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Number of functions f:\1,2,dots,100\rightarrow\0,1\, that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to \_\_\_\_.

Numerical Answer Type:
Enter a numerical value Answer: 392 +4 marks

Solution & Explanation

### Related Formula Fundamental Counting Principle: If an operation can be performed in n_1 ways, followed by a second operation in n_2 ways, the total configurations equal n_1 times n_2. ### Core Logic The domain set contains integers from 1 to 100. We divide the mapping requirements across distinct subsets of this domain[cite: 4064, 4068, 4069].
Function mapping grid for Q71 - JEE Main 2025 Evening
Function mapping grid for Q71 - JEE Main 2025 Evening
### Step 1: Choose the single element from \1, 2, dots, 98\ We must assign the image value 1 to exactly one positive integer from the \subset \1, 2, dots, 98\. The number of ways to pick this single element is : binom981 = 98 text ways ### Step 2: Mapping remaining elements The remaining 97 elements in the \1, 2, dots, 98\ \subset cannot map to 1, so they must map to 0. This leaves exactly 1 choice per remaining element. For the final two elements in the domain, 99 and 100, there are no structural constraints [cite: 4068, 4069]: - Element 99 can map to either 0 or 1 (2 options) . - Element 100 can map to either 0 or 1 (2 options). ### Step 3: Total functions combination Multiply the independent choices together : textTotal functions = 98 times 2 times 2 = 392 [cite: 4065, 4067] ### Pattern Recognition Separate domains tightly into restricted blocks vs completely free components. Realizing that elements 99 and 100 behave independently with full co-domain targets leaves a clear product formulation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions Class 11 Mathematics: Permutations and Combinations

Reference Study Guides

Relations Functions

More Relations and Functions Previous-Year Questions

Q54 2025 Relations
Let mathrmA = \1, 2, 3, dots, 100\ and mathrmR be a relation on mathrmA such that mathrmR = \(a, b) : a = 2b + 1\. Let (a_1, a_2), (a_2, a_3), (a_3, a_4), dots, (a_k, a_k+1) be a sequence of k elements of mathrmR 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, 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 ### Core Logic To find the longest sequence of connected pairs, we trace the relation backward starting from the smallest elements in A. ### Step 1: Trace the relations backward To maximize 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. Since a_k = 2 a_k+1 + 1: - If a_k+1 = 1 implies a_k = 3 - If a_k+1 = 2 implies a_k = 5 Let's test the chain starting with a_k+1 = 1: - a_k = 2(1) + 1 = 3 - a_k-1 = 2(3) + 1 = 7 - a_k-2 = 2(7) + 1 = 15 - a_k-3 = 2(15) + 1 = 31 - a_k-4 = 2(31) + 1 = 63 - a_k-5 = 2(63) + 1 = 127 (but 127 notin mathrmA!) Thus, the longest chain within the set A has 6 elements: \63, \, 31, \, 15, \, 7, \, 3, \, 1\ This chain corresponds to exactly 5 ordered pairs: (63, 31), \, (31, 15), \, (15, 7), \, (7, 3), \, (3, 1) So the maximum number of pairs in the sequence is k = 5. ### Step 2: Check alternative chains If we start with a_k+1 = 2: - a_k+1 = 2 - a_k = 5 - a_k-1 = 11 - a_k-2 = 23 - a_k-3 = 47 - a_k-4 = 95 - a_k-5 = 191 > 100 Again, the maximum number of pairs is k = 5. Thus, the largest integer k is 5. ### Pattern Recognition Recursive scaling: Tracing exponential chains of the form 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
Q62 2025 Domain of a Function
If the domain of the function f(x) = frac1sqrt10 + 3x - x^2 + frac1sqrtx + |x| is (a, b), then (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 ### 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 (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 - If x ge 0: x + x = 2x > 0 implies x > 0. - If x < 0: x - x = 0 ngtr 0. Thus, the domain of the second term is: 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) Comparing this with (a, b) gives a = 0 and b = 5. Now calculate the value: (1 + a)^2 + b^2 = (1 + 0)^2 + 5^2 = 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions
Q 2025 Types of Relations
Let A be the set of all functions fcolon mathbbZ to mathbbZ and R be a relation on A such that R = \(f, g): f(0) = g(1) text 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 iff f(0) = f(1) * Symmetric: (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 ### 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 mathbbZ to mathbbZ (e.g., f(x)=x), R is **not reflexive**. ### Step 2: Symmetry Audit Assume (f,g) in R implies 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 implies f(0) = g(1), f(1) = g(0). Assume (g,h) in R implies 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions
Q56 2025 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 = max\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 implies (y, x) in R. ### Core Logic Let's find the explicit set R using y = max\x, 1\: - x = -2 implies y = 1 implies (-2, 1) in R - x = -1 implies y = 1 implies (-1, 1) in R - x = 0 implies y = 1 implies (0, 1) in R - x = 1 implies y = 1 implies (1, 1) in R - x = 2 implies y = 2 implies (2, 2) 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)\ 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 neq y, then (y, x) must also be in R. - (-2, 1) in R implies need (1, -2) - (-1, 1) in R implies need (1, -1) - (0, 1) in R implies 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions
Q60 2025 Functional Equations
Let f be a function such that f(x) + 3fleft(frac24xright) = 4x, x neq 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 frac24x). ### Core Logic Given: f(x) + 3fleft(frac24xright) = 4x quad text--- (1) ### Step 1: Substitution of values Substitute x = 3: f(3) + 3f(8) = 12 quad text--- (2) Substitute x = 8: f(8) + 3f(3) = 32 quad text--- (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 to 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions

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