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

Solution & Explanation

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 + x = 2x, which is always even. Thus, (x, x) in R.
  • Symmetric: If x + y is even, then y + x must also be even due to commutative addition. Thus, if (x, y) in R (y, x) in R.
  • Transitive: If x + y is even and 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.
Step 1: Final Property Summary

Since all three criteria are satisfies simultaneously, 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

More Relations and Functions Previous-Year Questions — Page 10

Q53 jee_main_2025_24_jan_morning Functional Relations and Equations
Let f: R - 0 → R be a function such that f(x) - 6f((1)/(x)) = (35)/(3x) - (5)/(2) If x → 0 ( (1)/(α x) + f(x) ) = β for some α, β in R, then α + 2β is equal to :
  • A. 3
  • B. 5
  • C. 4
  • D. 6

Solution

Related Formula

For functional equations with inversion, substituting x → (1)/(x) establishes a solvable system of algebraic equations to isolate f(x) directly.

Core Logic

The given equation is:

f(x) - 6f((1)/(x)) = (35)/(3x) - (5)/(2) (1)

Substitute x → (1)/(x) in equation (1):

f((1)/(x)) - 6f(x) = (35x)/(3) - (5)/(2) (2)
Step 1: Eliminate f(1/x)

Multiply equation (2) by 6 and add it to equation (1):

[ f(x) - 6f((1)/(x)) ] + 6 [ f((1)/(x)) - 6f(x) ] = ( (35)/(3x) - (5)/(2) ) + 6 ( (35x)/(3) - (5)/(2) ) f(x) - 36f(x) = (35)/(3x) - (5)/(2) + 70x - 15 -35f(x) = 70x + (35)/(3x) - (35)/(2)

Divide across by -35:

f(x) = -2x - (1)/(3x) + (1)/(2)
Step 2: Evaluate the Limit

We are given that the following limit evaluates to a finite constant β:

x → 0 ( (1)/(α x) + f(x) ) = β x → 0 ( (1)/(α x) - 2x - (1)/(3x) + (1)/(2) ) = β x → 0 ( [ (1)/(α) - (1)/(3) ] (1)/(x) - 2x + (1)/(2) ) = β

For the limit to be a finite value, the coefficient of (1)/(x) must vanish completely:

(1)/(α) - (1)/(3) = 0 α = 3

When α = 3, the limit simplifies directly to the constant term:

β = x → 0 ( -2x + (1)/(2) ) = (1)/(2)
Step 3: Calculate Final Value

Substitute the determined parameters α and β:

α + 2β = 3 + 2((1)/(2)) = 3 + 1 = 4
Pattern Recognition

In limit problems involving fractional components where x → 0, any term like (1)/(x) or higher negative powers must have a net coefficient of zero to guarantee existence of a finite limit value.

Chapter Mix

Class 11 Mathematics: Functions Class 11 Mathematics: Limits and Derivatives

Q64 jee_main_2025_28_jan_evening Range and Onto Functions
Let f:[0,3]arrow A be defined by f(x)=2x³-15x²+36x+7 and g:[0,∞)arrow B be defined by g(x)= x²⁰²⁵x²⁰²⁵+1. If both the functions are onto and S=xin Z:xin A or xin B, then n(S) is equal to:
  • A. 30
  • B. 36
  • C. 29
  • D. 31

Solution

Related Formula

For a function to be onto, its Codomain must equal its Range.

Core Logic

First, find range A for f(x) = 2x³ - 15x² + 36x + 7 over [0, 3]. Differentiating f(x):

f'(x) = 6x² - 30x + 36 = 6(x² - 5x + 6) = 6(x-2)(x-3)

Critical points are x=2 and x=3. Evaluate f(x) at boundary and critical points:

  • f(0) = 7
  • f(2) = 2(8) - 15(4) + 36(2) + 7 = 16 - 60 + 72 + 7 = 35
  • f(3) = 2(27) - 15(9) + 36(3) + 7 = 54 - 135 + 108 + 7 = 34
  • Thus, Range A = [7, 35].

Step 1: Find Range B for g(x)

Now look at g(x) = x²⁰²⁵x²⁰²⁵+1 = 1 - 1x²⁰²⁵+1 over [0, ∞).

  • At x = 0, g(0) = 0.
  • As x → ∞, g(x) → 1.
  • Since g(x) is continuous and monotonically strictly increasing, Range B = [0, 1).

Step 2: Find the Integer Count of Union Set S

S = x in Z : x in A or x in B = Z (A B)

A B = [0, 1) [7, 35]

The integers in this set are:

  • From [0, 1): x = 0
  • From [7, 35]: x = 7, 8, 9, , 35
  • Total number of integers n(S):

n(S) = 1 + (35 - 7 + 1) = 1 + 29 = 30
Pattern Recognition

The condition 'or' means union. Be careful not to include integers between 1 and 6 since they are not present in either continuous range segment.

