Let mathrmA = \1, 2, 3, dots, 100\$\mathrm{A} = \{1, 2, 3, \dots, 100\}$ and mathrmR$\mathrm{R}$ be a relation on mathrmA$\mathrm{A}$ such that mathrmR = \(a, b) : a = 2b + 1\$\mathrm{R} = \{(a, b) : a = 2b + 1\}$. Let (a_1, a_2), (a_2, a_3), (a_3, a_4), dots, (a_k, a_k+1)$(a_1, a_2), (a_2, a_3), (a_3, a_4), \dots, (a_k, a_{k+1})$ be a sequence of k$k$ elements of mathrmR$\mathrm{R}$ such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k$k$, for which such a sequence exists, is equal to:

### Related Formula textTotal number of symmetric relations on set with n text elements = 2^fracn(n+1)2$$\text{Total number of symmetric relations on set with } n \text{ elements} = 2^{\frac{n(n+1)}{2}}$$ textTotal number of relations both symmetric and reflexive = 2^fracn(n-1)2$$\text{Total number of relations both symmetric and reflexive} = 2^{\frac{n(n-1)}{2}}$$ ### Core Logic A relation is represented by an n times n$n \times n$ matrix. For symmetry, the upper triangle determines the lower triangle completely. The diagonal elements (a_ii$a_{ii}$) can be either 0 or 1. Number of independent elements (upper triangle + diagonal) = fracn^2 - n2 + n = fracn(n+1)2$\frac{n^2 - n}{2} + n = \frac{n(n+1)}{2}$. Total symmetric relations = 2^fracn(n+1)2$2^{\frac{n(n+1)}{2}}$. For a relation to be *both* symmetric and reflexive, all diagonal elements must be 1$1$. The remaining independent elements (upper triangle) are fracn(n-1)2$\frac{n(n-1)}{2}$. Total symmetric AND reflexive relations = 2^fracn(n-1)2$2^{\frac{n(n-1)}{2}}$. ### Step 1: Calculating for n = 4 For a set of n = 4$n = 4$ elements: Total number of symmetric relations: N_textsymm = 2^frac4(4+1)2 = 2^10 = 1024$$N_{\text{symm}} = 2^{\frac{4(4+1)}{2}} = 2^{10} = 1024$$ Total number of symmetric and reflexive relations: N_textsymm+refl = 2^frac4(4-1)2 = 2^6 = 64$$N_{\text{symm+refl}} = 2^{\frac{4(4-1)}{2}} = 2^6 = 64$$ ### Step 2: Finding Requested Number We need the number of symmetric relations that are NOT reflexive. N = N_textsymm - N_textsymm+refl$$N = N_{\text{symm}} - N_{\text{symm+refl}}$$ N = 1024 - 64 = 960$$N = 1024 - 64 = 960$$ ### Pattern Recognition Condition 'not' acts as a simple set subtraction. Count the entire universe of symmetric relations (diagonal is free) and subtract the subset where reflexivity locks the diagonal elements to 1. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sets, Relations and Functions

