CUET 2026 Mathematics Paper Analysis, Check Key Topics Asked

CUET 2026 Mathematics Paper Analysis is here and it highlights important asked topics such as functions, inverse trigonometry, continuity, differentiation, and probability. It also stresses key exam strategies like time management, strong conceptual understanding, and keeping challenging topics like integration for the end. Candidates who have their exams in upcoming dates are advised to check this information detailed here about the kind of questions asked and how to tackle these types of questions. 

CUET 2026 Mathematics Paper Analysis

The CUET 2026 Mathematics Paper Analysis covers all the essential asked  topics, including functions, inverse trigonometry, continuity, differentiation, and probability. It also recommends smart preparation strategies such as managing time well, focusing on concepts, and attempting difficult topics like integration at the end. Candidates who have their exams in upcoming dates are advised to check this information detailed here about the kind of questions asked and how to tackle these types of questions. 

Functions: One-one, Onto, and Bijective

Understanding function classifications is fundamental.

• One-one (Injective): Each domain element maps to a unique codomain element. (Memory Tip: 'One-one' implies unique pairing.)

• Onto (Surjective): Every codomain element has a corresponding domain element. (Memory Tip: 'Onto' means the entire codomain is 'covered'.)

• Bijective: Both one-one and onto.

Also Check: CUET 2026 Paper Easy or Tough

Case 1: Linear Function f(x) = ax + b (Domain and Codomain R)

• One-one: f'(x) = a ≠ 0, thus always increasing or decreasing.

• Onto: Range is R, matching codomain.

• Conclusion: Bijective.

Case 2: Greatest Integer Function f(x) = [x]

• Not One-one: Multiple inputs map to same output (e.g., f(1.3) = 1, f(1.4) = 1).

• Not Onto: Range is Z, not codomain R.

• Conclusion: Neither one-one nor onto.

Domain of Inverse Trigonometric Functions

Identify the expression within the inverse function and set it within its standard domain. (Memory Tip: For sin⁻¹(expression), expression must be in [-1, 1].)

Problem: Find the domain of sin⁻¹(2x).

Solution: -1 ≤ 2x ≤ 1 => x ∈ [-1/2, 1/2].

Inverse Trigonometric Simplification

Convert the inner inverse trigonometric function to match the outer using a right-angled triangle. (Memory Tip: Use a right-angled triangle for conversion.)

Problem: Simplify tan(cos⁻¹x).

Solution: Let y = cos⁻¹x => cos y = x/1. A triangle yields tan y = √(1 - x²) / x. So, tan(cos⁻¹x) = **√(1 - x²) / x**.

Continuity and Discontinuity of a Function

A function f(x) is continuous at x = c if lim (x→c) f(x) = f(c).

Problem: Determine continuity of f(x) at x = 1, where f(x) = (x² - 4x + 3) / (x - 1) for x ≠ 1; f(x) = 2 for x = 1.

Solution:

1. f(1): From definition, f(1) = 2.

2. lim (x→1) f(x): lim (x→1) (x - 1)(x - 3) / (x - 1) = lim (x→1) (x - 3) = 1 - 3 = -2.

3. Comparison: Since f(1) = 2 ≠ lim (x→1) f(x) = -2, the function is discontinuous at x = 1.

Derivative of Infinite Series

For infinite series problems where y is defined recursively, use the "ocean analogy": removing one term does not change the overall value. (Memory Tip: For y = √(f(x) + y), the "ocean analogy" applies.)

Problem: Find dy/dx for y = √(log x + √(log x + √(log x + ...))).

Solution:

1. Rewrite as y = √(log x + y).

2. Square both sides: y² = log x + y.

3. Differentiate implicitly w.r.t. x: 2y (dy/dx) = (1/x) + (dy/dx).

4. Solve for dy/dx: dy/dx = **1 / [x(2y - 1)]**.

Calculus Concepts: Extrema and Critical Points

A strong understanding of fundamental function properties and shortcuts is crucial for efficient problem-solving. (Memory Tip: Knowledge of fundamental function properties and practice of "hacks" (shortcuts) are essential for competitive exams.)

