Open App

JEE Exam > JEE Notes > Mock Tests for JEE Main and Advanced 2025 > JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions

JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


 
 
18
th
 March. 2021 | Shift 2
SECTION – A 
 
 
1. Let the system of linear equations 
 4x y 2z 0 ? ? ? ? 
 2x y z 0 ? ? ? 
 x 2y 3z 0, , R ? ? ? ? ? ? ? 
 Has a non-trivial solution. Then which of the following is true? 
(1)  6, R ? ? ? ? 
(2) 2, R ? ? ? ? 
(3) ? ? ? ? 3, R 
(4) 6, R ? ? ? ? ? 
Ans. (1) 
Sol. For non trivial solution 
  ? ? 0 
  
?
? ?
?
4 2
2 1 1 0
2 3
 
  4( 3 2) (6 ) 2(4 ) ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? 20 6 8 2 
  = ? ? ? ? ? ? ? 12 6 2 
 ? ? ? ? ? ? ? ? ? ? 12 6 2 
  ? ? ? ? 6, R 
 
2. A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from 
each corner of the park be 
?
3
. If the radius of the circumcircle of ?ABC is 2, then the height of the pole is 
equal to: 
 (1) 
1
3
 
 (2) 3 
 (3) 2 3 
Page 2


 
 
18
th
 March. 2021 | Shift 2
SECTION – A 
 
 
1. Let the system of linear equations 
 4x y 2z 0 ? ? ? ? 
 2x y z 0 ? ? ? 
 x 2y 3z 0, , R ? ? ? ? ? ? ? 
 Has a non-trivial solution. Then which of the following is true? 
(1)  6, R ? ? ? ? 
(2) 2, R ? ? ? ? 
(3) ? ? ? ? 3, R 
(4) 6, R ? ? ? ? ? 
Ans. (1) 
Sol. For non trivial solution 
  ? ? 0 
  
?
? ?
?
4 2
2 1 1 0
2 3
 
  4( 3 2) (6 ) 2(4 ) ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? 20 6 8 2 
  = ? ? ? ? ? ? ? 12 6 2 
 ? ? ? ? ? ? ? ? ? ? 12 6 2 
  ? ? ? ? 6, R 
 
2. A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from 
each corner of the park be 
?
3
. If the radius of the circumcircle of ?ABC is 2, then the height of the pole is 
equal to: 
 (1) 
1
3
 
 (2) 3 
 (3) 2 3 
 
 
 (4) 
2 3
3
 
 
Ans. (3) 
Sol. 
   
 ? ? ? ?
h
tan60 h 2 3
2
 
 
3.  Let in a series of 2n observations, half of them are equal to a and remaining half are equal to –a. Also by 
adding a constant b in each of these observations, the mean and standard deviation of new set become 5 
and 20, respectively. Then the value of a
2
 + b
2
 is equal to: 
 (1) 250 
 (2) 925 
 (3) 650 
 (4) 425 
Ans. (4) 
Sol. Given series 
 (a,a,a…….n times), (–a, –a, –a,…… n times) 
 Now 
i
x
x 0
2n
? ?
?
 
 as x i ? x i + b 
 then x x b ? ? 
 So , x + b = 5 ? b = 5 
 No change in S.D. due to change in origin 
 ? = 
2
2
i 2
x
2na
– (x) – 0
2n 2n
?
?
 
 20 = 
2
a ? a = 20 
 a
2
 + b
2
 = 425 
 
 
 
A 
B C 
2 
• 
Page 3


 
 
18
th
 March. 2021 | Shift 2
SECTION – A 
 
 
1. Let the system of linear equations 
 4x y 2z 0 ? ? ? ? 
 2x y z 0 ? ? ? 
 x 2y 3z 0, , R ? ? ? ? ? ? ? 
 Has a non-trivial solution. Then which of the following is true? 
(1)  6, R ? ? ? ? 
(2) 2, R ? ? ? ? 
(3) ? ? ? ? 3, R 
(4) 6, R ? ? ? ? ? 
Ans. (1) 
Sol. For non trivial solution 
  ? ? 0 
  
?
? ?
?
4 2
2 1 1 0
2 3
 
  4( 3 2) (6 ) 2(4 ) ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? 20 6 8 2 
  = ? ? ? ? ? ? ? 12 6 2 
 ? ? ? ? ? ? ? ? ? ? 12 6 2 
  ? ? ? ? 6, R 
 
2. A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from 
each corner of the park be 
?
3
. If the radius of the circumcircle of ?ABC is 2, then the height of the pole is 
equal to: 
 (1) 
1
3
 
 (2) 3 
 (3) 2 3 
 
 
 (4) 
2 3
3
 
 
Ans. (3) 
Sol. 
   