### Related Formula -1 leq textargument of cos^-1 leq 1$$-1 \leq \text{argument of } \cos^{-1} \leq 1$$ log_a x text is defined for x > 0 text and base a > 0, a neq 1$$\log_a x \text{ is defined for } x > 0 \text{ and base } a > 0, a \neq 1$$ f^-1 text signifies denominator neq 0$$f^{-1} \text{ signifies denominator } \neq 0$$ ### Core Logic For f(x)$f(x)$ to be defined, both terms must be defined independently. Term 1: cos^-1left(frac2 - |x|4right)$\cos^{-1}\left(\frac{2 - |x|}{4}\right)$ The domain requires: -1 leq frac2 - |x|4 leq 1$$-1 \leq \frac{2 - |x|}{4} \leq 1$$ -4 leq 2 - |x| leq 4$$-4 \leq 2 - |x| \leq 4$$ -6 leq -|x| leq 2$$-6 \leq -|x| \leq 2$$ -2 leq |x| leq 6$$-2 \leq |x| \leq 6$$ Since |x|$|x|$ is always non-negative, the valid condition reduces to |x| leq 6$|x| \leq 6$. This implies x in [-6, 6] quad dots (1)$x \in [-6, 6] \quad \dots (1)$ ### Step 1: Domain of logarithmic term Term 2: (log_e(3 - x))^-1 = frac1log_e(3 - x)$(\log_e(3 - x))^{-1} = \frac{1}{\log_e(3 - x)}$ The argument of the logarithm must be positive: 3 - x > 0 Rightarrow x < 3 quad dots (2)$$3 - x > 0 \Rightarrow x < 3 \quad \dots (2)$$ Also, the denominator cannot be zero: log_e(3 - x) neq 0 Rightarrow 3 - x neq 1 Rightarrow x neq 2 quad dots (3)$$\log_e(3 - x) \neq 0 \Rightarrow 3 - x \neq 1 \Rightarrow x \neq 2 \quad \dots (3)$$ ### Step 2: Combining intersection regions Taking the intersection of (1), (2), and (3): x in [-6, 6] cap (-infty, 3) cap x neq 2$$x \in [-6, 6] \cap (-\infty, 3) \cap x \neq 2$$ x in [-6, 3) - \2\$$x \in [-6, 3) - \{2\}$$ Comparing this with the given domain [-alpha, beta) - \gamma\$[-\alpha, \beta) - \{\gamma\}$: alpha = 6$\alpha = 6$, beta = 3$\beta = 3$, gamma = 2$\gamma = 2$. ### Step 3: Calculating target value alpha + beta + gamma = 6 + 3 + 2 = 11$\alpha + \beta + \gamma = 6 + 3 + 2 = 11$ ### Pattern Recognition Isolate the constraints of piecewise function composition: inner constraints (argument bounds), denominator zero-checks, and fundamental range boundaries. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Relations and Functions

### Related Formula n(M cup P cup C) = n(M) + n(P) + n(C) - n(M cap P) - n(P cap C) - n(M cap C) + n(M cap P cap C)$$n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C)$$ ### Core Logic

Let x$x$ be the number of students who passed in all three subjects, so n(M cap P cap C) = x$n(M \cap P \cap C) = x$. We are given: n(M) = 20, n(P) = 25, n(C) = 16$n(M) = 20, n(P) = 25, n(C) = 16$ n(M cup P cup C) = 40$n(M \cup P \cup C) = 40$ Constraints: n(M cap P) le 11$n(M \cap P) \le 11$ n(P cap C) le 15$n(P \cap C) \le 15$ n(M cap C) le 15$n(M \cap C) \le 15$ If we let n(M cap P) = 11$n(M \cap P) = 11$, then the region representing only M cap P$M \cap P$ is 11 - x$11 - x$. For this to be non-negative, x le 11$x \le 11$. ### Step 1: Testing x = 11

Assume x = 11$x = 11$. To maximize, let's fix n(M cap P) = 11$n(M \cap P) = 11$. This leaves 0$0$ students in (M cap P) - (M cap P cap C)$(M \cap P) - (M \cap P \cap C)$. Let n(M cap C) = 11 + z$n(M \cap C) = 11 + z$ and n(P cap C) = 11 + y$n(P \cap C) = 11 + y$. Constraints limit these subsets: 11 + z le 15 Rightarrow z le 4$11 + z \le 15 \Rightarrow z \le 4$ 11 + y le 15 Rightarrow y le 4$11 + y \le 15 \Rightarrow y \le 4$ Filling the Venn diagram nodes: Only M = 20 - 11 - z - 0 = 9 - z$M = 20 - 11 - z - 0 = 9 - z$ Only P = 25 - 11 - y - 0 = 14 - y$P = 25 - 11 - y - 0 = 14 - y$ Only C = 16 - 11 - y - z = 5 - y - z$C = 16 - 11 - y - z = 5 - y - z$ Summing all disjoint areas: Total = (9 - z) + 0 + (14 - y) + z + 11 + y + (5 - y - z) = 40$= (9 - z) + 0 + (14 - y) + z + 11 + y + (5 - y - z) = 40$ 39 - y - z = 40 Rightarrow y + z = -1$39 - y - z = 40 \Rightarrow y + z = -1$ Since subsets cannot be negative, x=11$x=11$ is impossible. ### Step 2: Testing x = 10