1. Exponential Function Property: f(x) = eˣ

• Property: An exponential function eˣ is always increasing because its derivative f'(x) = eˣ, which is always positive (> 0).

2. Finding Minima for a Quadratic Function: f(x) = x² - 4x

• Solution: Find the first derivative f'(x) = 2x - 4. Setting f'(x) = 0 gives x = 2. The second derivative f''(x) = 2 > 0, confirming x = **2** as the point of minima.

3. Maximum Value of (1/x)ˣ

• Task: Find x at which the maximum value occurs.

• Solution: Using logarithmic differentiation, log y = -x log x. Differentiating and setting dy/dx = 0 yields log x = -1, so x = **1/e**.

4. Critical Points of x + 1/x

• Solution: Find f'(x) = 1 - 1/x². Setting f'(x) = 0 gives 1 - 1/x² = 0 => x² = 1 => x = **±1**.

Probability

Probability is a scoring topic; mastering various problem types is beneficial.

1. Probability of Not a Black Card

• Problem: One card is drawn from a pack of 52 cards. Find the probability that the chosen card is not a black card.

• Solution: Total cards = 52. Number of black cards = 26. Number of red cards = 26. The probability of the card not being black is the probability of it being a red card, so P(red) = 26/52 = 1/2.

2. Probability of (A' ∩ B')

• Problem: Find P(A' ∩ B') given P(A) = 2/3, P(B) = 4/9, and P(A ∩ B) = 14/15.

• Solution: Using De Morgan's Law, P(A' ∩ B') = P((A ∪ B)'). This simplifies to 1 - P(A ∪ B).

• First, calculate P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 2/3 + 4/9 - 14/15 = (30 + 20 - 42) / 45 = 8/45.

• Then, P(A' ∩ B') = 1 - 8/45 = 37/45.

3. Probability of (B'|A')

• Problem: Find P(B'|A') using P(A) = 1/2, P(B) = 1/4, and P(A ∩ B) = 1/5.

• Solution: The formula for conditional probability is P(B'|A') = P(B' ∩ A') / P(A'). This can be rewritten as (1 - P(A ∪ B)) / (1 - P(A)).

• First, calculate P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 1/2 + 1/4 - 1/5 = (10 + 5 - 4) / 20 = 11/20.

• Then, P(B'|A') = (1 - 11/20) / (1 - 1/2) = (9/20) / (1/2) = 9/10.

Linear Programming Problems (LPP)

When an LPP has multiple optimal solutions, the objective function value is identical at all optimal corner points.

Worked Example:

• Problem: The objective function Z = ax + by (where a, b > 0) has an optimal solution at two corner points, (30, 0) and (20, 30). Find the relation between 'a' and 'b'.

• Solution: Set the objective function value equal at both points:
  a(30) + b(0) = a(20) + b(30)
  30a = 20a + 30b
  10a = 30b
  Therefore, a = 3b.

3D Geometry: Perpendicular Distance and Image

These problems involve finding a point on a line (foot of perpendicular) and its reflection (image) from another point.

Method for Foot of Perpendicular (Q) and Perpendicular Distance (PQ)

1. Represent any point Q on the line parametrically (e.g., from symmetric form of line).

2. Calculate the direction ratios of the vector PQ from the given point P to Q.

3. Use the perpendicularity condition: the dot product of PQ's direction ratios and the line's direction ratios is zero. This allows finding the parameter (e.g., λ).

4. Substitute λ back into the parametric form to find the coordinates of the foot of the perpendicular Q.

5. Calculate the distance PQ using the distance formula.

• Example: For P(1, 2, 3) and a line (with direction ratios (3, 2, -2)), setting the dot product to zero yields λ = -1.

• Foot Q: Substituting λ = -1 gives Q(3, 5, 9).

• Distance PQ: √[(3-1)² + (5-2)² + (9-3)²] = √[2² + 3² + 6²] = √4 + 9 + 36 = √49 = **7 units**.

Finding the Image of Point P (P')

• The foot of the perpendicular (Q) is the midpoint between the original point (P) and its image (P').

• Example: Given P(1, 2, 3) and Q(3, 5, 9), if P' is (a, b, c):

• (1 + a)/2 = 3 => a = 5

• (2 + b)/2 = 5 => b = 8

• (3 + c)/2 = 9 => c = 15

• The image P' is (5, 8, 15).

