All Courses
Get the App Signup / Login
Sign in Open App
JEE Exam  >  JEE Notes  >  Mock Tests for JEE Main and Advanced 2025  >  JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions

JEE Main 2020 Mathematics September 5 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


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 1
Date : 5
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 If x=1 is a critical point of the function f(x)=(3x
2
+ax–2–a)e
x
, then:
(1) x=1 is a local minima and 
2
x
3
? ? is a local maxima of f. .
(2) x=1 is a local maxima and 
2
x
3
? ? is a local minima of f. .
(3) x=1 and 
2
x
3
? ? are local minima of f. .
(4) x=1 and 
2
x
3
? ? are local maxima of f. .
Sol. 1
f(x) = (3x
2
+ax–2–a)e
x
f
’
(x) = (3x
2
+ax–2–a)e
x
 + (6x+a)e
x
= 0
? ?
x 2
e 3x a 6 x 2 ? ? ? ? ?
? ?
=0
at x = 1, 3 +a+6 - 2 = 0
a=–7
f(x) = (3x
2
 - 7x + 5 )e
x
f
’
(x) = (6x—7)e
x 
+(3x
2
-7x+5)e
x
= e
x
(3x
2
-x-2) = 0
= 3x
2
 —3x + 2x - 2 = 0
= (3x+2) (x—1) = 0
x = 1, -2/3
+ – +
–2/3
1
x = 1 is point of local minima.
x = –2/3 is point of local maxima.
Q.2
? ?
2 4
1 x x 1 x
1
x 0 2 4
x e
1 x x 1
/
lim
? ? ?
?
?
? ?
? ?
? ?
? ? ?
(1) is equal to 
e
(2) is equal to 1 (3) is equal to 0 (4) does not exist
Sol. 2
? ?
? ?
2 4
1 x x 1 /x
x 0 2 4
x e 1
lim
1 x x 1
? ? ?
?
? ?
?
? ?
? ?
? ? ?
Page 2


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 1
Date : 5
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 If x=1 is a critical point of the function f(x)=(3x
2
+ax–2–a)e
x
, then:
(1) x=1 is a local minima and 
2
x
3
? ? is a local maxima of f. .
(2) x=1 is a local maxima and 
2
x
3
? ? is a local minima of f. .
(3) x=1 and 
2
x
3
? ? are local minima of f. .
(4) x=1 and 
2
x
3
? ? are local maxima of f. .
Sol. 1
f(x) = (3x
2
+ax–2–a)e
x
f
’
(x) = (3x
2
+ax–2–a)e
x
 + (6x+a)e
x
= 0
? ?
x 2
e 3x a 6 x 2 ? ? ? ? ?
? ?
=0
at x = 1, 3 +a+6 - 2 = 0
a=–7
f(x) = (3x
2
 - 7x + 5 )e
x
f
’
(x) = (6x—7)e
x 
+(3x
2
-7x+5)e
x
= e
x
(3x
2
-x-2) = 0
= 3x
2
 —3x + 2x - 2 = 0
= (3x+2) (x—1) = 0
x = 1, -2/3
+ – +
–2/3
1
x = 1 is point of local minima.
x = –2/3 is point of local maxima.
Q.2
? ?
2 4
1 x x 1 x
1
x 0 2 4
x e
1 x x 1
/
lim
? ? ?
?
?
? ?
? ?
? ?
? ? ?
(1) is equal to 
e
(2) is equal to 1 (3) is equal to 0 (4) does not exist
Sol. 2
? ?
? ?
2 4
1 x x 1 /x
x 0 2 4
x e 1
lim
1 x x 1
? ? ?
?
? ?
?
? ?
? ?
? ? ?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 2
? ?
? ?
? ?
2
2 4
1 x x 1
x 2
2 4
2 4
x 0
x e 1 1 x x 1
lim
x x
? ?
? ? ? ? ?
? ?
? ? ?
? ?
? ?
?
? ?
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
? ?
?
? ?
?
? ?
? ?
? ?
?
?
?
? ? ?
?
? ?
? ?
3
x x
2
3 x 0
e 1
lim 2
x x
2
2
 = 1
Q.3 The statement p q p p p q ( ( )) ( ( )) ? ? ? ? ? is:
(1) equivalent to p q p ( ) (~ ) ? ?
(2) equivalent to p q p ( ) (~ ) ? ?
(3) a contradiction
(4) a tautology
Sol. 4
? ? ?
? ? ? ? ? ?
? ?
(p (q p))
p q q p p (q p) p q p p q
(p (p q))
T T T T T T T
T F T T T T T
F T F T T T T
F F T T F T T
Q.4 If 
2 2
L
16 8
sin sin
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
 and 
2 2
M
16 8
cos sin
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
, then:
(1) 
1 1
M
2 8
2 2
cos
?
? ?
(2) 
1 1
M
4 8
4 2
cos
?
? ?
(3) 
1 1
L
2 8
2 2
cos
?
? ? ?
(4) 
1 1
L
4 8
4 2
cos
?
? ?
Sol. 1
L = 
? ? ?
? ?
? ?
3
s in
1 6
sin
16
? ? ? ?
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
1
L cos cos
2 8 4
?
1
L
2 2
—
1
cos
2 8
?
Page 3


