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JEE Main Preparation Tips for Mathematics: Important Topics and Strategies
Mathematics is probably the lifesaver for a large number of aspirants of JEE Main. Preparation for JEE Main Mathematics is comparatively easier than the other two sections and due to a foundation based on Sets, Relations and Functions, the learning does not require to start everything from scratch. The first step that comes under the JEE Main Preparation Tips for Mathematics is to be updated with the syllabus of JEE Main Mathematics 2020. Being well informed with the syllabus will definitely help you work upon the most important topics.
Mathematics is yet considered to be hard because of the fear created out of it. In order to overcome your fear, you are advised to start preparing Mathematics pretty well in advance. Candidates are advised to practice a sufficient number of IIT JEE previous year question papers. When you will start regularly solving previous years papers, you will be able to know the type of questions, the patterns that might be present in the upcoming examination and will also be able to figure out your weaknesses and strengths. Let us start with some very basic JEE Main Preparation Tips for Mathematics.
JEE Main Preparation Tips for Mathematics: Basic Rules to Follow
- Revision is Important – Organise your study time and revision time. Revision ensures you are on track with the topics that are not under purview but must be worked on. Take regular breaks and practice previous years’ JEE Main question papers. Revise the topics every alternate day or at least twice a week.
- Topic-wise Mocks– Attempting mock tests and practice papers can actually tell what level of preparation is needed. Solving the question papers enables the candidates to know difficulty levels and their own preparation against it so they can work on their weak areas. With online mocks, a deep analysis is easily available online, to understand accuracy, strong topics, time-consuming topics and weak topics.
- Speedy Calculations – With better practice, students will build speed in core topics. This speed must be maintained during the examination as well. Since there is no sectional timing, ensure that you solve enough questions from each section within 1 and a half hour (90 minutes) such that the score reaches the expected cutoff.
- Accurate Answers – If you follow the speed rule in the first 90 minutes, follow the accuracy rule in the next 90 minutes. Instead of solving maximum questions, try solving what you can, but with accurate answers. This will help build your score and probably help in overall rank considered for IITs.
JEE Main Preparation Tips for Mathematics: Key Pointers
- The trend of questions from previous year question papers can be analysed. Questions come from almost every chapter in the exam.
- Apply for the official JEE Main Mock Test. It will help you overcome the nervousness of exam halls.
- You cannot learn a formula by reading but only through practice. The more you practice, the better you perform.
- Solve online mock tests on a table and chair. It gives an exam hall viewpoint and helps with time management.
- The graphical approach helps a lot in solving Mathematics questions. If you can learn how to create a graph, many questions of calculus and trigonometry could be solved in a blink.
- Hit and trial method works in some questions where reverse calculations might be time-consuming. Use it with care.
- While solving questions, make sure the most common errors of addition, subtraction, multiplication or division do not occur. Marks in JEE Main are only for the right answer, none for steps.
- Strengthen your grip on function theory because it is the foundation for almost each and every topic of class 11th and 12th.
- If solving questions on a rough sheet, use as many sheets as required. Do not scribble over and over on the same sheet. One may never know when the answer might get mixed up with another question.
- As soon as the timer starts on the computer, spend around 5 to 10 minutes searching for easy questions from all the 25 questions. 40% to 50% of JEE Main Mathematics questions are easy and scorable. Around 20% to 30% are moderate and almost 10% to 20% difficult.
- Instead of getting entangled in tough questions, you must try to solve easy questions while giving two to three minutes towards each question. Approximately, 20 questions can be solved in 60 minutes if each question takes a maximum duration of 3 minutes. Every single mark counts. Don’t waste time.
- Focus on the topics you have a good grip over. Pay attention to them. Memorise formulas and concepts of all the major topics.
JEE Main Preparation Tips for Mathematics: Crucial Topics
- Calculus: Calculus is one of the largest topics of mathematics. It is divided into two major studies – Differential and Integral calculus. Key topics to study in differential calculus are – limits and continuity, application of derivatives, differentiation. Key topics from integral calculus are – Indefinite integral, definite integral, the area under curves.