 ? ? ? ?
h
tan60 h 2 3
2
 
 
3.  Let in a series of 2n observations, half of them are equal to a and remaining half are equal to –a. Also by 
adding a constant b in each of these observations, the mean and standard deviation of new set become 5 
and 20, respectively. Then the value of a
2
 + b
2
 is equal to: 
 (1) 250 
 (2) 925 
 (3) 650 
 (4) 425 
Ans. (4) 
Sol. Given series 
 (a,a,a…….n times), (–a, –a, –a,…… n times) 
 Now 
i
x
x 0
2n
? ?
?
 
 as x i ? x i + b 
 then x x b ? ? 
 So , x + b = 5 ? b = 5 
 No change in S.D. due to change in origin 
 ? = 
2
2
i 2
x
2na
– (x) – 0
2n 2n
?
?
 
 20 = 
2
a ? a = 20 
 a
2
 + b
2
 = 425 
 
 
 
A 
B C 
2 
• 
 
 
18
th
 March. 2021 | Shift 2
4. Let g(x) =
?
x
0
f(t)dt, where f is continuous function in [0, 3] such that  ? ?
1
f(t) 1
3
for all ? ? ?
? ?
t 0,1 and 
? ?
1
0 f(t)
2
 for all t (1,3] ? . The largest possible interval in which g(3) lies is: 
 (1) [1,3]   (2) 
? ?
? ?
? ?
? ?
1
1,
2
  (3) 
? ?
? ?
? ?
? ?
3
, 1
2
  (4) 
? ?
? ?
? ?
1
,2
3
 
 
 
Ans. (4) 
Sol. 
1 3 1 3
0 1 0 1
1 1
dt 0.dt g(3) 1.dt dt
3 2
? ? ? ?
? ? ? ?
 
1
3
? g(3) ? 2 
 
5. If ? ? ? ?
4 4
15sin 10cos 6 , for some ? ? R , then the value of ? ? ?
6 6
27sec 8cosec is equal to: 
(1) 250 
(2) 500 
(3) 400 
(4) 350 
Ans. (1) 
Sol. 15 sin
4
 ? + 10 cos
4
 ? = 6 
 ? 15 sin
4
? + 10(1–sin
2
?)
2
 = 6 
 ? 25 sin
4
? – 20sin
2
? + 4 = 0 
 ? 
? ?
? ? ? ? ? ? ? ?
2
2 2 2
2 3
5sin 2 0 sin ,cos
5 5
 
 Now 
6 6
125 125
27cosec 8sec 27 8 250
27 8
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
 
 
6. Let ? ? ? ? ? ? ? f :R 3 R 1 be defind by 
?
?
?
x 2
f(x)
x 3
. 
 Let ? g:R R be given as ? ? g(x) 2x 3 . The, the sum of all the values of x for which 
? ?
? ?
1 1
13
f (x) g (x)
2
 is 
equal to  
(1) 7 
(2) 5 
(3) 2 
(4) 3 
 
 
 
 
 
Page 4


 
 
18
th
 March. 2021 | Shift 2
SECTION – A 
 
 
1. Let the system of linear equations 
 4x y 2z 0 ? ? ? ? 
 2x y z 0 ? ? ? 
 x 2y 3z 0, , R ? ? ? ? ? ? ? 
 Has a non-trivial solution. Then which of the following is true? 
(1)  6, R ? ? ? ? 
(2) 2, R ? ? ? ? 
(3) ? ? ? ? 3, R 
(4) 6, R ? ? ? ? ? 
Ans. (1) 
Sol. For non trivial solution 
  ? ? 0 
  
?
? ?
?
4 2
2 1 1 0
2 3
 
  4( 3 2) (6 ) 2(4 ) ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? 20 6 8 2 
  = ? ? ? ? ? ? ? 12 6 2 
 ? ? ? ? ? ? ? ? ? ? 12 6 2 
  ? ? ? ? 6, R 
 
2. A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from 
each corner of the park be 
?
3
. If the radius of the circumcircle of ?ABC is 2, then the height of the pole is 
equal to: 
 (1) 
1
3
 
 (2) 3 
 (3) 2 3 
 
 
 (4) 
2 3
3
 
 
Ans. (3) 
Sol. 
   
