Keywords:#domain of the function#JEE Main 2026 Morning Q1#Functions JEE Main 2026#Domain of Inverse Trigonometric Functions JEE Main 2026
More Functions Previous-Year Questions — Page 6
Q71jee_main_2025_24_jan_eveningNumber of Functions under Constraints
Number of functions f:\1,2,dots,100\rightarrow\0,1\$f:\{1,2,\dots,100\}\rightarrow\{0,1\}$, that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to \_\_\_\_.
Numerical Answer.Answer: 392
Solution
### Related Formula
Fundamental Counting Principle: If an operation can be performed in n_1$n_1$ ways, followed by a second operation in n_2$n_2$ ways, the total configurations equal n_1 times n_2$n_1 \times n_2$.
### 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
### Step 1: Choose the single element from \1, 2, dots, 98\$\{1, 2, \dots, 98\}$
We must assign the image value 1 to exactly one positive integer from the \subset \1, 2, dots, 98\$\{1, 2, \dots, 98\}$. The number of ways to pick this single element is :
binom981 = 98 text ways$$\binom{98}{1} = 98 \text{ ways}$$
### Step 2: Mapping remaining elements
The remaining 97 elements in the \1, 2, dots, 98\$\{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$$\text{Total functions} = 98 \times 2 \times 2 = 392$$ [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
Q53jee_main_2025_24_jan_morningFunctional Relations and Equations
Let f: mathbbR - \0\ to mathbbR$f: \mathbb{R} - \{0\} \to \mathbb{R}$ be a function such that
f(x) - 6fleft(frac1xright) = frac353x - frac52$$f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$$
If lim_x to 0 left( frac1alpha x + f(x) right) = beta$\lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta$ for some alpha, beta in mathbbR$\alpha, \beta \in \mathbb{R}$, then alpha + 2beta$\alpha + 2\beta$ is equal to :
A.3$3$
B.5$5$
C.4$4$
D.6$6$
Solution
### Related Formula
For functional equations with inversion, substituting x to frac1x$x \to \frac{1}{x}$ establishes a solvable system of algebraic equations to isolate f(x)$f(x)$ directly.
### Core Logic
The given equation is:
f(x) - 6fleft(frac1xright) = frac353x - frac52 quad dots (1)$$f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2} \quad \dots (1)$$
Substitute x to frac1x$x \to \frac{1}{x}$ in equation (1):
fleft(frac1xright) - 6f(x) = frac35x3 - frac52 quad dots (2)$$f\left(\frac{1}{x}\right) - 6f(x) = \frac{35x}{3} - \frac{5}{2} \quad \dots (2)$$
### Step 1: Eliminate f(1/x)$f(1/x)$
Multiply equation (2) by 6$6$ and add it to equation (1):
left[ f(x) - 6fleft(frac1xright) right] + 6 left[ fleft(frac1xright) - 6f(x) right] = left( frac353x - frac52 right) + 6 left( frac35x3 - frac52 right)$$\left[ f(x) - 6f\left(\frac{1}{x}\right) \right] + 6 \left[ f\left(\frac{1}{x}\right) - 6f(x) \right] = \left( \frac{35}{3x} - \frac{5}{2} \right) + 6 \left( \frac{35x}{3} - \frac{5}{2} \right)$$f(x) - 36f(x) = frac353x - frac52 + 70x - 15$$f(x) - 36f(x) = \frac{35}{3x} - \frac{5}{2} + 70x - 15$$-35f(x) = 70x + frac353x - frac352$$-35f(x) = 70x + \frac{35}{3x} - \frac{35}{2}$$
Divide across by -35$-35$:
f(x) = -2x - frac13x + frac12$$f(x) = -2x - \frac{1}{3x} + \frac{1}{2}$$
### Step 2: Evaluate the Limit
We are given that the following limit evaluates to a finite constant beta$\beta$:
lim_x to 0 left( frac1alpha x + f(x) right) = beta$$\lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta$$lim_x to 0 left( frac1alpha x - 2x - frac13x + frac12 right) = beta$$\lim_{x \to 0} \left( \frac{1}{\alpha x} - 2x - \frac{1}{3x} + \frac{1}{2} \right) = \beta$$lim_x to 0 left( left[ frac1alpha - frac13 right] frac1x - 2x + frac12 right) = beta$$\lim_{x \to 0} \left( \left[ \frac{1}{\alpha} - \frac{1}{3} \right] \frac{1}{x} - 2x + \frac{1}{2} \right) = \beta$$
For the limit to be a finite value, the coefficient of frac1x$\frac{1}{x}$ must vanish completely:
frac1alpha - frac13 = 0 implies alpha = 3$$\frac{1}{\alpha} - \frac{1}{3} = 0 \implies \alpha = 3$$
When alpha = 3$\alpha = 3$, the limit simplifies directly to the constant term:
beta = lim_x to 0 left( -2x + frac12 right) = frac12$$\beta = \lim_{x \to 0} \left( -2x + \frac{1}{2} \right) = \frac{1}{2}$$
### Step 3: Calculate Final Value
Substitute the determined parameters alpha$\alpha$ and beta$\beta$:
alpha + 2beta = 3 + 2left(frac12right) = 3 + 1 = 4$$\alpha + 2\beta = 3 + 2\left(\frac{1}{2}\right) = 3 + 1 = 4$$
### Pattern Recognition
In limit problems involving fractional components where x to 0$x \to 0$, any term like frac1x$\frac{1}{x}$ or higher negative powers must have a net coefficient of zero to guarantee existence of a finite limit value.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Functions
Class 11 Mathematics: Limits and Derivatives
Q55jee_main_2025_24_jan_morningSymmetric Property of Functions
Let f(x) = frac2^x + 2 + 162^2x + 1 + 2^x + 4 + 32$f(x) = \frac{2^{x + 2} + 16}{2^{2x + 1} + 2^{x + 4} + 32}$ . Then the value of 8left(fleft(frac115
ight) + fleft(frac215
ight) + ldots + fleft(frac5915
ight)
ight)$8\left(f\left(\frac{1}{15}
ight) + f\left(\frac{2}{15}
ight) + \ldots + f\left(\frac{59}{15}
ight)
ight)$ is equal to :
A.118$118$
B.92$92$
C.102$102$
D.108$108$
Solution
### Related Formula
Many finite fractional sum questions involving functional terms rely on identifying an underlying symmetric summation invariant, typically of the form f(x) + f(k-x) = textconstant$f(x) + f(k-x) = \text{constant}$.
### Core Logic
First simplify the expression for f(x)$f(x)$ algebraically:
f(x) = frac4 cdot 2^x + 162 cdot (2^x)^2 + 16 cdot 2^x + 32$$f(x) = \frac{4 \cdot 2^x + 16}{2 \cdot (2^x)^2 + 16 \cdot 2^x + 32}$$
Factor out 4$4$ from the numerator and 2$2$ from the denominator:
f(x) = frac4(2^x + 4)2[(2^x)^2 + 8 cdot 2^x + 16] = frac2(2^x + 4)(2^x + 4)^2 = frac22^x + 4$$f(x) = \frac{4(2^x + 4)}{2[(2^x)^2 + 8 \cdot 2^x + 16]} = \frac{2(2^x + 4)}{(2^x + 4)^2} = \frac{2}{2^x + 4}$$
### Step 1: Establish Symmetry Pairings
Let's check the value of f(x) + f(4-x)$f(x) + f(4-x)$:
f(4-x) = frac22^4-x + 4 = frac2frac162^x + 4 = frac2 cdot 2^x16 + 4 cdot 2^x = frac2 cdot 2^x4(2^x + 4) = frac2^x2(2^x + 4)$$f(4-x) = \frac{2}{2^{4-x} + 4} = \frac{2}{\frac{16}{2^x} + 4} = \frac{2 \cdot 2^x}{16 + 4 \cdot 2^x} = \frac{2 \cdot 2^x}{4(2^x + 4)} = \frac{2^x}{2(2^x + 4)}$$
Now compute the sum directly:
f(x) + f(4-x) = frac22^x + 4 + frac2^x2(2^x + 4) = frac4 + 2^x2(2^x + 4) = frac12$$f(x) + f(4-x) = \frac{2}{2^x + 4} + \frac{2^x}{2(2^x + 4)} = \frac{4 + 2^x}{2(2^x + 4)} = \frac{1}{2}$$
Hence, whenever two input arguments sum up to 4$4$, the sum of their functional values is exactly frac12$\frac{1}{2}$.
