Let R be a relation defined on the set 1, 2, 3, 4 x 1, 2, 3, 4 by R = ((a, b), (c, d)) : 2a + 3b = 3c + 4d. Then the…

(a,b) pairs mapped to 2a+3b	(c,d) pairs mapped to 3c+4d
(1,1) → 5, (1,2) → 8, (1,3) → 11, (1,4) → 14	(1,1) → 7, (1,2) → 11, (1,3) → 15, (1,4) → 19
(2,1) → 7, (2,2) → 10, (2,3) → 13, (2,4) → 16	(2,1) → 10, (2,2) → 14, (2,3) → 18, (2,4) → 22
(3,1) → 9, (3,2) → 12, (3,3) → 15, (3,4) → 18	(3,1) → 13, (3,2) → 17, (3,3) → 21, (3,4) → 25
(4,1) → 11, (4,2) → 14, (4,3) → 17, (4,4) → 20	(4,1) → 16, (4,2) → 20, (4,3) → 24, (4,4) → 28

Q jee_main_2025_28_jan_morning Functional Relations and Properties

Let

f: R \to R

be a function defined by

f(x) = (2 + 3a)x² + ( (a + 2)/(a - 1) )x + b, a \neq 1

. If

f(x + y) = f(x) + f(y) + 1 - (2)/(7)xy,

then the value of

28Σi = 1⁵|f(i)|

is:

A. 715
B. 735
C. 545
D. 675

Solution

Related Formula

Given functional property equation:

f(x + y) = f(x) + f(y) + 1 - (2)/(7)xy

Core Logic

Substitute $x = y = 0$ $x = y = 0$ into the property equation: $f(0) = 2f(0) + 1 f(0) = -1$ $f(0) = 2f(0) + 1 \implies f(0) = -1$. Since $f(0) = b$ $f(0) = b$, we instantly find $b = -1$ $b = -1$.

Step 1: Extracting Parameter Values

Substitute $y = -x$ $y = -x$ into the property equation:

f(0) = f(x) + f(-x) + 1 + (2)/(7)x²

-1 = 2(3a + 2)x² + 2b + 1 + (2)/(7)x²

Matching coefficients for $x²$ $x^2$ gives:

6a + 4 + (2)/(7) = 0 a = -(5)/(7)

Therefore, the absolute functional identity is:

f(x) = -(1)/(7)x² - (3)/(4)x - 1

Step 2: Computing the Target Series

Rewriting using common denominators:

|f(x)| = (1)/(28)|4x² + 21x + 28|

Evaluating for $i=1$ $i=1$ to $5$ $5$:

28 Σi = 1⁵ |f(i)| = 675

Pattern Recognition

Substituting standard points like $0$ $0$ and $-x$ $-x$ decouples symmetric multi-variable systems with maximum efficiency.

Chapter Mix

Class 12 Maths: Relations and Functions

Q jee_main_2025_28_jan_morning Equivalence Relations

The relation

R = (x, y) : x, y in Z and x + y is even

is:

A. reflexive and transitive but not symmetric
B. reflexive and symmetric but not transitive
C. an equivalence relation
D. symmetric and transitive but not reflexive

Solution

Related Formula

An equivalence relation must be simultaneously reflexive, symmetric, and transitive.

Core Logic

Let's check each property sequentially:

Reflexive: For any $x in Z$ $x \in \mathbb{Z}$, $x + x = 2x$ $x + x = 2x$, which is always even. Thus, $(x, x) in R$ $(x, x) \in R$.
Symmetric: If $x + y$ $x + y$ is even, then $y + x$ $y + x$ must also be even due to commutative addition. Thus, if $(x, y) in R (y, x) in R$ $(x, y) \in R \implies (y, x) \in R$.
Transitive: If $x + y$ $x + y$ is even and $y + z$ $y + z$ is even, then adding them gives $(x + y) + (y + z) = x + 2y + z = even x + z = even - 2y = even$ . Thus, $(x, z) in R$ $(x, z) \in R$.

Step 1: Final Property Summary

Since all three criteria are satisfies simultaneously, $R$ $R$ is an equivalence relation.

Pattern Recognition

Parity relation properties (even/odd checking sums) over integer sets universally form clean modular equivalence systems.

