)\} \). Then R is
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) Equivalence
Answer: (b) Symmetric
We are given a set \( X = \{1, 2, 3, 4\} \) and a relation R on X. We need to check the properties of R: reflexivity, symmetry, and transitivity.

1. **Reflexivity:** For R to be reflexive, every element \( x \in X \) must be related to itself, i.e., \( (x, x) \in R \) for all \( x \in \{1, 2, 3, 4\} \). We check the pairs: \( (1, 1) \in R \), \( (2, 2) \in R \), \( (3, 3) \in R \). However, \( (4, 4) \notin R \). Since \( (4, 4) \) is not in R, the relation R is not reflexive.

2. **Symmetry:** For R to be symmetric, if \( (x, y) \in R \), then \( (y, x) \) must also be in R. Let's check the pairs: \( (1, 1) \in R \), and \( (1, 1) \in R \). (OK) \( (1, 2) \in R \), and \( (2, 1) \in R \). (OK) \( (1, 3) \in R \), and \( (3, 1) \in R \). (OK) \( (2, 2) \in R \), and \( (2, 2) \in R \). (OK) \( (3, 3) \in R \), and \( (3, 3) \in R \). (OK) \( (1, 4) \in R \), and \( (4, 1) \in R \). (OK) Since for every pair \( (x, y) \) in R, the pair \( (y, x) \) is also in R, the relation R is symmetric.

3. **Transitivity:** For R to be transitive, if \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \) must also be in R. Let's test an example: We have \( (1, 4) \in R \) and \( (4, 1) \in R \). For transitivity, \( (1, 1) \) must be in R, which it is. However, consider \( (1, 4) \in R \) and \( (4, 1) \in R \). If R were transitive, then \( (4, 4) \) must be in R. But we already established that \( (4, 4) \notin R \). Since \( (1, 4) \in R \) and \( (4, 1) \in R \) but \( (4, 4) \notin R \), the relation R is not transitive. An equivalence relation must be reflexive, symmetric, and transitive. Since R is not reflexive and not transitive, it is not an equivalence relation.

Therefore, the relation R is symmetric. This example shows that a relation can have some properties without having all of them. Each property must be checked independently.
In simple words: We check if the relation R has three main features. First, 'reflexive' means every number relates to itself, but 4 does not relate to 4, so it's not reflexive. Second, 'symmetric' means if A relates to B, then B relates to A. This is true for all pairs in R. Third, 'transitive' means if A relates to B and B relates to C, then A relates to C. This doesn't hold for 1, 4, and 1, because (1,4) and (4,1) are in R, but (4,4) is not. So, R is only symmetric.

🎯 Exam Tip: When checking relation properties, explicitly list the conditions required for each (reflexivity: all (x,x) present; symmetry: if (x,y) then (y,x); transitivity: if (x,y) and (y,z) then (x,z)). If even one instance fails, the property does not hold.

Question 15. The range of the function \( f(x) = \frac{1}{1-2 \sin x} \) is
(a) \( (-\infty, -1) \cup (\frac{1}{3}, \infty) \)
(b) \( (-1, \frac{1}{3}) \)
(c) \( [-1, \frac{1}{3}] \)
(d) \( (-\infty, -1] \cup [\frac{1}{3}, \infty) \)
Answer: (d) \( (-\infty, -1] \cup [\frac{1}{3}, \infty) \)
To find the range of \( f(x) = \frac{1}{1-2 \sin x} \), we start with the known range of \( \sin x \).
We know that for all real values of \( x \), \( -1 \leq \sin x \leq 1 \).

Now, let's manipulate this inequality to get \( 1-2 \sin x \):
First, multiply by -2. When multiplying an inequality by a negative number, reverse the inequality signs:
\( -1 \times (-2) \geq -2 \sin x \geq 1 \times (-2) \)
\( 2 \geq -2 \sin x \geq -2 \)
Rearrange this to put the smaller number on the left:
\( -2 \leq -2 \sin x \leq 2 \)

Next, add 1 to all parts of the inequality:
\( 1 - 2 \leq 1 - 2 \sin x \leq 1 + 2 \)
\( -1 \leq 1 - 2 \sin x \leq 3 \)

Now, we need to find the reciprocal of \( 1 - 2 \sin x \). When taking the reciprocal of an inequality, we must be careful, especially if the interval includes zero. In this case, \( 1 - 2 \sin x \) can take any value between -1 and 3, including 0 if \( \sin x = \frac{1}{2} \). Since the denominator cannot be zero, \( 1 - 2 \sin x \neq 0 \). This means \( \sin x \neq \frac{1}{2} \). So, the term \( 1 - 2 \sin x \) will never actually be zero. Therefore, we can split the interval \( [-1, 3] \) into two parts where 0 is excluded: \( [-1, 0) \) and \( (0, 3] \).

For the interval \( -1 \leq 1 - 2 \sin x < 0 \):
Taking reciprocals reverses the inequality and also changes the sign. For example, if \( -1 \leq A < 0 \), then \( \frac{1}{A} \leq \frac{1}{-1} \) is not correct. Instead, as A approaches 0 from the negative side, \( \frac{1}{A} \) approaches \( -\infty \). And if \( A = -1 \), then \( \frac{1}{A} = -1 \).
So, if \( -1 \leq 1 - 2 \sin x < 0 \), then \( \frac{1}{1-2 \sin x} \leq -1 \). This implies \( \frac{1}{1-2 \sin x} \in (-\infty, -1] \).

For the interval \( 0 < 1 - 2 \sin x \leq 3 \):
Taking reciprocals also reverses the inequality. For example, if \( 0 < A \leq 3 \), then \( \frac{1}{A} \geq \frac{1}{3} \). So, if \( 0 < 1 - 2 \sin x \leq 3 \), then \( \frac{1}{1-2 \sin x} \geq \frac{1}{3} \). This implies \( \frac{1}{1-2 \sin x} \in [\frac{1}{3}, \infty) \).

Combining these two parts, the range of \( f(x) \) is \( (-\infty, -1] \cup [\frac{1}{3}, \infty) \). This method ensures that the range is correctly derived by handling the reciprocal function's behavior around zero. This is a typical approach for finding the range of rational functions involving trigonometric terms.
In simple words: We know \( \sin x \) goes from -1 to 1. We use this to find the range of \( 1 - 2 \sin x \), which goes from -1 to 3. Then, to find the range of \( \frac{1}{1-2 \sin x} \), we flip these numbers. Because the number in the bottom can't be zero, we split the range into two parts. When you flip numbers between -1 and 0, they become numbers from negative infinity to -1. When you flip numbers between 0 and 3, they become numbers from \( \frac{1}{3} \) to positive infinity.

🎯 Exam Tip: When finding the range of a function of the form \( \frac{1}{g(x)} \), first find the range of \( g(x) \). Then, carefully consider how the reciprocal function behaves, especially around zero and when the interval includes negative numbers.

Question 16. The range of the function \( f(x) = |[x] - x|, x \in R \) is
(a) \( [0, 1] \)
(b) \( [0, \infty) \)
(c) \( [0, 1) \)
(d) \( (0, 1) \)
Answer: (c) \( [0, 1) \)
The function is \( f(x) = |[x] - x| \), where \( [x] \) denotes the greatest integer less than or equal to \( x \) (also known as the floor function).
Let \( [x] = n \), where \( n \) is an integer. By definition of the floor function, we know that \( n \leq x < n+1 \).
We want to find the range of \( |[x] - x| \). Let's rewrite \( [x] - x \) in terms of \( n \):
\( n \leq x < n+1 \)
Subtract \( x \) from all parts of the inequality:
\( n - x \leq 0 < n+1 - x \)
Rearrange the first part to get \( x - n \geq 0 \).
Rearrange the second part to get \( x - (n+1) < 0 \), which implies \( x - n - 1 < 0 \).

From \( n \leq x < n+1 \), we can also say \( 0 \leq x - n < 1 \).
Since \( [x] = n \), the expression \( [x] - x \) can be rewritten as \( n - x \).
From \( 0 \leq x - n < 1 \), multiplying by -1 and reversing the inequality signs gives:
\( -1 < -(x - n) \leq 0 \)
So, \( -1 < n - x \leq 0 \).

Now, we need to take the absolute value of \( n - x \): \( |n - x| \).
Since \( -1 < n - x \leq 0 \), the values of \( n - x \) are negative or zero. Taking the absolute value will make them positive or zero.
If \( n - x = 0 \), then \( |n - x| = 0 \). (This happens when \( x \) is an integer, e.g., if \( x=1 \), \( [1]-1 = 1-1 = 0 \)).
If \( n - x \) is a value like \( -0.5 \), then \( |n - x| = |-0.5| = 0.5 \).
As \( n - x \) approaches -1 (but never reaches it), its absolute value \( |n - x| \) approaches 1 (but never reaches it).
Therefore, the range of \( |n - x| \) is \( [0, 1) \).

Let's check with some specific values: * If \( x = 1 \), \( [x] = 1 \). \( f(1) = |1 - 1| = 0 \). * If \( x = 1.5 \), \( [x] = 1 \). \( f(1.5) = |1 - 1.5| = |-0.5| = 0.5 \). * If \( x = 2.5 \), \( [x] = 2 \). \( f(2.5) = |2 - 2.5| = |-0.5| = 0.5 \). * If \( x = -2.5 \), \( [x] = -3 \). \( f(-2.5) = |-3 - (-2.5)| = |-3 + 2.5| = |-0.5| = 0.5 \). The function produces values between 0 (inclusive) and 1 (exclusive). This behavior is characteristic of the fractional part of a number, \( x - [x] \), except it's its absolute value. This function essentially measures the distance from x to the largest integer less than or equal to x.
In simple words: The function \( f(x) = |[x] - x| \) means taking a number, subtracting the largest whole number less than or equal to it, and then taking the positive value of the result. For any number, this difference will always be 0 (if \( x \) is a whole number) or a positive fraction less than 1. So, the answers will be from 0 up to, but not including, 1.

🎯 Exam Tip: Understand the definition and properties of the floor function \( [x] \). Remember that \( x - [x] \) represents the fractional part of \( x \), which always lies in \( [0, 1) \). The expression \( [x] - x \) is \( -(x - [x]) \), so its absolute value will also be in \( [0, 1) \).

Question 17. The rule \( f(x) = x^2 \) is a bijection if the domain and the co-domain are given by
(a) R, R
(b) R, \( (0, \infty) \)
(c) \( (0, \infty) \), R
(d) \( [0, \infty), [0, \infty) \)
Answer: (d) \( [0, \infty), [0, \infty) \)
For a function to be a bijection, it must be both one-to-one (injective) and onto (surjective). Let's analyze the function \( f(x) = x^2 \) with different domains and codomains.

**Case 1: Domain = R, Codomain = R**
* **One-to-one?** No. For example, \( f(1) = 1^2 = 1 \) and \( f(-1) = (-1)^2 = 1 \). Since \( 1 \neq -1 \) but \( f(1) = f(-1) \), the function is not one-to-one. * **Onto?** No. The range of \( f(x) = x^2 \) for \( x \in R \) is \( [0, \infty) \). Since the codomain is R, and not all real numbers (e.g., negative numbers) are in the range, it's not onto.

**Case 2: Domain = R, Codomain = \( (0, \infty) \)**
* **One-to-one?** Still not one-to-one (same reason as above, \( f(1) = f(-1) = 1 \)). * **Onto?** The range is \( [0, \infty) \). The codomain is \( (0, \infty) \). Since 0 is in the range but not in the codomain, it's not onto. Also, the problem of \( f(x) = 0 \) for \( x=0 \) means it doesn't map to `(0, inf)`.

**Case 3: Domain = \( (0, \infty) \), Codomain = R**
* **One-to-one?** Yes. If \( x_1, x_2 \in (0, \infty) \) and \( f(x_1) = f(x_2) \), then \( x_1^2 = x_2^2 \). Since \( x_1, x_2 \) are positive, \( x_1 = x_2 \). So, it's one-to-one. * **Onto?** No. The range of \( f(x) = x^2 \) for \( x \in (0, \infty) \) is \( (0, \infty) \). Since the codomain is R, and not all real numbers are in the range, it's not onto.

**Case 4: Domain = \( [0, \infty) \), Codomain = \( [0, \infty) \)**
* **One-to-one?** Yes. If \( x_1, x_2 \in [0, \infty) \) and \( f(x_1) = f(x_2) \), then \( x_1^2 = x_2^2 \). Since \( x_1, x_2 \) are non-negative, \( x_1 = x_2 \). So, it's one-to-one. * **Onto?** Yes. For any \( y \) in the codomain \( [0, \infty) \), we need to find an \( x \) in the domain \( [0, \infty) \) such that \( f(x) = y \). Let \( x^2 = y \). Since \( y \geq 0 \), \( x = \sqrt{y} \) is a real number. Since \( y \geq 0 \), \( \sqrt{y} \geq 0 \), so \( x = \sqrt{y} \) is in the domain \( [0, \infty) \). Thus, for every element in the codomain, there is a corresponding element in the domain. So, it's onto.

Since \( f(x) = x^2 \) is both one-to-one and onto when the domain is \( [0, \infty) \) and the codomain is \( [0, \infty) \), it is a bijection in this case. This careful selection of domain and codomain is crucial for classifying functions.
In simple words: For a function to be a bijection, it needs two things: (1) each input must give a unique output (one-to-one), and (2) every possible output in the codomain must be produced by some input (onto). The function \( f(x) = x^2 \) works as a bijection only if we limit its inputs to positive numbers (including zero) and its outputs to positive numbers (including zero). If we include negative inputs, different inputs give the same output (like -1 and 1 both give 1), so it's not one-to-one.

🎯 Exam Tip: To determine if a function is a bijection, always check for both injectivity (one-to-one, using the horizontal line test or \( f(x_1)=f(x_2) \implies x_1=x_2 \)) and surjectivity (onto, ensuring the range equals the codomain) based on the given domain and codomain.

Question 18. The number of constant functions from a set containing \( m \) elements to a set containing \( n \) elements is
(a) \( mn \)
(b) \( m \)
(c) \( n \)
(d) \( m + n \)
Answer: (c) \( n \)
Let A be a set with \( m \) elements, \( A = \{a_1, a_2, ..., a_m\} \).
Let B be a set with \( n \) elements, \( B = \{b_1, b_2, ..., b_n\} \).
A function \( f: A \to B \) is called a constant function if every element in the domain A maps to the *same* single element in the codomain B.

Consider the elements of B:
1. We can define a constant function where every element in A maps to \( b_1 \). This is one constant function.
Example: \( f_1(a_i) = b_1 \) for all \( a_i \in A \). 2. We can define another constant function where every element in A maps to \( b_2 \). This is a second constant function.
Example: \( f_2(a_i) = b_2 \) for all \( a_i \in A \).
We can continue this process for all elements in B. Since there are \( n \) elements in the set B, we can define \( n \) distinct constant functions.
Each element in the codomain B acts as a potential single target for all elements in the domain A. Thus, the number of choices for this common target element determines the number of constant functions. The number of elements in the domain (m) does not affect the count of constant functions, as long as it's non-empty. This concept simplifies counting problems in basic set theory. For instance, if you have 5 friends (set A) and 3 ice cream flavors (set B), you can have 3 constant functions (everyone gets vanilla, everyone gets chocolate, or everyone gets strawberry).
In simple words: A constant function means all the inputs from the first set (with \( m \) elements) point to just one single output in the second set (with \( n \) elements). Since there are \( n \) different elements in the second set, there are \( n \) different choices for that single output. So, there are \( n \) possible constant functions.

🎯 Exam Tip: For constant functions \( f: A \to B \), the domain \( A \) must map all its elements to a single element in the codomain \( B \). The number of such functions is simply the cardinality of the codomain, \( n(B) \).

Question 19. The function \( f:[0, 2\pi] \rightarrow [-1, 1] \) defined by \( f(x) = \sin x \) is
(1) one to one
(2) onto
(3) bijection
(4) cannot be defined
Answer: (2) onto
Answer: For a function to be onto, its range must cover the entire co-domain. The sine function, \( f(x) = \sin x \), when defined over the interval \( [0, 2\pi] \), takes on all values between -1 and 1. This means its range is \( [-1, 1] \). Since the co-domain given for this function is also \( [-1, 1] \), every value in the co-domain has a corresponding input in the domain.
In simple words: The function \( \sin x \) in the given range covers all possible output values exactly as specified by the problem. Because its actual output matches the expected output range, it is an "onto" function.

🎯 Exam Tip: To check if a function is 'onto', always compare its actual range (all possible output values) with its defined co-domain. If they are identical, the function is onto.

