JEE Exam > JEE Notes > JEE Main & Advanced Previous Year Papers > JEE Main 2020 January 9 Question Paper Shift 2
JEE Main 2020 January 9 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
9
th
January 2020 (Shift 2), Mathematics Page | 1
Date: 9
th
January 2020 (Shift 2)
Time: 2:30 P.M. to 5:30 P.M.
Subject: Mathematics
1. If ?? (?? )= |
?? +?? ?? +2 ?? +1
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
| and ?? -2?? +?? =1 then
a. ?? (-50)=-1 b. ?? (50)= 1
c. ?? (50)=-501 d. ?? (50)= 501
Answer: (?? )
Solution:
Given ?? (?? )= |
?? +?? ?? +2 ?? +1
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
|
?? -2?? +?? =1
Applying ?? 1
??? 1
-2?? 2
+?? 3
?? (?? )= |
?? -2?? +?? 0 0
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
|
Using ?? -2?? +?? =1
??? (?? )= (?? +3)
2
-(?? +2)(?? +4)
??? (?? )=1
??? (50)=1
??? (-50)=1
2. If ?? (?? )=
{
?? 0<?? <
1
2
1
2
?? =
1
2
1-??
1
2
<?? <1
?? (?? )= (?? -
1
2
)
2
then find the area bounded by ?? (?? ) and ?? (?? ) from ?? =
1
2
to ?? =
v3
2
.
a.
v3
2
-
1
3
b.
v3
4
+
1
3
c. 2v3 d. 3v3
Page 2
9
th
January 2020 (Shift 2), Mathematics Page | 1
Date: 9
th
January 2020 (Shift 2)
Time: 2:30 P.M. to 5:30 P.M.
Subject: Mathematics
1. If ?? (?? )= |
?? +?? ?? +2 ?? +1
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
| and ?? -2?? +?? =1 then
a. ?? (-50)=-1 b. ?? (50)= 1
c. ?? (50)=-501 d. ?? (50)= 501
Answer: (?? )
Solution:
Given ?? (?? )= |
?? +?? ?? +2 ?? +1
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
|
?? -2?? +?? =1
Applying ?? 1
??? 1
-2?? 2
+?? 3
?? (?? )= |
?? -2?? +?? 0 0
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
|
Using ?? -2?? +?? =1
??? (?? )= (?? +3)
2
-(?? +2)(?? +4)
??? (?? )=1
??? (50)=1
??? (-50)=1
2. If ?? (?? )=
{
?? 0<?? <
1
2
1
2
?? =
1
2
1-??
1
2
<?? <1
?? (?? )= (?? -
1
2
)
2
then find the area bounded by ?? (?? ) and ?? (?? ) from ?? =
1
2
to ?? =
v3
2
.
a.
v3
2
-
1
3
b.
v3
4
+
1
3
c. 2v3 d. 3v3
9
th
January 2020 (Shift 2), Mathematics Page | 2
Answer: (?? )
Solution:
Given ?? (?? )=
{
?? 0<?? <
1
2
1
2
?? =
1
2
1-??
1
2
<?? <1
?? (?? )= (?? -
1
2
)
2
The area between ?? (?? ) and ?? (?? ) from ?? =
1
2
to =
v3
2
:
Points of intersection of ?? (?? ) and (?? ) :
1-?? = (?? -
1
2
)
2
??? =
v3
2
,-
v3
2
Required area =?
(?? (?? )-?? (?? ))????
v3
2
1
2
=? (1-?? - (?? -
1
2
)
2
)????
v3
2
1
2
=?? -
?? 2
2
-
1
3
(?? -
1
2
)
3
|
1
2
v3
2
=
v3
4
-
1
3
Page 3
9
th
January 2020 (Shift 2), Mathematics Page | 1
Date: 9
th
January 2020 (Shift 2)
Time: 2:30 P.M. to 5:30 P.M.