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 1
Date : 5
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 If x=1 is a critical point of the function f(x)=(3x
2
+ax–2–a)e
x
, then:
(1) x=1 is a local minima and 
2
x
3
? ? is a local maxima of f. .
(2) x=1 is a local maxima and 
2
x
3
? ? is a local minima of f. .
(3) x=1 and 
2
x
3
? ? are local minima of f. .
(4) x=1 and 
2
x
3
? ? are local maxima of f. .
Sol. 1
f(x) = (3x
2
+ax–2–a)e
x
f
’
(x) = (3x
2
+ax–2–a)e
x
 + (6x+a)e
x
= 0
? ?
x 2
e 3x a 6 x 2 ? ? ? ? ?
? ?
=0
at x = 1, 3 +a+6 - 2 = 0
a=–7
f(x) = (3x
2
 - 7x + 5 )e
x
f
’
(x) = (6x—7)e
x 
+(3x
2
-7x+5)e
x
= e
x
(3x
2
-x-2) = 0
= 3x
2
 —3x + 2x - 2 = 0
= (3x+2) (x—1) = 0
x = 1, -2/3
+ – +
–2/3
1
x = 1 is point of local minima.
x = –2/3 is point of local maxima.
Q.2
? ?
2 4
1 x x 1 x
1
x 0 2 4
x e
1 x x 1
/
lim
? ? ?
?
?
? ?
? ?
? ?
? ? ?
(1) is equal to 
e
(2) is equal to 1 (3) is equal to 0 (4) does not exist
Sol. 2
? ?
? ?
2 4
1 x x 1 /x
x 0 2 4
x e 1
lim
1 x x 1
? ? ?
?
? ?
?
? ?
? ?
? ? ?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 2
? ?
? ?
? ?
2
2 4
1 x x 1
x 2
2 4
2 4
x 0
x e 1 1 x x 1
lim
x x
? ?
? ? ? ? ?
? ?
? ? ?
? ?
? ?
?
? ?
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
? ?
?
? ?
?
? ?
? ?
? ?
?
?
?
? ? ?
?
? ?
? ?
3
x x
2
3 x 0
e 1
lim 2
x x
2
2
 = 1
Q.3 The statement p q p p p q ( ( )) ( ( )) ? ? ? ? ? is:
(1) equivalent to p q p ( ) (~ ) ? ?
(2) equivalent to p q p ( ) (~ ) ? ?
(3) a contradiction
(4) a tautology
Sol. 4
? ? ?
? ? ? ? ? ?
? ?
(p (q p))
p q q p p (q p) p q p p q
(p (p q))
T T T T T T T
T F T T T T T
F T F T T T T
F F T T F T T
Q.4 If 
2 2
L
16 8
sin sin
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
 and 
2 2
M
16 8
cos sin
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
, then:
(1) 
1 1
M
2 8
2 2
cos
?
? ?
(2) 
1 1
M
4 8
4 2
cos
?
? ?
(3) 
1 1
L
2 8
2 2
cos
?
? ? ?
(4) 
1 1
L
4 8
4 2
cos
?
? ?
Sol. 1
L = 
? ? ?
? ?
? ?
3
s in
1 6
sin
16
? ? ? ?
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
1
L cos cos
2 8 4
?
1
L
2 2
—
1
cos
2 8
?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 3
M = 
3
cos
16
? ? ?
? ?
? ?
cos
16
? ? ?
? ?
? ?
M = 
1
cos cos
2 4 8
? ? ? ?
?
? ?
? ?
M = 
1
2 2
+
1
2
cos
8
?
Q.5 If the sum of the first 20 terms of the series 
? ? ? ? ? ?
1 2 1 3 1 4
7 7 7
x x x
/ / /
log log log ... ? ? ? is 460,
then x is equal to:
(1) 7
1/2
(2) 7
2
(3) e
2
(4) 7
46/21
Sol. 2
(2 + 3 + 4 +... + 21)log
7
x = 460
? ?
? ?
7
20 21 2
log 460
2
? ?
? x
? ?230 log
7
x = 460 ? log
7
x = 2 ? ?x = 7
2
Q.6 There are 3 sections in a question paper and each section contains 5 questions. A
candidate has to answer a total of 5 questions, choosing at least one question from
each section. Then the number of ways, in which the candidate can choose the questions,
is:
(1) 2250 (2) 2255 (3) 1500 (4) 3000
Sol. 1
S—1 S—2 S—3
1,2,3,4,5 1,2,3,4,5 1,2,3,4,5
6 case
1
1
1
2
2
3
1
2
3
2
1
1
3
2
1
1
2
1
= 3(
5
C
1
 × 
5
C
2 
× 
5
C
2
) + 3(
5
C
1 
×
5
C
1
×
5
C
3
)
= 3(5 × 5 × 2 × 5 × 2) + 3(5 × 5 × 10)
= 750 + 1500 = 2250
Q.7 If the mean and the standard deviation of the data 3,5,7,a,b are 5 and 2 respectively,
then a and b are the roots of the equation:
(1) x
2
–20x+18=0 (2) x
2
–10x+19=0 (3) 2x
2
–20x+19=0 (4) x
2
–10x+18=0
Sol. 2
S.D. =
? ?
2
2
i
x
x
n
?
?
Page 4


