Tricks To Solve JEE Main Maths MCQs Within Seconds

Tricks To Solve JEE Main Maths MCQs Within Seconds

Vishal kumarUpdated on 19 Apr 2024, 12:43 PM IST

We all are familiar with Multiple Choice Questions (MCQs). These are the questions which give us some options, usually 4, out of which one, or sometimes more than one, is correct. The Joint Entrance Examination Main, or JEE Main, has 30 Questions per subject out of which 20 are MCQs with only one correct option. Generally in mathematics for some questions, you will find the options in integer values like chapters trigonometry, sequence and series and others for which you can easily get the answer by using the values given in options directly to solve questions and verify it in less time. In this article, we are going to explain some ticks and trips with the help of examples which you can implement in various questions. Let us understand the trick with the help of some examples.

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Tricks To Solve JEE Main Maths MCQs Within Seconds
Tricks to solve MCQs within seconds

Example 1

If A + B = 45°, then find the value of \frac{(1+tanA)}{(1+tanB)}equals

a)1

b)0

c)2

d)1

Let us first try to understand the question. It means that if A + B = 45°, then the value of \frac{(1+tanA)}{(1+tanB)}is always constant, which is equal to one of the options. So, for all combinations of angles A and B whose sum is 45°, the value of (\frac{(1+tanA)}{(1+tanB)}will always be the same, which is equal to one of the four options.

So, whether (A = 0° and B = 45°) or (A = 1 ° and B = 44°) or (A = 10° and B = 35°), they all will give the same value of (1+tanA)(1+tanB). So, we can easily replace A and B with some convenient values of angles and get the required value of (1+tanA) * (1+tanB). Let us put A = 0° and B = 45°,

\frac{(1+tanA)}{(1+tanB)} = (1+tan0°)(1+tan45°) = (1 + 0)(1 + 1) = 1*2 = 2

So, option C is correct

Example 2

\text{Q: The value of} \ \sum_{r=1}^{n} \frac{r}{1+r^2+r^4} \text{ equals}

\\a.\ \frac{n^2+n}{1+n+n^2} \ \\b.\ \frac{n^2+n}{2(1+n+n^2)} \\c.\ \frac{1}{n+2} \ \\d. \ \frac{1}{n^2+n-1}

Now we have to select the option that gives the correct sum for all natural number values of n. So, the correct option has to be true for n = 1, 2, 3, …and all other natural number values of n. So, if any option is giving the wrong sum for n = 1, we can be sure that that option is incorrect.

Also Read,

Let us try to find the options that are incorrect.

For n = 1, the actual sum of the given series is \frac{1}{1+1+1}= \frac{1}{3}

Checking what value option A for n = 1: \frac{1+1}{1+1+1}=\frac{2}{3}. So, it gives the wrong answer for n = 1, and hence it cannot give the correct answer for all values of n. Hence this option is wrong.

Checking what value option B for n = 1: (\frac{1+1}{2(1+1+1)}=\frac{1}{3}. So, it gives the correct answer for n = 1. But this might give wrong answers for higher values of n. So, we will not mark this as correct right now.

Checking what value option C for n = 1: \frac{1}{2(1+2)}=\frac{1}{3} So, it gives the correct answer for n = 1. But this might give wrong answers for higher values of n. So, we will not mark this as correct right now.

Checking what value option D for n = 1: \frac{1}{1+1-1}=1 So, it gives the wrong answer for n = 1, and hence it cannot give the correct answer for all values of n. Hence this option is wrong.

Now both options A and D are eliminated, and one of option B or C is correct. Let us now see which of these two is giving the correct answer for n = 2.

For n = 2, the actual sum of the given series is

\frac{1}{1+1+1}+\frac{2}{1+4+16}=\frac{9}{21}=\frac{3}{7}

Checking what value option B for n = 2: \frac{4+2}{2(1+2+4)}=\frac{3}{7}. So, it gives the correct answer for n = 2 as well. But this might give wrong answers for higher values of n. So, we will not mark this as correct right now.

Checking what value option C for n = 2: \frac{1}{2+2}=\frac{1}{4}. So, it gives the wrong answer for n = 2, and hence it cannot give the correct answer for all values of n. Hence this option is wrong.

Now we know that 3 out of 4 options are wrong, and only option B is left. So, we can safely mark option B as correct.

Example 3

1645770629771

If the value of the determinant given equals ka3b3c3, then the value of k is

a)1

b)0

c)-1

d)2

This question means that the value of the determinant always equals ka3b3c3 for all sets of values of a, b and c. In such questions, we can substitute some values of a, b and c and check to see the values of the determinant and the value of ka3b3c3. Also try to keep values of a, b and c such that ka3b3c3 does not become 0. So we will keep non-zero values of a, b and c. Let us put a = b = c = 1.

The value of the determinant is

1645770629606

which equals 0 (As two columns are the same) and the value of ka3b3c3 is k. So comparing these, we get k = 0.

Know when to use this trick and when not to. Don't use it for questions with more than one right answer or for multiple-choice questions where 'None of these' or 'All of these' is the only correct option.

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Questions related to JEE Main

On Question asked by student community

Have a question related to JEE Main ?

Hello

Here is the short guidance for you to score well and prepare for JEE

  • Stick to a smart routine – Prioritize high-weightage topics and revise daily with a focused timetable.

  • Solve PYQs & mock tests – Practice like it’s the real thing; analyze mistakes without panic.

  • Quality over quantity – Understand concepts deeply rather than rushing through chapters.

  • Stay healthy & rested – Your brain needs sleep, water, and breaks to perform at its best.

  • Trust your journey – Stay calm, believe in your preparation, and avoid comparing with others.

Hello

Since you’re reappearing for your Class 12 boards in 2026, that will be counted as your official passing year. So, you will be eligible to write JEE Advanced in 2026 and 2027. Although you submitted the boards in 2025, that attempt won’t be considered due to the compartment. JEE Advanced rules allow 2 attempts in 2 consecutive years after passing 12th. So you still have both chances left, which is great! Just make sure you meet the other eligibility conditions too. Keep your focus strong, you’ve got this!

Hello aspirant,

Like all competitive exams, JEE MAINS requires careful preparation and strategy to pass.  The JEE Main exam is extremely competitive, so it's critical to carefully plan and strategy.  Students can increase their speed, accuracy, and confidence by adhering to a systematic study schedule and practicing frequently.

To know complete preparation tips, you can visit our site through following link:

https://engineering.careers360.com/articles/jee-main-preparation-tips

Thank you

Hello aspirant,

Start with NCERT in every subject for JEE Main.  For physics, use H.C. Verma and D.C. Pandey; for chemistry, use R.C. Mukherjee and O.P. Tandon; and for mathematics, use R.D. Sharma or Cengage.  J.D. Lee assists with inorganic chemistry, while Morrison & Boyd assists with organic chemistry.  Practice chapter-by-chapter Arihant or MTG past year's papers.  For efficient preparation, concentrate on clear concepts, frequent revision, and passing practice exams.

Thank you

Hello,

Generally an income certificate isn't required for the JEE Main registration, but if you want to claim the EWS quota, then you need this. You must provide the certificate, issued by a government authority, as proof of your family's income being below the specified limit for the reservation category you wish to apply under.

I hope it will clear your query!!