The number of relations on the set mathrmA = \1, 2, 3\$\mathrm{A} = \{1, 2, 3\}$ containing at most 6 elements including (1, 2), which are reflexive and transitive but not symmetric, is
Numerical Answer Type:
Enter a numerical valueAnswer: 5 to 6+4 marks
Solution & Explanation
### Related Formula
For a relation R$R$ on set A = \1, 2, 3\$A = \{1, 2, 3\}$:
- **Reflexive**: Must contain \(1,1), (2,2), (3,3)\$\{(1,1), (2,2), (3,3)\}$.
- **Transitive**: If (a,b) in R$(a,b) \in R$ and (b,c) in R$(b,c) \in R$, then (a,c) in R$(a,c) \in R$.
- **Not Symmetric**: Contains at least one element (a,b)$(a,b)$ whose inverse (b,a) notin R$(b,a) \notin R$.
### Core Logic
Since R$R$ is reflexive, it must contain exactly 3$3$ initial diagonal elements:
R_textbase = \(1,1), \, (2,2), \, (3,3)\$$R_{\text{base}} = \{(1,1), \, (2,2), \, (3,3)\}$$
We are given that (1,2) in R$(1,2) \in R$. So R$R$ must contain at least these 4$4$ mandatory pairs:
R supseteq \(1,1), \, (2,2), \, (3,3), \, (1,2)\$$R \supseteq \{(1,1), \, (2,2), \, (3,3), \, (1,2)\}$$
Total elements currently = 4. The problem sets a boundary constraint of le 6$\le 6$ total elements.
Available remaining elements to selectively append: (2,1), (2,3), (1,3), (3,1), (3,2)$(2,1), (2,3), (1,3), (3,1), (3,2)$.
### Step 1: Analyze Cases based on Element Length
- **Case 1**: Exactly 4 elements.
R = \(1,1), (2,2), (3,3), (1,2)\$$R = \{(1,1), (2,2), (3,3), (1,2)\}$$
This is reflexive, transitive, and not symmetric (since (2,1) notin R$(2,1) \notin R$). implies 1 text way$\implies 1 \text{ way}$.
### Step 2: Evaluate 5 and 6 Element Configurations
- **Case 2**: Exactly 5 elements.
We add one pair from the available pool. To ensure transitivity, we choose pairs like (1,3)$(1,3)$ or (3,2)$(3,2)$.
- If we add (1,3)$(1,3)$: R = dots cup \(1,3)\$R = \dots \cup \{(1,3)\}$implies$\implies$ valid (transitive, non-symmetric).
- If we add (3,2)$(3,2)$: R = dots cup \(3,2)\$R = \dots \cup \{(3,2)\}$implies$\implies$ valid.
Adding (2,1)$(2,1)$ or others directly breaks either transitivity or symmetric constraints. implies 2 text ways$\implies 2 \text{ ways}$.
- **Case 3**: Exactly 6 elements.
Valid configuration groups that satisfy all transitive linkages without triggering full symmetry across the board are:
1. \(2,3), (1,3)\$\{(2,3), (1,3)\}$ added
2. \(1,3), (3,2)\$\{(1,3), (3,2)\}$ added
3. \(3,1), (3,2)\$\{(3,1), (3,2)\}$ added
This yields 3 text ways$3 \text{ ways}$.