 ? ? ? ?
h
tan60 h 2 3
2
 
 
3.  Let in a series of 2n observations, half of them are equal to a and remaining half are equal to –a. Also by 
adding a constant b in each of these observations, the mean and standard deviation of new set become 5 
and 20, respectively. Then the value of a
2
 + b
2
 is equal to: 
 (1) 250 
 (2) 925 
 (3) 650 
 (4) 425 
Ans. (4) 
Sol. Given series 
 (a,a,a…….n times), (–a, –a, –a,…… n times) 
 Now 
i
x
x 0
2n
? ?
?
 
 as x i ? x i + b 
 then x x b ? ? 
 So , x + b = 5 ? b = 5 
 No change in S.D. due to change in origin 
 ? = 
2
2
i 2
x
2na
– (x) – 0
2n 2n
?
?
 
 20 = 
2
a ? a = 20 
 a
2
 + b
2
 = 425 
 
 
 
A 
B C 
2 
• 
 
 
18
th
 March. 2021 | Shift 2
4. Let g(x) =
?
x
0
f(t)dt, where f is continuous function in [0, 3] such that  ? ?
1
f(t) 1
3
for all ? ? ?
? ?
t 0,1 and 
? ?
1
0 f(t)
2
 for all t (1,3] ? . The largest possible interval in which g(3) lies is: 
 (1) [1,3]   (2) 
? ?
? ?
? ?
? ?
1
1,
2
  (3) 
? ?
? ?
? ?
? ?
3
, 1
2
  (4) 
? ?
? ?
? ?
1
,2
3
 
 
 
Ans. (4) 
Sol. 
1 3 1 3
0 1 0 1
1 1
dt 0.dt g(3) 1.dt dt
3 2
? ? ? ?
? ? ? ?
 
1
3
? g(3) ? 2 
 
5. If ? ? ? ?
4 4
15sin 10cos 6 , for some ? ? R , then the value of ? ? ?
6 6
27sec 8cosec is equal to: 
(1) 250 
(2) 500 
(3) 400 
(4) 350 
Ans. (1) 
Sol. 15 sin
4
 ? + 10 cos
4
 ? = 6 
 ? 15 sin
4
? + 10(1–sin
2
?)
2
 = 6 
 ? 25 sin
4
? – 20sin
2
? + 4 = 0 
 ? 
? ?
? ? ? ? ? ? ? ?
2
2 2 2
2 3
5sin 2 0 sin ,cos
5 5
 
 Now 
6 6
125 125
27cosec 8sec 27 8 250
27 8
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
 
 
6. Let ? ? ? ? ? ? ? f :R 3 R 1 be defind by 
?
?
?
x 2
f(x)
x 3
. 
 Let ? g:R R be given as ? ? g(x) 2x 3 . The, the sum of all the values of x for which 
? ?
? ?
1 1
13
f (x) g (x)
2
 is 
equal to  
(1) 7 
(2) 5 
(3) 2 
(4) 3 
 
 
 
 
 
 
 
Ans. (2) 
Sol. 
? ?
? ?
1 1
13
f (x) g (x)
2
 
 ? 
? ?
? ?
?
3x 2 x 3 13
x 1 2 2
 
 ? ? ? ? ? ? ? 2(3x 2) (x 1)(x 3) 13(x 1) 
 ? ? ? ?
2
x 5x 6 0 
 ? ? x 2 or 3 
 
7. Let S 1 be the sum of frist 2n terms of an arithmetic progression. Let S 2 be the sum of first 4n terms of the 
same arithmetic progression. If (S 2–S 1) is 1000, then the sum of the first 6n terms of the arithmetic 
progression is equal to : 
 (1) 3000  (2) 7000  (3) 5000  (4) 1000 
 
Ans (1) 
Sol. S 4n – S 2n = 1000 
 ? 
4n
2
(2a + (4n–1)d)–
2n
2
(2a+(2n–1)d)=1000 
 ? ?2an + 6n
2
d–nd = 1000 
 ? ?
6n
2
?2a + (6n–1)d) = 3000 
 ? S 6n = 3000 
 
 
8. Let ? ?
2 2
1
S : x y 9 and ? ? ?
2 2
2
S :(x 2) y 1. Then the locus of center of a variable circle S which touches 
S 1 internally and S 2 externally always passes through the points: 
 (1) 
? ?
? ? ?
? ?
? ?
1 5
,
2 2
 
 (2) 
? ?
?
? ?
? ?
3
2,
2
 
 (3) ? ? ? 1, 2 
 (4) 
? ?
? 0, 3 
Ans. (2) 
Sol. C 1 : (0,0) , r 1 = 3 
 C 2 : (2, 0), r 2 = 1 
 Let centre of variable circle be C 3(h,k) and radius be r. 
Page 5


 
 
