### Related Formula A function is one-one if its derivative is strictly monotonic (always positive or always negative) across its domain. It is onto if its range equals its co-domain. ### Core Logic Rewrite the function by multiplying the numerator and denominator by 2^x$2^x$: f(x) = frac2^2x - 12^2x + 1 = 1 - frac22^2x + 1$f(x) = \frac{2^{2x} - 1}{2^{2x} + 1} = 1 - \frac{2}{2^{2x} + 1}$ ### Step 1: Check One-One property Differentiating f(x)$f(x)$ with respect to x$x$ : f'(x) = frac2(2^2x + 1)^2 cdot 2 cdot 2^2x cdot ln 2 = frac4 cdot 2^2x cdot ln 2(2^2x + 1)^2$f'(x) = \frac{2}{(2^{2x} + 1)^2} \cdot 2 \cdot 2^{2x} \cdot \ln 2 = \frac{4 \cdot 2^{2x} \cdot \ln 2}{(2^{2x} + 1)^2}$ Since 2^2x > 0$2^{2x} > 0$ and ln 2 > 0$\ln 2 > 0$, f'(x) > 0$f'(x) > 0$ always. Thus, f(x)$f(x)$ is strictly increasing, confirming it is a one-one function. ### Step 2: Check Onto property Analyze the limits at boundaries [cite: 3880, 3881]: lim_x to -infty f(x) = 1 - frac20 + 1 = -1$\lim_{x \to -\infty} f(x) = 1 - \frac{2}{0 + 1} = -1$ lim_x to infty f(x) = 1 - 0 = 1$\lim_{x \to \infty} f(x) = 1 - 0 = 1$ Thus, the range of the function is (-1, 1)$(-1, 1)$. Since the given co-domain is (-infty, 1)$(-\infty, 1)$ and textRange neq textCo-domain$\text{Range} \neq \text{Co-domain}$ , the function is not onto. ### Pattern Recognition The expression given is a shifted form of the hyperbolic tangent function tanh(x ln 2)$\tanh(x \ln 2)$. Hyperbolic tangent always maps to (-1, 1)$(-1, 1)$, making its restriction against (-infty, 1)$(-\infty, 1)$ non-surjective (not onto). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions

### Related Formula Fundamental Counting Principle: If an operation can be performed in n_1$n_1$ ways, followed by a second operation in n_2$n_2$ ways, the total configurations equal n_1 times n_2$n_1 \times n_2$. ### Core Logic The domain set contains integers from 1 to 100. We divide the mapping requirements across distinct subsets of this domain[cite: 4064, 4068, 4069].

### Step 1: Choose the single element from \1, 2, dots, 98\$\{1, 2, \dots, 98\}$ We must assign the image value 1 to exactly one positive integer from the \subset \1, 2, dots, 98\$\{1, 2, \dots, 98\}$. The number of ways to pick this single element is : binom981 = 98 text ways$\binom{98}{1} = 98 \text{ ways}$ ### Step 2: Mapping remaining elements The remaining 97 elements in the \1, 2, dots, 98\$\{1, 2, \dots, 98\}$ \subset cannot map to 1, so they must map to 0. This leaves exactly 1 choice per remaining element. For the final two elements in the domain, 99 and 100, there are no structural constraints [cite: 4068, 4069]: - Element 99 can map to either 0 or 1 (2 options) . - Element 100 can map to either 0 or 1 (2 options). ### Step 3: Total functions combination Multiply the independent choices together : textTotal functions = 98 times 2 times 2 = 392$\text{Total functions} = 98 \times 2 \times 2 = 392$ [cite: 4065, 4067] ### Pattern Recognition Separate domains tightly into restricted blocks vs completely free components. Realizing that elements 99 and 100 behave independently with full co-domain targets leaves a clear product formulation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions Class 11 Mathematics: Permutations and Combinations

### 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