### Step 2: Group the Finite Series Terms
Consider the terms inside the requested sequence:
frac115 + frac5915 = frac6015 = 4 implies fleft(frac115
ight) + fleft(frac5915
ight) = frac12$$\frac{1}{15} + \frac{59}{15} = \frac{60}{15} = 4 \implies f\left(\frac{1}{15}
ight) + f\left(\frac{59}{15}
ight) = \frac{1}{2}$$frac215 + frac5815 = frac6015 = 4 implies fleft(frac215
ight) + fleft(frac5815
ight) = frac12$$\frac{2}{15} + \frac{58}{15} = \frac{60}{15} = 4 \implies f\left(\frac{2}{15}
ight) + f\left(\frac{58}{15}
ight) = \frac{1}{2}$$
This complementary pairing continues up to:
fleft(frac2915
ight) + fleft(frac3115
ight) = frac12$$f\left(\frac{29}{15}
ight) + f\left(\frac{31}{15}
ight) = \frac{1}{2}$$
This yields exactly 29$29$ distinct pairs. The single middle term left unpaired corresponds to:
textMiddle Term = fleft(frac3015
ight) = f(2) = frac22^2 + 4 = frac28 = frac14$$\text{Middle Term} = f\left(\frac{30}{15}
ight) = f(2) = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}$$
### Step 3: Evaluate Final Expression
Compute the total value by multiplying the grouped sum by 8:
textTotal = 8 cdot left[ 29 cdot left(frac12right) + frac14 right]$$\text{Total} = 8 \cdot \left[ 29 \cdot \left(\frac{1}{2}\right) + \frac{1}{4} \right]$$textTotal = 8 cdot frac292 + 8 cdot frac14 = 116 + 2 = 118$$\text{Total} = 8 \cdot \frac{29}{2} + 8 \cdot \frac{1}{4} = 116 + 2 = 118$$
### Pattern Recognition
Whenever a symmetric set of arguments is presented inside a summation matching frackn + fracN-kn = textconstant$\frac{k}{n} + \frac{N-k}{n} = \text{constant}$, look for an algebraic reduction of f(x)$f(x)$ that yields a uniform constant sum for symmetric pairs.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Functions
Q64jee_main_2025_28_jan_eveningRange and Onto Functions
Let f:[0,3]rightarrow A$f:[0,3]\rightarrow A$ be defined by f(x)=2x^3-15x^2+36x+7$f(x)=2x^{3}-15x^{2}+36x+7$ and g:[0,infty)rightarrow B$g:[0,\infty)\rightarrow B$ be defined by g(x)=fracx^2025x^2025+1$g(x)=\frac{x^{2025}}{x^{2025}+1}$. If both the functions are onto and S=\xin Z:xin Atext or xin B\$S=\{x\in Z:x\in A\text{ or }x\in B\}$, then n(S)$n(S)$ is equal to:
A.30$30$
B.36$36$
C.29$29$
D.31$31$
Solution
### Related Formula
For a function to be onto, its Codomain must equal its Range.
### Core Logic
First, find range A$A$ for f(x) = 2x^3 - 15x^2 + 36x + 7$f(x) = 2x^3 - 15x^2 + 36x + 7$ over [0, 3]$[0, 3]$.
Differentiating f(x)$f(x)$:
f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$$f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$$
Critical points are x=2$x=2$ and x=3$x=3$.
Evaluate f(x)$f(x)$ at boundary and critical points:
- f(0) = 7$f(0) = 7$
- f(2) = 2(8) - 15(4) + 36(2) + 7 = 16 - 60 + 72 + 7 = 35$f(2) = 2(8) - 15(4) + 36(2) + 7 = 16 - 60 + 72 + 7 = 35$
- f(3) = 2(27) - 15(9) + 36(3) + 7 = 54 - 135 + 108 + 7 = 34$f(3) = 2(27) - 15(9) + 36(3) + 7 = 54 - 135 + 108 + 7 = 34$
Thus, Range A = [7, 35]$A = [7, 35]$.
### Step 1: Find Range B for g(x)
Now look at g(x) = fracx^2025x^2025+1 = 1 - frac1x^2025+1$g(x) = \frac{x^{2025}}{x^{2025}+1} = 1 - \frac{1}{x^{2025}+1}$ over [0, infty)$[0, \infty)$.
- At x = 0$x = 0$, g(0) = 0$g(0) = 0$.
- As x to infty$x \to \infty$, g(x) to 1$g(x) \to 1$.
Since g(x)$g(x)$ is continuous and monotonically strictly increasing, Range B = [0, 1)$B = [0, 1)$.
### Step 2: Find the Integer Count of Union Set S
S = \x in mathbbZ : x in A text or x in B\ = mathbbZ cap (A cup B)$S = \{x \in \mathbb{Z} : x \in A \text{ or } x \in B\} = \mathbb{Z} \cap (A \cup B)$A cup B = [0, 1) cup [7, 35]$$A \cup B = [0, 1) \cup [7, 35]$$
The integers in this set are:
- From [0, 1)$[0, 1)$: x = 0$x = 0$
- From [7, 35]$[7, 35]$: x = 7, 8, 9, dots, 35$x = 7, 8, 9, \dots, 35$
Total number of integers n(S)$n(S)$:
n(S) = 1 + (35 - 7 + 1) = 1 + 29 = 30$$n(S) = 1 + (35 - 7 + 1) = 1 + 29 = 30$$
### Pattern Recognition
The condition 'or' means union. Be careful not to include integers between 1$1$ and 6$6$ since they are not present in either continuous range segment.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Functions
Class 12 Mathematics: Application of Derivatives
Q65jee_main_2025_28_jan_eveningDomain of Inverse Trigonometric Functions
Let [x] denote the greatest integer less than or equal to x. Then domain of f(x)=sec^-1(2[x]+1)$f(x)=\sec^{-1}(2[x]+1)$ is:
### Related Formula
The domain of sec^-1(y)$\sec^{-1}(y)$ is given by |y| ge 1$|y| \ge 1$, which means:
y le -1 quad textor quad y ge 1$$y \le -1 \quad \text{or} \quad y \ge 1$$
### Core Logic
For f(x) = sec^-1(2[x]+1)$f(x) = \sec^{-1}(2[x]+1)$ to be defined:
2[x] + 1 le -1 quad textor quad 2[x] + 1 ge 1$$2[x] + 1 \le -1 \quad \text{or} \quad 2[x] + 1 \ge 1$$
### Step 1: Solve individual inequalities
Case 1:
2[x] + 1 le -1 implies 2[x] le -2 implies [x] le -1$$2[x] + 1 \le -1 \implies 2[x] \le -2 \implies [x] \le -1$$
This holds true for all x < 0$x < 0$, i.e., x in (-infty, 0)$x \in (-\infty, 0)$.
Case 2:
2[x] + 1 ge 1 implies 2[x] ge 0 implies [x] ge 0$$2[x] + 1 \ge 1 \implies 2[x] \ge 0 \implies [x] \ge 0$$
This holds true for all x ge 0$x \ge 0$, i.e., x in [0, infty)$x \in [0, \infty)$.
### Step 2: Take Union of the Solutions
textDomain = (-infty, 0) cup [0, infty) = (-infty, infty)$$\text{Domain} = (-\infty, 0) \cup [0, \infty) = (-\infty, \infty)$$
### Pattern Recognition
Since [x]$[x]$ covers all integer values and 2[x]+1$2[x]+1$ forms all odd integer values, the expression inside sec^-1$\sec^{-1}$ is always a non-zero integer. Non-zero integers always have absolute value ge 1$\ge 1$. Hence, it is valid for all real numbers.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Functions
Class 12 Mathematics: Inverse Trigonometric Functions
More Functions Questions — jee_main_2026_21_jan_morning
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