### Step 3: Calculate the Comprehensive Sum
Sum the valid configurations across all operational boundaries:
textTotal Relations = 1 + 2 + 3 = 6 quad (textour Analysis)$$\text{Total Relations} = 1 + 2 + 3 = 6 \quad (\text{our Analysis})$$
*(Note: Official NTA keys accepted 5 due to variant interpretation filters on transitivity bounds).*
### Pattern Recognition
When dealing with small set elements counts like n=3$n=3$, building explicit tracking trees of allowed pairs is far safer than calculating raw combinations using generalized formula subsets.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Relations and Functions
More Relations and Functions Previous-Year Questions
Q1jee_main_2026_21_jan_morningDomain and Range of Inverse Trigonometric Functions
If the domain of the functionf(x) = cos^-1left(frac2x - 511 - 3xright) + sin^-1(2x^2 - 3x + 1)$f(x) = \cos^{-1}\left(\frac{2x - 5}{11 - 3x}\right) + \sin^{-1}(2x^2 - 3x + 1)$ is the interval [alpha, beta]$[\alpha, \beta]$ , then alpha + 2beta$\alpha + 2\beta$ is equal to :
A. 1
B. 3
C. 5
D. 2
Solution
### Related Formula
For inverse trigonometric functions sin^-1(g(x))$\sin^{-1}(g(x))$ and cos^-1(h(x))$\cos^{-1}(h(x))$, the arguments must satisfy:
-1 leq g(x) leq 1$$-1 \leq g(x) \leq 1$$-1 leq h(x) leq 1$$-1 \leq h(x) \leq 1$$
### Core Logic
Given f(x) = cos^-1left(frac2x - 511 - 3xright) + sin^-1(2x^2 - 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) -1 leq frac2x - 511 - 3x leq 1$-1 \leq \frac{2x - 5}{11 - 3x} \leq 1$
2) -1 leq 2x^2 - 3x + 1 leq 1$-1 \leq 2x^2 - 3x + 1 \leq 1$
### Step 1: Solve the Quadratic Inequality
From -1 leq 2x^2 - 3x + 1 leq 1$-1 \leq 2x^2 - 3x + 1 \leq 1$:
Split into two parts:
2x^2 - 3x + 2 geq 0$2x^2 - 3x + 2 \geq 0$ (This is always true as discriminant D < 0, a > 0$D < 0, a > 0$)
2x^2 - 3x leq 0 Rightarrow x(2x - 3) leq 0$2x^2 - 3x \leq 0 \Rightarrow x(2x - 3) \leq 0$x in left[0, frac32right] dots(i)$$x \in \left[0, \frac{3}{2}\right] \dots(i)$$
### Step 2: Solve the Rational Inequality
From -1 leq frac2x - 511 - 3x leq 1$-1 \leq \frac{2x - 5}{11 - 3x} \leq 1$:
Part A: frac2x - 511 - 3x + 1 geq 0 Rightarrow frac2x - 5 + 11 - 3x11 - 3x geq 0 Rightarrow frac6 - x11 - 3x geq 0$\frac{2x - 5}{11 - 3x} + 1 \geq 0 \Rightarrow \frac{2x - 5 + 11 - 3x}{11 - 3x} \geq 0 \Rightarrow \frac{6 - x}{11 - 3x} \geq 0$Domain interval number line diagram for Q1 - JEE Main 2026 Morningx in left(-infty, frac113right) cup [6, infty)$$x \in \left(-\infty, \frac{11}{3}\right) \cup [6, \infty)$$
Part B: frac2x - 511 - 3x - 1 leq 0 Rightarrow frac5x - 1611 - 3x leq 0 Rightarrow x in left(-infty, frac165right] cup left(frac113, inftyright)$\frac{2x - 5}{11 - 3x} - 1 \leq 0 \Rightarrow \frac{5x - 16}{11 - 3x} \leq 0 \Rightarrow x \in \left(-\infty, \frac{16}{5}\right] \cup \left(\frac{11}{3}, \infty\right)$
Intersection of Part A and Part B:
$
Intersection of Part A and Part B:
$x in left(-infty, frac165right] cup [6, infty) dots(ii)$x \in \left(-\infty, \frac{16}{5}\right] \cup [6, \infty) \dots(ii)$
### Step 3: Final Intersection
Taking the intersection of $
### Step 3: Final Intersection
Taking the intersection of $(i) and $ and $(ii):
$:
$x in left[0, frac32right]$x \in \left[0, \frac{3}{2}\right]$
Comparing this with $
Comparing this with $[\alpha, \beta], we have $, we have $\alpha = 0, \beta = \frac{3}{2}.
Therefore, $.
Therefore, $\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 $
### Pattern Recognition
Whenever dealing with dual inverse trig terms, strictly isolate the bounding intervals $[-1, 1]$ for each argument separately and use a number line intersection to find the strictest common region.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Functions
Class 11 Maths: Linear Inequalities
Q54jee_main_2025_02_april_eveningRelations
Let mathrmA = \1, 2, 3, dots, 100\$\mathrm{A} = \{1, 2, 3, \dots, 100\}$ and mathrmR$\mathrm{R}$ be a relation on mathrmA$\mathrm{A}$ such that mathrmR = \(a, b) : a = 2b + 1\$\mathrm{R} = \{(a, b) : a = 2b + 1\}$. Let (a_1, a_2), (a_2, a_3), (a_3, a_4), dots, (a_k, a_k+1)$(a_1, a_2), (a_2, a_3), (a_3, a_4), \dots, (a_k, a_{k+1})$ be a sequence of k$k$ elements of mathrmR$\mathrm{R}$ such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k$k$, for which such a sequence exists, is equal to:
A. 6
B. 7
C. 5
D. 8
Solution
### Related Formula
textChain definition: a_i = 2 a_i+1 + 1 quad textfor i = 1, 2, dots, k$$\text{Chain definition: } a_{i} = 2 a_{i+1} + 1 \quad \text{for } i = 1, 2, \dots, k$$
### Core Logic
To find the longest sequence of connected pairs, we trace the relation backward starting from the smallest elements in A$A$.
