Let A=1,2,3,.....20$A=\{1,2,3,.....20\}$ Let R₁$R_{1}$ and R₂$R_{2}$ two relation on A such that R₁=(a,b):b is divisible by a$R_{1}=\{(a,b):b \text{ is divisible by } a\}$ and R₂=(a,b) a is an integral multiple of b$R_{2}=\{(a,b) \text{ a is an integral multiple of } b\}$. Then, number of elements in R₁-R₂$R_{1}-R_{2}$ is equal to

Related Formula

For inverse trigonometric functions ⁻¹(g(x))$\sin^{-1}(g(x))$ and ⁻¹(h(x))$\cos^{-1}(h(x))$, the arguments must satisfy:

Core Logic

Given f(x) = ⁻¹((2x - 5)/(11 - 3x)) + ⁻¹(2x² - 3x + 1)$f(x) = \cos^{-1}\left(\frac{2x - 5}{11 - 3x}\right) + \sin^{-1}(2x^2 - 3x + 1)$

We establish two simultaneous inequalities for the domain:

• -1 ≤ (2x - 5)/(11 - 3x) ≤ 1$-1 \leq \frac{2x - 5}{11 - 3x} \leq 1$
• -1 ≤ 2x² - 3x + 1 ≤ 1$-1 \leq 2x^2 - 3x + 1 \leq 1$

Step 1: Solve the Quadratic Inequality

From -1 ≤ 2x² - 3x + 1 ≤ 1$-1 \leq 2x^2 - 3x + 1 \leq 1$: Split into two parts: 2x² - 3x + 2 ≥ 0$2x^2 - 3x + 2 \geq 0$ (This is always true as discriminant D < 0, a > 0$D < 0, a > 0$) 2x² - 3x ≤ 0 ⇒ x(2x - 3) ≤ 0$2x^2 - 3x \leq 0 \Rightarrow x(2x - 3) \leq 0$

Step 2: Solve the Rational Inequality

From -1 ≤ (2x - 5)/(11 - 3x) ≤ 1$-1 \leq \frac{2x - 5}{11 - 3x} \leq 1$: Part A: (2x - 5)/(11 - 3x) + 1 ≥ 0 ⇒ (2x - 5 + 11 - 3x)/(11 - 3x) ≥ 0 ⇒ (6 - x)/(11 - 3x) ≥ 0$\frac{2x - 5}{11 - 3x} + 1 \geq 0 \Rightarrow \frac{2x - 5 + 11 - 3x}{11 - 3x} \geq 0 \Rightarrow \frac{6 - x}{11 - 3x} \geq 0$

Part B: (2x - 5)/(11 - 3x) - 1 ≤ 0 ⇒ (5x - 16)/(11 - 3x) ≤ 0 ⇒ x in (-∞, (16)/(5)] ((11)/(3), ∞)$\frac{2x - 5}{11 - 3x} - 1 \leq 0 \Rightarrow \frac{5x - 16}{11 - 3x} \leq 0 \Rightarrow x \in \left(-\infty, \frac{16}{5}\right] \cup \left(\frac{11}{3}, \infty\right)$

Intersection of Part A and Part B:

Step 3: Final Intersection

Taking the intersection of (i)$(i)$ and (ii)$(ii)$:

Comparing this with [α, β]$[\alpha, \beta]$, we have α = 0, β = (3)/(2)$\alpha = 0, \beta = \frac{3}{2}$.

Therefore, α + 2β = 0 + 2((3)/(2)) = 3$\alpha + 2\beta = 0 + 2\left(\frac{3}{2}\right) = 3$

Pattern Recognition

Whenever dealing with dual inverse trig terms, strictly isolate the bounding intervals [-1, 1]$[-1, 1]$ for each argument separately and use a number line intersection to find the strictest common region.

Chapter Mix

Class 12 Maths: Functions Class 11 Maths: Linear Inequalities

Related Formula

For a set with n$n$ elements, the number of relations that are both reflexive and symmetric is given by:

Core Logic

A reflexive relation must contain all diagonal pairs (x,x)$(x,x)$. There is only 1$1$ way to assign these elements (they MUST be present). A symmetric relation requires that if (x,y)$(x,y)$ is present, (y,x)$(y,x)$ must also be present. Thus, we only have the freedom to choose whether to include the unordered pairs x, y$\{x, y\}$ where x ≠ y$x \neq y$.

