Consider two sets A = xin Z:|(|x - 3| - 3)|≤ 1$A = \{x\in \mathbb{Z}:|(|x - 3| - 3)|\leq 1\}$ and B = xin R - 1,2 :((x - 2)(x - 4))/(x - 1) ₑ(|x - 2|) = 0$B = \left\{x\in \mathbb{R} - \{1,2\} :\frac{(x - 2)(x - 4)}{x - 1}\log_{e}(|x - 2|) = 0\right\}$. Then the number of onto functions f: A → B$f: A \to B$ is equal to:

Related Formula

A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

Core Logic

First, partition set A$A$ into equivalence classes modulo 3$3$: Numbers of form 3K$3K$: 0, 3, 6, 9$\{0, 3, 6, 9\}$ $\implies$ 4$4$ elements Numbers of form 3K+1$3K+1$: 1, 4, 7$\{1, 4, 7\}$ $\implies$ 3$3$ elements Numbers of form 3K+2$3K+2$: 2, 5, 8$\{2, 5, 8\}$ $\implies$ 3$3$ elements

Any pair (x, y)$(x, y)$ inside the same group will have a difference |x - y|$|x - y|$ that is a multiple of 3$3$. The total number of elements in the relation R$R$ is the sum of pairs possible from each group:

Statement I claims n(R) = 36$n(R) = 36$, so it is false.

Step 1: Check Statement II

Reflexive: |x-x| = 0$|x-x| = 0$, which is a multiple of 3$3$. (True) Symmetric: If |x-y|$|x-y|$ is a multiple of 3$3$, then |y-x|$|y-x|$ is also a multiple of 3$3$. (True) Transitive: If (x-y) = 3λ$(x-y) = 3\lambda$ and (y-z) = 3μ$(y-z) = 3\mu$, then (x-z) = (x-y) + (y-z) = 3(λ + μ)$(x-z) = (x-y) + (y-z) = 3(\lambda + \mu)$, which is a multiple of 3$3$. (True) Thus, R$R$ is an equivalence relation. Statement II is true.

Pattern Recognition

Relations defined by divisibility of differences inherently form congruence classes. Size of relation = sum of squares of equivalence class sizes.

Chapter Mix

Class 11 Maths: Sets, Relations and Functions

Related Formula

Core Logic

For the domain, three conditions must be satisfied:

• Argument > 0$> 0$: 18x² - 11x + 1 > 0$18x^2 - 11x + 1 > 0$
• Base > 0$> 0$: 10x² - 17x + 7 > 0$10x^2 - 17x + 7 > 0$
• Base ≠ 1$\neq 1$: 10x² - 17x + 7 ≠ 1$10x^2 - 17x + 7 \neq 1$

Step 1: Argument Condition

x < (1)/(9)$x < \frac{1}{9}$ or x > (1)/(2)$x > \frac{1}{2}$

Step 2: Base Positivity

x < (7)/(10)$x < \frac{7}{10}$ or x > 1$x > 1$

Step 3: Base Not Equal to 1

Step 4: Intersection and Value Evaluation

Taking the intersection of all conditions: x in (-∞, (1)/(9)) ((1)/(2), (7)/(10)) (1, ∞) - (6)/(5)$x \in \left(-\infty, \frac{1}{9}\right) \cup \left(\frac{1}{2}, \frac{7}{10}\right) \cup (1, \infty) - \left\{\frac{6}{5}\right\}$ Comparing with (-∞, a) (b, c) (d, ∞) - e$(-\infty, a) \cup (b, c) \cup (d, \infty) - \{e\}$: a = (1)/(9)$a = \frac{1}{9}$, b = (1)/(2)$b = \frac{1}{2}$, c = (7)/(10)$c = \frac{7}{10}$, d = 1$d = 1$, e = (6)/(5)$e = \frac{6}{5}$

Pattern Recognition

A standard combined inequality problem. Always factorize quadratics early and map the boundaries on a number line to prevent overlapping interval errors.

Chapter Mix

Class 11 Maths: Relations and Functions Class 11 Maths: Linear Inequalities

Related Formula

Count combinations (a, b)$(a, b)$ and (c, d)$(c, d)$ from the set S = 1, 2, 3, 4$S = \{1, 2, 3, 4\}$ fulfilling 2a + 3b = 3c + 4d$2a + 3b = 3c + 4d$.

Core Logic

Evaluate possible values for LHS = 2a + 3b$2a + 3b$ where a,b in 1, 2, 3, 4$a,b \in \{1, 2, 3, 4\}$. Min value = 5$5$, Max value = 20$20$. Evaluate RHS = 3c + 4d$3c + 4d$ where c,d in 1, 2, 3, 4$c,d \in \{1, 2, 3, 4\}$. Find exact matches.