Subject: Mathematics
1. If ?? (?? )= |
?? +?? ?? +2 ?? +1
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
| and ?? -2?? +?? =1 then
a. ?? (-50)=-1 b. ?? (50)= 1
c. ?? (50)=-501 d. ?? (50)= 501
Answer: (?? )
Solution:
Given ?? (?? )= |
?? +?? ?? +2 ?? +1
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
|
?? -2?? +?? =1
Applying ?? 1
??? 1
-2?? 2
+?? 3
?? (?? )= |
?? -2?? +?? 0 0
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
|
Using ?? -2?? +?? =1
??? (?? )= (?? +3)
2
-(?? +2)(?? +4)
??? (?? )=1
??? (50)=1
??? (-50)=1
2. If ?? (?? )=
{
?? 0<?? <
1
2
1
2
?? =
1
2
1-??
1
2
<?? <1
?? (?? )= (?? -
1
2
)
2
then find the area bounded by ?? (?? ) and ?? (?? ) from ?? =
1
2
to ?? =
v3
2
.
a.
v3
2
-
1
3
b.
v3
4
+
1
3
c. 2v3 d. 3v3
9
th
January 2020 (Shift 2), Mathematics Page | 2
Answer: (?? )
Solution:
Given ?? (?? )=
{
?? 0<?? <
1
2
1
2
?? =
1
2
1-??
1
2
<?? <1
?? (?? )= (?? -
1
2
)
2
The area between ?? (?? ) and ?? (?? ) from ?? =
1
2
to =
v3
2
:
Points of intersection of ?? (?? ) and (?? ) :
1-?? = (?? -
1
2
)
2
??? =
v3
2
,-
v3
2
Required area =?
(?? (?? )-?? (?? ))????
v3
2
1
2
=? (1-?? - (?? -
1
2
)
2
)????
v3
2
1
2
=?? -
?? 2
2
-
1
3
(?? -
1
2
)
3
|
1
2
v3
2
=
v3
4
-
1
3
9
th
January 2020 (Shift 2), Mathematics Page | 3
3. If ?? ?(?? ?~ ?? ) is false. Truth value of ?? and ?? will be
a. TF b. FT
c. TT d. FF
Answer: (?? )
Solution:
Given ?? ?(?? ?~ ?? )
Truth table:
?? ?? ~?? (?? ?~ ?? ) ?? ?(?? ?~ ?? )
T T F F F
T F T T T
F T F F T
F F T F T
?? ?(?? ?~ ?? ) is false when ?? is true and ?? is true.
4. ?
????
cos
2
??
(sec2?? +tan2?? )
=?? tan?? +2log?? (?? )+?? then ordered pair (?? , ?? (?? )) is
a. (1,1+tan?? ) b. (1,1-tan?? )
c. (-1,1+tan?? ) d. (-1,1-tan?? )
Answer: (?? )
Solution:
Let ?? = ?
????
cos
2
?? (sec2?? +tan2?? )
?? = ?
sec
2
?? ????
(
1+tan
2
?? 1-tan
2
?? )+ (
2tan?? 1- tan
2
?? )
?? = ?
(1- tan
2
?? )(sec
2
?? )????
(1+tan?? )
2
Let tan?? =?? ? sec
2
?? ???? =d??
?? = ?
(1- ?? 2
)
(1+?? )
2
???? = ?
(1- ?? )
(1+?? )
????
?? = (
2
1+?? -1)????
?? =2ln|1+?? |-?? +??
?? =2ln|1+tan?? |-tan?? +??
Page 4
9
th
January 2020 (Shift 2), Mathematics Page | 1
Date: 9
th
January 2020 (Shift 2)
Time: 2:30 P.M. to 5:30 P.M.
Subject: Mathematics
1. If ?? (?? )= |
?? +?? ?? +2 ?? +1
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
| and ?? -2?? +?? =1 then
a. ?? (-50)=-1 b. ?? (50)= 1
c. ?? (50)=-501 d. ?? (50)= 501
Answer: (?? )
Solution:
Given ?? (?? )= |
?? +?? ?? +2 ?? +1
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
|
?? -2?? +?? =1
Applying ?? 1
??? 1
-2?? 2
+?? 3
?? (?? )= |
?? -2?? +?? 0 0
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
|
Using ?? -2?? +?? =1
??? (?? )= (?? +3)
2
-(?? +2)(?? +4)
??? (?? )=1
??? (50)=1
??? (-50)=1
2. If ?? (?? )=
{
?? 0<?? <
1
2
1
2
?? =
1
2
1-??
