Relations And Functions

The given problem is to determine the number of elements in the relation from the set to itself. The relation is defined such that , where and are prime numbers ≥ 3.

To solve this, we need to find the values of which can be expressed as a product of two prime numbers ≥ 3 and also lie within the set .

First, let's identify the prime numbers in this range: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, and 59.

Now, we need to form products of these primes (b = pq) and ensure that each product is ≤ 60.

• : Possible values are 3, 5, 7, 11, 13, 17 (all products are ≤ 60).
• : Possible values are 3, 5, 7, 11 (all products are ≤ 60).
• : Possible values are 3, 5 (all products are ≤ 60).
• : Possible value is 3 (product is ≤ 60).

Any higher prime number and lower prime pair that results in a product > 60 are omitted:

Now, count the distinct products:

• For : 3*3, 3*5, 3*7, 3*11, 3*13, 3*17 (6 products).
• For : 5*3, 5*5, 5*7, 5*11 (includes new products 5*9 distinct from 5*3, etc., 4 products) .
• For : 7*3, 7*5 (2 products).
• For : 11*3 (1 product).

Total distinct products = 6 (from when ) + 4 (from ) + 2 (from ) + 1 (from ) = 13 distinct values.

Therefore, the number of elements in is calculated by multiplying each pair's count with the total 60 values (from all permutations in the set): .

So, the correct answer is 660.

To determine the number of points where the function is not differentiable in the interval (–20, 20), we need to analyze where each component causes non-differentiability.
1. **Absolute Value Function**: The term is not differentiable when its argument is zero. Solving gives .
2. **Floor Functions**: The terms and are not differentiable at integer values of their arguments.
For : Non-differentiability occurs at , where is an integer. Thus, . In the range , ranges from to , giving points .
For : Non-differentiability occurs at , integer. So . In the interval for , ranges from to , giving points .
3. **Combining Points**: Align distinct points of non-differentiability: . The combinations correspond to distinct values of .
4. **Count**: Count unique points, totaling 79 non-differentiable points within the specified range.

The number of points in the interval (–20, 20) where is not differentiable is , verified by checking the distinct values of such .

To find the domain of the given function , we need to determine when the expression inside the inverse sine function is valid. The domain of is valid for values in the range . Therefore, we need to establish the values of for which:

We will break this down into two inequalities:

**Step 1:** Solve the inequality

Rewriting, we get:

Simplifying gives:

This is a quadratic inequality. Solving using the quadratic formula:

\)

where .

Calculating the discriminant:

Thus, the roots are:

This gives valid intervals where , which is outside the roots.

**Step 2:** Solve the inequality

Rewriting, we get:

Simplifying gives:

So, or . This inequality can only hold when is less than or equal to 2.

**Combine Both Results:**

Combining both inequalities, the domain of the function where satisfies both inequalities is .

Thus, the correct answer is .

The function is given as:

First Iteration:

Second Iteration:

Notice how the numerator and denominator coefficients evolve as the function is iterated.

Third Iteration:

The pattern becomes clearer as we proceed further. Observe the symmetry in the coefficients.

Fifth Iteration:

This results from applying the function iteratively, maintaining the structure of coefficients in the numerator and denominator.

Given Condition:

The condition provided is:

Here, and are the coefficients of and the constant term in the numerator of , respectively.

Conclusion:

We have and , so:

Thus, the given condition is satisfied.

The correct answer is 72.

2y−x=15
A=0∫3​(2x+15​−x2)dx+21​×215​×15
4x2​+215x​−3x3​∣∣​03​+4225​
=49​+245​−9+4225​=499−36+225​
=4288​=72

The correct answer is (B) : 180
Total no. of onto functions

So , when f(a) = 1

Required functions :
= 240 -36 -24
=180

We are given Notice that wherever we see in an expression for , we can treat that as a single variable, say . Then because substituting into recovers the given formula: exactly matching the right‐hand side. Hence, . Therefore, Note on domain: Strictly speaking, for to be defined at , must lie in the domain of . Here , which poses no difficulty for the form . Thus the final result is .

The correct answer is option (C): 582

n(A)=5, n(B)=2

Number of subsets having 3 elements = ¹⁰C₃

Number of subsets having 3 elements = ¹⁰C₄

Number of subsets having 3 elements = ¹⁰C₅

¹⁰C₃+¹⁰C₄+¹⁰C₅

=120+210+252

=582

We are given that for all , and .
Step 1: Differentiating the functional equation. To solve this, let's differentiate the given functional equation with respect to : This simplifies to: Thus, we find that for all . This implies that is a constant function. Let , where is a constant.
Step 2: Determining the constant . We are given that . Since , it follows that . Therefore, for all .
Step 3: Finding the general form of . Since , we integrate to find : where is a constant.

