If f(x) = 2^x2^x + √(2), x in R, then Σk=1⁸¹ f((k)/(82)) is equal to:

Solution & Explanation

Related Formula

Symmetric identity wrapper for matching indices:

f(x) + f(1-x) = 1
Core Logic

Let's evaluate f(x) + f(1-x):

f(x) + f(1-x) = 2^x2^x + √(2) + 21-x21-x + √(2) = 2^x2^x + √(2) + 22 + √(2)· 2^x = 2^x + √(2)2^x + √(2) = 1
Step 1: Expanding the Series

Pairing matching terms from opposite ends of the summation:

Σk=1⁸¹ f((k)/(82)) = [f((1)/(82)) + f((81)/(82))] + + f((41)/(82))

There are 40 complete pairs matching the f(x) + f(1-x) = 1 identity, plus one lone center term f((1)/(2)).

Step 2: Computing Final Valuation
Sum = 40 + f((1)/(2)) = 40 + √(2)√(2) + √(2) = 40 + (1)/(2) = (81)/(2)
Pattern Recognition

When encountering fractional summation bounds, always check the sum of components x + (1-x) to find linear reduction templates.

Chapter Mix

Class 11 Maths: Sequences and Series Class 12 Maths: Relations and Functions

More Functions Previous-Year Questions — Page 5

Q62 jee_main_2025_02_april_evening Domain of a Function
If the domain of the function f(x) = 1√(10 + 3x - x²) + 1√(x + |x|) is (a, b), then (1 + a)² + b² is equal to:
  • A. 26
  • B. 29
  • C. 25
  • D. 30

Solution

Related Formula
For 1√(g(x)) 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 - x² > 0 x² - 3x - 10 < 0 (x - 5)(x + 2) < 0 x in (-2, 5) --- (1)
Step 2: Find the domain of the second term

For the second term to be defined:

x + |x| > 0

  • If x ≥ 0: x + x = 2x > 0 x > 0.
  • If x < 0: x - x = 0 0.
  • Thus, the domain of the second term is:

x in (0, ∞) --- (2)
Step 3: Find intersection and calculate the final expression

Intersecting domains (1) and (2):

x in (-2, 5) (0, ∞) x in (0, 5)

Comparing this with (a, b) gives a = 0 and b = 5.

Now calculate the value:

(1 + a)² + b² = (1 + 0)² + 5² = 1 + 25 = 26
Pattern Recognition

Modulus domain constraint: The function x + |x| is non-zero only for positive values of x. This is a standard math trick that collapses complex domains down to x > 0 instantly.

Chapter Mix

Class 11 Mathematics: Relations and Functions

Q jee_main_2025_02_april_morning Types of Relations
Let A be the set of all functions f Z → Z and R be a relation on A such that R = (f, g): f(0) = g(1) and f(1) = g(0). Then 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 f(0) = f(1)
  • Symmetric: (f, g) in R (g, f) in R
  • Transitive: (f, g) in R and (g, h) in R (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, we require f(0) = f(1) and f(1) = f(0). This holds true only for functions whose values at 0 and 1 are identical. Since it does not hold true for all possible functions mapping Z → Z (e.g., f(x)=x), R is not reflexive.

Step 2: Symmetry Audit

Assume (f,g) in R f(0) = g(1) and f(1) = g(0). To check if (g,f) in R, check its matching constraints: g(0) = f(1) and g(1) = f(0). Both statements are perfectly identical to our assumption. Therefore, R is symmetric.

Step 3: Transitivity Audit

Assume (f,g) in R f(0) = g(1), f(1) = g(0). Assume (g,h) in R g(0) = h(1), g(1) = h(0). For (f,h) in R, we need f(0) = h(1) and f(1) = h(0). From assumptions: f(0) = g(1) = h(0) and f(1) = g(0) = h(1). This means f(0) = h(0) and f(1) = h(1), which does not necessarily satisfy f(0)=h(1). Hence, R is not transitive.

Pattern Recognition

The relation swaps indices 0 and 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.

Chapter Mix

Class 12 Mathematics: Relations and Functions

Q jee_main_2025_03_april_evening Domain of Functions
If the domain of the function f(x) = ₇(1 - ₄(x² - 9x + 18)) is (α, β) (γ, δ), then the sum α + β + γ + δ is equal to
  • A. 18
  • B. 16
  • C. 15
  • D. 17

Solution

Related Formula

For a logarithmic term b(g(x)) to be defined:

  • g(x) > 0
  • b > 0, b ≠ 1
Core Logic

Let's set defining inequalities sequentially:

  • Inside the outer logarithm:
1 - ₄(x² - 9x + 18) > 0 ₄(x² - 9x + 18) < 1

Since base is 4 > 1:

x² - 9x + 18 < 4 x² - 9x + 14 < 0 (x-2)(x-7) < 0 x in (2, 7) --- (1)
Step 1: Finding bounds for inner logarithmic term
  • Inside the inner logarithm:
x² - 9x + 18 > 0 (x-3)(x-6) > 0 x in (-∞, 3) (6, ∞) --- (2)
Step 2: Intersection of regions

Taking the intersection of (1) and (2):

x in (2, 3) (6, 7)

This gives:

α = 2, β = 3, γ = 6, δ = 7

Calculating the sum:

α + β + γ + δ = 2 + 3 + 6 + 7 = 18
Pattern Recognition

Logarithmic domains must check arguments from the innermost level to the outermost level. Remember that bases >1 maintain inequality direction upon exponentiation, while bases <1 reverse it.

Chapter Mix

Class 11 Mathematics: Relations and Functions

Q56 jee_main_2025_03_april_evening Types of Relations
Let A = -2, -1, 0, 1, 2, 3. Let R be a relation on A defined by xRy if and only if y = x, 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 + m + n is equal to
  • A. 12
  • B. 11
  • C. 13
  • D. 14

Solution

Related Formula

A relation R on set A is:

  • Reflexive: If (x, x) in R for all x in A.
  • Symmetric: If (x, y) in R (y, x) in R.
Core Logic

Let's find the explicit set R using y = x, 1:

  • x = -2 y = 1 (-2, 1) in R
  • x = -1 y = 1 (-1, 1) in R
  • x = 0 y = 1 (0, 1) in R
  • x = 1 y = 1 (1, 1) in R
  • x = 2 y = 2 (2, 2) in R
  • x = 3 y = 3 (3, 3) in R
R = (-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)

Thus, l = 6 elements.

Step 1: Making the Relation Reflexive

For R to be reflexive, it must contain all elements (x, x) where x in A = -2, -1, 0, 1, 2, 3. Currently, R has (1,1), (2,2), (3,3). Missing elements: (-2, -2), (-1, -1), (0, 0).

Thus, minimum number of elements to add: m = 3

Step 2: Making the Relation Symmetric

For R to be symmetric, if (x, y) in R and x ≠ y, then (y, x) must also be in R.

  • (-2, 1) in R need (1, -2)
  • (-1, 1) in R need (1, -1)
  • (0, 1) in R need (1, 0)
  • Missing elements to form symmetric pairs: (1, -2), (1, -1), (1, 0).

    Thus, minimum number of elements to add: n = 3

    Calculating total:

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). Count the elements already satisfying standard relations, then subtract from |A| to find missing diagonal terms for reflexivity.

Chapter Mix

Class 12 Mathematics: Relations and Functions

Q60 jee_main_2025_03_april_evening Functional Equations
Let f be a function such that f(x) + 3f((24)/(x)) = 4x, x ≠ 0. Then f(3) + f(8) is equal to
  • A. 11
  • B. 10
  • C. 12
  • D. 13

Solution

Related Formula

A functional equation relates the values of a function at different arguments. We can find values by substituting symmetric inputs that map to each other (e.g., x and (24)/(x)).

Core Logic

Given:

f(x) + 3f((24)/(x)) = 4x --- (1)
Step 1: Substitution of values

Substitute x = 3:

f(3) + 3f(8) = 12 --- (2)

Substitute x = 8:

f(8) + 3f(3) = 32 --- (3)
Step 2: Linear combination of equations

Add equations (2) and (3) directly:

(f(3) + 3f(8)) + (f(8) + 3f(3)) = 12 + 32 4(f(3) + f(8)) = 44 f(3) + f(8) = 11
Pattern Recognition

Instead of solving for the general function f(x) (which is also easy by substitution: replace x → 24/x), look at the symmetric nature of the target expression f(3) + f(8). Direct addition of symmetric systems avoids resolving the individual values and saves time.

Chapter Mix

Class 11 Mathematics: Relations and Functions

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