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 Page 1


                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 1  
Date of Exam: 9
th
 January 2020 (Shift 1) 
Time: 9:30 A.M. to 12:30 P.M. 
Subject: Mathematics 
 
 
1. A sphere of 10 cm radius has a uniform thickness of ice around it. If the ice is melting at 
the rate of 50 cm
3
/min when thickness is 5 cm, then the rate of change of thickness is 
a. 
1
12?? b. 
1
18?? 
c. 
1
9?? d. 
1
36?? Answer: (?? ) 
Solution: 
 Let thickness of ice be ?? cm. 
 Therefore, net radius of sphere =(10+?? ) cm 
 Volume of sphere ?? =
4
3
?? (10+?? )
3
 
 ?
????
????
=4?? (10+?? )
2
????
????
  
 At ?? =5,
????
????
=50 cm
3
/min 
 ?50=4?? ×225×
????
????
 
 
????
????
=
1
18?? cm/min 
 
2. The number of real roots of ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 is  
a. 1 b. 2 
c. 3 d. 4 
Answer: (?? ) 
Solution: 
 ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 
 ??? 2?? +?? ?? -4+
1
?? ?? +
1
?? 2?? =0 
 ?(?? 2?? +
1
?? 2?? )+(?? ?? +
1
?? ?? )-4=0 
 ?(?? ?? +
1
?? ?? )
2
-2+(?? ?? +
1
?? ?? )-4=0 
 Let ?? ?? +
1
?? ?? =?? 
 Then, ?? 2
+?? -6=0 
Page 2


                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 1  
Date of Exam: 9
th
 January 2020 (Shift 1) 
Time: 9:30 A.M. to 12:30 P.M. 
Subject: Mathematics 
 
 
1. A sphere of 10 cm radius has a uniform thickness of ice around it. If the ice is melting at 
the rate of 50 cm
3
/min when thickness is 5 cm, then the rate of change of thickness is 
a. 
1
12?? b. 
1
18?? 
c. 
1
9?? d. 
1
36?? Answer: (?? ) 
Solution: 
 Let thickness of ice be ?? cm. 
 Therefore, net radius of sphere =(10+?? ) cm 
 Volume of sphere ?? =
4
3
?? (10+?? )
3
 
 ?
????
????
=4?? (10+?? )
2
????
????
  
 At ?? =5,
????
????
=50 cm
3
/min 
 ?50=4?? ×225×
????
????
 
 
????
????
=
1
18?? cm/min 
 
2. The number of real roots of ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 is  
a. 1 b. 2 
c. 3 d. 4 
Answer: (?? ) 
Solution: 
 ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 
 ??? 2?? +?? ?? -4+
1
?? ?? +
1
?? 2?? =0 
 ?(?? 2?? +
1
?? 2?? )+(?? ?? +
1
?? ?? )-4=0 
 ?(?? ?? +
1
?? ?? )
2
-2+(?? ?? +
1
?? ?? )-4=0 
 Let ?? ?? +
1
?? ?? =?? 
 Then, ?? 2
+?? -6=0 
                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 2  
 ??? =2,-3 
 ?? ?-3 as ?? >0 (??? ?? >0) 
 ??? ?? +
1
?? ?? =2?(?? ?? -1)
2
=0??? ?? =1??? =0 
 Hence, only one real solution is possible. 
  
3. If ?? '(?? )=tan
-1
(sec?? +tan?? ),?? ?(-
?? 2
,
?? 2
) and ?? (0)=0, then the value of ?? (1) is 
a. 
?? -1
4
  b. 
?? +1
4
 
c. 
?? +1
2
  
d. 0 
Answer: (?? ) 
Solution: 
?? '(?? )=tan
-1
(sec?? +tan?? )=tan
-1
(
1+sin?? cos?? ) 
?? '(?? )=tan
-1
(
sin
2
?? 2
+cos
2
?? 2
+2sin
?? 2
cos
?? 2
cos
2
?? 2
-sin
2
?? 2
) 
?? '(?? )=tan
-1
[
(cos
?? 2
+sin
?? 2
)
2
(cos
?? 2
+sin
?? 2
)(cos
?? 2
-sin
?? 2
)
] 
?? '(?? )=tan
-1
[
1+tan
?? 2
1-tan
?? 2
] 
?? '(?? )=tan
-1
[tan(
?? 4
+
?? 2
)] 
?? '(?? )=
?? 4
+
?? 2
 
??? (?? )=
?? 4
?? +
?? 2
4
+?? 
?? (0)=0??? =0 
?? (1)=
?? 4
+
1
4
=
?? +1
4
 
 
4. The number of solutions of log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ] is   
a. 2 b. 4 
c. 8 d. 6 
Answer: (?? ) 
Page 3


                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 1  
Date of Exam: 9
th
 January 2020 (Shift 1) 
Time: 9:30 A.M. to 12:30 P.M. 
Subject: Mathematics 
 
 
1. A sphere of 10 cm radius has a uniform thickness of ice around it. If the ice is melting at 
the rate of 50 cm
3
/min when thickness is 5 cm, then the rate of change of thickness is 
a. 
1
12?? b. 
1
18?? 
c. 
1
9?? d. 
1
36?? Answer: (?? ) 
Solution: 
 Let thickness of ice be ?? cm. 
 Therefore, net radius of sphere =(10+?? ) cm 
 Volume of sphere ?? =
4
3
?? (10+?? )
3
 
 ?
????
????
=4?? (10+?? )
2
????
????
  
 At ?? =5,
????
????
=50 cm
3
/min 
 ?50=4?? ×225×
????
????
 
 
????
????
=
1
18?? cm/min 
 
2. The number of real roots of ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 is  
a. 1 b. 2 
c. 3 d. 4 
Answer: (?? ) 
Solution: 
 ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 
 ??? 2?? +?? ?? -4+
1
?? ?? +
1
?? 2?? =0 
 ?(?? 2?? +
1
?? 2?? )+(?? ?? +
1
?? ?? )-4=0 
 ?(?? ?? +
1
?? ?? )
2
-2+(?? ?? +
1
?? ?? )-4=0 
 Let ?? ?? +
1
?? ?? =?? 
 Then, ?? 2
+?? -6=0 
                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 2  
 ??? =2,-3 
 ?? ?-3 as ?? >0 (??? ?? >0) 
 ??? ?? +
1
?? ?? =2?(?? ?? -1)
2
=0??? ?? =1??? =0 
 Hence, only one real solution is possible. 
  
3. If ?? '(?? )=tan
-1
(sec?? +tan?? ),?? ?(-
?? 2
,
?? 2
) and ?? (0)=0, then the value of ?? (1) is 
a. 
?? -1
4
  b. 
?? +1
4
 
c. 
?? +1
2
  
d. 0 
Answer: (?? ) 
Solution: 
?? '(?? )=tan
-1
(sec?? +tan?? )=tan
-1
(
1+sin?? cos?? ) 
?? '(?? )=tan
-1
(
sin
2
?? 2
+cos
2
?? 2
+2sin
?? 2
cos
?? 2
cos
2
?? 2
-sin
2
?? 2
) 
?? '(?? )=tan
-1
[
(cos
?? 2
+sin
?? 2
)
2
(cos
?? 2
+sin
?? 2
)(cos
?? 2
-sin
?? 2
)
] 
?? '(?? )=tan
-1
[
1+tan
?? 2
1-tan
?? 2
] 
?? '(?? )=tan
-1
[tan(
?? 4
+
?? 2
)] 
?? '(?? )=
?? 4
+
?? 2
 
??? (?? )=
?? 4
?? +
?? 2
4
+?? 
?? (0)=0??? =0 
?? (1)=
?? 4
+
1
4
=
?? +1
4
 
 
4. The number of solutions of log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ] is   
a. 2 b. 4 
c. 8 d. 6 
Answer: (?? ) 
                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 3  
Solution: 
log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ]  
?log1
2
|sin?? ||cos?? |=2  
?|sin?? cos?? |=
1
4
  
?sin2?? =±
1
2
  
  
? We have 8 solutions for ?? ?[0,2?? ] 
 
 
5. If ?? 1
and ?? 2
are the eccentricities of  
?? 2
18
+
?? 2
4
=1 and  
?? 2
9
-
?? 2
4
=1, respectively. If the 
points  (?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=?? . Then the value of ?? is 
a. 16 b. 14  
c. 15 d. 17 
Answer: (?? ) 
Solution: 
 e
1
=
v
1-
4
18
=
v7
3
 & ?? 2
=
v
1+
4
9
=
v13
3
   
?(?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=?? 
?15?? 1
2
+3?? 2
2
=?? 
?15×
7
9
+3×
13
9
=?? ??? =16  
 
 
 
 
Page 4


                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 1  
Date of Exam: 9
th
 January 2020 (Shift 1) 
Time: 9:30 A.M. to 12:30 P.M. 
Subject: Mathematics 
 
 
1. A sphere of 10 cm radius has a uniform thickness of ice around it. If the ice is melting at 
the rate of 50 cm
3
/min when thickness is 5 cm, then the rate of change of thickness is 
a. 
1
12?? b. 
1
18?? 
c. 
1
9?? d. 
1
36?? Answer: (?? ) 
Solution: 
 Let thickness of ice be ?? cm. 
 Therefore, net radius of sphere =(10+?? ) cm 
 Volume of sphere ?? =
4
3
?? (10+?? )
3
 
 ?
????
????
=4?? (10+?? )
2
????
????
  
 At ?? =5,
????
????
=50 cm
3
/min 
 ?50=4?? ×225×
????
????
 
 
????
????
=
1
18?? cm/min 
 
2. The number of real roots of ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 is  
a. 1 b. 2 
c. 3 d. 4 
Answer: (?? ) 
Solution: 
 ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 
 ??? 2?? +?? ?? -4+
1
?? ?? +
1
?? 2?? =0 
 ?(?? 2?? +
1
?? 2?? )+(?? ?? +
1
?? ?? )-4=0 
 ?(?? ?? +
1
?? ?? )
2
-2+(?? ?? +
1
?? ?? )-4=0 
 Let ?? ?? +
1
?? ?? =?? 
 Then, ?? 2
+?? -6=0 
                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 2  
 ??? =2,-3 
 ?? ?-3 as ?? >0 (??? ?? >0) 
 ??? ?? +
1
?? ?? =2?(?? ?? -1)
2
=0??? ?? =1??? =0 
 Hence, only one real solution is possible. 
  
3. If ?? '(?? )=tan
-1
(sec?? +tan?? ),?? ?(-
?? 2
,
?? 2
) and ?? (0)=0, then the value of ?? (1) is 
a. 
?? -1
4
  b. 
?? +1
4
 
c. 
?? +1
2
  
d. 0 
Answer: (?? ) 
Solution: 
?? '(?? )=tan
-1
(sec?? +tan?? )=tan
-1
(
1+sin?? cos?? ) 
?? '(?? )=tan
-1
(
sin
2
?? 2
+cos
2
?? 2
+2sin
?? 2
cos
?? 2
cos
2
?? 2
-sin
2
?? 2
) 
?? '(?? )=tan
-1
[
(cos
?? 2
+sin
?? 2
)
2
(cos
?? 2
+sin
?? 2
)(cos
?? 2
-sin
?? 2
)
] 
?? '(?? )=tan
-1
[
1+tan
?? 2
1-tan
?? 2
] 
?? '(?? )=tan
-1
[tan(
?? 4
+
?? 2
)] 
?? '(?? )=
?? 4
+
?? 2
 
??? (?? )=
?? 4
?? +
?? 2
4
+?? 
?? (0)=0??? =0 
?? (1)=
?? 4
+
1
4
=
?? +1
4
 
 
4. The number of solutions of log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ] is   
a. 2 b. 4 
c. 8 d. 6 
Answer: (?? ) 
                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 3  
Solution: 
log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ]  
?log1
2
|sin?? ||cos?? |=2  
?|sin?? cos?? |=
1
4
  
?sin2?? =±
1
2
  
  
? We have 8 solutions for ?? ?[0,2?? ] 
 
 
5. If ?? 1
and ?? 2
are the eccentricities of  
?? 2
18
+
?? 2
4
=1 and  
?? 2
9
-
?? 2
4
=1, respectively. If the 
points  (?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=?? . Then the value of ?? is 
a. 16 b. 14  
c. 15 d. 17 
Answer: (?? ) 
Solution: 
 e
1
=
v
1-
4
18
=
v7
3
 & ?? 2
=
v
1+
4
9
=
v13
3
   
?(?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=?? 
?15?? 1
2
+3?? 2
2
=?? 
?15×
7
9
+3×
13
9
=?? ??? =16  
 
 
 
 
                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 4  
6. The value of ?
????
(?? -3)
6
7
×(?? +4)
8
7
 is –  
 
a. 7(
?? -3
?? +4
)
1
7
+?? b. 7(
?? -3
?? +4
)
6
7
+?? 
c. (
?? -3
?? +4
)
1
7
+?? d. 7(
?? +4
?? -3
)
6
7
+?? 
Answer: (?? ) 
Solution: 
?? =?
????
(?? -3)
6
7
×(?? +4)
8
7
  
??? =?
(?? +4)
6
7
????
(?? -3)
6
7
×(?? +4)
2
=?(
?? -3
?? +4
)
- 
6
7
× 
????
(?? +4)
2
  
Put 
?? -3
?? +4
=?? ????? =7(
1
(?? +4)
2
)????    
??? =
??? -
6
7
7
???? =?? 1
7
+?? =(
?? -3
?? +4
)
1
7
+??   
 