Question 20. If the function \( f : [-3, 3] \rightarrow S \) defined by \( f(x) = x^2 \) is onto, then \( S \) is
(1) \( [-9, 9] \)
(2) \( R \)
(3) \( [-3, 3] \)
(4) \( [0, 9] \)
Answer: (4) [0, 9]
Answer: For a function to be "onto", its range must be exactly equal to its co-domain. Here, the function is \( f(x) = x^2 \) and the domain is \( [-3, 3] \). To find the range, we look at the minimum and maximum values of \( x^2 \) within this domain. When \( x=0 \), \( f(x) = 0^2 = 0 \), which is the smallest value. When \( x = -3 \) or \( x = 3 \), \( f(x) = (-3)^2 = 9 \) or \( f(x) = 3^2 = 9 \), which is the largest value. Thus, the range of \( f(x) = x^2 \) on \( [-3, 3] \) is \( [0, 9] \). Since the function is onto, the co-domain \( S \) must be equal to this range.
In simple words: If a function is "onto", its actual output values (the range) must cover every single value in the target set (co-domain). For \( x^2 \) when \( x \) is between -3 and 3, the smallest output is 0 and the largest is 9, so the target set must be all numbers from 0 to 9.

🎯 Exam Tip: When finding the range of a quadratic function like \( x^2 \) over an interval, always check the function's value at \( x=0 \) (if in the interval) and at the interval's endpoints.

Question 21. Let \( X = \{1, 2, 3, 4\}, Y = \{ a, b, c, d \} \) and \( f = \{(1, a), (4, b), (2, c), (3, d), (2, d)\} \). Then \( f \) is
(1) an one – to – one function
(2) an onto function
(3) a function which is not one to one
(4) not a function
Answer: (4) not a function
Answer: For a relation to be considered a function, each element in the domain (set X) must map to exactly one element in the co-domain (set Y). In the given set of ordered pairs, \( f \), the element \( 2 \in X \) is mapped to two different elements in Y: \( c \) and \( d \) (from \( (2, c) \) and \( (2, d) \)). Because one input has more than one output, this relation is not a function.
In simple words: A relation is only a function if each starting number goes to only one ending number. Here, the number '2' tries to go to two different ending numbers ('c' and 'd'). Because of this, it cannot be called a function.

🎯 Exam Tip: The 'vertical line test' is a visual way to check if a graph represents a function: if any vertical line intersects the graph more than once, it is not a function. For sets, look for any element in the domain mapping to multiple elements in the co-domain.