Step 4: Using the given functional equation to find . Substitute into the original functional equation : Equating the two expressions: This simplifies to:

Step 5: Final form of . Thus, the function is:

Step 6: Finding . Substitute into the function : Thus, .

Step 1: Understanding the functional equation. The functional equation for all suggests that is a linear function. So, let: where is some constant.

Step 2: Using the given information. We are given that . Substituting this into the linear form: Thus, the function becomes:

Step 3: Substituting into the sum. We are given the equation: Substitute into this sum: Simplifying the expression: Factor out : Multiply both sides by 5:

Step 4: Simplifying the sum. The sum can be simplified using partial fractions. We decompose: Thus, the sum becomes: This is a telescoping series, and when we expand it, most terms cancel out: The remaining terms are: We are given that this sum equals :

Step 5: Solving for . Solve for : Thus:

Solution:
First, observe that replacing by in the given functional equation yields two simultaneous equations: We can solve this system for . Multiply (1) by 3 and (2) by 2: Subtracting the second from the first:

Step (i): Evaluate .
Substitute into :

Step (ii): Find and then .
Differentiate with respect to : Now substitute :

Step (iii): Combine to get the final answer:
If the problem denotes as an absolute value or final numeric value, we get . From the official result and context

Given: The piecewise function:

• Step 1: Check for discontinuity:

Clearly, is discontinuous at . It is also non-differentiable at this point.

Thus, .

• Step 2: Check for differentiability:

Differentiate :

is non-differentiable at .

• Step 3: Check :

The absolute value remains the same.

Thus, .

• Step 4: Calculate :

.

Final Answer: .

Step 1: Analyze the given conditions for and

We are given the following choices:

• choices of
• choices of
• choices of
• choices of
• choice of

Total number of ways for :

5 + 4 + 4 + 2 + 1 = 16 ways.

Step 2: Analyze choices for and

We are given the following choices:

• choices of
• choices of
• choices of
• choice of

Total number of ways for :

4 + 3 + 2 + 1 = 10 ways.

Step 3: Calculate the total number of required elements

Combining the above results, the total number of required elements in is:

Final Answer

The required number of elements in is 160.

Step 1: Define the relation.
- .
- .
Step 2: Check for symmetry.
- For each pair , ensure to make symmetric.
Step 3: Count the missing pairs.
- Add 15 pairs for odd positive differences and 4 pairs for .
Final Answer: A minimum of 19 pairs must be added.

Solution: Step 1: Rewriting the function.

The given function is:

We can rewrite it as:

Step 2: Analyzing the function.

The function has the form:

By differentiating , we get:

We analyze the sign of :

• For , , indicating that the function is increasing and thus one-one in this interval.
• For , , indicating that the function is decreasing and thus not one-one in this interval.

Step 3: Conclusion.

Thus, is one-one in but not in .

Step 4: Conclusion from the graph.

The graph of the function confirms that is one-one in and not in .

Given A = {1, 2, 3, . . . , 10} and B = {0, 1, 2, 3, 4}, the relation R is defined as:

R = {(a, b) ∈ A × A : 2(a − b)² + 3(a − b) ∈ B}.

Let a − b = k. Then 2k² + 3k ∈ B. Since a, b ∈ A, a − b can take integer values from -9 to 9.

We need to find the number of pairs (a, b) such that 2(a − b)² + 3(a − b) ∈ {0, 1, 2, 3, 4}.

Solution:Click to view the solution

Let a − b = k. Then we have 2k² + 3k ∈ B. This means 0 ≤ 2k² + 3k ≤ 4. Since a, b ∈ {1, 2, ..., 10}, we have −9 ≤ k ≤ 9.

Case 1: If k = 0, then 2(0)² + 3(0) = 0 ∈ B. This corresponds to a = b. Since there are 10 possible values for a, there are 10 such ordered pairs.

Case 2: If k = −2, then 2(−2)² + 3(−2) = 8 − 6 = 2 ∈ B. This corresponds to a = b − 2. Since 1 ≤ a ≤ 10 and 1 ≤ b ≤ 10, we have 1 ≤ b − 2 ≤ 10. Therefore, 3 ≤ b ≤ 10, which gives 8 possible values for b. Thus, there are 8 such ordered pairs.

Case 3: If k = −1, then 2(−1)² + 3(−1) = 2 − 3 = −1, which is not in B.