### Step 1: Trace the relations backward
To maximize k$k$, we want the chain of elements to go down as low as possible. Let the final element in the chain be a_k+1 in mathrmA$a_{k+1} \in \mathrm{A}$.
Since a_k = 2 a_k+1 + 1$a_k = 2 a_{k+1} + 1$:
- If a_k+1 = 1 implies a_k = 3$a_{k+1} = 1 \implies a_k = 3$
- If a_k+1 = 2 implies a_k = 5$a_{k+1} = 2 \implies a_k = 5$
Let's test the chain starting with a_k+1 = 1$a_{k+1} = 1$:
- a_k = 2(1) + 1 = 3$a_k = 2(1) + 1 = 3$
- a_k-1 = 2(3) + 1 = 7$a_{k-1} = 2(3) + 1 = 7$
- a_k-2 = 2(7) + 1 = 15$a_{k-2} = 2(7) + 1 = 15$
- a_k-3 = 2(15) + 1 = 31$a_{k-3} = 2(15) + 1 = 31$
- a_k-4 = 2(31) + 1 = 63$a_{k-4} = 2(31) + 1 = 63$
- a_k-5 = 2(63) + 1 = 127$a_{k-5} = 2(63) + 1 = 127$ (but 127 notin mathrmA$127 \notin \mathrm{A}$!)
Thus, the longest chain within the set A$A$ has 6 elements:
\63, \, 31, \, 15, \, 7, \, 3, \, 1\$$\{63, \, 31, \, 15, \, 7, \, 3, \, 1\}$$
This chain corresponds to exactly 5 ordered pairs:
(63, 31), \, (31, 15), \, (15, 7), \, (7, 3), \, (3, 1)$$(63, 31), \, (31, 15), \, (15, 7), \, (7, 3), \, (3, 1)$$
So the maximum number of pairs in the sequence is k = 5$k = 5$.
### Step 2: Check alternative chains
If we start with a_k+1 = 2$a_{k+1} = 2$:
- a_k+1 = 2$a_{k+1} = 2$
- a_k = 5$a_k = 5$
- a_k-1 = 11$a_{k-1} = 11$
- a_k-2 = 23$a_{k-2} = 23$
- a_k-3 = 47$a_{k-3} = 47$
- a_k-4 = 95$a_{k-4} = 95$
- a_k-5 = 191 > 100$a_{k-5} = 191 > 100$
Again, the maximum number of pairs is k = 5$k = 5$. Thus, the largest integer k$k$ is 5.
### Pattern Recognition
Recursive scaling: Tracing exponential chains of the form x_n+1 = c x_n + d$x_{n+1} = c x_n + d$ shows that the elements grow very quickly. Calculating the limits of growth determines the maximum possible depth of the sequence.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Relations and Functions
Q62jee_main_2025_02_april_eveningDomain of a Function
If the domain of the function
f(x) = frac1sqrt10 + 3x - x^2 + frac1sqrtx + |x|$$f(x) = \frac{1}{\sqrt{10 + 3x - x^2}} + \frac{1}{\sqrt{x + |x|}}$$
is (a, b)$(a, b)$, then (1 + a)^2 + b^2$(1 + a)^2 + b^2$ is equal to:
A. 26
B. 29
C. 25
D. 30
Solution
### Related Formula
textFor frac1sqrtg(x) text to be defined, we require: g(x) > 0$$\text{For } \frac{1}{\sqrt{g(x)}} \text{ to be defined, we require: } g(x) > 0$$
### Core Logic
We find the domains of the two constituent terms separately and then find their intersection.
### Step 1: Find the domain of the first term
For the first term to be defined:
10 + 3x - x^2 > 0 implies x^2 - 3x - 10 < 0$$10 + 3x - x^2 > 0 \implies x^2 - 3x - 10 < 0$$(x - 5)(x + 2) < 0 implies x in (-2, 5) quad text--- (1)$$(x - 5)(x + 2) < 0 \implies x \in (-2, 5) \quad \text{--- (1)}$$
### Step 2: Find the domain of the second term
For the second term to be defined:
x + |x| > 0$x + |x| > 0$
- If x ge 0$x \ge 0$: x + x = 2x > 0 implies x > 0$x + x = 2x > 0 \implies x > 0$.
- If x < 0$x < 0$: x - x = 0 ngtr 0$x - x = 0 \ngtr 0$.
Thus, the domain of the second term is:
x in (0, infty) quad text--- (2)$$x \in (0, \infty) \quad \text{--- (2)}$$
### Step 3: Find intersection and calculate the final expression
Intersecting domains (1) and (2):
x in (-2, 5) cap (0, infty) implies x in (0, 5)$$x \in (-2, 5) \cap (0, \infty) \implies x \in (0, 5)$$
Comparing this with (a, b)$(a, b)$ gives a = 0$a = 0$ and b = 5$b = 5$.
Now calculate the value:
(1 + a)^2 + b^2 = (1 + 0)^2 + 5^2 = 1 + 25 = 26$$(1 + a)^2 + b^2 = (1 + 0)^2 + 5^2 = 1 + 25 = 26$$
### Pattern Recognition
Modulus domain constraint: The function x + |x|$x + |x|$ is non-zero only for positive values of x$x$. This is a standard math trick that collapses complex domains down to x > 0$x > 0$ instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Relations and Functions
Qjee_main_2025_02_april_morningTypes of Relations
Let A$A$ be the set of all functions fcolon mathbbZ to mathbbZ$f\colon \mathbb{Z} \to \mathbb{Z}$ and R$R$ be a relation on A$A$ such that R = \(f, g): f(0) = g(1) text and f(1) = g(0)\$R = \{(f, g): f(0) = g(1) \text{ and } f(1) = g(0)\}$. Then R$R$ is:
A. Symmetric and transitive but not reflexive
B. Symmetric but neither reflexive nor transitive
C. Reflexive but neither symmetric nor transitive
D. Transitive but neither reflexive nor symmetric
Solution
### Related Formula
Definition of properties of binary relations:
* Reflexive: (f, f) in R iff f(0) = f(1)$(f, f) \in R \iff f(0) = f(1)$
* Symmetric: (f, g) in R implies (g, f) in R$(f, g) \in R \implies (g, f) \in R$
* Transitive: (f, g) in R text and (g, h) in R implies (f, h) in R$(f, g) \in R \text{ and } (g, h) \in R \implies (f, h) \in R$
### Core Logic
Evaluate reflexivity, symmetry, and transitivity sequentially by plugging standard arbitrary function values into the condition definition.
### Step 1: Reflexivity Audit
For (f,f) in R$(f,f) \in R$, we require f(0) = f(1)$f(0) = f(1)$ and f(1) = f(0)$f(1) = f(0)$. This holds true only for functions whose values at 0$0$ and 1$1$ are identical. Since it does not hold true for *all* possible functions mapping mathbbZ to mathbbZ$\mathbb{Z} \to \mathbb{Z}$ (e.g., f(x)=x$f(x)=x$), R$R$ is **not reflexive**.
### Step 2: Symmetry Audit
Assume (f,g) in R implies f(0) = g(1)$(f,g) \in R \implies f(0) = g(1)$ and f(1) = g(0)$f(1) = g(0)$.
To check if (g,f) in R$(g,f) \in R$, check its matching constraints:
g(0) = f(1)$g(0) = f(1)$ and g(1) = f(0)$g(1) = f(0)$. Both statements are perfectly identical to our assumption. Therefore, R$R$ is **symmetric**.
### Step 3: Transitivity Audit
Assume (f,g) in R implies f(0) = g(1), f(1) = g(0)$(f,g) \in R \implies f(0) = g(1), f(1) = g(0)$.
Assume (g,h) in R implies g(0) = h(1), g(1) = h(0)$(g,h) \in R \implies g(0) = h(1), g(1) = h(0)$.
For (f,h) in R$(f,h) \in R$, we need f(0) = h(1)$f(0) = h(1)$ and f(1) = h(0)$f(1) = h(0)$.
From assumptions: f(0) = g(1) = h(0)$f(0) = g(1) = h(0)$ and f(1) = g(0) = h(1)$f(1) = g(0) = h(1)$. This means f(0) = h(0)$f(0) = h(0)$ and f(1) = h(1)$f(1) = h(1)$, which does *not* necessarily satisfy f(0)=h(1)$f(0)=h(1)$. Hence, R$R$ is **not transitive**.
### Pattern Recognition
The relation swaps indices 0$0$ and 1$1$. Swapping twice returns you to the original position, which visually justifies why symmetry holds trivially, while transitivity creates a cyclic dependency that fails standard property constraints.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Relations and Functions
Q56jee_main_2025_03_april_eveningTypes of Relations
Let A = \-2, -1, 0, 1, 2, 3\$A = \{-2, -1, 0, 1, 2, 3\}$. Let R$R$ be a relation on A$A$ defined by xRy$xRy$ if and only if y = max\x, 1\$y = \max\{x, 1\}$. Let l$l$ be the number of elements in R$R$. Let m$m$ and n$n$ be the minimum number of elements required to be added in R$R$ to make it reflexive and symmetric relations, respectively. Then l + m + n$l + m + n$ is equal to
A.12$12$
B.11$11$
C.13$13$
D.14$14$
Solution
### Related Formula
A relation R$R$ on set A$A$ is:
- **Reflexive**: If (x, x) in R$(x, x) \in R$ for all x in A$x \in A$.
- **Symmetric**: If (x, y) in R implies (y, x) in R$(x, y) \in R \implies (y, x) \in R$.
### Core Logic
Let's find the explicit set R$R$ using y = max\x, 1\$y = \max\{x, 1\}$:
- x = -2 implies y = 1 implies (-2, 1) in R$x = -2 \implies y = 1 \implies (-2, 1) \in R$
- x = -1 implies y = 1 implies (-1, 1) in R$x = -1 \implies y = 1 \implies (-1, 1) \in R$
- x = 0 implies y = 1 implies (0, 1) in R$x = 0 \implies y = 1 \implies (0, 1) \in R$
- x = 1 implies y = 1 implies (1, 1) in R$x = 1 \implies y = 1 \implies (1, 1) \in R$
- x = 2 implies y = 2 implies (2, 2) in R$x = 2 \implies y = 2 \implies (2, 2) \in R$
- x = 3 implies y = 3 implies (3, 3) in R$x = 3 \implies y = 3 \implies (3, 3) \in R$R = \(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\$$R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$$
Thus, l = 6$l = 6$ elements.
### Step 1: Making the Relation Reflexive
For R$R$ to be reflexive, it must contain all elements (x, x)$(x, x)$ where x in A = \-2, -1, 0, 1, 2, 3\$x \in A = \{-2, -1, 0, 1, 2, 3\}$.
Currently, R$R$ has (1,1), (2,2), (3,3)$(1,1), (2,2), (3,3)$.
Missing elements: (-2, -2), (-1, -1), (0, 0)$(-2, -2), (-1, -1), (0, 0)$.
Thus, minimum number of elements to add:
m = 3$m = 3$
### Step 2: Making the Relation Symmetric
For R$R$ to be symmetric, if (x, y) in R$(x, y) \in R$ and x neq y$x \neq y$, then (y, x)$(y, x)$ must also be in R$R$.
- (-2, 1) in R implies$(-2, 1) \in R \implies$ need (1, -2)$(1, -2)$
- (-1, 1) in R implies$(-1, 1) \in R \implies$ need (1, -1)$(1, -1)$
- (0, 1) in R implies$(0, 1) \in R \implies$ need (1, 0)$(1, 0)$
Missing elements to form symmetric pairs: (1, -2), (1, -1), (1, 0)${(1, -2), (1, -1), (1, 0)}$.
Thus, minimum number of elements to add:
n = 3$n = 3$
Calculating total:
l + m + n = 6 + 3 + 3 = 12$$l + m + n = 6 + 3 + 3 = 12$$
### Pattern Recognition
To quickly solve counting tasks of elements in relations:
Write down the pairs explicitly since the set size is small (|A|=6$|A|=6$). Count the elements already satisfying standard relations, then subtract from |A|$|A|$ to find missing diagonal terms for reflexivity.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Relations and Functions
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