Assume x = 10$x = 10$. Let's construct a valid distribution with x = 10$x = 10$. We can choose intersection values: n(M cap P) = 10$n(M \cap P) = 10$, so only M cap P$M \cap P$ is 0$0$. n(M cap C) = 11$n(M \cap C) = 11$, so only M cap C$M \cap C$ is 1$1$. n(P cap C) = 10$n(P \cap C) = 10$, so only P cap C$P \cap C$ is 0$0$. This gives: Only M = 20 - 10 - 0 - 1 = 9$M = 20 - 10 - 0 - 1 = 9$ Only P = 25 - 10 - 0 - 0 = 15$P = 25 - 10 - 0 - 0 = 15$ Only C = 16 - 10 - 1 - 0 = 5$C = 16 - 10 - 1 - 0 = 5$ Summing up = 9 + 15 + 5 + 0 + 1 + 0 + 10 = 40$9 + 15 + 5 + 0 + 1 + 0 + 10 = 40$. All conditions are satisfied perfectly. Hence, the maximum number is 10. ### Pattern Recognition In multi-constraint Venn diagram maximization, test the theoretical upper bound sequentially downward until a non-negative configuration for all disjoint subsets is achieved. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sets

### Related Formula textSize of Power Set |P(A)| = 2^|A|$$\text{Size of Power Set } |P(A)| = 2^{|A|}$$ ### Core Logic For a function f: A to P(A)$f: A \to P(A)$, each element a in A$a \in A$ maps to a subset of A$A$. The condition is a in f(a)$a \in f(a)$. This means that the subset assigned to a$a$ must contain the element a$a$. How many such subsets exist in P(A)$P(A)$? Since |A| = 7$|A| = 7$, there are 2^7$2^7$ total subsets. Exactly half of these contain a specific element a$a$. Number of subsets containing 'a' is 2^7-1 = 2^6 = 64$2^{7-1} = 2^6 = 64$. ### Step 1: Calculating total functions For each element in A$A$ (which has 7 elements), there are independently 2^6$2^6$ choices for its image in P(A)$P(A)$. Total options for 1 will be 2^6$2^6$. Similarly, for every other element up to 7. Hence, total number of functions is: (2^6) times (2^6) times (2^6) times (2^6) times (2^6) times (2^6) times (2^6) = (2^6)^7 = 2^42$$(2^6) \times (2^6) \times (2^6) \times (2^6) \times (2^6) \times (2^6) \times (2^6) = (2^6)^7 = 2^{42}$$ ### Step 2: Extracting m and n We need to express this as m^n$m^n$ where m$m$ is the least natural number. The least base possible is m = 2$m = 2$. Thus, m = 2$m = 2$ and n = 42$n = 42$. m + n = 2 + 42 = 44$m + n = 2 + 42 = 44$. ### Pattern Recognition Permutations into power sets with inclusion constraints always isolate the constrained element and leave the remaining elements free to permute in 2^k-1$2^{k-1}$ subsets. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sets Class 12 Maths: Relations and Functions

### Related Formula textA relation R_1 text is symmetric if (x,y) in R_1 implies (y,x) in R_1$$\text{A relation } R_1 \text{ is symmetric if } (x,y) \in R_1 \implies (y,x) \in R_1$$ ### Core Logic Given 2x = 3y$2x = 3y$ for (x,y) in A times A$(x,y) \in A \times A$. Elements in A$A$ range up to 100. Thus y = frac2x3$y = \frac{2x}{3}$. For y$y$ to be an integer, x$x$ must be a multiple of 3. x in \3, 6, 9, dots, 99\$x \in \{3, 6, 9, \dots, 99\}$ are valid inputs. There are frac993 = 33$\frac{99}{3} = 33$ such elements. The maximum value of x$x$ is 99$99$, which gives y = frac2(99)3 = 66 le 100$y = \frac{2(99)}{3} = 66 \le 100$ (valid). The relation R$R$ explicitly contains 33 elements: R = \(3,2), (6,4), dots, (99,66)\$R = \{(3,2), (6,4), \dots, (99,66)\}$. None of the pairs in R$R$ have x=y$x=y$ because 2x=3x implies x=0$2x=3x \implies x=0$, which is not in A$A$. For mathbfR_1$\mathbf{R}_1$ to be a symmetric relation containing mathbfR$\mathbf{R}$, it must contain all elements of mathbfR$\mathbf{R}$ and their inverse pairs (y,x)$(y,x)$. Minimum number of elements in mathbfR_1 = 2 times |mathbfR| = 2 times 33 = 66$\mathbf{R}_1 = 2 \times |\mathbf{R}| = 2 \times 33 = 66$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Relations and Functions