Case 4: If k = 1, then 2(1)² + 3(1) = 2 + 3 = 5, which is not in B.

Therefore, the only valid cases are a = b and a = b − 2. When a = b, there are 10 pairs: (1,1), (2,2), ..., (10,10). When a = b − 2, there are 8 pairs: (1,3), (2,4), (3,5), ..., (8,10).

Final Answer: The total number of elements in R is 10 + 8 = 18.

Let .

Step 1: Reflexive Property

For the relation to be reflexive:

Step 2: Transitive Property

For the relation to be transitive:

Step 3: Symmetric Property

The relation is not symmetric because:

Relations:

Step 1 : Let

Taking the natural logarithm on both sides:

Simplify:

Step 2: Differentiating with respect to
Differentiating both sides with respect to :

Multiply through by :

Step 3: Behavior of the function
For , the function is decreasing.
Thus, we can establish the following inequalities:

Given the sets:

The relation is defined by:

We list all possible valid pairs and satisfying the condition:

× 2

Thus, the total number of such relations is:

To determine the nature of the relation defined on the cartesian product , we are given that if and only if is an even integer. Let's analyze whether this relation is reflexive, symmetric, and transitive:

1. Reflexivity:
   • For reflexivity, each ordered pair with itself should satisfy the condition .
   • Substituting into the relation condition, we have .
   • Clearly, is an even integer (since any integer multiplied by an even number is even).
   • Therefore, the relation is reflexive on .
2. Symmetry:
   • For symmetry, if then should also hold.
   • If , we have is even.
   • We must check if is also even.
   • Since integer addition/subtraction preserves parity, implies symmetry.
   • Therefore, the relation is symmetric.
3. Transitivity:
   • For transitivity, if and , then should hold as well.
   • Given and are even, nothing guarantees is even without additional conditions.
   • Thus, the relation is not transitive in general.

Based on the analysis, the relation is reflexive and symmetric but not transitive. Thus, the correct answer is:

• Reflexive and symmetric but not transitive.

To solve this problem, we need to analyze the conditions under which each given relation becomes symmetric. A relation on a set is symmetric if whenever , then as well. We have two relations defined as:

We determine how many elements need to be added to each set to achieve symmetry.

1. For :
   • The relation implies . To check symmetry, we also need or satisfied for each pair in , if not, add to .
   • By solving, we find that:
     satisfies both conditions.
     For , the reverse needs to be added.
     Continue similarly.
   • Compute: Needs exactly 5 additional elements to make symmetric covering up the instances where symmetry fails.
2. For :
   • The relation implies . To check symmetry, similarly test if or holds for all pairings.
   • Upon examination, if a pair satisfies, can be cross-verified.
   • Compute: Find needing also 5 additional elements for symmetry.

Thus, the minimum number of elements to be added to make symmetric is and for is . Therefore, .

Hence, the correct answer is .

Given:

Step 1: Define the relation

Therefore,

Number of elements in :

Since

Step 2: Define

Step 3: Minimum number of elements

Final Answer:

Given the relations, let's evaluate whether and are equivalence relations by checking the properties: reflexivity, symmetry, and transitivity.

Relation :

• Reflexivity: For , we need , which simplifies to . However, this doesn't hold for all . Thus, is not reflexive.
• Symmetry: Suppose , which implies . Clearly, if , then , so . Hence, is symmetric.
• Transitivity: Suppose and . We have and . Adding these gives . This doesn't imply unless , which fails for all . Thus, is not transitive.

fails the reflexivity and transitivity tests, so it is not an equivalence relation.

Relation :

• Reflexivity: We need to check if . In this case, , which is true for all . Hence, is reflexive.
• Symmetry: If , then . This implies , so . Therefore, is symmetric.
• Transitivity: Suppose and . Then and . Adding these gives . Hence, , showing is transitive.

satisfies reflexivity, symmetry, and transitivity, so it is an equivalence relation.

Based on these analyses, the correct answer is: Only is an equivalence relation.

Given:

The functional equation is given by:

Task: We are asked to calculate the values of and .

Expression for : We are given the integral expression for :

Symmetry Consideration: By observing the symmetry of the problem, we deduce the relationship between and . From the symmetry, we find that: which simplifies to:

Given Values: We are also given the values and . The least value of is computed as:

To solve the problem, we need to construct an equivalence relation on set where is a subset of . An equivalence relation must be reflexive, symmetric, and transitive.

Step 1: Reflexivity
For reflexivity, every element in must relate to itself. Therefore, we add , , , and to .

Step 2: Symmetry
For symmetry, if , then must also be in . Using , we add the pairs , , and .

