Whenever dealing with dual inverse trig terms, strictly isolate the bounding intervals [-1, 1]$[-1, 1]$ for each argument separately and use a number line intersection to find the strictest common region.
Chapter Mix
Class 12 Maths: Functions
Class 11 Maths: Linear Inequalities
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 9
Q54jee_main_2025_07_april_eveningRange of Rational Functions
If the range of the function f(x) = (5 - x)/(x² - 3x + 2)$f(x) = \frac{5 - x}{x^2 - 3x + 2}$, x ≠ 1, 2$x \neq 1, 2$, is (-∞ , α ] [ β , ∞)$(-\infty , \alpha ] \cup [ \beta , \infty)$, then α² +β²$\alpha^2 +\beta^2$ is equal to :
A.190$190$
B.192$192$
C.188$188$
D.194$194$
Solution
Related Formula
For a quadratic equation Ax² + Bx + C = 0$Ax^2 + Bx + C = 0$ to yield real roots, its discriminant must satisfy:
D = B² - 4AC ≥ 0$$D = B^2 - 4AC \ge 0$$
Core Logic
Set y = (5 - x)/(x² - 3x + 2)$y = \frac{5 - x}{x^2 - 3x + 2}$:
Case I: If y = 0$y = 0$, the equation simplifies to x - 5 = 0 x = 5$x - 5 = 0 \implies x = 5$, which is a valid part of the domain. Thus, 0$0$ belongs to the range.
Case II: If y ≠ 0$y \neq 0$, for x$x$ to be real, D ≥ 0$D \ge 0$:
For rational expressions of the form LinearQuadratic$\frac{\text{Linear}}{\text{Quadratic}}$, converting to a quadratic in x$x$ and forcing D ≥ 0$D \ge 0$ establishes the range boundaries elegantly.
Chapter Mix
Class 11 Mathematics: Sets, Relations and Functions
Q52jee_main_2025_24_jan_eveningOne-One and Onto Functions
The function f:(-∞,∞)arrow(-∞,1)$f:(-\infty,\infty)\rightarrow(-\infty,1)$, defined by f(x)= 2x-2-x2x+2-x$f(x)=\frac{2^{x}-2^{-x}}{2^{x}+2^{-x}}$ is:
A.One-one but not onto$\text{One-one but not onto}$
B.Onto but not one-one$\text{Onto but not one-one}$
C.Both one-one and onto$\text{Both one-one and onto}$
D.Neither one-one nor onto$\text{Neither one-one nor onto}$
Solution
Related Formula
A function is one-one if its derivative is strictly monotonic (always positive or always negative) across its domain. It is onto if its range equals its co-domain.
Core Logic
Rewrite the function by multiplying the numerator and denominator by 2^x$2^x$:
Thus, the range of the function is (-1, 1)$(-1, 1)$. Since the given co-domain is (-∞, 1)$(-\infty, 1)$ and Range ≠ Co-domain$\text{Range} \neq \text{Co-domain}$ , the function is not onto.
Pattern Recognition
The expression given is a shifted form of the hyperbolic tangent function (x ln 2)$\tanh(x \ln 2)$. Hyperbolic tangent always maps to (-1, 1)$(-1, 1)$, making its restriction against (-∞, 1)$(-\infty, 1)$ non-surjective (not onto).
Chapter Mix
Class 12 Mathematics: Relations and Functions
Q58jee_main_2025_24_jan_eveningLinear Programming and Inequalities in Two Variables
Let the points ((11)/(2),α)$(\frac{11}{2},\alpha)$ lie on or inside the $\triangle$ with sides x+y=11$x+y=11$, x+2y=16$x+2y=16$ and 2x+3y=29$2x+3y=29$ Then the product of the smallest and the largest values of a is equal to:
A.22$22$
B.44$44$
C.33$33$
D.55$55$
Solution
Related Formula
For a vertical line x = x₀$x = x_0$ crossing a bounded region, the valid coordinates of y$y$ sit between the boundary lines intersecting that specific line vertical plane.
Core Logic
The point given is fixed at x = (11)/(2) = 5.5$x = \frac{11}{2} = 5.5$. We evaluate the values of y$y$ along this vertical line segment across each boundary edge.
Linear Programming region graph for Q58 - JEE Main 2025 Evening
Step 1: Evaluate Intersections
Substitute x = (11)/(2)$x = \frac{11}{2}$ into the three linear constraints:
From x + y = 11$x + y = 11$:
(11)/(2) + y = 11 ⇒ y = 11 - 5.5 = 5.5$$\frac{11}{2} + y = 11 \Rightarrow y = 11 - 5.5 = 5.5$$
Checking the internal region of the $\triangle$ bounded by these lines, the valid range for α$\alpha$ along the section line is delimited by y = 5.5$y = 5.5$ and y = 6$y = 6$:
Instead of drawing full coordinate diagrams or computing all three vertex points, evaluating values directly at the fixed coordinate constraint x = 5.5$x = 5.5$ saves time during multi-line area problems.
Chapter Mix
Class 11 Mathematics: Linear Inequalities
Class 11 Mathematics: Straight Lines
Q71jee_main_2025_24_jan_eveningNumber of Functions under Constraints
Number of functions f:1,2, ,100arrow0,1$f:\{1,2,\dots,100\}\rightarrow\{0,1\}$, 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 n₁$n_1$ ways, followed by a second operation in n₂$n_2$ ways, the total configurations equal n₁ × n₂$n_1 \times n_2$.
Core Logic
The domain set contains integers from 1 to 100. We divide the mapping requirements across distinct subsets of this domain.
Function mapping grid for Q71 - JEE Main 2025 Evening
Step 1: Choose the single element from 1, 2, , 98$\{1, 2, \dots, 98\}$
We must assign the image value 1 to exactly one positive integer from the subset 1, 2, , 98$\{1, 2, \dots, 98\}$. The number of ways to pick this single element is :
981 = 98 ways$$\binom{98}{1} = 98 \text{ ways}$$
Step 2: Mapping remaining elements
The remaining 97 elements in the 1, 2, , 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:
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.
Chapter Mix
Class 12 Mathematics: Relations and Functions
Class 11 Mathematics: Permutations and Combinations
Qjee_main_2025_24_jan_morningSymmetric Property of Functions
Let f(x) = 2x + 2 + 1622x + 1 + 2x + 4 + 32$f(x) = \frac{2^{x + 2} + 16}{2^{2x + 1} + 2^{x + 4} + 32}$. Then the value of 8(f((1)/(15)) + f((2)/(15)) + … + f((59)/(15)))$8\left(f\left(\frac{1}{15}\right) + f\left(\frac{2}{15}\right) + \ldots + f\left(\frac{59}{15}\right)\right)$ 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) = constant$f(x) + f(k-x) = \text{constant}$.
Core Logic
First simplify the expression for f(x)$f(x)$ algebraically:
Whenever a symmetric set of arguments is presented inside a summation matching (k)/(n) + (N-k)/(n) = constant$\frac{k}{n} + \frac{N-k}{n} = \text{constant}$, look for an algebraic reduction of f(x)$f(x)$ that yields a uniform constant sum for symmetric pairs.
Chapter Mix
Class 11 Mathematics: Functions
More Functions Questions — jee_main_2026_21_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.