7. If |
?? -?? ?? +2?? | =1 ,|?? |=
5
2
 then the value of |?? +3?? | is    
a. v10 b. v5 
c. 
7
2
  
d. v3 
Answer: (?? ) 
Solution: 
If |
?? -?? ?? +2?? | =1 & |?? |=
5
2
 
?|?? -?? |=|?? +2?? |  
??? 2
+(?? -1)
2
=?? 2
+(?? +2)
2
  
??? -1=±(?? +2)  
??? -1=-?? -2  
??? =-
1
2
  
?|?? |=
5
2
??? 2
+?? 2
=
25
4
  
??? 2
+
1
4
=
25
4
  
??? =±v6  
Page 5


                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 1  
Date of Exam: 9
th
 January 2020 (Shift 1) 
Time: 9:30 A.M. to 12:30 P.M. 
Subject: Mathematics 
 
 
1. A sphere of 10 cm radius has a uniform thickness of ice around it. If the ice is melting at 
the rate of 50 cm
3
/min when thickness is 5 cm, then the rate of change of thickness is 
a. 
1
12?? b. 
1
18?? 
c. 
1
9?? d. 
1
36?? Answer: (?? ) 
Solution: 
 Let thickness of ice be ?? cm. 
 Therefore, net radius of sphere =(10+?? ) cm 
 Volume of sphere ?? =
4
3
?? (10+?? )
3
 
 ?
????
????
=4?? (10+?? )
2
????
????
  
 At ?? =5,
????
????
=50 cm
3
/min 
 ?50=4?? ×225×
????
????
 
 
????
????
=
1
18?? cm/min 
 
2. The number of real roots of ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 is  
a. 1 b. 2 
c. 3 d. 4 
Answer: (?? ) 
Solution: 
 ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 
 ??? 2?? +?? ?? -4+
1
?? ?? +
1
?? 2?? =0 
 ?(?? 2?? +
1
?? 2?? )+(?? ?? +
1
?? ?? )-4=0 
 ?(?? ?? +
1
?? ?? )
2
-2+(?? ?? +
1
?? ?? )-4=0 
 Let ?? ?? +
1
?? ?? =?? 
 Then, ?? 2
+?? -6=0 
                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 2  
 ??? =2,-3 
 ?? ?-3 as ?? >0 (??? ?? >0) 
 ??? ?? +
1
?? ?? =2?(?? ?? -1)
2
=0??? ?? =1??? =0 
 Hence, only one real solution is possible. 
  
3. If ?? '(?? )=tan
-1
(sec?? +tan?? ),?? ?(-
?? 2
,
?? 2
) and ?? (0)=0, then the value of ?? (1) is 
a. 
?? -1
4
  b. 
?? +1
4
 
c. 
?? +1
2
  
d. 0 
Answer: (?? ) 
Solution: 
?? '(?? )=tan
-1
(sec?? +tan?? )=tan
-1
(
1+sin?? cos?? ) 
?? '(?? )=tan
-1
(
sin
2
?? 2
+cos
2
?? 2
+2sin
?? 2
cos
?? 2
cos
2
?? 2
-sin
2
?? 2
) 
?? '(?? )=tan
-1
[
(cos
?? 2
+sin
?? 2
)
2
(cos
?? 2
+sin
?? 2
)(cos
?? 2
-sin
?? 2
)
] 
?? '(?? )=tan
-1
[
1+tan
?? 2
1-tan
?? 2
] 
?? '(?? )=tan
-1
[tan(
?? 4
+
?? 2
)] 
?? '(?? )=
?? 4
+
?? 2
 
??? (?? )=
?? 4
?? +
?? 2
4
+?? 
?? (0)=0??? =0 
?? (1)=
?? 4
+
1
4
=
?? +1
4
 
 
4. The number of solutions of log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ] is   
a. 2 b. 4 
c. 8 d. 6 
Answer: (?? ) 
                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 3  
Solution: 
log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ]  
?log1
2
|sin?? ||cos?? |=2  
?|sin?? cos?? |=
1
4
  
?sin2?? =±
1
2
  
  
? We have 8 solutions for ?? ?[0,2?? ] 
 
 
5. If ?? 1
and ?? 2
are the eccentricities of  
?? 2
18
+
?? 2
4
=1 and  
?? 2
9
-
?? 2
4
=1, respectively. If the 
points  (?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=?? . Then the value of ?? is 
a. 16 b. 14  
c. 15 d. 17 
Answer: (?? ) 
Solution: 
 e
1
=
v
1-
4
18
=
v7
3
 & ?? 2
=
v
1+
4
9
=
v13
3
   
?(?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=?? 
?15?? 1
2
+3?? 2
2
=?? 
?15×
7
9
+3×
13
9
=?? ??? =16  
 
 
 
 
                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 4  
6. The value of ?
????
(?? -3)
6
7
×(?? +4)
8
7
 is –  
 
a. 7(
?? -3
?? +4
)
1
7
+?? b. 7(
?? -3
?? +4
)
6
7
+?? 
c. (
?? -3
?? +4
)
1
7
+?? d. 7(
?? +4
?? -3
)
6
7
+?? 
Answer: (?? ) 
Solution: 
?? =?
????
(?? -3)
6
7
×(?? +4)
8
7
  
??? =?
(?? +4)
6
7
????
(?? -3)
6
7
×(?? +4)
2
=?(
?? -3
?? +4
)
- 
6
7
× 
????
(?? +4)
2
  
Put 
?? -3
?? +4
=?? ????? =7(
1
(?? +4)
2
)????    
??? =
??? -
6
7
7
???? =?? 1
7
+?? =(
?? -3
?? +4
)
1
7
+??   
 
7. If |
?? -?? ?? +2?? | =1 ,|?? |=
5
2
 then the value of |?? +3?? | is    
a. v10 b. v5 
c. 
7
2
  
d. v3 
Answer: (?? ) 
Solution: 
If |
?? -?? ?? +2?? | =1 & |?? |=
5
2
 
?|?? -?? |=|?? +2?? |  
??? 2
+(?? -1)
2
=?? 2
+(?? +2)
2
  
??? -1=±(?? +2)  
??? -1=-?? -2  
??? =-
1
2
  
?|?? |=
5
2
??? 2
+?? 2
=
25
4
  
??? 2
+
1
4
=
25
4
  
??? =±v6  
                                               9
th
 January 2020 (Shift 1), Mathematics                                              Page | 5  
|?? +3?? |=v?? 2
+(?? +3)
2
  
? |?? +3?? |=
7
2
  
8. The value of  2
1
4
×4
1
16
×8
1
48
…8 is  
a. 2 b. 1 
c. v2  
d. 2
1
4
Answer: (?? )  
Solution: 
 2
1
4
×4
1
16
×8
1
48
…8= 2
1
4
×2
2
16
×2
4
48
…8  
?2
1
4
×2
1
8
×2
1
16
…8= 2
1
4
+
1
8
+
1
16
+?8
  
? 2
(
1
4
1-
1
2
)
  =v2  
 
 
9. The value of cos
3
?? 8
cos
3?? 8
+sin
3
?? 8
sin
3?? 8
  is –   
a. 
1
2
 b. -
1
2
 
c. 
1
v2
 d. 
1
2v2
 
       Answer: (?? ) 
       Solution: 
cos
3
?? 8
cos
3?? 8
+sin
3
?? 8
sin
3?? 8
 =cos
3
?? 8
[4cos
3
?? 8
-3cos
?? 8
]+sin
3
?? 8
[3sin
?? 8
-4sin
3
?? 8
]   
=4[cos
6
?? 8
-sin
6
?? 8
]+3[sin
4
?? 8
-cos
4
?? 8
 ]  
=4[cos
2
?? 8
-sin
2
?? 8
][cos
4
?? 8
+sin
4
?? 8
+cos
2
?? 8
sin
2
?? 8
]-3[cos
2
?? 8
-sin
2
?? 8
]  
=[cos
2
?? 8
-sin
2
?? 8
][4(1-cos
2
?? 8
sin
2
?? 8
)-3]  
=cos
?? 4
[1-sin
2
?? 4
]= 
1
2v2
 
 
10. The value of ?
?? sin
8
?? sin
8
?? +cos
8
?? ???? 2?? 0
 is  
a. 4?? 2
 b. 2?? 2
 
c. ?? 2
 d. 3?? 2
 
 
Answer: (?? )
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Document Description: JEE Main 2020 January 9 Question Paper Shift 1 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 1 have been prepared according to the JEE exam syllabus. Information about JEE Main 2020 January 9 Question Paper Shift 1 covers topics like and JEE Main 2020 January 9 Question Paper Shift 1 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 1.
Introduction of JEE Main 2020 January 9 Question Paper Shift 1 in English is available as part of our JEE Main & Advanced Previous Year Papers for JEE & JEE Main 2020 January 9 Question Paper Shift 1 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 1 | JEE Main & Advanced Previous Year Papers
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