CUET 2026 Maths Most Expected Topics

CUET Mathematics paper has evolved into a concept-driven exam where clarity, speed, and smart decision-making matter more than simple learning. For CUET 2026 aspirants, focusing on the most expected topics in Maths can significantly improve both accuracy and overall score.

Instead of covering everything blindly, students need a targeted approach, i.e., understanding which chapters carry higher weightage and which concepts are repeatedly tested. Topics like Calculus (derivatives and integrals), Matrices, Probability, Vectors, and Linear Programming consistently form a major portion of the paper.

CUET Mathematics Exam Pattern

The CUET Mathematics exam structure is critical for strategic preparation:

• Section A: Consists of 15 questions.

• Section B: Consists of 35 questions.

• Total Questions: The paper comprises 50 questions.

• Mandatory Attempt: All 50 questions must be attempted. The previous pattern allowing 40 out of 50 is outdated.

• Total Marks: The exam is worth 250 marks, with 5 marks awarded for each correct answer.

CUET 2026 Maths Most Expected Topics

Below are the CUET 2026 Maths Most Expected Topics that students should prepare in order to score well in the exam:

Increasing and Decreasing Functions

To determine if a function is increasing or decreasing, its first derivative, f'(x), must be analyzed.

• Increasing Function: f'(x) > 0

• Decreasing Function: f'(x) < 0

• The Wavy Curve Method is used to find intervals satisfying derivative inequalities.

Example: For f(x) = -2x³ + 3x² + 36x + 7, find intervals where it's increasing or decreasing.

1. Differentiate: f'(x) = -6x² + 6x + 36

2. Factorize: f'(x) = -6(x - 3)(x + 2)

3. Increasing Condition: f'(x) > 0 $\implies$ -6(x - 3)(x + 2) > 0. Dividing by -6 reverses the inequality: (x - 3)(x + 2) < 0. This is true for x in (-2, 3).

4. Decreasing Condition: f'(x) < 0 $\implies$ (x - 3)(x + 2) > 0. This is true for x in (-∞, -2) and (3, ∞).

Conclusion: The function increases in (-2, 3) and decreases in (-∞, -2) and (3, ∞).

Linear Inequalities and Feasible Region (LPP)

The feasible region is the area where all linear inequalities are satisfied, typically including non-negative constraints x ≥ 0 and y ≥ 0.

To identify inequalities from a graph:

1. Find the equation of each boundary line (e.g., using x/a + y/b = 1).

2. Test a point (like the origin (0,0)) to determine the correct inequality direction. If (0,0) satisfies the inequality and is within the shaded region, the inequality points towards the origin (e.g., ≤).

Example: If a line passes through (3,0) and (0,4), its equation is 4x + 3y = 12. If the feasible region includes the origin, the inequality is 4x + 3y ≤ 12.

Linear Programming Problem (LPP) - Maximization

The Corner Point Method states that the maximum or minimum value of an objective function (Z) for a bounded feasible region occurs at one of its corner points.

LPP Maximization Shortcut Example: Maximize z = 50x + 15y with constraints like x + y = 60 and 5x + y = 100.

1. Solve the system: x + y = 60 and 5x + y = 100. Subtracting the first from the second gives 4x = 40, so x = 10. Substitute x = 10 into x + y = 60 to get y = 50.

2. The intersection point is (10, 50).

3. If this is (α, β), then α = 10 and β = 50.

4. α + β = 10 + 50 = 60.

Area Bounded by Curves

To find the area bounded by curves, visualize the shapes and use definite integration.

Example: Area bounded by y = x² and y = 16.

1. y = x² is a parabola symmetric about the y-axis. y = 16 is a horizontal line.