Chapter Mix

Class 12 Mathematics: Functions Class 12 Mathematics: Application of Derivatives

Q65 jee_main_2025_28_jan_evening Domain of Inverse Trigonometric Functions
Let [x] denote the greatest integer less than or equal to x. Then domain of f(x)= ⁻¹(2[x]+1) is:
  • A. (-∞,-1] [0,∞)
  • B. (-∞,∞)
  • C. (-∞,-1] [1,∞)
  • D. (-∞,∞)-0

Solution

Related Formula

The domain of ⁻¹(y) is given by |y| ≥ 1, which means:

y ≤ -1 or y ≥ 1
Core Logic

For f(x) = ⁻¹(2[x]+1) to be defined:

2[x] + 1 ≤ -1 or 2[x] + 1 ≥ 1
Step 1: Solve individual inequalities

Case 1:

2[x] + 1 ≤ -1 2[x] ≤ -2 [x] ≤ -1

This holds true for all x < 0, i.e., x in (-∞, 0).

Case 2:

2[x] + 1 ≥ 1 2[x] ≥ 0 [x] ≥ 0

This holds true for all x ≥ 0, i.e., x in [0, ∞).

Step 2: Take Union of the Solutions
Domain = (-∞, 0) [0, ∞) = (-∞, ∞)
Pattern Recognition

Since [x] covers all integer values and 2[x]+1 forms all odd integer values, the expression inside ⁻¹ is always a non-zero integer. Non-zero integers always have absolute value ≥ 1. Hence, it is valid for all real numbers.

Chapter Mix

Class 11 Mathematics: Functions Class 12 Mathematics: Inverse Trigonometric Functions

Q70 jee_main_2025_28_jan_evening Functional Equations
Let f:R-0arrow(-∞,1] be a polynomial of degree 2, satisfying f(x)f((1)/(x))=f(x)+f((1)/(x)). If f(K)=-2K then the sum of squares of all possible values of K is:
  • A. 1
  • B. 6
  • C. 7
  • D. 9

Solution

Related Formula

Standard result for functional equation of a polynomial satisfying f(x)f(1/x) = f(x) + f(1/x):

f(x) = 1 ± xⁿ
Core Logic

Given that f(x) is a polynomial of degree 2, the identity implies:

f(x) = 1 + x² or f(x) = 1 - x²

We are given the range is bounded above: (-∞, 1].

  • For 1 + x², the range is [1, ∞).
  • For 1 - x², the range is (-∞, 1].
  • Therefore, the correct functional form is f(x) = 1 - x².

Step 1: Solve for K

Given condition: f(K) = -2K

1 - K² = -2K K² - 2K - 1 = 0

Let the roots of this equation be K₁ and K₂. From quadratic properties (Vieta's formulas): K₁ + K₂ = 2

K₁ · K₂ = -1
Step 2: Calculate Sum of Squares

We need the sum of squares of the values of K:

K₁² + K₂² = (K₁ + K₂)² - 2K₁K₂ K₁² + K₂² = (2)² - 2(-1) = 4 + 2 = 6
Pattern Recognition

The functional equation f(x)f(1/x)=f(x)+f(1/x) uniquely forces polynomials to be 1 ± xⁿ. Remembering this shortcut saves valuable time required to derive the template from general coefficients.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations Class 12 Mathematics: Functions

Q61 jee_main_2025_29_jan_morning Types of Relations
Define a relation R on the interval [0,(π)/(2)) by x R y if and only if ²x - ²y = 1 . Then R is:
  • A. an equivalence relation
  • B. both reflexive and transitive but not symmetric
  • C. both reflexive and symmetric but not transitive
  • D. reflexive but neither symmetric nor transitive

Solution

Related Formula
² θ - ² θ = 1
Core Logic

To show R is an equivalence relation, verify reflexive, symmetric, and transitive properties sequentially.

Step 1: Reflexive Property

For any x in [0, π/2):

² x - ² x = 1 xRx (Reflexive)
Step 2: Symmetric Property

If xRy ² x - ² y = 1. Using identities: (1 + ² x) - ( ² y - 1) = 1 ² y - ² x = 1 yRx (Symmetric)

Step 3: Transitive Property

If

Step 3: Transitive Property

If $xRyandyRz \implies \sec^2 x - \tan^2 y = 1and ² y - ² z = 1. Adding both equations:

² x - ² y + ² y - ² z = 2² x + ( ² y - ² y) - ² z = 2 ² x + 1 - ² z = 2² x - ² z = 1 xRz (Transitive)

Hence,

Hence, $Ris an equivalence relation.

Pattern Recognition

Converting the relation constraint to

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.

Chapter Mix

Class 12 Mathematics: Relations and Functions

Rankbit System
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