### Related Formula sec^2 theta - tan^2 theta = 1$\sec^2 \theta - \tan^2 \theta = 1$ ### Core Logic To show R$R$ is an equivalence relation, verify reflexive, symmetric, and transitive properties sequentially. ### Step 1: Reflexive Property For any x in [0, pi/2)$x \in [0, \pi/2)$: sec^2 x - tan^2 x = 1 implies xRx quad text(Reflexive)$\sec^2 x - \tan^2 x = 1 \implies xRx \quad \text{(Reflexive)}$ ### Step 2: Symmetric Property If xRy implies sec^2 x - tan^2 y = 1$xRy \implies \sec^2 x - \tan^2 y = 1$. Using identities: (1 + tan^2 x) - (sec^2 y - 1) = 1 implies sec^2 y - tan^2 x = 1 implies yRx quad text(Symmetric)$(1 + \tan^2 x) - (\sec^2 y - 1) = 1 \implies \sec^2 y - \tan^2 x = 1 \implies yRx \quad \text{(Symmetric)}$ ### Step 3: Transitive Property If $ ### Step 3: Transitive Property If $xRy and $ and $yRz \implies \sec^2 x - \tan^2 y = 1 and sec^2 y - tan^2 z = 1. Adding both equations: $ and \sec^2 y - \tan^2 z = 1. Adding both equations: $sec^2 x - tan^2 y + sec^2 y - tan^2 z = 2$\sec^2 x - \tan^2 y + \sec^2 y - \tan^2 z = 2$ $ $sec^2 x + (sec^2 y - tan^2 y) - tan^2 z = 2 implies sec^2 x + 1 - tan^2 z = 2$\sec^2 x + (\sec^2 y - \tan^2 y) - \tan^2 z = 2 \implies \sec^2 x + 1 - \tan^2 z = 2$ $ $sec^2 x - tan^2 z = 1 implies xRz quad text(Transitive)$\sec^2 x - \tan^2 z = 1 \implies xRz \quad \text{(Transitive)}$ Hence, $ Hence, $R is an equivalence relation. ### Pattern Recognition Converting the relation constraint to $ is an equivalence relation. ### Pattern Recognition Converting the relation constraint to $\sec^2 x - 1 = \tan^2 y \implies \tan^2 x = \tan^2 y$ makes the equivalence property obvious by basic equality comparison rules. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions

### Related Formula For composite functions, g(f(x))$g(f(x))$ is determined by substituting the range of f(x)$f(x)$ into the appropriate domain intervals of g(y)$g(y)$: g(f(x)) = begincases f(x) & , & f(x) ge 0 \\ e^f(x) & , & f(x) < 0 endcases$g(f(x)) = \begin{cases} f(x) & , & f(x) \ge 0 \\ e^{f(x)} & , & f(x) < 0 \end{cases}$ ### Core Logic Let us analyze the definition of g(f(x))$g(f(x))$ branch-by-branch based on the domain of x$x$: 1. **Case 1: x le 0$x \le 0$** Here, f(x) = e^-x$f(x) = e^{-x}$. Since x le 0$x \le 0$, -x ge 0 implies e^-x ge 1 > 0$-x \ge 0 \implies e^{-x} \ge 1 > 0$. Since f(x) ge 0$f(x) \ge 0$, we use the upper branch of g(y)$g(y)$: g(f(x)) = f(x) = e^-x$g(f(x)) = f(x) = e^{-x}$ 2. **Case 2: x > 0$x > 0$** Here, f(x) = log_e x$f(x) = \log_e x$. - Subcase (a): If f(x) ge 0 implies log_e x ge 0 implies x ge 1$f(x) \ge 0 \implies \log_e x \ge 0 \implies x \ge 1$. Then, g(f(x)) = f(x) = log_e x$g(f(x)) = f(x) = \log_e x$. - Subcase (b): If f(x) < 0 implies log_e x < 0 implies 0 < x < 1$f(x) < 0 \implies \log_e x < 0 \implies 0 < x < 1$. Then, g(f(x)) = e^f(x) = e^log_e x = x$g(f(x)) = e^{f(x)} = e^{\log_e x} = x$. ### Step 1: Constructing the Composition Function Combining the branches obtained, the composite function is: g(f(x)) = begincases e^-x & , & x le 0 \\ x & , & 0 < x < 1 \\ log_e x & , & x ge 1 endcases$g(f(x)) = \begin{cases} e^{-x} & , & x \le 0 \\ x & , & 0 < x < 1 \\ \log_e x & , & x \ge 1 \end{cases}$

### Step 2: Injectivity and Surjectivity Analysis - **Injectivity (One-One Check):** Let's test two different inputs: x_1 = 0$x_1 = 0$ and x_2 = e$x_2 = e$. g(f(0)) = e^-0 = 1$g(f(0)) = e^{-0} = 1$ g(f(e)) = log_e e = 1$g(f(e)) = \log_e e = 1$ Since distinct inputs yield identical outputs (g(f(0)) = g(f(e)) = 1$g(f(0)) = g(f(e)) = 1$), the function is **many-one** (not one-one). - **Surjectivity (Onto Check):** Evaluating the range across the branches: - For x le 0$x \le 0$, e^-x in [1, infty)$e^{-x} \in [1, \infty)$. - For 0 < x < 1$0 < x < 1$, x in (0, 1)$x \in (0, 1)$. - For x ge 1$x \ge 1$, \log_e x in [0, infty)$\log_e x \in [0, \infty)$. The union of these sets gives the total range as [0, infty)$[0, \infty)$. Since the codomain is given as mathbbR$\mathbb{R}$, textRange neq textCodomain$\text{Range} \neq \text{Codomain}$, so the function is **into** (not onto). Therefore, the function is neither one-one nor onto. ### Pattern Recognition Sees: Piecewise branch composition. Shortcut: Sketching the graph quickly shows that a horizontal line at y=1$y=1$ intersects the function multiple times (not one-one) and no part of the graph goes below the x-axis (not onto). Trap: Always determine the range of the inner function first to select the correct branch of the outer function. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions

### Related Formula Set Difference Cardinality Identity: n(R_1 - R_2) = n(R_1) - n(R_1 cap R_2)$n(R_1 - R_2) = n(R_1) - n(R_1 \cap R_2)$ ### Core Logic Let's first determine the number of pairs in relation R_1$R_1$, where b$b$ is divisible by a$a$ (b = k cdot a$b = k \cdot a$): For each element a in \1, 2, dots, 20\$a \in \{1, 2, \dots, 20\}$, the number of multiples b le 20$b \le 20$ is equal to leftlfloor frac20a rightrfloor$\left\lfloor \frac{20}{a} \right\rfloor$. ### Step 1: Calculate Cardinality of R1 Summing the total possible pairings for each distinct a$a$: - a=1 implies 20$a=1 \implies 20$ - a=2 implies 10$a=2 \implies 10$ - a=3 implies 6$a=3 \implies 6$ - a=4 implies 5$a=4 \implies 5$ - a=5 implies 4$a=5 \implies 4$ - a=6 implies 3$a=6 \implies 3$ - a=7, 8, 9, 10 implies 2 times 4 = 8$a=7, 8, 9, 10 \implies 2 \times 4 = 8$ - a=11 text to 20 implies 1 times 10 = 10$a=11 \text{ to } 20 \implies 1 \times 10 = 10$ n(R_1) = 20 + 10 + 6 + 5 + 4 + 3 + 8 + 10 = 66$n(R_1) = 20 + 10 + 6 + 5 + 4 + 3 + 8 + 10 = 66$ ### Step 2: Calculate Cardinality of Intersection The intersection R_1 cap R_2$R_1 \cap R_2$ requires both b$b$ to be divisible by a$a$ and a$a$ to be divisible by b$b$. Since all elements are positive integers within the set domain, this statement holds true if and only if: a = b$a = b$ Thus, the matching intersections are all reflexive pairs: \(1,1), (2,2), dots, (20,20)\$\{(1,1), (2,2), \dots, (20,20)\}$, giving: n(R_1 cap R_2) = 20$n(R_1 \cap R_2) = 20$ ### Step 3: Evaluate Final Set Difference Applying the set difference relation: n(R_1 - R_2) = n(R_1) - n(R_1 cap R_2) = 66 - 20 = 46$n(R_1 - R_2) = n(R_1) - n(R_1 \cap R_2) = 66 - 20 = 46$ ### Pattern Recognition Sees: Divisibility relations matched via sets concepts. Shortcut: Recognizing that mutual divisibility between positive integers implies absolute equality (a=b$a=b$) eliminates the need to detail individual intersection pairs manually. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions Class 11 Mathematics: Sets

### Related Formula Relation definitions: Reflexive: (X, X) in R$(X, X) \in R$ Symmetric: (X, Y) in R Rightarrow (Y, X) in R$(X, Y) \in R \Rightarrow (Y, X) \in R$ Transitive: (X, Y) in R text and (Y, Z) in R Rightarrow (X, Z) in R$(X, Y) \in R \text{ and } (Y, Z) \in R \Rightarrow (X, Z) \in R$ ### Core Logic Testing Reflexivity: M$M$ is the set of all subsets, which strictly includes the empty set phi$\phi$. For reflexivity, every element A in M$A \in M$ must satisfy A cap A ne phi$A \cap A \ne \phi$. However, for A = phi in M$A = \phi \in M$, phi cap phi = phi$\phi \cap \phi = \phi$, which violates the given relation condition. Hence, R$R$ is NOT reflexive. ### Step 1: Testing Symmetry Assume (A, B) in R$(A, B) \in R$. This implies A cap B ne phi$A \cap B \ne \phi$. By the commutative property of intersections, B cap A ne phi$B \cap A \ne \phi$. This means (B, A) in R$(B, A) \in R$. Hence, R$R$ is symmetric. ### Step 2: Testing Transitivity Take specific subsets to test condition leakage. Let A = \1, 2\$A = \{1, 2\}$, B = \2, 3\$B = \{2, 3\}$, C = \3, 4\$C = \{3, 4\}$. A cap B = \2\ ne phi Rightarrow (A, B) in R$A \cap B = \{2\} \ne \phi \Rightarrow (A, B) \in R$ B cap C = \3\ ne phi Rightarrow (B, C) in R$B \cap C = \{3\} \ne \phi \Rightarrow (B, C) \in R$ However, A cap C = phi$A \cap C = \phi$, which means (A, C) notin R$(A, C) \notin R$. Hence, R$R$ is NOT transitive. ### Step 3: Final Conclusion The relation is symmetric only. ### Pattern Recognition The empty set is a subset of every set. Since phi cap phi = phi$\phi \cap \phi = \phi$, any set intersection relation bounded over a universal powerset will automatically fail reflexivity unless the empty set is explicitly excluded from the domain. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Relations and Functions Class 11 Maths: Sets