### Related Formula One-one (Injective): f(x) = f(y) Rightarrow x = y$f(x) = f(y) \Rightarrow x = y$ Onto (Surjective): Range = Codomain. ### Core Logic Evaluate injectivity (one-one property): Check standard small inputs. f(2) = texthighest prime factor of 2 = 2$f(2) = \text{highest prime factor of } 2 = 2$. f(4) = texthighest prime factor of 4 = 2$f(4) = \text{highest prime factor of } 4 = 2$. Since f(2) = f(4) = 2$f(2) = f(4) = 2$ but 2 ne 4$2 \ne 4$, the function is many-to-one. ### Step 1: Evaluating Surjectivity Evaluate surjectivity (onto property): The codomain is defined as N$N$ (Natural numbers). Consider an element in the codomain that is NOT prime, for example, 4 in N$4 \in N$. The highest prime factor of any integer will ALWAYS be a prime number. It is impossible for the output of f(n)$f(n)$ to be a composite number like 4. Since 4 is not the image of any element in the domain, Range neq$\neq$ Codomain. Thus, the function is into (not onto). ### Step 2: Final Conclusion The function is neither one-one nor onto. ### Pattern Recognition Functions extracting "highest prime factors" or "digit sums" are universally many-to-one because multiple numbers share these properties. If the codomain is ALL natural numbers, but the output is strictly restricted to primes, it universally fails surjectivity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Relations and Functions

### Related Formula f(g(x)) = f(y) quad textwhere y = g(x)$f(g(x)) = f(y) \quad \text{where } y = g(x)$ The domain restrictions of f(y)$f(y)$ must be cross-checked against the range evaluated for g(x)$g(x)$. ### Core Logic First, analyze the range of the inner function g(x)$g(x)$: g(x) = -x$g(x) = -x$ for -3 le x le 0$-3 \le x \le 0$. The range here is [0, 3]$[0, 3]$. g(x) = x$g(x) = x$ for 0 lt x le 1$0 \lt x \le 1$. The range here is (0, 1]$(0, 1]$. Therefore, the complete range of g(x)$g(x)$ across its entire domain [-3, 1]$[-3, 1]$ is [0, 3]$[0, 3]$. Note that g(x) ge 0$g(x) \ge 0$ for all valid x$x$.

Now, plug g(x)$g(x)$ into f(x)$f(x)$: f(g(x)) = begincases2+2g(x),& -1 le g(x) lt 0 quad dots (1) \\ 1-fracg(x)3,& 0 le g(x) le 3 quad dots (2)endcases$f(g(x)) = \begin{cases}2+2g(x),& -1 \le g(x) \lt 0 \quad \dots (1) \\ 1-\frac{g(x)}{3},& 0 \le g(x) \le 3 \quad \dots (2)\end{cases}$

### Step 1: Evaluate the Branches Check branch (1): -1 le g(x) lt 0$-1 \le g(x) \lt 0$. Since we already established the range of g(x)$g(x)$ is entirely non-negative (g(x) ge 0$g(x) \ge 0$), this branch condition is never satisfied. No values of x$x$ map here (x in phi$x \in \phi$). Check branch (2): 0 le g(x) le 3$0 \le g(x) \le 3$. This perfectly aligns with the entire range of g(x)$g(x)$. Hence, this branch is active for all x in [-3, 1]$x \in [-3, 1]$. We evaluate f(g(x)) = 1 - fracg(x)3$f(g(x)) = 1 - \frac{g(x)}{3}$. Since g(x)$g(x)$ spans all values continuously from 0$0$ to 3$3$: When g(x) = 0 Rightarrow f(g(x)) = 1 - 0 = 1$g(x) = 0 \Rightarrow f(g(x)) = 1 - 0 = 1$. When g(x) = 3 Rightarrow f(g(x)) = 1 - 1 = 0$g(x) = 3 \Rightarrow f(g(x)) = 1 - 1 = 0$. Since g(x)$g(x)$ is continuous and covers [0, 3]$[0, 3]$, 1 - fracg(x)3$1 - \frac{g(x)}{3}$ covers [0, 1]$[0, 1]$ continuously. ### Step 2: Final Conclusion The range of f(g(x))$f(g(x))$ is purely evaluated from branch (2), leading to [0, 1]$[0, 1]$. ### Pattern Recognition When finding the range of f(g(x))$f(g(x))$, evaluate the global range of g(x)$g(x)$ first. Use that resulting interval as the "domain" input for f(x)$f(x)$ to trace the final output bounds, eliminating unneeded piecewise branches. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sets, Relations and Functions

### Related Formula textReflexive: (x,x) in R text for all x$\text{Reflexive: } (x,x) \in R \text{ for all } x$ textSymmetric: (x,y) in R implies (y,x) in R$\text{Symmetric: } (x,y) \in R \implies (y,x) \in R$ textTransitive: (x,y) in R text and (y,z) in R implies (x,z) in R$\text{Transitive: } (x,y) \in R \text{ and } (y,z) \in R \implies (x,z) \in R$ ### Core Logic Check Reflexivity: Let (a,b) in Z times Z$(a,b) \in Z \times Z$. (a,b) R (a,b) iff ab - ba$(a,b) R (a,b) \iff ab - ba$ is divisible by 5. ab - ba = 0$ab - ba = 0$. Since 0 is divisible by 5, the relation is strictly Reflexive. Check Symmetry: Let (a,b) R (c,d) implies ad - bc = 5k$(a,b) R (c,d) \implies ad - bc = 5k$ for some integer k$k$. We need to check (c,d) R (a,b)$(c,d) R (a,b)$, which evaluates to cb - da$cb - da$. Notice that cb - da = -(ad - bc) = -5k = 5(-k)$cb - da = -(ad - bc) = -5k = 5(-k)$. Since -k$-k$ is an integer, cb - da$cb - da$ is divisible by 5. The relation is Symmetric. ### Step 1: Test Transitivity To test transitivity, attempt to construct a counter-example where (a,b)R(c,d)$(a,b)R(c,d)$ and (c,d)R(e,f)$(c,d)R(e,f)$, but (a,b)$(a,b)$ does not relate to (e,f)$(e,f)$. Let's pick numbers carefully. We need ad-bc equiv 0 pmod 5$ad-bc \equiv 0 \pmod 5$ and cf-de equiv 0 pmod 5$cf-de \equiv 0 \pmod 5$. Try (a,b) = (3,1)$(a,b) = (3,1)$ and (c,d) = (10,5)$(c,d) = (10,5)$: ad-bc = (3)(5) - (1)(10) = 15 - 10 = 5$ad-bc = (3)(5) - (1)(10) = 15 - 10 = 5$ (Divisible by 5. Thus (3,1)R(10,5)$(3,1)R(10,5)$ holds). Try (c,d) = (10,5)$(c,d) = (10,5)$ and (e,f) = (1,1)$(e,f) = (1,1)$: cf-de = (10)(1) - (5)(1) = 10 - 5 = 5$cf-de = (10)(1) - (5)(1) = 10 - 5 = 5$ (Divisible by 5. Thus (10,5)R(1,1)$(10,5)R(1,1)$ holds). Now check (a,b)R(e,f)$(a,b)R(e,f)$, mapping (3,1)R(1,1)$(3,1)R(1,1)$: af-be = (3)(1) - (1)(1) = 3 - 1 = 2$af-be = (3)(1) - (1)(1) = 3 - 1 = 2$. 2 is NOT divisible by 5. Therefore, the relation is Not Transitive. ### Pattern Recognition Divisibility relations on cross-multiplied pairs (ad-bc$ad-bc$) are almost always reflexive and symmetric, but typically fail transitivity because the intermediate pairing scales elements in non-linear ways that break the modulo arithmetic chain. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sets, Relations and Functions

### 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