1
2
<?? <1
?? (?? )= (?? -
1
2
)
2
then find the area bounded by ?? (?? ) and ?? (?? ) from ?? =
1
2
to ?? =
v3
2
.
a.
v3
2
-
1
3
b.
v3
4
+
1
3
c. 2v3 d. 3v3
9
th
January 2020 (Shift 2), Mathematics Page | 2
Answer: (?? )
Solution:
Given ?? (?? )=
{
?? 0<?? <
1
2
1
2
?? =
1
2
1-??
1
2
<?? <1
?? (?? )= (?? -
1
2
)
2
The area between ?? (?? ) and ?? (?? ) from ?? =
1
2
to =
v3
2
:
Points of intersection of ?? (?? ) and (?? ) :
1-?? = (?? -
1
2
)
2
??? =
v3
2
,-
v3
2
Required area =?
(?? (?? )-?? (?? ))????
v3
2
1
2
=? (1-?? - (?? -
1
2
)
2
)????
v3
2
1
2
=?? -
?? 2
2
-
1
3
(?? -
1
2
)
3
|
1
2
v3
2
=
v3
4
-
1
3
9
th
January 2020 (Shift 2), Mathematics Page | 3
3. If ?? ?(?? ?~ ?? ) is false. Truth value of ?? and ?? will be
a. TF b. FT
c. TT d. FF
Answer: (?? )
Solution:
Given ?? ?(?? ?~ ?? )
Truth table:
?? ?? ~?? (?? ?~ ?? ) ?? ?(?? ?~ ?? )
T T F F F
T F T T T
F T F F T
F F T F T
?? ?(?? ?~ ?? ) is false when ?? is true and ?? is true.
4. ?
????
cos
2
??
(sec2?? +tan2?? )
=?? tan?? +2log?? (?? )+?? then ordered pair (?? , ?? (?? )) is
a. (1,1+tan?? ) b. (1,1-tan?? )
c. (-1,1+tan?? ) d. (-1,1-tan?? )
Answer: (?? )
Solution:
Let ?? = ?
????
cos
2
?? (sec2?? +tan2?? )
?? = ?
sec
2
?? ????
(
1+tan
2
?? 1-tan
2
?? )+ (
2tan?? 1- tan
2
?? )
?? = ?
(1- tan
2
?? )(sec
2
?? )????
(1+tan?? )
2
Let tan?? =?? ? sec
2
?? ???? =d??
?? = ?
(1- ?? 2
)
(1+?? )
2
???? = ?
(1- ?? )
(1+?? )
????
?? = (
2
1+?? -1)????
?? =2ln|1+?? |-?? +??
?? =2ln|1+tan?? |-tan?? +??
9
th
January 2020 (Shift 2), Mathematics Page | 4
Given ?? =?? tan?? +2log?? (?? )+??
??? = -1 , ?? (?? )=|1+???????? |
5. Let ?? ?? is a positive term of GP and ? ?? 2?? +1
=200
100
?? =1
, ? ?? 2?? =100
100
?? =1
, then the value of ? ?? ?? 200
?? =1
is
a. 150 b. 225
c. 300 d. 175
Answer: (?? )
Solution:
?? ?? is a positive term of GP.
Let GP be ?? ,???? ,????
2
,.....
? ?? 2?? +1
100
?? =1
=?? 3
+ ?? 5
+ .......+?? 201
200=????
2
+ ????
4
+ .......+ ????
201
200=
????
2
(?? 200
-1)
?? 2
-1
. . . (1)
Also, ? ?? 2?? 100
?? =1
=100
100=?? 2
+ ?? 4
+ ... .....+ ?? 200
100=???? + ????
3
+ ........+????