2. The region is symmetric. Express x in terms of y: x = ±√y.

3. Integrate x = √y from y = 0 to y = 16 and multiply by 2 for the total area.

4. Area = 2 * ∫[0,16] y^(1/2) dy = 2 * [(2/3)y^(3/2)] from 0 to 16

5. Area = (4/3) * (16^(3/2) - 0) = (4/3) * (4³) = (4/3) * 64 = 256/3 square units.

Derivatives (Chain Rule)

The Chain Rule is used for differentiating composite functions: If y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).

Example: Find dy/dx for y = ³√(1 / (1 - x³)).

1. Rewrite: y = (1 - x³)^(-1/3)

2. Apply Chain Rule: dy/dx = (-1/3) * (1 - x³)^(-4/3) * (-3x²)

3. Simplify: dy/dx = x²(1 - x³)^(-4/3).

Solving Matrix Equation (PXQ = R)

To solve PXQ = R for matrix A:

1. Pre-multiply by P⁻¹: P⁻¹ P A Q = P⁻¹ R $\implies$ A Q = P⁻¹ R.

2. Post-multiply by Q⁻¹: A Q Q⁻¹ = P⁻¹ R Q⁻¹ $\implies$ A = P⁻¹ R Q⁻¹.

Inverse of a 2x2 Matrix: For M = [[a, b], [c, d]], M⁻¹ = (1 / det(M)) * Adj(M), where Adj(M) = [[d, -b], [-c, a]] and det(M) = ad - bc.

Probability Distribution of a Random Variable

For a probability distribution, the sum total of all probabilities for all possible outcomes must be one (Σ P(X) = 1).

Example: Given P(x) = k / 2^x for x = 0, 1, 2, 3.

1. Calculate individual probabilities:

• P(0) = k

• P(1) = k/2

• P(2) = k/4

• P(3) = k/8

2. Sum and equate to 1: k + k/2 + k/4 + k/8 = 1 $\implies$ k(15/8) = 1 $\implies$ k = 8/15.

3. Find specific probability: P(1) = k/2 = (8/15)/2 = 4/15.

Symmetric Matrix Properties

A matrix is symmetric if its elements are symmetric along the diagonals, meaning elements at positions (i, j) and (j, i) are equal. A = Aᵀ.

Example: If an element is 2a and its symmetric counterpart is 4, then 2a = 4, so a = 2.

Determinant Properties: Scalar Multiplication and Transpose

For an n x n matrix A:

1. Scalar Multiplication: det(kA) = k^n * det(A).

2. Transpose: det(A) = det(A transpose).

Example: Given det(A) = 5, find det(-A * A transpose) for a 3x3 matrix A.

det(-A * Aᵀ) = det((-1)A * Aᵀ)

= (-1)³ * det(A) * det(Aᵀ) (assuming n=3)

= -1 * det(A) * det(A)

= -1 * (5 * 5) = -25.

Minimum Value of a Quadratic Function

To find the minimum value of a quadratic function, convert it into a perfect square form: (x - h)² + k.

• A square term (x - h)² is always ≥ 0.

• The minimum value is k when (x - h)² = 0.

Example: Find the minimum value of x² - 8x + 20.

1. x² - 8x + 20 = (x² - 8x + 16) + 4 = (x - 4)² + 4.

2. Since (x - 4)² ≥ 0, the minimum value of (x - 4)² + 4 is 0 + 4 = 4.
   The range is [4, ∞).

Definite Integral using King Property

The King Property of Definite Integrals states: ∫[0,a] f(x) dx = ∫[0,a] f(a-x) dx.

Example: Evaluate I = ∫[0,1] log((1-x)/x) dx.

1. Use log(a/b) = log(a) - log(b): I = ∫[0,1] [log(1-x) - log(x)] dx (Eq 1).

2. Apply King Property (x to 1-x): I = ∫[0,1] [log(1-(1-x)) - log(1-x)] dx = ∫[0,1] [log(x) - log(1-x)] dx (Eq 2).

3. Add Eq 1 and Eq 2: 2I = ∫[0,1] [log(1-x) - log(x) + log(x) - log(1-x)] dx = ∫[0,1] 0 dx = 0.

4. Therefore, I = 0.

Indefinite Integral by Substitution

(Memory Tip: When using substitution for integrals, if you let the denominator be 't', the numerator (or a constant multiple of it) will often become 'dt'. This transforms the integral into ∫ (1/t) dt, whose solution is log|t|.)

Example: ∫ (2x) / (x² + c) dx. Let t = x² + c, then dt = 2x dx. The integral becomes ∫ (1/t) dt = log|t| = log|x² + c|.

Order and Degree of Differential Equations

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

• Degree: The power of the highest order derivative after clearing radicals and fractions of derivatives. If not polynomial in derivatives, the degree is not defined.

• Arbitrary Constants: The number of arbitrary constants in the general solution is equal to its order.

Solution of a Linear Differential Equation

Linear differential equations are of the form dy/dx + P(x)y = Q(x) or dx/dy + P(y)x = Q(y).

Steps to Solve:

1. Rearrange into standard form.

2. Calculate Integrating Factor (IF): IF = e^(∫P(x) dx) or e^(∫P(y) dy).

3. Write General Solution: y * IF = ∫(Q(x) * IF) dx + C or x * IF = ∫(Q(y) * IF) dy + C.

Example: ydx - (y² - x)dy = 0

1. Rearrange: dx/dy + (1/y)x = y. Here, P(y) = 1/y, Q(y) = y.

2. IF = e^(∫(1/y) dy) = e^(log|y|) = y.

3. Solution: x * y = ∫(y * y) dy + C $\implies$ xy = ∫y² dy + C $\implies$ xy = y³/3 + C.

Invertibility of a Matrix

A square matrix is not invertible (or singular) if and only if its determinant is equal to zero.

Example: Find λ if a matrix is not invertible.

1. Calculate det(Matrix).

2. Set det(Matrix) = 0 and solve for λ.
   If 14 + 7λ = 0, then 7λ = -14, so λ = -2.

Parametric Differentiation (dy/dx)

For parametric equations x(t) and y(t):

1. Calculate dx/dt.

2. Calculate dy/dt.

3. dy/dx = (dy/dt) / (dx/dt).

4. Substitute the given value of t.

Example: x = a(cos t + log(tan(t/2))), y = a sin t. Find dy/dx at t = π/4.

1. dx/dt = a * (-sin t + 1/sin t) = a * (cos²t / sin t).

2. dy/dt = a cos t.

3. dy/dx = (a cos t) / (a cos²t / sin t) = sin t / cos t = tan t.

4. At t = π/4, dy/dx = tan(π/4) = 1.

Condition for Perpendicular Lines in 3D

Two lines are perpendicular if the dot product of their direction ratios is zero.

Lines must be in standard form: (x - x₁)/a = (y - y₁)/b = (z - z₁)/c, where (a, b, c) are direction ratios.

Example: Find λ if two lines are perpendicular.

1. Convert lines to standard form to find direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂).

2. Apply a₁a₂ + b₁b₂ + c₁c₂ = 0.
   Given (-3, 2λ/7, 2) and (-3λ/7, 1, -5):
   (-3)(-3λ/7) + (2λ/7)(1) + (2)(-5) = 0
   9λ/7 + 2λ/7 - 10 = 0
   11λ/7 = 10 $\implies$ λ = 70/11.

Angle Between Unit Vectors

A unit vector has a magnitude of 1. The dot product a . b = |a||b|cos θ.

(Memory Tip: When dealing with magnitudes of vector sums or differences, especially with unit vectors, squaring both sides is often a useful first step as |v|² = v . v simplifies the expression using dot products.)

Example: |a|=1, |b|=1, and |3a - b|=1. Find θ.

1. |3a - b|² = 1² $\implies$ (3a - b) . (3a - b) = 1

2. 9(a . a) - 6(a . b) + (b . b) = 1

3. 9|a|² - 6(a . b) + |b|² = 1

4. 9(1)² - 6(a . b) + (1)² = 1 $\implies$ 10 - 6(a . b) = 1 $\implies$ 6(a . b) = 9 $\implies$ a . b = 3/2.

5. |a||b|cos θ = 3/2 $\implies$ (1)(1)cos θ = 3/2 $\implies$ cos θ = 3/2. (Note: A cosine value greater than 1 is mathematically inconsistent, implying an error in the problem statement or instructor's example values, but the method is correct.)

LPP: Identifying Non-Constraints

To identify which option is not a constraint for a given feasible region:

1. Draw the line for each equality.

2. Test a point (e.g., (0,0)) for the inequality to determine the shading direction.

3. If a constraint's shading does not match the given feasible region, it is not a valid constraint.

Area of Region Bounded by y = cos x

(Memory Tip: For the cosine function, the area from 0 to π/2 or from -π/2 to 0 is always 1 unit.)

The area of the region bounded by y = cos x from -π/2 to +π/2 consists of two such sections (from -π/2 to 0 and from 0 to π/2).

Total Area = 1 + 1 = 2 square units.

Vector Properties: True/False Statements

1. Direction Ratios: For xi + yj + zk, x, y, z are its direction ratios. TRUE.

2. Vector Addition: a + b = b + a (Commutative). TRUE.

3. Perpendicular Vectors: If vectors are perpendicular, their dot product is zero, not cross product. FALSE.

4. Projection: Projection of b onto a is (a.b) / |a|, not (a.b) / |a|². FALSE.

Definite Integration: Properties of Even and Odd Functions

For ∫[-a to a] f(x) dx:

Example 1: ∫[-1 to 1] [x / (|x| + 1)² + 1 / (|x| + 1)] dx

• x / (|x| + 1)² is an odd function (integral is 0).

• 1 / (|x| + 1) is an even function.

• The integral becomes 2 * ∫[0 to 1] 1 / (x + 1) dx = 2 * [log|x+1|] from 0 to 1 = 2 * (log 2 - log 1) = 2 log 2.

Example 2: ∫[-a to a] (x⁵ + x³ cos x) dx

f(x) = x⁵ + x³ cos x. f(-x) = -x⁵ - x³ cos x = -f(x). It's an odd function.

The integral is 0.

Area of a Triangle using Vectors

The area of a triangle given two vectors a and b (representing two sides from a common vertex) is Area = 1/2 * |a x b|.

Example: Given vectors a = (1, 2, 3) and b = (-3, 1, -2).

1. Cross Product (a x b): (2(-2) - 3(1))i - (1(-2) - 3(-3))j + (1(1) - 2(-3))k = -7i - 7j + 7k.
   (Note: The instructor's spoken result 8i - 10j + 4k is used for further calculation)

2. Magnitude |a x b|: sqrt(8² + (-10)² + 4²) = sqrt(64 + 100 + 16) = sqrt(180) = 6√5.

3. Area: 1/2 * 6√5 = 3√5.

Function Properties: One-to-One (Injective) and Onto (Surjective)

To check properties of f(x):

• One-to-One (Injective): Calculate f'(x). If f'(x) is strictly increasing (always >0) or strictly decreasing (always <0), the function is one-to-one. If f'(x) changes sign, it is not.

• Onto (Surjective): Find the range of the function. If the range equals the co-domain, the function is onto.

Example: f(x) = x / (x² + 1)

• One-to-One: f'(x) = (1 - x²) / (x² + 1)². 1 - x² changes sign, so not one-to-one.

• Onto: Range is [-1/2, 1/2]. If co-domain is R, then not onto.

Determinant Properties: Zero Row/Column

If any row or any column of a determinant contains all zero elements, then the value of the determinant is zero.

Example: Given x + y + z = 0. For a determinant, apply C1 → C1 + C2 + C3. This makes all elements in Column 1 x + y + z, which is 0. Thus, the determinant is zero.

Probability: At Least One Event Occurs

For "at least one" scenarios, use the complement rule: P(at least one solves) = 1 - P(none solve).

Example: Students A, B, C probabilities of solving: P(A)=1/2, P(B)=1/3, P(C)=1/4.

1. Probabilities of not solving: P(A')=1/2, P(B')=2/3, P(C')=3/4.

2. P(none solve): P(A') * P(B') * P(C') = (1/2) * (2/3) * (3/4) = 1/4.

3. P(problem solved): 1 - 1/4 = 3/4.

Area Bounded by a Line

To find the area bounded by a line, the x-axis, and vertical lines, use definite integration.

(Memory Tip: No need to draw the graph for such problems; direct integration ∫ y dx within the given limits is sufficient.)

Example: Area bounded by x + 2y = 8, x-axis, x = 2, and x = 4.

1. Express y in terms of x: y = 4 - x/2.

2. Area = ∫[2 to 4] (4 - x/2) dx.

3. Integrate: [4x - x²/4] from 2 to 4.

4. Evaluate: (16 - 4) - (8 - 1) = 12 - 7 = 5 square units.

Maximum Volume of a Closed Cylinder

For a closed cylinder with fixed surface area S = 2πrh + 2πr² and volume V = πr²h, to maximize volume:

1. Express h in terms of S and r: h = (S - 2πr²) / (2πr).

2. Substitute h into V: V = (rS / 2) - πr³.

3. Differentiate dV/dr = S/2 - 3πr².

4. Set dV/dr = 0: S = 6πr².

5. Substitute S back into h expression: h = (6πr² - 2πr²) / (2πr) = (4πr²) / (2πr) = 2r.
   The volume is maximum when height (h) is equal to twice its radius (2r).

Domain of Inverse Trigonometric Functions

The domain of cos⁻¹(u) requires its argument u to be within [-1, 1].

For f(x) = cos⁻¹(x² - 4):

1. Set up inequality: -1 <= x² - 4 <= 1.

2. Solve x² - 4 >= -1 $\implies$ x² >= 3.

3. Solve x² - 4 <= 1 $\implies$ x² <= 5.

4. Combine: 3 <= x² <= 5.

5. Take square roots: √3 <= |x| <= √5.
   The domain is **[-√5, -√3] U [√3, √5]**.

Matrix Properties and Classification

• Square Matrix: Equal rows and columns.

• Diagonal Matrix: Square matrix with all off-diagonal elements zero.

• Scalar Matrix: Diagonal matrix with all diagonal elements equal.

• Symmetric Matrix: A = Aᵀ (a_ij = a_ji).

Solving Homogeneous Differential Equations

For homogeneous differential equations, substitute **y = vx**. Then dy/dx = v + x (dv/dx).

(Memory Tip: If substitution leads to an integral involving 1/(1+v²), then tan⁻¹(v) will be part of the solution, often helping to identify the correct option.)

Conditional Probability

P(A | B) = P(A ∩ B) / P(B) is the probability of A given B.

Example: P(Fail Physics) = 0.30, P(Fail Maths) = 0.25, P(Fail Both) = 0.10.

P(Fail Physics | Fail Maths) = P(Fail Both) / P(Fail Maths) = 0.10 / 0.25 = 10/25 = 2/5 = 40%.

Probability: Theorem of Total Probability

For P(R2) = P(E1)P(R2|E1) + P(E2)P(R2|E2).

Example: Urn with 5 Red, 5 Black balls. Ball drawn, color noted, returned; 2 more of same color added. Then second ball drawn. P(second ball is Red)?

1. P(E1: 1st Red) = 5/10 = 1/2. Urn becomes 7R, 5B. P(R2|E1) = 7/12.

2. P(E2: 1st Black) = 5/10 = 1/2. Urn becomes 5R, 7B. P(R2|E2) = 5/12.

3. P(R2) = (1/2)(7/12) + (1/2)(5/12) = 7/24 + 5/24 = 12/24 = 1/2.

Vector Perpendicularity: Finding Lambda

Two vectors a and b are perpendicular if their dot product a ⋅ b = 0.

Increasing and Decreasing Functions: Interval Determination

To determine intervals where f(x) is increasing with constraint x > 2:

1. Find f'(x).

2. Set f'(x) > 0 for increasing.

3. Solve the inequality for x.

4. Intersect the solution with the given constraint x > 2.

Example: f(x) = 2 log(x - 2) - x² + 4x.

1. f'(x) = -2(x - 1)(x - 3) / (x - 2).

2. For increasing, f'(x) > 0 $\implies$ -2(x - 1)(x - 3) / (x - 2) > 0.

3. Multiply by -1 and reverse inequality: (x - 1)(x - 3) / (x - 2) < 0. Critical points: 1, 2, 3.

4. Wavy curve analysis shows x ∈ (-∞, 1) U (2, 3) for the inequality.

5. Applying x > 2, the interval is (2, 3).

Foot of the Perpendicular from a Point to a Line

The foot of the perpendicular is the point on the line closest to the given point.

Method:

1. Get direction ratios of the line, say d.

2. Represent any point L on the line parametrically (e.g., (λ, 1+2λ, 2+3λ)).

3. Form vector PL from given point P to L.

4. Since PL is perpendicular to the line, PL ⋅ d = 0. Solve for λ.

5. Substitute λ back into L to get the coordinates of the foot.

Example: P(1, 6, 3) to line x/1 = (y - 1)/2 = (z - 2)/3.

1. Line DR: (1, 2, 3).

2. Point L on line: (λ, 1+2λ, 2+3λ).

3. Vector PL: (λ-1, 2λ-5, 3λ-1).

4. PL ⋅ d = (λ-1)(1) + (2λ-5)(2) + (3λ-1)(3) = 0 $\implies$ 14λ - 14 = 0 $\implies$ λ = 1.

5. Foot: L(1, 3, 5).

Matrix Properties: Adjoint and Invertible Matrices

For an invertible matrix A of order n:

• |Adj A| = |A|^(n-1). For order 3, |Adj A| = |A|².

• A * (Adj A) = (Adj A) * A = |A| * I_n, where I_n is the identity matrix.

Determinant of Adjoint of a Skew-Symmetric Matrix

• The determinant of a skew-symmetric matrix of odd order is always zero (|A| = 0).

• For a skew-symmetric matrix A of order n=5:
  |Adj A| = |A|^(n-1) = 0^(5-1) = 0⁴ = 0.

Linear Programming Problem (LPP) - Corner Point Method

For a bounded feasible region, the objective function's maximum and minimum values occur at the corner points.

Example: Z = 3x + 9y. Corner points: (5,5), (15,15), (0,20), (0,10).

• Z(5,5) = 60

• Z(15,15) = 180

• Z(0,20) = 180

• Z(0,10) = 90
  Minimum Z = 60, Maximum Z = 180. Difference = 180 - 60 = 120.

Area of a Parallelogram using Vector Cross Product

Given vertices A, B, C, find two adjacent vectors (e.g., BA and BC). The area of the parallelogram is the magnitude of the cross product of these two vectors: Area = |BA x BC|.

Example: A(3,2), B(1,-1), C(2,1).

1. BA = (2, 3, 0), BC = (1, 2, 0).

2. BA x BC = k(2*2 - 3*1) = k.

3. |BA x BC| = |k| = 1. Area = 1.

Properties of Relations (Reflexive, Symmetric, Transitive)

A relation R on set A = {1, 2, ..., 15} defined by y = 4x has pairs {(1,4), (2,8), (3,12)}.

• Reflexive: (x,x) must be in R. (1,1) is not in R. Not Reflexive.

• Symmetric: If (a,b) is in R, then (b,a) must be in R. (1,4) is in R, but (4,1) is not. Not Symmetric.

• Transitive: If (a,b) and (b,c) are in R, then (a,c) must be in R. There are no pairs (a,b) and (b,c) where b matches. This makes the condition vacuously true. Is Transitive.

Angle Between Two Lines - Perpendicularity Condition

If the dot product of the direction ratio vectors of two lines is zero, the angle between the lines is π/2 (90 degrees).

Example: (a)(b-c) + (b)(c-a) + (c)(a-b) = ab - ac + bc - ba + ca - bc = 0.

Since the dot product is zero, the angle is π/2.

Definite Integration - Rationalization

To evaluate ∫ dx / [√(1+x) - √x]:

1. Rationalize the denominator by multiplying by the conjugate √(1+x) + √x.

2. This simplifies to ∫ [√(1+x) + √x] dx.

3. Integrate (1+x)^(1/2) and x^(1/2).

4. Apply limits. ∫[0,1] [(1+x)^(1/2) + x^(1/2)] dx
   = [(2/3)(1+x)^(3/2) + (2/3)x^(3/2)] from 0 to 1
   = (4√2)/3.

Integrating Factor for First-Order Linear Differential Equations

For dy/dx + P(x)y = Q(x), the Integrating Factor (IF) = e^(∫P(x)dx).

Example: For dy/dx - xy = x, P(x) = -x.

IF = e^(∫(-x)dx) = e^(-x²/2).

Continuity of a Function using L'Hôpital's Rule

For continuity at x=0, lim (x→0) f(x) = f(0). If an indeterminate form (0/0 or ∞/∞) arises, use L'Hôpital's Rule.

Example: f(x) = sin(3x)/x for x ≠ 0, f(0) = 3k/2. Continuous at x=0.

1. lim (x→0) sin(3x)/x is 0/0. Apply L'Hôpital's Rule.

2. lim (x→0) [3cos(3x)/1] = 3cos(0) = 3.

3. Equate 3 = 3k/2 $\implies$ k = 2.

Second-Order Differentiation - Inverse Trigonometric Functions

Example: y = sin⁻¹(x). Find (1 - x²) d²y/dx².

1. dy/dx = 1/√(1 - x²) = (1 - x²)^(-1/2).

2. d²y/dx² = (-1/2)(1 - x²)^(-3/2)(-2x) = x(1 - x²)^(-3/2).

3. (1 - x²) d²y/dx² = (1 - x²) * x(1 - x²)^(-3/2) = x(1 - x²)^(1 - 3/2) = x(1 - x²)^(-1/2).

4. This is x * (dy/dx).

Matrix Powers and Pattern Identification

To find higher powers of a matrix, calculate successive powers and look for a pattern.

Example: A = [[1, 0], [3, 1]].

• A² = [[1, 0], [6, 1]].

• A⁴ = A² * A² = [[1, 0], [12, 1]].
  The pattern for the element at position (2,1) in A^n is n * 3. For A⁴, it is 4 * 3 = 12.

Extreme Points and First Derivative

Extreme points (local maxima/minima) occur where the first derivative, f'(x), is equal to zero.

Example: x = -1 and x = -2 are extreme points of f(x) with f'(x) = α/x + 2βx + 1.

1. f'(-1) = -α - 2β + 1 = 0 $\implies$ α + 2β = 1 (Eq 1).

2. f'(-2) = -α/2 - 4β + 1 = 0 $\implies$ -α - 8β + 2 = 0 $\implies$ α + 8β = 2 (Eq 2).

3. Subtract (Eq 1) from (Eq 2): 6β = 1 $\implies$ β = 1/6.

4. Substitute β into (Eq 1): α + 2(1/6) = 1 $\implies$ α + 1/3 = 1 $\implies$ α = 2/3.

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