Vector Algebra: Area of a Parallelogram

The area of a parallelogram can be found using the cross product of its diagonals.

Formula: If the diagonals of a parallelogram are represented by vectors d1 and d2, the area is 1/2 |d1 × d2|.

Worked Example:

• Problem: Find the area of a parallelogram with diagonals d1 = 3**i** + j - 2**k** and d2 = i + 3**j** - 4**k**.

• Solution:

1. Calculate cross product d1 × d2: (1*(-4) - (-2)*3)i - (3*(-4) - (-2)*1)j + (3*3 - 1*1)k = 2i + 10j + 8k.

2. Calculate magnitude |d1 × d2|: √[2² + 10² + 8²] = √[4 + 100 + 64] = √168 = 2√42.

3. Area: 1/2 * (2√42) = √42 square units.

Differential Equations: Order and Degree

These concepts define the structural properties of a differential equation.

• Order: The order of a differential equation is the order of the highest derivative present in the equation.

• Degree: The degree is the highest power of the highest order derivative, after the equation has been made free from radicals and fractions (as far as derivatives are concerned).

• Key Principle: If a derivative is inside a transcendental function (e.g., sin(dy/dx), log(d²y/dx²)), the degree is not defined.

Matrices

Fundamental definitions of different types of matrices.

• Orthogonal Matrix: A square matrix A is Orthogonal if its transpose is equal to its inverse (Aᵀ = A⁻¹).

• Diagonal Matrix: A square matrix where all the non-diagonal elements are zero.

• Skew-Symmetric Matrix: A square matrix A is skew-symmetric if Aᵀ = -A.

• Adjoint of a 2x2 Matrix: For a 2x2 matrix [[a, b], [c, d]], the adjoint is found by swapping the elements on the main diagonal (a and d) and changing the sign of the off-diagonal elements (b and c).

Determinants

Understanding properties of determinants with scalar multiplication is essential.

Property: If A is a square matrix of order 'n', and 'c' is a scalar, then the determinant of the scalar multiple (cA) is given by |cA| = cⁿ|A|.

Worked Example:

• Problem: If A is a 3x3 matrix and |3A| = K|A|, find the value of K.

• Solution: Using the property |cA| = cⁿ|A| with n = 3, we have |3A| = 3³|A| = 27|A|. Comparing this to |3A| = K|A|, we find K = 27.

Bayes' Theorem

Bayes' Theorem is frequently tested in conditional probability problems involving multiple events.

Formula: For mutually exclusive and exhaustive events E₁, E₂, …, Eₙ, and an event A, the conditional probability P(E₁|A) = [P(E₁) * P(A|E₁)] / [Σ P(Eᵢ) * P(A|Eᵢ)].

Worked Example (Factory Defectives):

• Problem: Given machine production percentages P(Eᵢ) and defective rates P(D|Eᵢ), if a randomly selected bolt is defective (D), find the probability it was from machine E1 (P(E1|D)).

• Solution:

• Numerator: P(E1)P(D|E1) = (25/100)(5/100) = 0.0125.

• Denominator (Total P(D)): P(E1)P(D|E1) + P(E2)P(D|E2) + P(E3)P(D|E3) = (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02) = 0.0125 + 0.0140 + 0.0080 = 0.0345.

• P(E1|D) = 0.0125 / 0.0345 = 125/345 = 25/69.

Exam Strategy: Approaching Integration Questions

Integration questions were noted to be extremely challenging in recent exams. A strategic approach is vital.

(Memory Tip: Always plan to solve integration problems at the very end of the exam. If an integration question does not immediately appear solvable or seems complex, skip it immediately to save time, and only return if time permits.)

• Prioritize Last: Never start your exam by attempting integration problems.

• Quick Assessment: Briefly check if an integration problem appears straightforward using familiar methods.

• Decision: If it looks complex or unfamiliar, skip it immediately.

• Time Management: Focus on completing other sections of the exam efficiently first.

Strategic Skipping: It is often better to accurately solve all other questions than to waste time on a few difficult integration problems and potentially miss easier marks.

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