199
100=
???? (?? 200
-1)
?? 2
-1
. . . (2)
From (1) and (2), r = 2
And ? ?? 2?? +1
100
?? =1
+ ? ?? 2?? 100
?? =1
=300
? ?? 2
+ ?? 3
+ ?? 4
... .....+ ?? 200
+?? 201
=300
? ???? + ????
2
+ ????
3
+ .......+ ????
200
=300
??? (?? +???? + ?? ?? 2
+ .....+ ????
199
)=300
?2(?? 1
+ ?? 2
+ ?? 3
+ ......+ ?? 200
)=300
? ?? ?? 200
?? =1
=150
6. ?? is a complex number such that |Re(?? )|+|Im(?? )|=4, then |?? | cannot be equal to
a. v8 b. v7
c. v
17
2
d. v10
Page 5
9
th
January 2020 (Shift 2), Mathematics Page | 1
Date: 9
th
January 2020 (Shift 2)
Time: 2:30 P.M. to 5:30 P.M.
Subject: Mathematics
1. If ?? (?? )= |
?? +?? ?? +2 ?? +1
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
| and ?? -2?? +?? =1 then
a. ?? (-50)=-1 b. ?? (50)= 1
c. ?? (50)=-501 d. ?? (50)= 501
Answer: (?? )
Solution:
Given ?? (?? )= |
?? +?? ?? +2 ?? +1
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
|
?? -2?? +?? =1
Applying ?? 1
??? 1
-2?? 2
+?? 3
?? (?? )= |
?? -2?? +?? 0 0
?? +?? ?? +3 ?? +2
?? +?? ?? +4 ?? +3
|
Using ?? -2?? +?? =1
??? (?? )= (?? +3)
2
-(?? +2)(?? +4)
??? (?? )=1
??? (50)=1
??? (-50)=1
2. If ?? (?? )=
{
?? 0<?? <
1
2
1
2
?? =
1
2
1-??
1
2
<?? <1
?? (?? )= (?? -
1
2
)
2
then find the area bounded by ?? (?? ) and ?? (?? ) from ?? =
1
2
to ?? =
v3
2
.
a.
v3
2
-
1
3
b.
v3
4
+
1
3
c. 2v3 d. 3v3
9
th
January 2020 (Shift 2), Mathematics Page | 2
Answer: (?? )
Solution:
Given ?? (?? )=
{
?? 0<?? <
1
2
1
2
?? =
1
2
1-??
1
2
<?? <1
?? (?? )= (?? -
1
2
)
2
The area between ?? (?? ) and ?? (?? ) from ?? =
1
2
to =
v3
2
:
Points of intersection of ?? (?? ) and (?? ) :
1-?? = (?? -
1
2
)
2
??? =
v3
2
,-
v3
2
Required area =?
(?? (?? )-?? (?? ))????
v3
2
1
2
=? (1-?? - (?? -
1
2
)
2
)????
v3
2
1
2
=?? -
?? 2
2
-
1
3
(?? -
1
2
)
3
|
1
2
v3
2
=
v3
4
-
1
3
9
th
January 2020 (Shift 2), Mathematics Page | 3
3. If ?? ?(?? ?~ ?? ) is false. Truth value of ?? and ?? will be
a. TF b. FT
c. TT d. FF
Answer: (?? )
Solution:
Given ?? ?(?? ?~ ?? )
Truth table:
?? ?? ~?? (?? ?~ ?? ) ?? ?(?? ?~ ?? )
T T F F F
T F T T T
F T F F T
F F T F T
?? ?(?? ?~ ?? ) is false when ?? is true and ?? is true.
4. ?
????
cos
2
??
(sec2?? +tan2?? )
=?? tan?? +2log?? (?? )+?? then ordered pair (?? , ?? (?? )) is
a. (1,1+tan?? ) b. (1,1-tan?? )
c. (-1,1+tan?? ) d. (-1,1-tan?? )
Answer: (?? )
Solution:
Let ?? = ?
????
cos
2
?? (sec2?? +tan2?? )
?? = ?
sec
2
?? ????
(
1+tan
2
?? 1-tan
2
?? )+ (
2tan?? 1- tan
2
?? )
?? = ?
(1- tan
2
?? )(sec
2
?? )????