- Coordinate Geometry: Comparatively easy and scorable, coordinate geometry is the study of shapes and curves on a graph. Major topics to be covered under coordinate geometry are – straight lines, hyperbola, parabola, ellipse, circle.
- Trigonometry: Trigonometry has been considered one of the most important topics since class 10th. Students must have a command over the basics of trigonometry. JEE Main and other exams have questions from trigonometric ratios and trigonometric functions. These are scorable and are moderately tough.
- Algebra: This is the vastest topic of Mathematics. Questions from algebra could vary from easy to very hard. It depends on your ability to decipher how long a question can take. Major topics to be covered under algebra are – sets, relations and functions, matrices and determinants, permutations and combinations, complex numbers, probability, differential equations, quadratic equations and expressions, progressions and series, binomial theorem.
How to prepare JEE Main Mathematics by studying at home?
It is a proven fact that when you are at your home, books are the best source of knowledge to study properly. But it is also very important to note that due to the changed exam mode into the computer-based environment, JEE Main is not exactly a pen and paper-based examination. Students also require to learn various tricks over how to increase productivity while attempting the full examination of 3 hours. Given below are some JEE Main Preparation Tips for Mathematics to Study from Home.
Question Papers of JEE Main Mathematics (by Allen Kota)
To begin with the preparation, it is often suggested that Computers and Smartphones are typically contributing to distractions. To end this struggle for students, download these question paper PDFs of JEE Main 2019 Mathematics section prepared by Allen Kota. If possible, it is also suggested to print these papers and solve them offline in a traditional pen and paper mode. Do remember to first use “Papers & Answers” as this pushes you to attempt the questions on your own, rather than copying the whole solution.
Date | Shift | Paper & Answer | Paper & Solutions |
January 9 | Shift 1 | Click Here | Click Here |
Shift 2 | Click Here | Click Here | |
January 10 | Shift 1 | Click Here | Click Here |
Shift 2 | Click Here | Click Here | |
January 11 | Shift 1 | Click Here | Click Here |
Shift 2 | Click Here | Click Here | |
January 12 | Shift 1 | Click Here | Click Here |
Shift 2 | Click Here | Click Here |
Free Online Mock Tests Provided by NTA
The online mock tests provided by NTA (National Testing Agency) are a simulation of the actual exam which will help students understand how to attempt the examination in a computer-based environment. The only disadvantage here is that these are full mock tests and not subject-wise mock tests. Although, students can utilise this portal if they have completed the syllabus for JEE Main Mathematics and now are under revision mode.
- To access the mock tests head over to this website – https://nta.ac.in/quiz.
- After this, just select the examination and paper from 2020, 2019, 2018 and 2017 sessions.
- Once selected, start your mock test.
Latest News Updates
- NTA Re-opens JEE Main Application Form For Candidates, Last Date To Apply Is 24th May
- Turn Lockdown Period Into An Opportunity”, says Anand Kumar of Super 30 to JEE Aspirants
- JEE Main 2020 Syllabus not reduced, says NTA
- JEE Main 2020: Dates announced by the MHRD, Examination to be conducted from July 18 to 23, 2020
- New Batches for IITs and NITs to start in September 2020
NTA Content-Based Lectures (by IIT-PAL)
National Testing Agency has provided video lectures recorded by a few reputed IIT Professors under the nemesis of IIT-PAL on Youtube. These lectures are divided into sub-topics for each chapter to focus only on main topics. There are more than 200 videos of approximately 50 to 60-minute runtime for each video. To access these videos-
- Head over to the official website of NTA – www.nta.ac.in
- Click on the ‘Students’ tab.
- Under the ‘Test Practice’ section, select ‘Content Based Lectures’.
E-Learning Platforms
There are various e-learning platforms which are the best source for preparation of almost each and every competitive examination. In the past few years, the use of smartphones and online courses have increased multiple folds due to decreased cost of the internet in India. A lot of Ed-Tech companies started growing and now provide video lectures, mock tests, study material and a lot more stuff which helps students improve their preparation levels. Given below are some E-learning portals which can help students boost their JEE Main Mathematics Preparation.
