JEE Main Maths 2025 Formulas - Topic wise Important Mathematics Formulas

JEE Main Maths 2025 Formulas- Students preparing for JEE Main 2025 should keep a list of important formulas handy. This list of JEE Main Maths formulas 2025 will aid students to solve questions quickly. When creating short notes, it's essential to include these important formulas for JEE Mains 2025. The list of JEE Main Maths 2025 formulas is created after considering the NTA JEE Mains new syllabus. NTA will conduct the JEE Mains 2025 in two sessions, January & April.

There will be 14 chapters in Mathematics including the Class 11 and 12 topics. Hence, it is suggested to follow the NCERT books to prepare for JEE Mains. To help the aspirants, we have created an ebook on JEE Main Maths 2025 formulas. Students can download the JEE Main Maths all formulas PDF from the link provided below.
JEE Main 2025 Maths Formulas PDF Free Download

Mastery of these concepts not only enhances problem-solving skills but also boosts confidence in tackling the diverse range of questions expected in JEE Main. By focusing on these formulas and practising their applications, aspirants can significantly improve their readiness for the upcoming JEE Main sessions, thereby increasing their chances of achieving excellent results and securing admission to their desired engineering programs.

JEE Main Subject-wise Important Formulas

S.NoSubject-wise Important Formula Links
1Important for Physics Formula
2Important for Chemistry Formula

JEE Main Maths 2025 Formula List

Probability Formula

Trigonometric Limits

Some important JEE formulas for trigonometric limit are

Also, JEE Main Chemistry Formulas

To get radius and centre if only the equation of the circle (ii) is given:

Compare eq (i) and eq (ii)

h = - g, k = - h and c = h2 + k2 - r2

Coordinates of the centre (-g,-f)

Radius =

The equation of a circle with centre at (h,k) and radius r is

This is known as the general equation of the circle.

Also Read- JEE Main Physics Formulas

Find the equation of the locus of the point which is at a constant distance of 5 units from a point (2, 3)

Solution

Let A = (2, 3) and B = (h, k)

AB = constant = 5

(AB)2 = 52 = 25

(AB)2 = (h - 2)2 + (k - 3)2 = 25

h2 - 4h + 4 + k2 - 6k + 9 = 25

The equation of locus is

x2 + y2 - 4x - 6y - 12 = 0

Exponential Limits

To solve the limit of the function involving exponential function, we use the following standard results

Proof:


In General, if , then we have