(1+tan?? )
2
Let tan?? =?? ? sec
2
?? ???? =d??
?? = ?
(1- ?? 2
)
(1+?? )
2
???? = ?
(1- ?? )
(1+?? )
????
?? = (
2
1+?? -1)????
?? =2ln|1+?? |-?? +??
?? =2ln|1+tan?? |-tan?? +??
9
th
January 2020 (Shift 2), Mathematics Page | 4
Given ?? =?? tan?? +2log?? (?? )+??
??? = -1 , ?? (?? )=|1+???????? |
5. Let ?? ?? is a positive term of GP and ? ?? 2?? +1
=200
100
?? =1
, ? ?? 2?? =100
100
?? =1
, then the value of ? ?? ?? 200
?? =1
is
a. 150 b. 225
c. 300 d. 175
Answer: (?? )
Solution:
?? ?? is a positive term of GP.
Let GP be ?? ,???? ,????
2
,.....
? ?? 2?? +1
100
?? =1
=?? 3
+ ?? 5
+ .......+?? 201
200=????
2
+ ????
4
+ .......+ ????
201
200=
????
2
(?? 200
-1)
?? 2
-1
. . . (1)
Also, ? ?? 2?? 100
?? =1
=100
100=?? 2
+ ?? 4
+ ... .....+ ?? 200
100=???? + ????
3
+ ........+????
199
100=
???? (?? 200
-1)
?? 2
-1
. . . (2)
From (1) and (2), r = 2
And ? ?? 2?? +1
100
?? =1
+ ? ?? 2?? 100
?? =1
=300
? ?? 2
+ ?? 3
+ ?? 4
... .....+ ?? 200
+?? 201
=300
? ???? + ????
2
+ ????
3
+ .......+ ????
200
=300
??? (?? +???? + ?? ?? 2
+ .....+ ????
199
)=300
?2(?? 1
+ ?? 2
+ ?? 3
+ ......+ ?? 200
)=300
? ?? ?? 200
?? =1
=150
6. ?? is a complex number such that |Re(?? )|+|Im(?? )|=4, then |?? | cannot be equal to
a. v8 b. v7
c. v
17
2
d. v10
9
th
January 2020 (Shift 2), Mathematics Page | 5
Answer: (?? )
Solution:
|Re(?? )|+|Im(?? )|=4
Let ?? =?? +????
?|?? |+|?? |=4
??? lies on the rhombus.
Maximum value of |?? |=4 when ?? =4,-4,4?? ,-4??
Minimum value of |?? |=2v2 when ?? =2±2??, ±2+2??
|?? |?[2v2 ,4]
|?? |?[v8 ,v16]
|?? |?v7
7. ?? (?? ):[0,5]??? ,?? (?? )=? ?? 2
?? (?? )???? ,?? (1)=3,
?? 0
?? (?? )=? ?? (?? )????
?? 1
then correct choice is
a. ?? (?? ) has no critical point
b. ?? (?? ) has local minimum at ?? =1
c. ?? (?? ) has local maximum at ?? =1
d. ?? (?? ) has point of inflection at ?? =1
Answer: (b)
Solution:
?? (?? )=?? 2
?? (?? )
Put ?? =1
Read More
|
254 docs|1 tests
|
FAQs on JEE Main 2020 January 9 Question Paper Shift 2 - JEE Main & Advanced Previous Year Papers
| 1. क्या JEE Main 2020 January 9 Question Paper Shift 2 में किसी खास विषय पर प्रश्न पूछे गए थे? |  |
Ans. हां, JEE Main 2020 January 9 Question Paper Shift 2 में विभिन्न विषयों पर प्रश्न पूछे गए थे जैसे कि गणित, भौतिक विज्ञान और रसायन विज्ञान।
| 2. JEE Main 2020 January 9 Question Paper Shift 2 में कितने प्रश्न पूछे गए थे? | |
Ans. JEE Main 2020 January 9 Question Paper Shift 2 में कुल मिलाकर 75 प्रश्न पूछे गए थे।
| 3. क्या JEE Main 2020 January 9 Question Paper Shift 2 में किसी विषय पर अधिक प्रश्न पूछे गए थे? | |
Ans. हां, JEE Main 2020 January 9 Question Paper Shift 2 में गणित विषय पर अधिक प्रश्न पूछे गए थे।
| 4. JEE Main 2020 January 9 Question Paper Shift 2 की तैयारी के लिए सर्वश्रेष्ठ संदेश क्या होगा? | |
Ans. JEE Main 2020 January 9 Question Paper Shift 2 की तैयारी के लिए प्रैक्टिस सेट्स और मॉक टेस्ट करना बेहतर हो सकता है।
| 5. JEE Main 2020 January 9 Question Paper Shift 2 के नतीजे कब जारी होंगे? | |
Ans. JEE Main 2020 January 9 Question Paper Shift 2 के नतीजे जल्द ही ऑफिशियल वेबसाइट पर जारी किए जाएंगे।
Related Exams
About this Document
|
Oct 24, 2024 Last updated |
Document Description: JEE Main 2020 January 9 Question Paper Shift 2 for JEE 2024 is part of JEE Main & Advanced Previous Year Papers preparation.