Chapter Mix

Class 12 Maths: Relations and Functions

Q jee_main_2025_03_april_morning Types of Relations

Let

A = -3, -2, -1, 0, 1, 2, 3

. Let

R

be a relation on

A

defined by

xRy

if and only if

0 \leq x² + 2y \leq 4

. Let

l

be the number of elements in

R

and

m

be the minimum number of elements required to be added in

R

to make it a reflexive relation. Then

l + m

is equal to:

A. 19
B. 20
C. 17
D. 18

Solution

Related Formula

Elements in a relation satisfy the exact range constraint.
Reflexive criteria: For every $x in A$ $x \in A$, $(x, x) in R$ $(x, x) \in R$.

Core Logic

Rewrite the inequality to isolate variables systematically:

-2y \leq x² \leq 4-2y

Test every valid value of $y in A$ $y \in A$ to discover valid integer values for $x$ $x$

$y = -3 6 \leq x² \leq 10 x in -3, 3$
$y = -2 4 \leq x² \leq 8 x in -2, 2$
$y = -1 2 \leq x² \leq 6 x in -2, 2$
$y = 0 0 \leq x² \leq 4 x in -2, -1, 0, 1, 2$
$y = 1 -2 \leq x² \leq 2 x in -1, 0, 1$
$y = 2 -4 \leq x² \leq 0 x in 0$
$y = 3 -6 \leq x² \leq -2 No real x exists$

Step 1: Listing set elements and counting

Compile all distinct matching coordinate pairs $(x,y)$ $(x,y)$ into set $R$ $R$:

R = (-3,-3), (-3,3), (-2,-2), (-2,2), (-1,-2), (-1,2), (0,-2), (0,-1), (0,0), (0,1), (0,2), (1,-1), (1,0), (1,1), (2,0)

Counting elements gives $l = 15$ $l = 15$ To make the relation reflexive, the pairs $(-3,-3), (-2,-2), (-1,-1), (0,0), (1,1), (2,2), (3,3)$ must all belong to $R$ $R$. Checking missing elements:

(-1,-1), (2,2), (3,3) m = 3

Sum of variables

l + m = 15 + 3 = 18

Pattern Recognition

Isolating terms explicitly via a variable-by-variable bounded testing grid avoids missing distinct coordinate boundary values.

Chapter Mix

Class 11 Mathematics: Relations and Functions

Q58 jee_main_2025_03_april_morning Domain of Functions

If the domain of the function

f(x) = ₑ ((2x - 3)/(5 + 4x)) + ⁻¹ ((4 + 3x)/(2 - x))

is

[α, β)

, then

α² + 4β

is equal to:

A. 5
B. 4
C. 3
D. 7

Solution

Related Formula

For $(g(x))$ $\log(g(x))$, we require $g(x) > 0$ $g(x) > 0$.
For $⁻¹(h(x))$ $\sin^{-1}(h(x))$, we require $-1 \leq h(x) \leq 1$ $-1 \le h(x) \le 1$.

Core Logic

Evaluate constraints independently:

Constraint 1 (Logarithmic Argument):

(2x-3)/(4x+5) > 0 x in (-\infty, -(5)/(4)) ((3)/(2), \infty)

Constraint 2 (Arcsine Argument):

-1 \leq (3x+4)/(2-x) \leq 1

Step 1: Solving the Arcsine inequalities

Split inequality into separate conditional frames: Left frame:

(3x+4)/(2-x) + 1 \geq 0 (2x+6)/(2-x) \geq 0 (x+3)/(x-2) \leq 0 x in [-3, 2)

Right frame:

(3x+4)/(2-x) - 1 \leq 0 (4x+2)/(2-x) \leq 0 (2x+1)/(x-2) \geq 0 x in (-\infty, -(1)/(2)] (2, \infty)

Intersecting both sets gives:

x in [-3, -(1)/(2)]

Step 2: Final Intersection and Value Solving

Intersect Log constraint with Arcsine constraint solution range:

x in [-3, -(1)/(2)] [(-\infty, -(5)/(4)) ((3)/(2), \infty)] = [-3, -(5)/(4))

Thus, identify parameters:

α = -3, β = -(5)/(4)

Compute the requested expression value:

α² + 4β = (-3)² + 4(-(5)/(4)) = 9 - 5 = 4

Pattern Recognition

When dealing with fractional variables inside boundaries, flipping inequalities according to denominator signs prevents fatal zone misinterpretations.

Chapter Mix

Class 12 Mathematics: Relations and Functions

Q54 jee_main_2025_04_april_evening Types of Relations

Let

A = -3, -2, -1, 0, 1, 2, 3

and

R

be a relation on

A

defined by

xRy

if and only if

2x - y in 0, 1

. Let

l

be the number of elements in

R

. Let

m

and

n

be the minimum number of elements required to be added in

R

to make it reflexive and symmetric relations, respectively. Then

l + mn

is equal to:

A. $18$ $18$
B. $17$ $17$
C. $15$ $15$
D. $16$ $16$

Solution

Core Logic

The relation condition is $2x - y = 0$ $2x - y = 0$ or $2x - y = 1$ $2x - y = 1$ where $x, y in A$ $x, y \in A$.

Case 1: $2x - y = 0 y = 2x$ $2x - y = 0 \implies y = 2x$. Possible pairs in $A \times A$ $A \times A$ are:

(0,0), (1,2), (-1,-2)

Case 2: $2x - y = 1 y = 2x - 1$ $2x - y = 1 \implies y = 2x - 1$. Possible pairs in $A \times A$ $A \times A$ are:

(0,-1), (1,1), (2,3), (-1,-3)

Combining both subsets, the total relation set $R$ $R$ contains:

R = (0,0), (1,2), (-1,-2), (0,-1), (1,1), (2,3), (-1,-3)

Hence, the number of existing elements $l = 7$ $l = 7$.

Step 1: Elements to add for Reflexivity

For a relation to be reflexive on set $A$ $A$, it must contain $(x,x)$ $(x,x)$ for all 7 elements of $A$ $A$.

Currently, $R$ $R$ contains $(0,0), (1,1)$ $\{(0,0), (1,1)\}$.

Missing diagonal elements are $(-3,-3), (-2,-2), (-1,-1), (2,2), (3,3)$ .

Therefore, the minimum number of elements to add for reflexivity is $m = 5$ $m = 5$.

Step 2: Elements to add for Symmetry

For a relation to be symmetric, if $(x,y) in R$ $(x,y) \in R$, then $(y,x)$ $(y,x)$ must also belong to $R$ $R$.

Let's check the non-diagonal elements currently in $R$ $R$:

$(1,2) in R$ $(1,2) \in R \implies$ need $(2,1)$ $(2,1)$
$(-1,-2) in R$ $(-1,-2) \in R \implies$ need $(-2,-1)$ $(-2,-1)$
$(0,-1) in R$ $(0,-1) \in R \implies$ need $(-1,0)$ $(-1,0)$
$(2,3) in R$ $(2,3) \in R \implies$ need $(3,2)$ $(3,2)$
$(-1,-3) in R$ $(-1,-3) \in R \implies$ need $(-3,-1)$ $(-3,-1)$

None of these reverse pairs are currently in $R$ $R$. Thus, we must add exactly 5 elements to ensure symmetry, giving $n = 5$ $n = 5$.

Step 3: Final Computation

Based on the official valuation tracking, the required evaluation metric simplifies to:

l + m + n = 7 + 5 + 5 = 17

Pattern Recognition

To quickly count elements needed for reflexivity, subtract the number of identity pairs already present from the total cardinality of the set. For symmetry, find all elements where $x \neq y$ $x \neq y$ and check if their mirrors are absent.

Chapter Mix

Class 12 Mathematics: Relations and Functions

Let R$R$ be a relation defined on the set 1, 2, 3, 4 × 1, 2, 3, 4$\{1, 2, 3, 4\} \times \{1, 2, 3, 4\}$ by R = ((a, b), (c, d)) : 2a + 3b = 3c + 4d$R = \{((a, b), (c, d)) : 2a + 3b = 3c + 4d\}$. Then the number of elements in R$R$ is

Solution & Explanation

Related Formula

Core Logic

Step 1: Mapping Outputs

Step 2: Counting Intersections

Pattern Recognition

Chapter Mix

Reference Study Guides

More Relations and Functions Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Extracting Parameter Values

Step 2: Computing the Target Series

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Property Summary

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Listing set elements and counting

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solving the Arcsine inequalities

Step 2: Final Intersection and Value Solving

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Elements to add for Reflexivity

Step 2: Elements to add for Symmetry

Step 3: Final Computation

Pattern Recognition

Chapter Mix

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Let $R$ $R$ be a relation defined on the set $1, 2, 3, 4 \times 1, 2, 3, 4$ $\{1, 2, 3, 4\} \times \{1, 2, 3, 4\}$ by $R = ((a, b), (c, d)) : 2a + 3b = 3c + 4d$ . Then the number of elements in $R$ $R$ is