- Unacademy – This is the largest learning platform in India with a continuously growing user base. Students can watch completely free lectures on various subjects and topics directly streamed by the toppers of various examinations. Students can also interact with the educators or teachers through the comment section given below the video.
- EduRev – This application is a learning application which provides subject-wise notes and video lectures. It also houses a large number of Mock Tests and topic-wise tests for students to work on their weak topics. Students will also find a large database of questions from previous years provided within the application.
- Khan Academy – This application provides video lectures according to the curriculum of CBSE and for IIT JEE too. Video lectures of Khan Academy are very well explained to understand the topics of JEE Main Mathematics thoroughly. With each video lecture, students are also asked questions on the topic for which they earn points and badges as rewards.
- Vedantu – This application houses all the major resources required for a student. From Solutions of questions from books of major writers to All India Mock Tests and live video lectures, students will find all the necessary resources, completely free! The speciality of this application is their in-house WAVE technology which lets students (or users) interact with a question put-up on the screen by the educator.
Chapter-wise Weightage in JEE Main Mathematics
Given below is the chapter-wise weightage of JEE Main Mathematics which will help you understand what topics should be given more time. It is nonetheless a universal fact that about 30-40% of the syllabus is part of the Calculus which is divided into Limits and Derivatives, Differentiation, Integration, Applications of Calculus, Differential Equations and much more. Students who can complete Calculus build huge confidence to cover other chapters as well.
Topic | Number of Questions |
Sets, Relations and Functions | 1 |
Complex Numbers and Quadratic Equations | 2 |
Matrices and Determinants | 2 |
Permutations and Combinations | 1 |
Mathematical Induction | 1 |
Binomial Theorem and its applications | 1 |
Sequences and Series | 2 |
Limit, Continuity and Differentiability | 1 |
Differential Calculus | 4 |
Integral Calculus | 5 |
Differential Equations | 1 |
Coordinate Geometry | 2 |
Three Dimensional Geometry | 3 |
Vector Algebra | 1 |
Statistics and Probability | 2 |
Trigonometry | 2 |
Mathematical Reasoning | 1 |
Recommended Books for Preparing JEE Main Mathematics
Mathematics section requires comprehensive preparation. The following table represents some of the best books for JEE Main along with their author’s name that students can prefer for their Mathematical exam preparation:
Book | Author |
IIT Mathematics | M.L. Khanna |
Differential Calculus | Das Gupta |
Class XI and XII Mathematics | R.D. Sharma |
Trigonometry, Geometry Books | S.L. Loney |
Integral Calculus for IIT-JEE | Amit Agarwal |
Calculus and Analytic Geometry | Thomas and Finney |
Problems in Calculus of One Variable | I. A. Maron |
Higher Algebra | Hall and Knight |
Basic Sample Questions on Various Topics
Ques. The locus of the point of intersection of the lines xcost+(1-cost)y=asint and xsint-(1+cost)y+asint=0 is
1) x2-y2=a2 3) y2 = ax
2)x2+y2=a2 4) x2 = ay
Ans. Rearranging the above equations, we get
1-cost/sint=a-x/y and 1+cost/sint=a+x/y
multiplying the above equations ,we get x2+y2=a2
Ques. Lines ax+by+c=0, where 3a+2b+4c=0 & a,b,c all belong to the set of real numbers that are concurrent at the point?
1) (3,2) 3) (3,4)
2) (2,4) 4) (3/4, 1/2)
Ans. We know that,
3a+2b+4c =0
which equals (3/4)a+(1/2)b+c=0
So, the line passes through (3/4,1/2)
Ques. If the area of the triangle formed by the equation 8×2-6xy+y2=0 and the line 2x+3y=a is 7 then the value of a is?
1) 14 3) 7
2) 28 4) 17
Ans. Equation of the sides of the given triangle are y=2x, y=4x and 2x+3y=a
So, vertices of the triangle are (a/8,a/4);(a/14,2a/7) and (0,0)
By the determinant method, the area of the triangle formed by these coordinates comes out to be a2/112 which is equal to 7
This gives the value of a as 28.
Ques. If the line x=k; k= 1,2,3,…..,n meet the line y=3x+4 at the points Ak(xk,yk), k= 1,2,3…..,n then the ordinate of the centre of the mean position of points Ak, k= 1,2,3,…..,n is
1) n+1/2 3) 3(n+1)/2
2) 3n+11/2 4) None of the above
Ans. We have yk=3k+4, the ordinate of intersection of x=k and y=3x+4 . So the ordinate of the mean position of the points Ak k= 1,2,3,…..,n is
(1/n){sum of all yk’s} which comes out to be 3n+11/2
Ques. If the point (3,4) lies on the locus of the point of intersection of the lines xcost+ysint =a and xsint-ycost=b, the point (a,b) lies on the line 3x-4y=0 then |a+b|=?
1) 1 3) 7
2) 3 4) 12
Ans. Squaring and adding the given equations of the lines we get
x2+y2=a2+b2 as the locus of the point of intersection of these lines.
Since (3,4) lies on the locus we get
9+16=a2+b2
Also (a,b) lies on 3x-4y=0 so 3a-4b=0
Solving the two equations for a and b, we get |a+b|= 7
Ques. Equation of the circle with centre (-4,3) touching internally and containing the circle x2+y2=1is
1) x2+y2+8x-6y+9=0 3) x2+y2-8x+6y+9=0
2)x2+y2+8x-6y+11=0 4) x2+y2-8x+6y-11=0
Ans. Let the equation of the required circle be (x+4)2+(y-3)2=r2
If the above circle touches the circle x2+y2=1 internally, then the distance between the centres of the circles is equal to the difference of their radii
42+32=r-1
Which implies r=6
So the equation of the circle is x2+y2+8x-6y-11=0
Ques. If the normal chord at a point ‘t’ on the parabola y2=4axsubtends a right angle at the vertex, then the value of t is
1)4 3) 3
2)1 4) 2
Ans. Equation of the normal at ‘t’ to the parabola y2=4ax is y= -tx+2at+at3
The joint equation of the lines joining the vertex to the points of the intersection of parabola and the normal is
y2=4ax[y+tx/2at+t3]
4tx2-(2t+t3)y2+4xy=0
Since these lines are at right angles so the coefficient of x2+y2=0
So, t comes out to be 2.
Ques. The angle between a diagonal of a cube and one of its edges is,
1) cos-1(1/3) 3) /3
2) /4 4) /6
Ans. Let a= a1i, b=a1j ,c=a1k.
Then the vector d= a1(i+j+k) is a diagonal of the cube. The angle between one of the edges a, b or c and the diagonal d is given by,
cos=a.d/|a||d| which comes out to be cos-1(1/3).
Ques. Volume of the tetrahedron with vertices P(-1,2,0) ; Q(2,1,-3) ; R(1,0,1) and S (3,-2,3) is
1) 1/3 3) 1/4
2) 2/3 4) ¾
Ans. Volume of the tetrahedron is given by a scalar triple product,
Volume of tetrahedron= ⅙|PQ.(PR x PS)|
Here PQ, PR and PS are three vectors made from the above provided coordinates.
So, the volume of tetrahedron comes out to be ⅔ after solving the triple product.
Ques. If A,B,C and D are four points in space and |ABxCD+BCxAD+CAxBD|=(area of triangle ABC). Then the value of is-
1) 1 3) 3
2) 2 4) 4
Ans. Let D be the origin of reference and DA=a, DB= b, DC= c
So, |ABxCD+BCxAD+CAxBD|= |(b-a) x (-c) +(c-b) x (-a) + (a-c) x (-b)|
= 2|axb+bxc+cxa|
=2(2 area of ABC)
Hence equals 4
Ques. A unit tangent vector at t=2 on the curve x=t2+2,y=4t3-5,z=2t2-6t is?
Ans. The position vector of any point at t is r=(t2+2)i+(4t3-5)j+(2t2-6t) k
dr/dt=2ti+12t2j+(4t-6)k
At t=2, the above comes out to be 4i+48j+2k, and the unit vector comes out to be (1/580) 2i+24j+k
Ques. Let N be the foot of the perpendicular of length p from the origin to a plane and l,m,n be the direction cosines of ON, the equation of the plane is
1) px+my+nz=l 3) lx+my+pz=n
2) lx+py+nz =m 4) lx+my+nz=p
Ans. The coordinates of N are (pl,pm,pn) and let P(x,y,z) be any point on the plane. The direction cosines of PN are proportional to x-pl, y-pm and z-pm. Since ON is perpendicular to the plane, it is perp. To PN
Hence, l(x-pl)+m(y-pm)+n(z-pm)=0
lx+my+nz=p(l2+m2+n2)=p, which is the locus of P and is the required equation of the plane.
Ques. The image of the point (-1,3,4) in the plane x-2y = 0 is
1) 8,4,4 3) 15,11,4
2) 9/5, -13/5,4 4) 4,4,1
Ans. Required image of the line lies on the line through A(-1,3,4) and perpendicular to x-2y=0 that is on the line
x+1/1 = y-3/-2 = z-4/ 0 =t (say)
So, the coordinates of the image is (t-1,-2t+3,4)
This point also lies on the plane, so t comes out to be 14/5
So, the required image is (9/5, -13/5,4).
Ques. If (2,3,5) is one end of the diameter of the spherex2+y2+z2-6x-12y-2z+20=0 then the coordinates of the other end are
1) 4,9,-3 3) 4,3,3
2) 4,-3,3 4) 4,3,5
Ans. Let the other end of the diameter be (a,b,c) , then the equation of the sphere is (x-2)(x-a)+(y-3)(y-b)+(z-5)(z-c)=0
Which equals x2+y2+z2-(2+a)x-(3+b)y-(5+c)z+2a+3b+5c=0
Comparing the above equations of the sphere with the equation given and comparing the corresponding terms we get,
The required coordinates as (4,9,-3).
Ques. Four Persons independently solve a certain problem correctly with probabilities ½, ¾, ¼ and 1/8. Then the probability that the problem is solved correctly by at least one of them. [JEE Advanced 2013]
Ans. P(Problem solved by at least one of them)=1-P(Problem solved none of them)
=1-½*¼*¾*7/8
=1-21/256
=235/256
Ques. An unbiased coin is tossed. If the result is heads, a pair of unbiased dice is rolled, and the number obtained by adding the number on the two faces are noted. If the result is a tail, a card from a well-shuffled pack of 11 cards numbered 2, 3, 4…. .., 12 is picked & the number on the card is noted. What is the probability that the number noted is 7 or 8?
Ans. Let us define the events:
A : head appears.
B : Tail appears
C : 7 or 8 is noted.
We have to find the probability of C i.e. P (C)
P(C) = P(A) P (C/A) + P(B) P(C/B)
Now we calculate each of the constituents one by one P(A) = probability of appearing head=½
P(C/A) = Probability that event C takes place i.e. 7 or 8 being noted when head has already appeared. (If something has already happened then it becomes certain, i.e. now it is certain that head has appeared we have to certainly roll a pair of unbiased dice).
= 11/36 (since (6, 1) (1, 6) (5, 2) (2, 5) (3, 4) (4, 3) (6, 2) (2, 6) (3, 5) (5, 3) (4, 4) i.e. 11 favorable cases and of course 6 × 6 = 36 total number of cases)
Similarly, P(B) = 1/2
P(B/C) = 2/11 (Two favorable cases (7 and 8) and 11 total number of cases).
Hence, P(C) = ½ × 11/36 + ½ × 2/11 = 193/792 (Ans.)
Ques. Sixteen players P1, P2, ….. P16 play in a tournament. They are divided into eight pairs at random. From each pair, a winner is decided on the basis of a game played between the two players of the pair. Assuming that all the players are of equal strength, the probability that exactly one of the players P1 and P2 is among the eight winners is
(a) 4/15
(b) 5/9
(c) 3/8
(d) 8/15
Ans. Let E1 and E2 denote the event that P1 and P2 are paired or not paired together. Let A denote the event that one of the two players P1 and P2 is amongst the winners.
Since, P1 can be paired with any of the remaining 15 players, so P(E1) = 1/15
and P(E2) = 1 – P(E1) = 1 – 1/15 = 14/15
In case E1 occurs, it is certain that one of P1 and P2 will be among the winners. In case E2 occurs, the probability that exactly one of P1 and P2 is among the winners is
P[(P1 ∩ P2C) ∪ (P1C ∩ P2)] = P(P1 ∩ P2C) + P(P1C ∩ P2)
= P(P1) P(P2C) + P(P1C) P(P2)
= ½ (1 – 1/2) + (1 – 1/2)1/2
= ¼ + ¼
= ½
i.e. P(A/E1) = 1 and P(A/E2) = ½
By the total probability rule,
P(A) = P(E1). P(A/E1) + P(E2) P(A/E2)
= 1/15 (1) + 14/15(1/2)
= 8/15
Ques. In a test an examinee either guesses or copies or knows that answer to a multiple-choice question which has 4 choices. The probability that he makes a guess is 1/3 and the probability that he copies is 1/6. The probability that his answer is correct, given the copied it, is 1/8. Find the probability that he knew the answer to the question, given that he answered it correctly.
Ans.
P(g) = probability of guessing = 1/3
P(c) = probability of copying = 1/6
P(k) = probability of knowing = 1 – 1/3 – 1/6 = ½
(Since the three-event g, c and k are mutually exclusive and exhaustive)
P(w) = probability that answer is correct
P(k/w) = (P(w/k).P(k))/(P(w/c)P(c) + P(w/k)P(k) + P(w/g)P (g)) (using Bayes theorem)
= (1×1/2)/((1/8,1/6) + (1×1/2) + (1/4×1/3) )
= 24/29(Ans.)
Ques. A speaks truth 3 out of 4 times. He reported that Mohan Bagan had won the match. Find the probability that his report was correct.
Ans.
Method 1:
Let T : A speaks the truth
B : Mohan Bagan won the match
Given, P(T) = 3/4
.·. P(TC) = 1 – 1/3 = 1/4
A match can be won, drawn or loosen
.·. P(B/T) = 1/3 P(B/TC) = 2/3.
Using Bayes theorem we get
P(T/B) = (P(T).P(B/T))/(P(T).P(B/T) + P(TC)P(B/TC))
= 3/4×1/3)/(3/4×1/3 + 1/4×2/3) = (1/4)/(5/12)
=3/5
Method 2:
Let, T : The man speaks truth
A : Mohan Bagan won the match
B : He reported that Mohan Bagan has won.
P(A) = 1/3(the match may also end in a draw)
P(T) = ¾
P(B) = P(A) P(T) + P(AC) P(TC)
= 1/3×3/4 + 2/3×1/4
= ¼ + 1/6
= (3+2)/12 = 5/12
P(T/B) = (P(B/T).P(T))/(P(B)) = (1/3×3/4)/(5/12)
= 3/5 (Ans.)
JEE Main 2020 Syllabus
Mathematics: Sets, relations and functions, complex numbers and quadratic equations, matrices and determinants, permutations and combinations, mathematical signs, binomial theorems, etc.
Physics: Kinematics, Laws of Motion, Functions, Gravity, Solids and Liquids, Thermodynamics, Energy and Power, Rotating Motion etc.
Chemistry: Basic concepts of chemistry, chemical bonds and molecular structures, chemical kinetics, chemical thermodynamics, states of substances etc.
Mathematics can be easily scorable if you know the tricks and short methods for harder calculations. Remember that in a board examination marks are given for each step but in competitive exams like JEE Main, each question has 4 marks only. Now that question can take 10 steps or it can take 2 steps. That choice is the smart decision which brings out the toppers.
Keep revising your topics and maintain a small handbook for the basic formulae of trigonometry, calculus and other major sections. Mathematics is a mixture of hard work and smart work. You cannot just rely on one of them to ensure a good score. Work on your weak topics but even more on strong topics to give you an edge on them.