Step 1: Calculate the available independent pairs

Number of distinct elements n = 4$n = 4$. Total number of pairs in the cartesian product is n² = 16$n^2 = 16$. Number of diagonal pairs (reflexive necessity) = n = 4$= n = 4$. Remaining non-diagonal pairs = 16 - 4 = 12$= 16 - 4 = 12$. Since symmetry pairs them up (a,b) rightarrow (b,a)$(a,b) \leftrightarrow (b,a)$, there are exactly (12)/(2) = 6$\frac{12}{2} = 6$ independent choices.

Step 2: Calculate total relations

Each of the 6 independent pairs can either be included or excluded (2$2$ choices). Total relations = 1⁴ × 2⁶ = 64$= 1^4 \times 2^6 = 64$.

Pattern Recognition

Memorize the combinatorics of binary relations for n$n$ elements: Total = 2n²$= 2^{n^2}$, Reflexive = 2n(n-1)$= 2^{n(n-1)}$, Symmetric = 2n(n+1)/2$= 2^{n(n+1)/2}$, Reflexive & Symmetric = 2n(n-1)/2$= 2^{n(n-1)/2}$.

Chapter Mix

Class 12 Maths: Sets and Relations

Related Formula

Core Logic

Expand both set definitions onto the real number line to find explicit intervals for A$A$ and B$B$, then apply set operations to verify the options.

Step 1: Simplify Set A

This yields x in [-4, -2] [2, 4]$x \in [-4, -2] \cup [2, 4]$. So, A = [-4, -2] [2, 4]$A = [-4, -2] \cup [2, 4]$.

Step 2: Simplify Set B

|x - 2| > 1$|x - 2| > 1$

So, B = (-∞, 1) (3, ∞)$B = (-\infty, 1) \cup (3, \infty)$.

Step 3: Evaluate Options

A B = (-∞, 1) [2, ∞)$A \cup B = (-\infty, 1) \cup [2, \infty)$ (Option 1 is wrong, has 1]$1]$) A B = [-4, -2] (3, 4]$A \cap B = [-4, -2] \cup (3, 4]$ (Option 3 is wrong, has [3,4]$[3,4]$) A - B = A B^c$A - B = A \cap B^c$. B^c = [1, 3]$B^c = [1, 3]$. A [1, 3] = [2, 3]$A \cap [1, 3] = [2, 3]$. (Option 2 is wrong, has [2, 3)$[2, 3)$) B - A = B A^c$B - A = B \cap A^c$. A^c = (-∞, -4) (-2, 2) (4, ∞)$A^c = (-\infty, -4) \cup (-2, 2) \cup (4, \infty)$. B A^c = (-∞, -4) (-2, 1) (4, ∞)$B \cap A^c = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$. This matches Option 4 perfectly.

Pattern Recognition

Draw inequalities involving absolute values directly onto a single number line graph to perform union and intersection operations without logic gaps.

Chapter Mix

Class 11 Maths: Sets and Relations

Related Formula

Core Logic

Check condition y ≥ (2x)/(3)$y \geq \frac{2x}{3}$ for each x in A = 2, 3, 5, 7, 9$x \in A = \{2, 3, 5, 7, 9\}$ to generate ordered pairs (x,y)$(x,y)$. Count the pairs to get $\ell$. Then find asymmetric pairs to get m$m$.

Step 1: Enumerate elements of R

x = 2 y ≥ 4/3 = 1.33 y in 2, 3, 5, 7, 9$x = 2 \implies y \geq 4/3 = 1.33 \implies y \in \{2, 3, 5, 7, 9\}$ (5 elements) x = 3 y ≥ 6/3 = 2 y in 2, 3, 5, 7, 9$x = 3 \implies y \geq 6/3 = 2 \implies y \in \{2, 3, 5, 7, 9\}$ (5 elements) x = 5 y ≥ 10/3 = 3.33 y in 5, 7, 9$x = 5 \implies y \geq 10/3 = 3.33 \implies y \in \{5, 7, 9\}$ (3 elements) x = 7 y ≥ 14/3 = 4.66 y in 5, 7, 9$x = 7 \implies y \geq 14/3 = 4.66 \implies y \in \{5, 7, 9\}$ (3 elements) x = 9 y ≥ 18/3 = 6 y in 7, 9$x = 9 \implies y \geq 18/3 = 6 \implies y \in \{7, 9\}$ (2 elements)

Total elements in R$R$ is = 5 + 5 + 3 + 3 + 2 = 18$\ell = 5 + 5 + 3 + 3 + 2 = 18$.

Step 2: Determine Missing Symmetric Pairs

We need to check which reverse pairs (y,x)$(y,x)$ are missing. (2, 5) in R$(2, 5) \in R$, but (5, 2) R$(5, 2) \notin R$. (2, 7) in R$(2, 7) \in R$, but (7, 2) R$(7, 2) \notin R$. (2, 9) in R$(2, 9) \in R$, but (9, 2) R$(9, 2) \notin R$. (3, 5) in R$(3, 5) \in R$, but (5, 3) R$(5, 3) \notin R$. (3, 7) in R$(3, 7) \in R$, but (7, 3) R$(7, 3) \notin R$. (3, 9) in R$(3, 9) \in R$, but (9, 3) R$(9, 3) \notin R$. (5, 9) in R$(5, 9) \in R$, but (9, 5) R$(9, 5) \notin R$.

These are m = 7$m = 7$ pairs that need to be added.

Thus, + m = 18 + 7 = 25$\ell + m = 18 + 7 = 25$.

Pattern Recognition

When counting pairs for small sets, list them manually by rows. To make symmetric, any non-diagonal pair (x,y)$(x,y)$ that is present while (y,x)$(y,x)$ is absent counts as 1$1$ missing element.

Chapter Mix

Class 11 Maths: Sets and Relations

Related Formula

Core Logic

Given 4y = 5x - 3 y = (5x - 3)/(4)$4y = 5x - 3 \implies y = \frac{5x - 3}{4}$. We evaluate this for x in 1, 2, …, 16$x \in \{1, 2, \ldots, 16\}$ such that y$y$ is also an integer in the same set.

• If x = 3 y = (15 - 3)/(4) = 3 (3,3) in R$x = 3 \implies y = \frac{15 - 3}{4} = 3 \implies (3,3) \in R$
• If x = 7 y = (35 - 3)/(4) = 8 (7,8) in R$x = 7 \implies y = \frac{35 - 3}{4} = 8 \implies (7,8) \in R$
• If x = 11 y = (55 - 3)/(4) = 13 (11,13) in R$x = 11 \implies y = \frac{55 - 3}{4} = 13 \implies (11,13) \in R$
• If x = 15 y = (75 - 3)/(4) = 18 M$x = 15 \implies y = \frac{75 - 3}{4} = 18 \notin M$

Therefore, R = (3,3), (7,8), (11,13)$R = \{(3,3), (7,8), (11,13)\}$

Step 1: Identifying Missing Symmetric Elements

To make the relation symmetric, for every (x,y) in R$(x,y) \in R$, the pair (y,x)$(y,x)$ must also belong to R$R$.

• (3,3)$(3,3)$ is symmetric to itself.
• For (7,8)$(7,8)$, we must add (8,7)$(8,7)$.
• For (11,13)$(11,13)$, we must add (13,11)$(13,11)$.

Thus, the required elements to be added are (8,7)$(8,7)$ and (13,11)$(13,11)$, which totals 2 elements.

Pattern Recognition

When evaluating linear Diophantine equations over a small finite set, simply substitute modular values (here, 5x - 3 ≡ 0 4$5x - 3 \equiv 0 \pmod 4$) to find the explicit ordered pairs, then mechanically apply the equivalence property criteria.

Chapter Mix

Class 11 Maths: Sets, Relations and Functions