18
th
 March. 2021 | Shift 2
SECTION – A 
 
 
1. Let the system of linear equations 
 4x y 2z 0 ? ? ? ? 
 2x y z 0 ? ? ? 
 x 2y 3z 0, , R ? ? ? ? ? ? ? 
 Has a non-trivial solution. Then which of the following is true? 
(1)  6, R ? ? ? ? 
(2) 2, R ? ? ? ? 
(3) ? ? ? ? 3, R 
(4) 6, R ? ? ? ? ? 
Ans. (1) 
Sol. For non trivial solution 
  ? ? 0 
  
?
? ?
?
4 2
2 1 1 0
2 3
 
  4( 3 2) (6 ) 2(4 ) ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? 20 6 8 2 
  = ? ? ? ? ? ? ? 12 6 2 
 ? ? ? ? ? ? ? ? ? ? 12 6 2 
  ? ? ? ? 6, R 
 
2. A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from 
each corner of the park be 
?
3
. If the radius of the circumcircle of ?ABC is 2, then the height of the pole is 
equal to: 
 (1) 
1
3
 
 (2) 3 
 (3) 2 3 
 
 
 (4) 
2 3
3
 
 
Ans. (3) 
Sol. 
   
 ? ? ? ?
h
tan60 h 2 3
2
 
 
3.  Let in a series of 2n observations, half of them are equal to a and remaining half are equal to –a. Also by 
adding a constant b in each of these observations, the mean and standard deviation of new set become 5 
and 20, respectively. Then the value of a
2
 + b
2
 is equal to: 
 (1) 250 
 (2) 925 
 (3) 650 
 (4) 425 
Ans. (4) 
Sol. Given series 
 (a,a,a…….n times), (–a, –a, –a,…… n times) 
 Now 
i
x
x 0
2n
? ?
?
 
 as x i ? x i + b 
 then x x b ? ? 
 So , x + b = 5 ? b = 5 
 No change in S.D. due to change in origin 
 ? = 
2
2
i 2
x
2na
– (x) – 0
2n 2n
?
?
 
 20 = 
2
a ? a = 20 
 a
2
 + b
2
 = 425 
 
 
 
A 
B C 
2 
• 
 
 
18
th
 March. 2021 | Shift 2
4. Let g(x) =
?
x
0
f(t)dt, where f is continuous function in [0, 3] such that  ? ?
1
f(t) 1
3
for all ? ? ?
? ?
t 0,1 and 
? ?
1
0 f(t)
2
 for all t (1,3] ? . The largest possible interval in which g(3) lies is: 
 (1) [1,3]   (2) 
? ?
? ?
? ?
? ?
1
1,
2
  (3) 
? ?
? ?
? ?
? ?
3
, 1
2
  (4) 
? ?
? ?
? ?
1
,2
3
 
 
 
Ans. (4) 
Sol. 
1 3 1 3
0 1 0 1
1 1
dt 0.dt g(3) 1.dt dt
3 2
? ? ? ?
? ? ? ?
 
1
3
? g(3) ? 2 
 
5. If ? ? ? ?
4 4
15sin 10cos 6 , for some ? ? R , then the value of ? ? ?
6 6
27sec 8cosec is equal to: 
(1) 250 
(2) 500 
(3) 400 
(4) 350 
Ans. (1) 
Sol. 15 sin
4
 ? + 10 cos
4
 ? = 6 
 ? 15 sin
4
? + 10(1–sin
2
?)
2
 = 6 
 ? 25 sin
4
? – 20sin
2
? + 4 = 0 
 ? 
? ?
? ? ? ? ? ? ? ?
2
2 2 2
2 3
5sin 2 0 sin ,cos
5 5
 
 Now 
6 6
125 125
27cosec 8sec 27 8 250
27 8
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
 
 
6. Let ? ? ? ? ? ? ? f :R 3 R 1 be defind by 
?
?
?
x 2
f(x)
x 3
. 
 Let ? g:R R be given as ? ? g(x) 2x 3 . The, the sum of all the values of x for which 
? ?
? ?
1 1
13
f (x) g (x)
2
 is 
equal to  
(1) 7 
(2) 5 
(3) 2 
(4) 3 
 
 
 
 
 
 
 
Ans. (2) 
Sol. 
? ?
? ?
1 1
13
f (x) g (x)
2
 
 ? 
? ?
? ?
?
3x 2 x 3 13
x 1 2 2
 
 ? ? ? ? ? ? ? 2(3x 2) (x 1)(x 3) 13(x 1) 
 ? ? ? ?
2
x 5x 6 0 
 ? ? x 2 or 3 
 
7. Let S 1 be the sum of frist 2n terms of an arithmetic progression. Let S 2 be the sum of first 4n terms of the 
same arithmetic progression. If (S 2–S 1) is 1000, then the sum of the first 6n terms of the arithmetic 
progression is equal to : 
 (1) 3000  (2) 7000  (3) 5000  (4) 1000 
 
