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Tricks To Solve JEE Main Maths MCQs Within Seconds

Tricks To Solve JEE Main Maths MCQs Within Seconds

Edited By Vishal kumar | Updated on Apr 19, 2024 12:43 PM IST | #JEE Main

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.

Tricks To Solve JEE Main Maths MCQs Within Seconds
Tricks To Solve JEE Main Maths 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

Have a question related to JEE Main ?
my son got 96.9 in Jee mains with 47K rank. can you let me know the probable NITs and the course types.

Admissions to any universities or colleges are given based on Rank. As you mentioned total rank is 47000. It is possible to get admission into some good NITs.

To know more about NITs cutoff for 2024, click on the below link.

https://engineering.careers360.com/articles/jee-main-cutoff-for-top-nits


Read Complete Answer
Rishitha 30 April,2024
I got jee mains 87.6335322 category rank 62421 obc ncl will I get nit any branch

At this rank chances will be tough to get a seat in NIT, will suggest you to go for state counselling too. First where you do have a home state quota and other state where you do have a budget to go or also you can go for private colleges too.

To know in detail you can check your chances through our college predictor tool.

Read Complete Answer
Hritik.maheshwari 30 April,2024
I scored 52 percentile in jee mains 2024 and CRL rank is 713410 , so there is any govt. or private college for btech in cse with low fees and good placement

Dear Student,

At this score your chances will be tough to get admission into govt colleges but you can try for state counselling and private colleges, those who do accepts your JEE Main Score.

To know in detail you can check our college predictor and our latest articles.

Read Complete Answer
Hritik.maheshwari 30 April,2024
sir ,i have done my 12th in 2021 with pcb .now i wants to get into iisc banglore. so decided to repeat 12 with pcmb and give jee and neet in 2025 ,is am eligible for get into iisc banglore.

Hello khushabu

Whether You will be eligible to get a seat in iisc depends upon your jee mains score and cut off iisc jee in 2025.

For general category , the candidate must get around 350 rank in jee mains

Around 17000 rank for ews and obc  category

Around 20000 for obc and ews ( female )

Now a days there has been introduced a New method , iiser aptitude test

The eligibility criteria for getting a seat in iisc , the candidate must obtain minimum 60% marks in class xii.

In physics,  chemistry and mathematics.

10 % seats are reserved for women.

Read Complete Answer
Manisha 29 April,2024
i have got 3.5 lakh rank in jee mains which college will I get in cse

At this score you can go for private colleges, because govt colleges will be tough at this rank.

Even you can try for State counselling where private colleges do comes under that counselling.

To know in detail you can check your chances through our JEE Main College Predictor Tool.

Read Complete Answer
Hritik.maheshwari 30 April,2024
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 5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.820C.  If Na2SO4 is 81.5% ionised, the value of x (Kf for water=1.860C kg mol−1) is approximately : (molar mass of S=32 g mol−1 and that of Na=23 g mol−1)
Option: 1  15 g
Option: 2  25 g
Option: 3  45 g
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Option: 3

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Option: 2

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Option: 3

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Option: 2 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 + \frac{\alpha a }{4 d } \right )

Option: 3 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{\alpha a }{2 d } \right )

Option: 4 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{3 \alpha a }{2 d } \right )
 

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Option: 1  (I) < (II) < (IV) < (III)
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Option: 2  n=\frac{C-C_{p}}{C-C_{v}}


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Option: 1 \frac{\epsilon L }{R^{2}} \left ( 1 - \frac{1}{e} \right )
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