Question 22. The inverse of \( f(x) \) where \( f(x) = \begin{cases} x & \text{if } x < 1 \\ x^2 & \text{if } 1 \leq x \leq 4 \\ 8\sqrt{x} & \text{if } x > 4 \end{cases} \) is
(1) \( f^{-1}(x) = \begin{cases} x & \text{if } x < 1 \\ \sqrt{x} & \text{if } 1 \leq x \leq 16 \\ \frac{x^2}{64} & \text{if } x > 16 \end{cases} \)
(2) \( f^{-1}(x) = \begin{cases} -x & \text{if } x < 1 \\ \sqrt{x} & \text{if } 1 \leq x \leq 16 \\ \frac{x^2}{64} & \text{if } x > 16 \end{cases} \)
(3) \( f^{-1}(x) = \begin{cases} x^2 & \text{if } x < 1 \\ \sqrt{x} & \text{if } 1 \leq x \leq 16 \\ \frac{x^3}{64} & \text{if } x > 16 \end{cases} \)
(4) \( f^{-1}(x) = \begin{cases} 2x & \text{if } x < 1 \\ \sqrt{x} & \text{if } 1 \leq x \leq 16 \\ \frac{x^3}{8} & \text{if } x > 16 \end{cases} \)
Answer: (1) \( f^{-1}(x) = \begin{cases} x & \text{if } x < 1 \\ \sqrt{x} & \text{if } 1 \leq x \leq 16 \\ \frac{x^2}{64} & \text{if } x > 16 \end{cases} \)
Answer: To find the inverse of a piecewise function, we find the inverse for each part separately and adjust the domain for the inverse.
**Case 1:** \( y = x \) for \( x < 1 \)
If \( y = x \), then \( x = y \). The condition \( x < 1 \) becomes \( y < 1 \).
So, \( f^{-1}(y) = y \) for \( y < 1 \). When we replace \( y \) with \( x \), we get \( f^{-1}(x) = x \) for \( x < 1 \).
**Case 2:** \( y = x^2 \) for \( 1 \leq x \leq 4 \)
If \( y = x^2 \), then \( x = \sqrt{y} \) (we take the positive root since \( x \geq 1 \)).
The condition \( 1 \leq x \leq 4 \) means \( 1 \leq \sqrt{y} \leq 4 \). Squaring all parts, we get \( 1^2 \leq (\sqrt{y})^2 \leq 4^2 \), which is \( 1 \leq y \leq 16 \).
So, \( f^{-1}(y) = \sqrt{y} \) for \( 1 \leq y \leq 16 \). Replacing \( y \) with \( x \), we get \( f^{-1}(x) = \sqrt{x} \) for \( 1 \leq x \leq 16 \).
**Case 3:** \( y = 8\sqrt{x} \) for \( x > 4 \)
If \( y = 8\sqrt{x} \), then \( \sqrt{x} = \frac{y}{8} \). Squaring both sides, we get \( x = \left(\frac{y}{8}\right)^2 = \frac{y^2}{64} \).
The condition \( x > 4 \) means \( \frac{y^2}{64} > 4 \). Multiplying by 64 gives \( y^2 > 256 \). Taking the square root, \( \sqrt{y^2} > \sqrt{256} \), so \( |y| > 16 \). Since \( y = 8\sqrt{x} \) and \( \sqrt{x} \) is always positive, \( y \) must be positive, so \( y > 16 \).
So, \( f^{-1}(y) = \frac{y^2}{64} \) for \( y > 16 \). Replacing \( y \) with \( x \), we get \( f^{-1}(x) = \frac{x^2}{64} \) for \( x > 16 \).
Combining these three parts, the inverse function is:
\( f^{-1}(x) = \begin{cases} x & \text{if } x < 1 \\ \sqrt{x} & \text{if } 1 \leq x \leq 16 \\ \frac{x^2}{64} & \text{if } x > 16 \end{cases} \)
In simple words: To find the inverse function, you switch the roles of the input and output (x and y) for each part of the function. For example, if \( y=x^2 \), its inverse is \( x=\sqrt{y} \). You also need to change the conditions for when each rule applies, based on the new output values.

🎯 Exam Tip: Remember to adjust the domain of each piece of the inverse function to reflect the range of the corresponding piece of the original function. Always check for potential restrictions like positive square roots.

Question 23. Let \( f: R \rightarrow R \) be defined by \( f(x) = 1 - |x| \). Then the range of \( f \) is
(1) \( R \)
(2) \( (1,\infty) \)
(3) \( (-1,\infty) \)
(4) \( (-\infty, 1) \)
Answer: (4) \( (-\infty, 1) \)
Answer: We need to find all possible output values for \( f(x) = 1 - |x| \). We know that the absolute value of any real number, \( |x| \), is always greater than or equal to zero (i.e., \( |x| \geq 0 \)). This means that \( -|x| \) will always be less than or equal to zero (i.e., \( -|x| \leq 0 \)). If we add 1 to both sides of this inequality, we get \( 1 - |x| \leq 1 + 0 \), which simplifies to \( 1 - |x| \leq 1 \). This shows that the maximum possible value for \( f(x) \) is 1. Also, as \( |x| \) can become very large (approaching infinity), \( 1 - |x| \) will become a very large negative number (approaching negative infinity). Therefore, the set of all possible output values (the range) for \( f(x) \) is all numbers less than 1, going down forever.
In simple words: The absolute value of any number is always zero or positive. So, \( 1 \) minus a positive number will always be \( 1 \) or less. Since the positive number can get very big, \( 1 \) minus a very big number can be very small (a large negative number). So, the function can produce any number less than 1, and can go down to very low negative values.

🎯 Exam Tip: To find the range of functions involving absolute values, consider the minimum value of \( |x| \) (which is 0) and how increasing \( |x| \) affects the function's output. The output will either go up or down from that point.

Question 24. The function \( f : R \rightarrow R \) be defined by \( f(x) = \sin x + \cos x \) is
(1) an odd function
(2) neither an odd function nor an even function
(3) an even function
(4) both odd function and even function
Answer: (2) neither an odd function nor an even function
Answer: To determine if a function is odd or even, we examine \( f(-x) \).
For \( f(x) = \sin x + \cos x \):
\( f(-x) = \sin(-x) + \cos(-x) \)
Since \( \sin(-x) = -\sin x \) and \( \cos(-x) = \cos x \), we have:
\( f(-x) = -\sin x + \cos x \)
Now, we compare \( f(-x) \) with \( f(x) \) and \( -f(x) \):
\( f(-x) = -\sin x + \cos x \) is not equal to \( f(x) = \sin x + \cos x \), so it is not an even function.
Also, \( -f(x) = -(\sin x + \cos x) = -\sin x - \cos x \).
\( f(-x) = -\sin x + \cos x \) is not equal to \( -f(x) = -\sin x - \cos x \) (because \( \cos x \neq -\cos x \) in general), so it is not an odd function.
Since it is neither even nor odd, the correct answer is (2). This type of function often arises when mixing trigonometric terms.
In simple words: An even function stays the same when you swap \( x \) with \( -x \). An odd function flips its sign. For \( \sin x + \cos x \), when you swap \( x \) with \( -x \), the \( \sin x \) part flips its sign but the \( \cos x \) part stays the same. Because of this mix, the whole function is neither even nor odd.

🎯 Exam Tip: Remember the properties: \( \sin(-x) = -\sin x \) (odd) and \( \cos(-x) = \cos x \) (even). A sum of an odd and an even function is usually neither unless one of the parts is zero.

Question 25. The function \( f : R \rightarrow R \) be defined by \( f(x) = \frac{(x^2 + \cos x)(1 + x^4)}{(x - \sin x)(2x - x^3)} + e^{-|x|} \) is
(1) an odd function
(2) neither an odd function nor an even function
(3) an even function
(4) both odd function and even function.
Answer: (3) an even function
Answer: To find if the function \( f(x) \) is odd or even, we need to evaluate \( f(-x) \). We will examine each part of the expression:
**1. Numerator:** \( (x^2 + \cos x)(1 + x^4) \)
\( (-x)^2 = x^2 \) (even)
\( \cos(-x) = \cos x \) (even)
\( (-x)^4 = x^4 \) (even)
So, \( (x^2 + \cos x) \) is even, and \( (1 + x^4) \) is even. The product of two even functions is an even function.
**2. Denominator:** \( (x - \sin x)(2x - x^3) \)
For \( (x - \sin x) \):
\( (-x - \sin(-x)) = -x - (-\sin x) = -x + \sin x = -(x - \sin x) \). So, \( (x - \sin x) \) is an odd function.
For \( (2x - x^3) \):
\( (2(-x) - (-x)^3) = -2x - (-x^3) = -2x + x^3 = -(2x - x^3) \). So, \( (2x - x^3) \) is an odd function.
The product of two odd functions is an even function. So the entire denominator \( (x - \sin x)(2x - x^3) \) is an even function.
**3. The fraction part:** Since the numerator is an even function and the denominator is an even function, their ratio is also an even function.
**4. The exponential part:** \( e^{-|x|} \)
\( e^{-|-x|} = e^{-|x|} \). This is an even function.
**5. Total function:** \( f(x) \) is the sum of an even function (the fraction part) and another even function (the exponential part). The sum of two even functions is always an even function. Therefore, \( f(x) \) is an even function.
In simple words: We check if the function changes when \( x \) becomes \( -x \). If it stays the same, it's "even". If it completely flips its sign, it's "odd". For this problem, all the parts like \( x^2 \), \( \cos x \), \( x^4 \), and \( e^{-|x|} \) stay the same when \( x \) becomes \( -x \). The parts like \( x-\sin x \) and \( 2x-x^3 \) flip their sign, but when two such "odd" parts are multiplied together, they become "even". So, the whole big function is made up of only "even" parts, which means the whole function is even.

🎯 Exam Tip: When dealing with complex functions, break them down into smaller components. Remember that the product or quotient of two even functions is even, the product or quotient of two odd functions is even, and the product or quotient of an odd and an even function is odd.