Ans (1) 
Sol. S 4n – S 2n = 1000 
 ? 
4n
2
(2a + (4n–1)d)–
2n
2
(2a+(2n–1)d)=1000 
 ? ?2an + 6n
2
d–nd = 1000 
 ? ?
6n
2
?2a + (6n–1)d) = 3000 
 ? S 6n = 3000 
 
 
8. Let ? ?
2 2
1
S : x y 9 and ? ? ?
2 2
2
S :(x 2) y 1. Then the locus of center of a variable circle S which touches 
S 1 internally and S 2 externally always passes through the points: 
 (1) 
? ?
? ? ?
? ?
? ?
1 5
,
2 2
 
 (2) 
? ?
?
? ?
? ?
3
2,
2
 
 (3) ? ? ? 1, 2 
 (4) 
? ?
? 0, 3 
Ans. (2) 
Sol. C 1 : (0,0) , r 1 = 3 
 C 2 : (2, 0), r 2 = 1 
 Let centre of variable circle be C 3(h,k) and radius be r. 
 
 
18
th
 March. 2021 | Shift 2
 
 C 3C 1 = 3 – r  
 C 2C 1 = 1 + r 
 C 3C 1 + C 2C 1 = 4 
 So locus is ellipse whose focii are C 1 & C 2 
 And major axis is 2a = 4 and 2ae = C 1C 2 = 2 
 ? ? e = 
1
2
 
 ?  b
2
 = 4 
? ?
? ?
? ?
? ?
1
1 3
4
 
Centre of ellipse is midpoint of C 1 & C 2 is (1,0) 
Equation of ellipse is 
? ?
? ?
2
2
2 2
x 1
y
1
2
3
?
? ? 
Now by cross checking the option 
? ?
?
? ?
? ?
3
2,
2
 satisfied it.  
 
9. Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle 
be along the straight line x + y = 3. If R and r be the radius of circumcircle and incircle respectively of ?ABC, 
then (R+ r ) is equal to  
 (1) 2 2 
 (2) 3 2 
 (3) 7 2 
 (4) 
9
2
 
 
 
C 1 
C
2
 
C
3(h,k)

	Mock Tests for JEE Main and Advanced 2025 357 docs\|148 tests

Mock Tests for JEE Main and Advanced 2025

357 docs|148 tests

Join Course for Free

Top Courses for JEE

View all

Related Exams

JEE

About this Document

	4.72/5 Rating
	Dec 18, 2024 Last updated

Document Description: JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The notes and questions for JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions have been prepared according to the JEE exam syllabus. Information about JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions covers topics like and JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions.

Introduction of JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions in English is available as part of our Mock Tests for JEE Main and Advanced 2025 for JEE & JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions in Hindi for Mock Tests for JEE Main and Advanced 2025 course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025

Description

Full syllabus notes, lecture & questions for JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 - JEE | Plus excerises question with solution to help you revise complete syllabus for Mock Tests for JEE Main and Advanced 2025 | Best notes, free PDF download

Information about JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions

In this doc you can find the meaning of JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions defined & explained in the simplest way possible. Besides explaining types of JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions theory, EduRev gives you an ample number of questions to practice JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions tests, examples and also practice JEE tests

	Mock Tests for JEE Main and Advanced 2025 357 docs\|148 tests

Mock Tests for JEE Main and Advanced 2025

357 docs|148 tests

Join Course for Free

Download as PDF

Explore Courses for JEE exam

Top Courses for JEE

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

Previous Year Questions with Solutions

Free

Semester Notes

video lectures

shortcuts and tricks

Objective type Questions

Important questions

Sample Paper

practice quizzes

Extra Questions

Exam

Summary

study material

JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025

past year papers

mock tests for examination

ppt

MCQs

Viva Questions

JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025

pdf

;

Additional Information about JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions for JEE Preparation

JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions Free PDF Download

The JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions is an invaluable resource that delves deep into the core of the JEE exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions now and kickstart your journey towards success in the JEE exam.

Importance of JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions

The importance of JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions cannot be overstated, especially for JEE aspirants. This document holds the key to success in the JEE exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions Notes

JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions Notes on EduRev are your ultimate resource for success.

JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions JEE Questions

The "JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions JEE Questions" guide is a valuable resource for all aspiring students preparing for the JEE exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions on the App

Students of JEE can study JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of JEE Main 2021 Mathematics March 18 Shift 2 Paper & Solutions is prepared as per the latest JEE syllabus.

Education Revolution