Step 1: Mapping Outputs

Step 2: Counting Intersections

Matches found: Value 7: (2,1)$(2,1)$ matches (1,1)$(1,1)$ → 1 pair Value 10: (2,2)$(2,2)$ matches (2,1)$(2,1)$ → 1 pair Value 11: (1,3), (4,1)$(1,3), (4,1)$ match (1,2)$(1,2)$ → 2 pairs Value 13: (2,3)$(2,3)$ matches (3,1)$(3,1)$ → 1 pair Value 14: (1,4), (4,2)$(1,4), (4,2)$ match (2,2)$(2,2)$ → 2 pairs Value 15: (3,3)$(3,3)$ matches (1,3)$(1,3)$ → 1 pair Value 16: (2,4)$(2,4)$ matches (4,1)$(4,1)$ → 1 pair Value 17: (4,3)$(4,3)$ matches (3,2)$(3,2)$ → 1 pair Value 18: (3,4)$(3,4)$ matches (2,3)$(2,3)$ → 1 pair Value 20: (4,4)$(4,4)$ matches (4,2)$(4,2)$ → 1 pair

Total matches = 1+1+2+1+2+1+1+1+1+1 = 12$1+1+2+1+2+1+1+1+1+1 = 12$.

Pattern Recognition

For tiny finite sets, brute forcing the 4 × 4 = 16$4 \times 4 = 16$ mappings of LHS and RHS separately and tallying equal values is the quickest algorithm to avoid counting logic errors.

Chapter Mix

Class 11 Maths: Relations and Functions

Related Formula

Core Logic

For f(x) = ⁻¹((1)/(x² - 2x - 2))$f(x) = \sin^{-1}\left(\frac{1}{x^2 - 2x - 2}\right)$, the inner argument must lie in [-1, 1]$[-1, 1]$.

Step 1: Solving Right Inequality

Factorize numerator and denominator: Numerator roots: x² - 2x - 3 = (x - 3)(x + 1)$x^2 - 2x - 3 = (x - 3)(x + 1)$ Denominator roots: x² - 2x - 2 = 0 x = 2 ± √(4 + 8)2 = 1 ± √(3)$x^2 - 2x - 2 = 0 \implies x = \frac{2 \pm \sqrt{4 + 8}}{2} = 1 \pm \sqrt{3}$

Using the wavy curve method, the intervals are: x in (-∞, -1] (1 - √(3), 1 + √(3)) [3, ∞)$x \in (-\infty, -1] \cup (1 - \sqrt{3}, 1 + \sqrt{3}) \cup [3, \infty)$ --- (1)

Step 2: Solving Left Inequality

Factorize numerator roots: x² - 2x - 1 = 0 x = 2 ± √(4 + 4)2 = 1 ± √(2)$x^2 - 2x - 1 = 0 \implies x = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$

Using the wavy curve method, the intervals are: x in (-∞, 1 - √(3)) [1 - √(2), 1 + √(2)] (1 + √(3), ∞)$x \in (-\infty, 1 - \sqrt{3}) \cup [1 - \sqrt{2}, 1 + \sqrt{2}] \cup (1 + \sqrt{3}, \infty)$ --- (2)

Step 3: Intersection of Intervals

Take the intersection of (1) and (2). Note that: 1 - √(3) ≈ 1 - 1.732 = -0.732$1 - \sqrt{3} \approx 1 - 1.732 = -0.732$ 1 - √(2) ≈ 1 - 1.414 = -0.414$1 - \sqrt{2} \approx 1 - 1.414 = -0.414$ 1 + √(2) ≈ 1 + 1.414 = 2.414$1 + \sqrt{2} \approx 1 + 1.414 = 2.414$ 1 + √(3) ≈ 1 + 1.732 = 2.732$1 + \sqrt{3} \approx 1 + 1.732 = 2.732$

Intersection gives: x in (-∞, -1] [1 - √(2), 1 + √(2)] [3, ∞)$x \in (-\infty, -1] \cup [1 - \sqrt{2}, 1 + \sqrt{2}] \cup [3, \infty)$

Comparing this with (-∞, α] [β, γ] [δ, ∞)$(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$: α = -1$\alpha = -1$ β = 1 - √(2)$\beta = 1 - \sqrt{2}$ γ = 1 + √(2)$\gamma = 1 + \sqrt{2}$ δ = 3$\delta = 3$

Pattern Recognition

When dealing with rational polynomial inequalities for ⁻¹(g(x))$\sin^{-1}(g(x))$, always break into -1 ≤ g(x)$-1 \leq g(x)$ and g(x) ≤ 1$g(x) \leq 1$, and do not cross-multiply terms across the inequality if the sign of the denominator is undetermined.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Functions