Step 3: Transitivity
For transitivity, if and , then must be in . We now apply transitivity for the relations:

• and imply and .
• and imply , already included as part of reflexivity.
• and imply , already included.
• and imply , already included.
• and imply and .

By following reflexivity, symmetry, and transitivity, we find that the smallest equivalence relation is .

Thus, the minimal contains elements. Confirming this value fits within the given range: (16, 16).

The minimum value of is .

We are given the function , where denotes the greatest integer less than or equal to . We need to find the sum of all points of discontinuity of this function in the interval .

Concept Used:

The greatest integer function, , is discontinuous at every integer value of its argument . The function is a difference of two functions involving the greatest integer function. Therefore, will be discontinuous at a point if either or is discontinuous at , unless the jump discontinuities of the two functions cancel each other out.

A function is discontinuous at a point if the left-hand limit (LHL), right-hand limit (RHL), and the function's value at that point are not all equal. That is, if or if these limits are not equal to .

Step-by-Step Solution:

Step 1: Identify potential points of discontinuity from the first term, .

This term is discontinuous when its argument, , is an integer. Let , where is an integer. This implies . We need to find the values of in the interval .

The possible integer values for are 3, 4, 5, 6, and 7. The corresponding values for are:

• For :
• For :
• For :
• For :

The set of potential points of discontinuity from the first term in is .

Step 2: Identify potential points of discontinuity from the second term, .

This term is discontinuous when its argument, , is an integer. Let , where is an integer. This implies . We need to find the values of in the interval .

The possible integer values for are 1 and 2. (We check endpoints separately, ). The corresponding values for are:

• For :
• For :

The set of potential points of discontinuity from the second term in is .

Step 3: Combine the potential points and check for continuity at each one.

The combined set of potential points of discontinuity in is . We must test each point.

At :

LHL: . RHL: . Since LHL RHL, is discontinuous at .

At :

LHL: . RHL: . Since LHL RHL, is discontinuous at .

At :

LHL: . RHL: . Also, . Since LHL = RHL = , is continuous at .

At :

LHL: . RHL: . Since LHL RHL, is discontinuous at .

At : (Endpoint)

We check the left-hand limit against the function value.

LHL: . Value at : . Since , is not continuous from the left at , so it is discontinuous at .

Step 4: Sum the points of discontinuity.

The set of all points of discontinuity in the interval is .

The sum of these points is:

The final sum is 17.

To solve this problem, we need to analyze the relation defined on the set by the condition if and only if .

First, consider and list all pairs satisfying the condition :

Counting all these pairs, we have elements in .

Next, we need to make symmetric. A relation is symmetric if, whenever , then as well. Our task is to make this relation symmetric by adding the minimum number of pairs.

Let's analyze:

• For : All pairs are already symmetric because no extra is needed as they exist already or are the same pair.
• For : The pairs , , imply we need to add , , .
• For : Pairs needing symmetry are , .
• For : Pair needing symmetry is .

Summing the extra pairs needed: , we must add pairs to ensure all relations are symmetric.

Adding these pairs to the existing 15 elements in , we have .

The final answer is 25, which means:

• The relation has 15 original pairs satisfying .
• 10 additional pairs are needed to make it symmetric.

To find the value of , we first need to determine the domain of the function , which is the intersection of the domains of its two component functions.

Concept Used:

The domain of a function is the set of all possible input values for which the function is defined.

1. Domain of Logarithmic Function: For a function , the argument must be strictly positive, i.e., .

2. Domain of Inverse Cosine Function: For a function , the argument must be within the closed interval , i.e., .

3. Domain of Combined Function: The domain of is the intersection of the domains of and .

Step-by-Step Solution:

Step 1: Determine the domain of the logarithmic part .

The argument of the logarithm must be positive:

First, we factor the quadratic denominator: . The inequality becomes:

The critical points are and . Using the wavy curve method (sign analysis), we find the intervals where the expression is positive.

The solution to this inequality is . Let this be Domain .

Step 2: Determine the domain of the inverse cosine part .

The argument of the inverse cosine function must satisfy:

This can be split into two separate inequalities. Also, note that .

First inequality:

The solution for this part is .

Second inequality:

The solution for this part is .

The domain is the intersection of the solutions to these two inequalities: . This gives:

Step 3: Find the overall domain of by taking the intersection of and .

The intersection of and is empty.

The intersection of and is .

Thus, the domain of is .

Step 4: Identify and from the domain.

The given domain format is . Comparing this with our result , we get:

Final Computation & Result:

Now, we calculate the value of the expression .

The value of is 12.