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 1
Date : 5
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 If x=1 is a critical point of the function f(x)=(3x
2
+ax–2–a)e
x
, then:
(1) x=1 is a local minima and 
2
x
3
? ? is a local maxima of f. .
(2) x=1 is a local maxima and 
2
x
3
? ? is a local minima of f. .
(3) x=1 and 
2
x
3
? ? are local minima of f. .
(4) x=1 and 
2
x
3
? ? are local maxima of f. .
Sol. 1
f(x) = (3x
2
+ax–2–a)e
x
f
’
(x) = (3x
2
+ax–2–a)e
x
 + (6x+a)e
x
= 0
? ?
x 2
e 3x a 6 x 2 ? ? ? ? ?
? ?
=0
at x = 1, 3 +a+6 - 2 = 0
a=–7
f(x) = (3x
2
 - 7x + 5 )e
x
f
’
(x) = (6x—7)e
x 
+(3x
2
-7x+5)e
x
= e
x
(3x
2
-x-2) = 0
= 3x
2
 —3x + 2x - 2 = 0
= (3x+2) (x—1) = 0
x = 1, -2/3
+ – +
–2/3
1
x = 1 is point of local minima.
x = –2/3 is point of local maxima.
Q.2
? ?
2 4
1 x x 1 x
1
x 0 2 4
x e
1 x x 1
/
lim
? ? ?
?
?
? ?
? ?
? ?
? ? ?
(1) is equal to 
e
(2) is equal to 1 (3) is equal to 0 (4) does not exist
Sol. 2
? ?
? ?
2 4
1 x x 1 /x
x 0 2 4
x e 1
lim
1 x x 1
? ? ?
?
? ?
?
? ?
? ?
? ? ?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 2
? ?
? ?
? ?
2
2 4
1 x x 1
x 2
2 4
2 4
x 0
x e 1 1 x x 1
lim
x x
? ?
? ? ? ? ?
? ?
? ? ?
? ?
? ?
?
? ?
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
? ?
?
? ?
?
? ?
? ?
? ?
?
?
?
? ? ?
?
? ?
? ?
3
x x
2
3 x 0
e 1
lim 2
x x
2
2
 = 1
Q.3 The statement p q p p p q ( ( )) ( ( )) ? ? ? ? ? is:
(1) equivalent to p q p ( ) (~ ) ? ?
(2) equivalent to p q p ( ) (~ ) ? ?
(3) a contradiction
(4) a tautology
Sol. 4
? ? ?
? ? ? ? ? ?
? ?
(p (q p))
p q q p p (q p) p q p p q
(p (p q))
T T T T T T T
T F T T T T T
F T F T T T T
F F T T F T T
Q.4 If 
2 2
L
16 8
sin sin
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
 and 
2 2
M
16 8
cos sin
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
, then:
(1) 
1 1
M
2 8
2 2
cos
?
? ?
(2) 
1 1
M
4 8
4 2
cos
?
? ?
(3) 
1 1
L
2 8
2 2
cos
?
? ? ?
(4) 
1 1
L
4 8
4 2
cos
?
? ?
Sol. 1
L = 
? ? ?
? ?
? ?
3
s in
1 6
sin
16
? ? ? ?
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
1
L cos cos
2 8 4
?
1
L
2 2
—
1
cos
2 8
?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 3
M = 
3
cos
16
? ? ?
? ?
? ?
cos
16
? ? ?
? ?
? ?
M = 
1
cos cos
2 4 8
? ? ? ?
?
? ?
? ?
M = 
1
2 2
+
1
2
cos
8
?
Q.5 If the sum of the first 20 terms of the series 
? ? ? ? ? ?
1 2 1 3 1 4
7 7 7
x x x
/ / /
log log log ... ? ? ? is 460,
then x is equal to:
(1) 7
1/2
(2) 7
2
(3) e
2
(4) 7
46/21
Sol. 2
(2 + 3 + 4 +... + 21)log
7
x = 460
? ?
? ?
7
20 21 2
log 460
2
? ?
? x
? ?230 log
7
x = 460 ? log
7
x = 2 ? ?x = 7
2
Q.6 There are 3 sections in a question paper and each section contains 5 questions. A
candidate has to answer a total of 5 questions, choosing at least one question from
each section. Then the number of ways, in which the candidate can choose the questions,
is:
(1) 2250 (2) 2255 (3) 1500 (4) 3000
Sol. 1
S—1 S—2 S—3
1,2,3,4,5 1,2,3,4,5 1,2,3,4,5
6 case
1
1
1
2
2
3
1
2
3
2
1
1
3
2
1
1
2
1
= 3(
5
C
1
 × 
5
C
2 
× 
5
C
2
) + 3(
5
C
1 
×
5
C
1
×
5
C
3
)
= 3(5 × 5 × 2 × 5 × 2) + 3(5 × 5 × 10)
= 750 + 1500 = 2250
Q.7 If the mean and the standard deviation of the data 3,5,7,a,b are 5 and 2 respectively,
then a and b are the roots of the equation:
(1) x
2
–20x+18=0 (2) x
2
–10x+19=0 (3) 2x
2
–20x+19=0 (4) x
2
–10x+18=0
Sol. 2
S.D. =
? ?
2
2
i
x
x
n
?
?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 4
(2)
2
 = 
2 2
83 a b
5
? ?
—(5)
2
4 = 
2 2
83 a b
5
? ?
 — 25
29×5 — 83 = a
2
+b
2
 ? a
2
+b
2
 = 62
a b 15
5
? ?
= 5
? a b 10 ? ? ...(1)
2ab = 100 — 62 = 38
ab 19 ? ...(2)
from eq.(1) & (2)
x
2
 – 10x + 19 = 0
Q.8 The derivative of 
2
1
1 x 1
x
tan
?
? ?
? ?
? ?
? ?
? ?
 with respect to 
2
1
2
2x 1 x
1 2x
tan
?
? ?
?
? ?
? ?
?
? ?
 at 
1
x
2
? is:
(1) 
2 3
3
(2) 
2 3
5
(3) 
3
12
(4) 
3
10
Sol. 4
x = tan ?
u = 
1
sec 1
tan
tan
?
? ? ? ?
? ?
?
? ?
= ? ?
1
2
tan tan /
?
? =
2
?
 = 
1
tan
2
?
x
x = sin ?
v = 
1
2sin cos
tan
cos2
?
? ? ? ?
? ?
?
? ?
 = 2 ?
= 2sin
-1
x
du
dv
= 2
1
2(1 x ) ?
×
2
1 x
2
?
= 
3
2 2 ?
×
4
5 2 ?
= 
3
10
Q.9 If 
e 2
d A B C
5 7 2
cos
log | ( )|
sin cos
?
? ? ? ?
? ? ? ?
?
 where C is a constant of integration, then
B
A
( ) ?
 can be:
(1) 
5 2 1
3
( sin )
sin
? ?
? ?
(2) 
5 3
2 1
(sin )
sin
? ?
? ?
(3) 
2 1
3
sin
sin
? ?
? ?
(4) 
2 1
5 3
sin
(sin )
? ?
? ?
Sol. 1
2
cos
5 7sin 2 2sin
?
? ? ? ? ?
?
d ?
Put sin ? = t, cos ? d ? = dt
Page 5


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 1
Date : 5
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 If x=1 is a critical point of the function f(x)=(3x
2
+ax–2–a)e
x
, then:
(1) x=1 is a local minima and 
2
x
3
? ? is a local maxima of f. .
(2) x=1 is a local maxima and 
2
x
3
? ? is a local minima of f. .
(3) x=1 and 
2
x
3
? ? are local minima of f. .
(4) x=1 and 
2
x
3
? ? are local maxima of f. .
Sol. 1
f(x) = (3x
2
+ax–2–a)e
x
f
’
(x) = (3x
2
+ax–2–a)e
x
 + (6x+a)e
x
= 0
? ?
x 2
e 3x a 6 x 2 ? ? ? ? ?
? ?
=0
at x = 1, 3 +a+6 - 2 = 0
a=–7
f(x) = (3x
2
 - 7x + 5 )e
x
f
’
(x) = (6x—7)e
x 
+(3x
2
-7x+5)e
x
= e
x
(3x
2
-x-2) = 0
= 3x
2
 —3x + 2x - 2 = 0
= (3x+2) (x—1) = 0
x = 1, -2/3
+ – +
–2/3
1
x = 1 is point of local minima.
x = –2/3 is point of local maxima.
Q.2
? ?
2 4
1 x x 1 x
1
x 0 2 4
x e
1 x x 1
/
lim
? ? ?
?
?
? ?
? ?
? ?
? ? ?
(1) is equal to 
e
(2) is equal to 1 (3) is equal to 0 (4) does not exist
Sol. 2
? ?
? ?
2 4
1 x x 1 /x
x 0 2 4
x e 1
lim
1 x x 1
? ? ?
?
? ?
?
? ?
? ?
? ? ?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 2
? ?
? ?
? ?
2
2 4
1 x x 1
x 2
2 4
2 4
x 0
x e 1 1 x x 1
lim
x x
? ?
? ? ? ? ?
? ?
? ? ?
? ?
? ?
?
? ?
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
? ?
?
? ?
?
? ?
? ?
? ?
?
?
?
? ? ?
?
? ?
? ?
3
x x
2
3 x 0
e 1
lim 2
x x
2
2
 = 1
Q.3 The statement p q p p p q ( ( )) ( ( )) ? ? ? ? ? is:
(1) equivalent to p q p ( ) (~ ) ? ?
(2) equivalent to p q p ( ) (~ ) ? ?
(3) a contradiction
(4) a tautology
Sol. 4
? ? ?
? ? ? ? ? ?
? ?
(p (q p))
p q q p p (q p) p q p p q
(p (p q))
T T T T T T T
T F T T T T T
F T F T T T T
F F T T F T T
Q.4 If 
2 2
L
16 8
sin sin
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
 and 
2 2
M
16 8
cos sin
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
, then:
(1) 
1 1
M
2 8
2 2
cos
?
? ?
(2) 
1 1
M
4 8
4 2
cos
?
? ?
(3) 
1 1
L
2 8
2 2
cos
?
? ? ?
(4) 
1 1
L
4 8
4 2
cos
?
? ?
Sol. 1
L = 
? ? ?
? ?
? ?
3
s in
1 6
sin
16
? ? ? ?
? ?
? ?
? ? ? ? ?
? ?
? ?
? ?
1
L cos cos
2 8 4
?
1
L
2 2
—
1
cos
2 8
?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 3
M = 
3
cos
16
? ? ?
? ?
? ?
cos
16
? ? ?
? ?
? ?
M = 
1
cos cos
2 4 8
? ? ? ?
?
? ?
? ?
M = 
1
2 2
+
1
2
cos
8
?
Q.5 If the sum of the first 20 terms of the series 
? ? ? ? ? ?
1 2 1 3 1 4
7 7 7
x x x
/ / /
log log log ... ? ? ? is 460,
then x is equal to:
(1) 7
1/2
(2) 7
2
(3) e
2
(4) 7
46/21
Sol. 2
(2 + 3 + 4 +... + 21)log
7
x = 460
? ?
? ?
7
20 21 2
log 460
2
? ?
? x
? ?230 log
7
x = 460 ? log
7
x = 2 ? ?x = 7
2
Q.6 There are 3 sections in a question paper and each section contains 5 questions. A
candidate has to answer a total of 5 questions, choosing at least one question from
each section. Then the number of ways, in which the candidate can choose the questions,
is:
(1) 2250 (2) 2255 (3) 1500 (4) 3000
Sol. 1
S—1 S—2 S—3
1,2,3,4,5 1,2,3,4,5 1,2,3,4,5
6 case
1
1
1
2
2
3
1
2
3
2
1
1
3
2
1
1
2
1
= 3(
5
C
1
 × 
5
C
2 
× 
5
C
2
) + 3(
5
C
1 
×
5
C
1
×
5
C
3
)
= 3(5 × 5 × 2 × 5 × 2) + 3(5 × 5 × 10)
= 750 + 1500 = 2250
Q.7 If the mean and the standard deviation of the data 3,5,7,a,b are 5 and 2 respectively,
then a and b are the roots of the equation:
(1) x
2
–20x+18=0 (2) x
2
–10x+19=0 (3) 2x
2
–20x+19=0 (4) x
2
–10x+18=0
Sol. 2
S.D. =
? ?
2
2
i
x
x
n
?
?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 4
(2)
2
 = 
2 2
83 a b
5
? ?
—(5)
2
4 = 
2 2
83 a b
5
? ?
 — 25
29×5 — 83 = a
2
+b
2
 ? a
2
+b
2
 = 62
a b 15
5
? ?
= 5
? a b 10 ? ? ...(1)
2ab = 100 — 62 = 38
ab 19 ? ...(2)
from eq.(1) & (2)
x
2
 – 10x + 19 = 0
Q.8 The derivative of 
2
1
1 x 1
x
tan
?
? ?
? ?
? ?
? ?
? ?
 with respect to 
2
1
2
2x 1 x
1 2x
tan
?
? ?
?
? ?
? ?
?
? ?
 at 
1
x
2
? is:
(1) 
2 3
3
(2) 
2 3
5
(3) 
3
12
(4) 
3
10
Sol. 4
x = tan ?
u = 
1
sec 1
tan
tan
?
? ? ? ?
? ?
?
? ?
= ? ?
1
2
tan tan /
?
? =
2
?
 = 
1
tan
2
?
x
x = sin ?
v = 
1
2sin cos
tan
cos2
?
? ? ? ?
? ?
?
? ?
 = 2 ?
= 2sin
-1
x
du
dv
= 2
1
2(1 x ) ?
×
2
1 x
2
?
= 
3
2 2 ?
×
4
5 2 ?
= 
3
10
Q.9 If 
e 2
d A B C
5 7 2
cos
log | ( )|
sin cos
?
? ? ? ?
? ? ? ?
?
 where C is a constant of integration, then
B
A
( ) ?
 can be:
(1) 
5 2 1
3
( sin )
sin
? ?
? ?
(2) 
5 3
2 1
(sin )
sin
? ?
? ?
(3) 
2 1
3
sin
sin
? ?
? ?
(4) 
2 1
5 3
sin
(sin )
? ?
? ?
Sol. 1
2
cos
5 7sin 2 2sin
?
? ? ? ? ?
?
d ?
Put sin ? = t, cos ? d ? = dt
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Maths     Page | 5
2
dt
2t 7t 3 ? ?
?
 = 
1
2
2
dt
7t 3
t
2 2
? ?
?
= 
1
2
2
2
dt
7 7 49 24
t t
2 4 16 16
? ?
? ? ? ?
? ?
? ?
?
= 
1
2
? ? ? ?
2 2
dt
t 7 / 4 5 / 4 ? ?
?
1
2
×
1
5
2 .
4
ln
t 7 / 4 5 / 4
t 7 / 4 5 / 4
? ? ? ?
? ?
? ?
? ?
1
5
ln 
sin 1 / 2
sin 3
? ? ? ?
? ?
? ?
? ?
+C
B
A
( ) ?
 = 
2sin 1
5
sin 3
? ? ? ?
? ?
? ?
? ?
Q.10 If the length of the chord of the circle, x
2
+y
2
 = r
2
(r>0) along the line, y–2x=3 is r, then
r
2
 is equal to:
(1) 12 (2) 
24
5
(3) 
9
5
(4) 
12
5
Sol. 4
3
5
r
(0,0)
M
A
B
AB = 
2
2 r 9 / 5 ?
=r
r
2
 —9/5 = 
2
r
4
3r
2
/4 = 9/5
2
12
r
5
?
Read More
357 docs|148 tests
Join Course for Free

Top Courses for JEE

Related Exams
JEE
About this Document
4.82/5 Rating
Jan 03, 2025 Last updated
Document Description: JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions for JEE 2025 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The notes and questions for JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions have been prepared according to the JEE exam syllabus. Information about JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions covers topics like and JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions Example, for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions.
Introduction of JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions in English is available as part of our Mock Tests for JEE Main and Advanced 2025 for JEE & JEE Main 2020 Mathematics September 5 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 2020 Mathematics September 5 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025
Description
Full syllabus notes, lecture & questions for JEE Main 2020 Mathematics September 5 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 2020 Mathematics September 5 Shift 2 Paper & Solutions
In this doc you can find the meaning of JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions defined & explained in the simplest way possible. Besides explaining types of JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions theory, EduRev gives you an ample number of questions to practice JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions tests, examples and also practice JEE tests
Explore Courses for JEE exam

Top Courses for JEE

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

study material

,

Semester Notes

,

practice quizzes

,

Viva Questions

,

Objective type Questions

,

video lectures

,

Previous Year Questions with Solutions

,

Exam

,

MCQs

,

Free

,

JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025

,

Important questions

,

Extra Questions

,

Sample Paper

,

shortcuts and tricks

,

JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025

,

JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025

,

past year papers

,

pdf

,

Summary

,

mock tests for examination

,

ppt

;
Additional Information about JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions for JEE Preparation

JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions Free PDF Download

The JEE Main 2020 Mathematics September 5 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 2020 Mathematics September 5 Shift 2 Paper & Solutions now and kickstart your journey towards success in the JEE exam.

Importance of JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions

The importance of JEE Main 2020 Mathematics September 5 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 2020 Mathematics September 5 Shift 2 Paper & Solutions Notes

JEE Main 2020 Mathematics September 5 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 2020 Mathematics September 5 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 2020 Mathematics September 5 Shift 2 Paper & Solutions Notes on EduRev are your ultimate resource for success.

JEE Main 2020 Mathematics September 5 Shift 2 Paper & Solutions JEE Questions

The "JEE Main 2020 Mathematics September 5 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 2020 Mathematics September 5 Shift 2 Paper & Solutions on the App

Students of JEE can study JEE Main 2020 Mathematics September 5 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 2020 Mathematics September 5 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 2020 Mathematics September 5 Shift 2 Paper & Solutions is prepared as per the latest JEE syllabus.
© EduRev
Education Revolution
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup