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.
The NTA will use a normalisation method to convert raw scores in percentile. This method is used to prevent any student from gaining any advantage due to the varying difficulty level across multiple shifts. Candidates can check the expected JEE Main marks vs percentile 2026 based on past trends.
JEE Main Marks | JEE Main Percentile |
109-100 | 94.96737888 - 93.8020333 |
99-90 | 93.67910655 - 92.21882783 |
89-80 | 92.05811248 - 90.27631202 |
79-70 | 90.0448455 - 87.51810893 |
69-60 | 87.33654157 - 83.89085926 |
59-50 | 83.5119717 - 78.35114254 |
49-40 | 77.81927947 - 69.5797271 |
39-30 | 68.80219265 - 56.09102043 |
29-20 | 54.01037138 - 36.58463962 |
19-10 | 35.2885364 - 18.16647924 |
9-0 | 17.14582299 -5.71472799 |
Example 1
If A + B = 45°, then find the value of
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
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 (
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°,
= (1+tan0°)(1+tan45°) = (1 + 0)(1 + 1) = 1*2 = 2
So, option C is correct
Example 2


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
= 
Checking what value option A for n = 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.
Checking what value option B for n = 1: (
. 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:
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:
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

Checking what value option B for n = 2:
. 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:
. 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

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

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.
On Question asked by student community
Arnav Gautam & P.Mohith secured 300 out of 300 in JEE Mains session 1, as per the provisional answer key. The list of toppers to be released with JEE Main results
JEE Main 2026 session 1 result is not declared yet. JEE Main session 1 result will be declared on February 12.
A rank above 50,000 is considered good in JEE Mains. Some of the private colleges accepting JEE Mains are
The 69-77 marks in JEE Main examination are considered to be the average score and does not guarantee best branches in top tier NITs and IIITs. Some of the branches in which students with this much score can get admission are Electronics and Communication Engineering, Computer Science and Engineering, and
With a rank between 50,000 and 65,000 in JEE Main, you can get admission in some of the NITs, private and government colleges. Here is the list of some of the colleges.
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
Recognized as Institute of Eminence by Govt. of India | NAAC ‘A++’ Grade | Upto 75% Scholarships | Application Deadline: 15th Jan
India's youngest NAAC A++ accredited University | NIRF rank band 151-200 | 2200 Recruiters | 45.98 Lakhs Highest Package
Last Date to Apply: 25th Feb | Ranked #43 among Engineering colleges in India by NIRF | Highest Package 1.3 CR , 100% Placements
Top international universities | Know fees, location, courses offered.
98% Placement Record | Highest CTC 81.25 LPA | NAAC A++ Accredited | Ranked #62 in India by NIRF Ranking 2025 | JEE & JET Scores Accepted