The notes and questions for JEE Main 2020 January 9 Question Paper Shift 2 have been prepared according to the JEE exam syllabus. Information about JEE Main 2020 January 9 Question Paper Shift 2 covers topics
like and JEE Main 2020 January 9 Question Paper Shift 2 Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main 2020 January 9 Question Paper Shift 2.
Introduction of JEE Main 2020 January 9 Question Paper Shift 2 in English is available as part of our JEE Main & Advanced Previous Year Papers
for JEE & JEE Main 2020 January 9 Question Paper Shift 2 in Hindi for JEE Main & Advanced Previous Year Papers course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: JEE Main 2020 January 9 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers
Description
Full syllabus notes, lecture & questions for JEE Main 2020 January 9 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers - JEE | Plus excerises question with solution to help you revise complete syllabus for JEE Main & Advanced Previous Year Papers | Best notes, free PDF download
Information about JEE Main 2020 January 9 Question Paper Shift 2
In this doc you can find the meaning of JEE Main 2020 January 9 Question Paper Shift 2 defined & explained in the simplest way possible. Besides explaining types of
JEE Main 2020 January 9 Question Paper Shift 2 theory, EduRev gives you an ample number of questions to practice JEE Main 2020 January 9 Question Paper Shift 2 tests, examples and also practice JEE
tests
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Free
,JEE Main 2020 January 9 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers
,Viva Questions
,Summary
,Extra Questions
,Important questions
,Previous Year Questions with Solutions
,ppt
,Objective type Questions
,shortcuts and tricks
,past year papers
,study material
,video lectures
,mock tests for examination
,JEE Main 2020 January 9 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers
,JEE Main 2020 January 9 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers
,Semester Notes
,practice quizzes
,Exam
,Sample Paper
,MCQs
;
Additional Information about JEE Main 2020 January 9 Question Paper Shift 2 for JEE Preparation
JEE Main 2020 January 9 Question Paper Shift 2 Free PDF Download
The JEE Main 2020 January 9 Question Paper Shift 2 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 2020 January 9 Question Paper Shift 2 now and kickstart your journey towards success in the JEE exam.
Importance of JEE Main 2020 January 9 Question Paper Shift 2
The importance of JEE Main 2020 January 9 Question Paper Shift 2 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 2020 January 9 Question Paper Shift 2 Notes
JEE Main 2020 January 9 Question Paper Shift 2 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 2020 January 9 Question Paper Shift 2.
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 2020 January 9 Question Paper Shift 2 Notes on EduRev are your ultimate resource for success.
JEE Main 2020 January 9 Question Paper Shift 2 JEE Questions
The "JEE Main 2020 January 9 Question Paper Shift 2 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 2020 January 9 Question Paper Shift 2 on the App
Students of JEE can study JEE Main 2020 January 9 Question Paper Shift 2 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the JEE Main 2020 January 9 Question Paper Shift 2,
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 2020 January 9 Question Paper Shift 2 is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup