JEE Mains 12 January 2019 Question Paper Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


E 1
JEE (Main) Examination–2019/Morning Session/12-01-2019
1. For x >1, if (2x)
2y
 = 4e
2x?2y
, then
( )
2
e
dy
1 log 2x
dx
+
 is equal to :
(1) log
e
2x
(2) 
ee
x log 2x log 2
x
+
(3) xlog
e
2x
(4) 
ee
x log 2x log 2
x
-
Ans. (4)
Sol. (2x)
2y
 = 4e
2x?2y
2y ln2x = ln4 + 2x ? 2y
x n2
y
1 n2x
+
=
+
l
l
2
1
(1 n2x) (x n2)
x
y'
(1 n2x)
+ -+
=
+
ll
l
2
x n2x n2
y'(1 n2x)
x
- éù
+=
êú
ëû
ll
l
2. The sum of the distinct real values of m, for
which the vectors, 
? ??
i jk m++ , 
? ??
i j k, +m+
? ??
i jk + +m are co-planer, is :
(1) 2 (2) 0 (3) ?1 (4) 1
Ans. (3)
Sol.
11
1 10
11
m
m=
m
µ(µ
2
 ? 1)?1(µ?1) + 1(1?µ) = 0
µ
3
 ? µ ? µ + 1 + 1 µ = 0
µ
3
 ? 3µ + 2 = 0
µ
3
 ? 1 ? 3(µ?1) = 0
µ = 1, µ
2
 + µ ? 2 = 0
µ = 1, µ = ?2
sum of distinct solutions = ?1
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On SATURDAY 12
th
 JANUARY ., 2019) TIME : 09 : 30 AM To 12 : 30 PM
3. Let S be the set of all points in (?p,p) at which
the function, f(x) = min {sinx, cosx} is not
differentiable. Then S is a subset of which of
the following?
(1) 
{ }
33
, ,,
4 4 44
p p pp
--
(2) 
{ }
33
, ,,
4 2 24
p ppp
--
(3) 
{ }
, ,,
2 442
p p pp
--
(4) 
{ }
,0,
44
pp
-
Ans. (1)
Sol.
p
4
-3p
4
4. The product of three consecutive terms of a G.P.
is 512. If 4 is added to each of the first and the
second of these terms, the three terms now
from an A.P. Then the sum of the original three
terms of the given G.P. is
(1) 36 (2) 24
(3) 32 (4) 28
Ans. (4)
Sol. Let terms are 
a
,a,ar G.P
r
®
\ a
3
 = 512 Þ a = 8
8
4,12,8r A.P.
r
+®
8
24 4 8r
r
= ++
r = 2, 
1
r
2
=
r = 2 (4, 8, 16)
1
r (16,8,4)
2
=
Sum = 28
Page 2


E 1
JEE (Main) Examination–2019/Morning Session/12-01-2019
1. For x >1, if (2x)
2y
 = 4e
2x?2y
, then
( )
2
e
dy
1 log 2x
dx
+
 is equal to :
(1) log
e
2x
(2) 
ee
x log 2x log 2
x
+
(3) xlog
e
2x
(4) 
ee
x log 2x log 2
x
-
Ans. (4)
Sol. (2x)
2y
 = 4e
2x?2y
2y ln2x = ln4 + 2x ? 2y
x n2
y
1 n2x
+
=
+
l
l
2
1
(1 n2x) (x n2)
x
y'
(1 n2x)
+ -+
=
+
ll
l
2
x n2x n2
y'(1 n2x)
x
- éù
+=
êú
ëû
ll
l
2. The sum of the distinct real values of m, for
which the vectors, 
? ??
i jk m++ , 
? ??
i j k, +m+
? ??
i jk + +m are co-planer, is :
(1) 2 (2) 0 (3) ?1 (4) 1
Ans. (3)
Sol.
11
1 10
11
m
m=
m
µ(µ
2
 ? 1)?1(µ?1) + 1(1?µ) = 0
µ
3
 ? µ ? µ + 1 + 1 µ = 0
µ
3
 ? 3µ + 2 = 0
µ
3
 ? 1 ? 3(µ?1) = 0
µ = 1, µ
2
 + µ ? 2 = 0
µ = 1, µ = ?2
sum of distinct solutions = ?1
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On SATURDAY 12
th
 JANUARY ., 2019) TIME : 09 : 30 AM To 12 : 30 PM
3. Let S be the set of all points in (?p,p) at which
the function, f(x) = min {sinx, cosx} is not
differentiable. Then S is a subset of which of
the following?
(1) 
{ }
33
, ,,
4 4 44
p p pp
--
(2) 
{ }
33
, ,,
4 2 24
p ppp
--
(3) 
{ }
, ,,
2 442
p p pp
--
(4) 
{ }
,0,
44
pp
-
Ans. (1)
Sol.
p
4
-3p
4
4. The product of three consecutive terms of a G.P.
is 512. If 4 is added to each of the first and the
second of these terms, the three terms now
from an A.P. Then the sum of the original three
terms of the given G.P. is
(1) 36 (2) 24
(3) 32 (4) 28
Ans. (4)
Sol. Let terms are 
a
,a,ar G.P
r
®
\ a
3
 = 512 Þ a = 8
8
4,12,8r A.P.
r
+®
8
24 4 8r
r
= ++
r = 2, 
1
r
2
=
r = 2 (4, 8, 16)
1
r (16,8,4)
2
=
Sum = 28
E 2
JEE (Main) Examination–2019/Morning Session/12-01-2019
5. The integral 
e
cos(log x)dx
ò
 is equal to :
(where C is a constant of integration)
(1) 
ee
x
[sin(log x) ? cos(log x)] C
2
+
(2)  
ee
x
[cos(log x) sin(log x)] C
2
++
(3) 
ee
x[cos(log x) sin(log x)] C ++
(4) 
ee
x[cos(log x) sin(log x)] C -+
Ans. (2)
Sol. I cos( n x)dx =
ò
l
I cos(ln x).x sin( n x) dx =+
ò
l
cos( n x)x [sin( n x).x cos( n x)dx] +-
ò
l ll
x
I [sin( n x) cos( n x)] C
2
= ++ ll
6. Let S
k
 = 
1 2 3 .... k
k
+ + ++
. If
222
1 2 10
5
S S .... S A
12
+++ =
, then A is equal to :
(1) 303 (2) 283 (3) 156 (4) 301
Ans. (1)
Sol.
K
K1
S
2
+
=
2
k
5
SA
12
S=
2
222 10
K1
K1 2 3 115
A
2 4 12
=
+ + +--+ æö
==
ç÷
èø
å
11 12 23 5
1A
63
´´
-=
5
505 A
3
=
,   A = 303
7. Let S = {1,2,3, ...., 100}. The number of non-
empty subsets A of S such that the product of
elements in A is even is :-
(1) 2
50
(2
50
?1) (2) 2
100
?1
(3) 2
50
?1 (4) 2
50
+1
Ans. (1)
Sol. S = {1,2,3------100}
= Total non empty subsets-subsets with product
of element is odd
= 2
100
?1?1[(2
50
?1)]
= 2
100
 ? 2
50
= 2
50
(2
50
?1)
8. If the sum of the deviations of 50 observations
from 30 is 50, then the mean of these
observation is :
(1) 50 (2) 51 (3) 30 (4) 31
Ans. (4)
Sol.
50
i
i1
(x 30) 50
=
-=
å
i
x 50 30 50 S =´=
i
x 50 50 30 S = ++
Mean = 
i
x 50 30 50
x 30 1 31
n 50
S ´+
= = = +=
9. If a variable line, 3x+4y?l=0 is such that the
two circles x
2
 + y
2
 ? 2x ? 2y + 1 = 0 and
x
2
+y
2
?18x?2y+78 = 0 are on its opposite sides,
then the set of all values of l is the interval :-
(1) [12, 21] (2) (2, 17)
(3) (23, 31) (4) [13, 23]
Ans. (1)
Sol. Centre of circles are opposite side of line
(3 + 4 ? l) (27 + 4 ? l) < 0
(l ? 7) (l ? 31) < 0
l Î (7, 31)
distance from S
1
34
1
5
+ -l
³ Þ lÎ(?¥, 2] È[(12,¥]
distance from S
2
Page 3


E 1
JEE (Main) Examination–2019/Morning Session/12-01-2019
1. For x >1, if (2x)
2y
 = 4e
2x?2y
, then
( )
2
e
dy
1 log 2x
dx
+
 is equal to :
(1) log
e
2x
(2) 
ee
x log 2x log 2
x
+
(3) xlog
e
2x
(4) 
ee
x log 2x log 2
x
-
Ans. (4)
Sol. (2x)
2y
 = 4e
2x?2y
2y ln2x = ln4 + 2x ? 2y
x n2
y
1 n2x
+
=
+
l
l
2
1
(1 n2x) (x n2)
x
y'
(1 n2x)
+ -+
=
+
ll
l
2
x n2x n2
y'(1 n2x)
x
- éù
+=
êú
ëû
ll
l
2. The sum of the distinct real values of m, for
which the vectors, 
? ??
i jk m++ , 
? ??
i j k, +m+
? ??
i jk + +m are co-planer, is :
(1) 2 (2) 0 (3) ?1 (4) 1
Ans. (3)
Sol.
11
1 10
11
m
m=
m
µ(µ
2
 ? 1)?1(µ?1) + 1(1?µ) = 0
µ
3
 ? µ ? µ + 1 + 1 µ = 0
µ
3
 ? 3µ + 2 = 0
µ
3
 ? 1 ? 3(µ?1) = 0
µ = 1, µ
2
 + µ ? 2 = 0
µ = 1, µ = ?2
sum of distinct solutions = ?1
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On SATURDAY 12
th
 JANUARY ., 2019) TIME : 09 : 30 AM To 12 : 30 PM
3. Let S be the set of all points in (?p,p) at which
the function, f(x) = min {sinx, cosx} is not
differentiable. Then S is a subset of which of
the following?
(1) 
{ }
33
, ,,
4 4 44
p p pp
--
(2) 
{ }
33
, ,,
4 2 24
p ppp
--
(3) 
{ }
, ,,
2 442
p p pp
--
(4) 
{ }
,0,
44
pp
-
Ans. (1)
Sol.
p
4
-3p
4
4. The product of three consecutive terms of a G.P.
is 512. If 4 is added to each of the first and the
second of these terms, the three terms now
from an A.P. Then the sum of the original three
terms of the given G.P. is
(1) 36 (2) 24
(3) 32 (4) 28
Ans. (4)
Sol. Let terms are 
a
,a,ar G.P
r
®
\ a
3
 = 512 Þ a = 8
8
4,12,8r A.P.
r
+®
8
24 4 8r
r
= ++
r = 2, 
1
r
2
=
r = 2 (4, 8, 16)
1
r (16,8,4)
2
=
Sum = 28
E 2
JEE (Main) Examination–2019/Morning Session/12-01-2019
5. The integral 
e
cos(log x)dx
ò
 is equal to :
(where C is a constant of integration)
(1) 
ee
x
[sin(log x) ? cos(log x)] C
2
+
(2)  
ee
x
[cos(log x) sin(log x)] C
2
++
(3) 
ee
x[cos(log x) sin(log x)] C ++
(4) 
ee
x[cos(log x) sin(log x)] C -+
Ans. (2)
Sol. I cos( n x)dx =
ò
l
I cos(ln x).x sin( n x) dx =+
ò
l
cos( n x)x [sin( n x).x cos( n x)dx] +-
ò
l ll
x
I [sin( n x) cos( n x)] C
2
= ++ ll
6. Let S
k
 = 
1 2 3 .... k
k
+ + ++
. If
222
1 2 10
5
S S .... S A
12
+++ =
, then A is equal to :
(1) 303 (2) 283 (3) 156 (4) 301
Ans. (1)
Sol.
K
K1
S
2
+
=
2
k
5
SA
12
S=
2
222 10
K1
K1 2 3 115
A
2 4 12
=
+ + +--+ æö
==
ç÷
èø
å
11 12 23 5
1A
63
´´
-=
5
505 A
3
=
,   A = 303
7. Let S = {1,2,3, ...., 100}. The number of non-
empty subsets A of S such that the product of
elements in A is even is :-
(1) 2
50
(2
50
?1) (2) 2
100
?1
(3) 2
50
?1 (4) 2
50
+1
Ans. (1)
Sol. S = {1,2,3------100}
= Total non empty subsets-subsets with product
of element is odd
= 2
100
?1?1[(2
50
?1)]
= 2
100
 ? 2
50
= 2
50
(2
50
?1)
8. If the sum of the deviations of 50 observations
from 30 is 50, then the mean of these
observation is :
(1) 50 (2) 51 (3) 30 (4) 31
Ans. (4)
Sol.
50
i
i1
(x 30) 50
=
-=
å
i
x 50 30 50 S =´=
i
x 50 50 30 S = ++
Mean = 
i
x 50 30 50
x 30 1 31
n 50
S ´+
= = = +=
9. If a variable line, 3x+4y?l=0 is such that the
two circles x
2
 + y
2
 ? 2x ? 2y + 1 = 0 and
x
2
+y
2
?18x?2y+78 = 0 are on its opposite sides,
then the set of all values of l is the interval :-
(1) [12, 21] (2) (2, 17)
(3) (23, 31) (4) [13, 23]
Ans. (1)
Sol. Centre of circles are opposite side of line
(3 + 4 ? l) (27 + 4 ? l) < 0
(l ? 7) (l ? 31) < 0
l Î (7, 31)
distance from S
1
34
1
5
+ -l
³ Þ lÎ(?¥, 2] È[(12,¥]
distance from S
2
E 3
JEE (Main) Examination–2019/Morning Session/12-01-2019
274
2
5
+ -l
³
 Þ lÎ (?¥, 21] È[41, ¥)
so lÎ [12, 21]
10. A ratio of the 5
th
 term from the beginning to
the 5
th
 term from the end in the binomial
expansion of 
10
1
3
1
3
1
2
2(3)
æö
+
ç÷
ç÷
èø
 is :
(1) 1 : 
1
3
4(16) (2) 1 : 
1
3
2(6)
(3) 
1
3
2(36) : 1 (4) 
1
3
4(36) : 1
Ans. (4)
Sol.
4
10 1/ 3 10 4
4 1/3
1/3 5
104 1
5
10 1/34
4 1/3
1
C(2)
T 2(3)
4.(36)
T
1
C (2)
2(3)
-
-
æö
ç÷
èø
==
æö
ç÷
èø
11. let C
1
 and C
2
 be the centres of the circles
x
2
+y
2
?2x?2y?2 = 0 and x
2
+y
2
?6x?6y+14 = 0
respectively. If P and Q are the points of
intersection of these circles, then the area
(in sq. units) of the quadrilateral PC
1
QC
2
 is :
(1) 8 (2) 6 (3) 9 (4) 4
Ans. (4)
Sol.
2
(1,1)
(3,3)
2
Area = 2 ×  
1
.42
2
=
12. In a random experiment, a fair die is rolled until
two fours are obtained in succession. The
probability that the experiment will end in the
fifth throw of the die is equal to :
(1) 
5
150
6
(2) 
5
175
6
(3) 
5
200
6
(4) 
5
225
6
Ans. (2)
Sol. 4 4
32
1
23 35
1 5 2C .5 175
66 66
æö
+=
ç÷
èø
13. If the straight line, 2x?3y+17 = 0 is
perpendicular to the line passing through the
points (7, 17) and (15, b), then b equals :-
(1) ?5 (2) 
35
?
3
(3) 
35
3
(4) 5
Ans. (4)
Sol.
172
1
83
-b
´ =-
-
b = 5
14. Let f and g be continuous functions on [0, a]
such that f(x) = f(a?x) and g(x)+g(a?x)=4,
then
a
0
f(x)g(x)dx
ò
 is equal to :-
(1) 
a
0
4 f(x)dx
ò (2) 
a
0
2 f(x)dx
ò
(3) 
a
0
?3 f(x)dx
ò (4) 
a
0
f(x)dx
ò
Ans. (2)
Sol.
a
0
I f(x)g(x)dx =
ò
a
0
I f(a x)g(a x)dx = --
ò
a
0
I f(x)(4 g(x)dx =-
ò
a
0
I 4 f(x)dx I =-
ò
Þ
a
0
I 2 f(x)dx =
ò
15. The maximum area (in sq. units) of a rectangle
having its base on the x-axis and its other two
vertices on the parabola, y = 12?x
2
 such that
the rectangle lies inside the parabola, is :-
(1) 
202
(2) 
183
(3) 32 (4) 36
Page 4


E 1
JEE (Main) Examination–2019/Morning Session/12-01-2019
1. For x >1, if (2x)
2y
 = 4e
2x?2y
, then
( )
2
e
dy
1 log 2x
dx
+
 is equal to :
(1) log
e
2x
(2) 
ee
x log 2x log 2
x
+
(3) xlog
e
2x
(4) 
ee
x log 2x log 2
x
-
Ans. (4)
Sol. (2x)
2y
 = 4e
2x?2y
2y ln2x = ln4 + 2x ? 2y
x n2
y
1 n2x
+
=
+
l
l
2
1
(1 n2x) (x n2)
x
y'
(1 n2x)
+ -+
=
+
ll
l
2
x n2x n2
y'(1 n2x)
x
- éù
+=
êú
ëû
ll
l
2. The sum of the distinct real values of m, for
which the vectors, 
? ??
i jk m++ , 
? ??
i j k, +m+
? ??
i jk + +m are co-planer, is :
(1) 2 (2) 0 (3) ?1 (4) 1
Ans. (3)
Sol.
11
1 10
11
m
m=
m
µ(µ
2
 ? 1)?1(µ?1) + 1(1?µ) = 0
µ
3
 ? µ ? µ + 1 + 1 µ = 0
µ
3
 ? 3µ + 2 = 0
µ
3
 ? 1 ? 3(µ?1) = 0
µ = 1, µ
2
 + µ ? 2 = 0
µ = 1, µ = ?2
sum of distinct solutions = ?1
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On SATURDAY 12
th
 JANUARY ., 2019) TIME : 09 : 30 AM To 12 : 30 PM
3. Let S be the set of all points in (?p,p) at which
the function, f(x) = min {sinx, cosx} is not
differentiable. Then S is a subset of which of
the following?
(1) 
{ }
33
, ,,
4 4 44
p p pp
--
(2) 
{ }
33
, ,,
4 2 24
p ppp
--
(3) 
{ }
, ,,
2 442
p p pp
--
(4) 
{ }
,0,
44
pp
-
Ans. (1)
Sol.
p
4
-3p
4
4. The product of three consecutive terms of a G.P.
is 512. If 4 is added to each of the first and the
second of these terms, the three terms now
from an A.P. Then the sum of the original three
terms of the given G.P. is
(1) 36 (2) 24
(3) 32 (4) 28
Ans. (4)
Sol. Let terms are 
a
,a,ar G.P
r
®
\ a
3
 = 512 Þ a = 8
8
4,12,8r A.P.
r
+®
8
24 4 8r
r
= ++
r = 2, 
1
r
2
=
r = 2 (4, 8, 16)
1
r (16,8,4)
2
=
Sum = 28
E 2
JEE (Main) Examination–2019/Morning Session/12-01-2019
5. The integral 
e
cos(log x)dx
ò
 is equal to :
(where C is a constant of integration)
(1) 
ee
x
[sin(log x) ? cos(log x)] C
2
+
(2)  
ee
x
[cos(log x) sin(log x)] C
2
++
(3) 
ee
x[cos(log x) sin(log x)] C ++
(4) 
ee
x[cos(log x) sin(log x)] C -+
Ans. (2)
Sol. I cos( n x)dx =
ò
l
I cos(ln x).x sin( n x) dx =+
ò
l
cos( n x)x [sin( n x).x cos( n x)dx] +-
ò
l ll
x
I [sin( n x) cos( n x)] C
2
= ++ ll
6. Let S
k
 = 
1 2 3 .... k
k
+ + ++
. If
222
1 2 10
5
S S .... S A
12
+++ =
, then A is equal to :
(1) 303 (2) 283 (3) 156 (4) 301
Ans. (1)
Sol.
K
K1
S
2
+
=
2
k
5
SA
12
S=
2
222 10
K1
K1 2 3 115
A
2 4 12
=
+ + +--+ æö
==
ç÷
èø
å
11 12 23 5
1A
63
´´
-=
5
505 A
3
=
,   A = 303
7. Let S = {1,2,3, ...., 100}. The number of non-
empty subsets A of S such that the product of
elements in A is even is :-
(1) 2
50
(2
50
?1) (2) 2
100
?1
(3) 2
50
?1 (4) 2
50
+1
Ans. (1)
Sol. S = {1,2,3------100}
= Total non empty subsets-subsets with product
of element is odd
= 2
100
?1?1[(2
50
?1)]
= 2
100
 ? 2
50
= 2
50
(2
50
?1)
8. If the sum of the deviations of 50 observations
from 30 is 50, then the mean of these
observation is :
(1) 50 (2) 51 (3) 30 (4) 31
Ans. (4)
Sol.
50
i
i1
(x 30) 50
=
-=
å
i
x 50 30 50 S =´=
i
x 50 50 30 S = ++
Mean = 
i
x 50 30 50
x 30 1 31
n 50
S ´+
= = = +=
9. If a variable line, 3x+4y?l=0 is such that the
two circles x
2
 + y
2
 ? 2x ? 2y + 1 = 0 and
x
2
+y
2
?18x?2y+78 = 0 are on its opposite sides,
then the set of all values of l is the interval :-
(1) [12, 21] (2) (2, 17)
(3) (23, 31) (4) [13, 23]
Ans. (1)
Sol. Centre of circles are opposite side of line
(3 + 4 ? l) (27 + 4 ? l) < 0
(l ? 7) (l ? 31) < 0
l Î (7, 31)
distance from S
1
34
1
5
+ -l
³ Þ lÎ(?¥, 2] È[(12,¥]
distance from S
2
E 3
JEE (Main) Examination–2019/Morning Session/12-01-2019
274
2
5
+ -l
³
 Þ lÎ (?¥, 21] È[41, ¥)
so lÎ [12, 21]
10. A ratio of the 5
th
 term from the beginning to
the 5
th
 term from the end in the binomial
expansion of 
10
1
3
1
3
1
2
2(3)
æö
+
ç÷
ç÷
èø
 is :
(1) 1 : 
1
3
4(16) (2) 1 : 
1
3
2(6)
(3) 
1
3
2(36) : 1 (4) 
1
3
4(36) : 1
Ans. (4)
Sol.
4
10 1/ 3 10 4
4 1/3
1/3 5
104 1
5
10 1/34
4 1/3
1
C(2)
T 2(3)
4.(36)
T
1
C (2)
2(3)
-
-
æö
ç÷
èø
==
æö
ç÷
èø
11. let C
1
 and C
2
 be the centres of the circles
x
2
+y
2
?2x?2y?2 = 0 and x
2
+y
2
?6x?6y+14 = 0
respectively. If P and Q are the points of
intersection of these circles, then the area
(in sq. units) of the quadrilateral PC
1
QC
2
 is :
(1) 8 (2) 6 (3) 9 (4) 4
Ans. (4)
Sol.
2
(1,1)
(3,3)
2
Area = 2 ×  
1
.42
2
=
12. In a random experiment, a fair die is rolled until
two fours are obtained in succession. The
probability that the experiment will end in the
fifth throw of the die is equal to :
(1) 
5
150
6
(2) 
5
175
6
(3) 
5
200
6
(4) 
5
225
6
Ans. (2)
Sol. 4 4
32
1
23 35
1 5 2C .5 175
66 66
æö
+=
ç÷
èø
13. If the straight line, 2x?3y+17 = 0 is
perpendicular to the line passing through the
points (7, 17) and (15, b), then b equals :-
(1) ?5 (2) 
35
?
3
(3) 
35
3
(4) 5
Ans. (4)
Sol.
172
1
83
-b
´ =-
-
b = 5
14. Let f and g be continuous functions on [0, a]
such that f(x) = f(a?x) and g(x)+g(a?x)=4,
then
a
0
f(x)g(x)dx
ò
 is equal to :-
(1) 
a
0
4 f(x)dx
ò (2) 
a
0
2 f(x)dx
ò
(3) 
a
0
?3 f(x)dx
ò (4) 
a
0
f(x)dx
ò
Ans. (2)
Sol.
a
0
I f(x)g(x)dx =
ò
a
0
I f(a x)g(a x)dx = --
ò
a
0
I f(x)(4 g(x)dx =-
ò
a
0
I 4 f(x)dx I =-
ò
Þ
a
0
I 2 f(x)dx =
ò
15. The maximum area (in sq. units) of a rectangle
having its base on the x-axis and its other two
vertices on the parabola, y = 12?x
2
 such that
the rectangle lies inside the parabola, is :-
(1) 
202
(2) 
183
(3) 32 (4) 36
E 4
JEE (Main) Examination–2019/Morning Session/12-01-2019
Ans. (3)
Sol. f(a) = 2a(12 ? a)
2
(a, 0)
(a, 12 – a )
2
(0, 0)
f'(a) = 2(12 ? 3a
2
)
maximum at a = 2
maximum area = f(2) = 32
16. The Boolean expression
((p q) (p ~q)) (~p ~q) Ù ÚÚ ÙÙ is equivalent to:
(1) p (~ q) Ù (2) p (~ q) Ú
(3) (~ p) (~ q) Ù (4) pq Ù
Ans. (3)
17.
( )
3
x
4
cot x tanx
lim
cosx
4
p
®
-
p
+
 is :
(1) 4 (2) 8
2
(3) 8 (4) 
42
Ans. (3)
Sol.
3
x /4
cot x tan x
lim
cosx
4
®p
-
p æö
+
ç÷
èø
4
x /4
(1 tan x)
lim
cos(x / 4)
®p
-
+p
2
x /4
(1 tan x)
2 lim
cos(x / 4)
®p
-
+p
22
2
x /4
cosx sinx1
R lim
cosx sinx
cosx
2
®p
-
-
x /4
4 2 lim (cosx sin x) 8
®p
+=
18. Considering only the principal values
of inverse functions, the set
{ }
11
A x0:tan(2x)tan(3x)
4
--
p
=³ +=
(1) is an empty set
(2) Contains more than two elements
(3) Contains two elements
(4) is a singleton
Ans. (4)
Sol. tan
?1
(2x) + tan
?1
(3x) = p/4
Þ
2
5x
1
1 6x
=
-
Þ 6x
2
 + 5x ? 1 = 0
x = ?1 or 
1
x
6
=
1
x x0
6
=> Q
19. An ordered pair(a,b) for which the system of
linear equations
(1+a)x + by+z = 2
ax+(1+b)y+z = 3
ax+by+2z = 2   has a unique solution is
(1) (1,?3) (2) (?3,1)
(3) (2, 4) (4) (?4, 2)
Ans. (3)
Sol. For unique solution
11
0 11
2
+ab
D¹ Þ a +b ¹0
ab
1 10
0 1 10
2
-
- ¹ Þ a +b ¹ -2
ab
20. The area (in sq. units) of the region bounded
by the parabola, y = x
2
 + 2 and the lines,
y = x + 1, x = 0 and x = 3, is :
(1) 
15
4
(2) 
15
2
(3) 
21
2
(4) 
17
4
Ans. (2)
Sol.
1
0 3
4
Req. area = 
3
2
0
1 15 15
(x 2)dx .5.3 9 6
2 22
+ - =+- =
ò
Page 5


E 1
JEE (Main) Examination–2019/Morning Session/12-01-2019
1. For x >1, if (2x)
2y
 = 4e
2x?2y
, then
( )
2
e
dy
1 log 2x
dx
+
 is equal to :
(1) log
e
2x
(2) 
ee
x log 2x log 2
x
+
(3) xlog
e
2x
(4) 
ee
x log 2x log 2
x
-
Ans. (4)
Sol. (2x)
2y
 = 4e
2x?2y
2y ln2x = ln4 + 2x ? 2y
x n2
y
1 n2x
+
=
+
l
l
2
1
(1 n2x) (x n2)
x
y'
(1 n2x)
+ -+
=
+
ll
l
2
x n2x n2
y'(1 n2x)
x
- éù
+=
êú
ëû
ll
l
2. The sum of the distinct real values of m, for
which the vectors, 
? ??
i jk m++ , 
? ??
i j k, +m+
? ??
i jk + +m are co-planer, is :
(1) 2 (2) 0 (3) ?1 (4) 1
Ans. (3)
Sol.
11
1 10
11
m
m=
m
µ(µ
2
 ? 1)?1(µ?1) + 1(1?µ) = 0
µ
3
 ? µ ? µ + 1 + 1 µ = 0
µ
3
 ? 3µ + 2 = 0
µ
3
 ? 1 ? 3(µ?1) = 0
µ = 1, µ
2
 + µ ? 2 = 0
µ = 1, µ = ?2
sum of distinct solutions = ?1
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On SATURDAY 12
th
 JANUARY ., 2019) TIME : 09 : 30 AM To 12 : 30 PM
3. Let S be the set of all points in (?p,p) at which
the function, f(x) = min {sinx, cosx} is not
differentiable. Then S is a subset of which of
the following?
(1) 
{ }
33
, ,,
4 4 44
p p pp
--
(2) 
{ }
33
, ,,
4 2 24
p ppp
--
(3) 
{ }
, ,,
2 442
p p pp
--
(4) 
{ }
,0,
44
pp
-
Ans. (1)
Sol.
p
4
-3p
4
4. The product of three consecutive terms of a G.P.
is 512. If 4 is added to each of the first and the
second of these terms, the three terms now
from an A.P. Then the sum of the original three
terms of the given G.P. is
(1) 36 (2) 24
(3) 32 (4) 28
Ans. (4)
Sol. Let terms are 
a
,a,ar G.P
r
®
\ a
3
 = 512 Þ a = 8
8
4,12,8r A.P.
r
+®
8
24 4 8r
r
= ++
r = 2, 
1
r
2
=
r = 2 (4, 8, 16)
1
r (16,8,4)
2
=
Sum = 28
E 2
JEE (Main) Examination–2019/Morning Session/12-01-2019
5. The integral 
e
cos(log x)dx
ò
 is equal to :
(where C is a constant of integration)
(1) 
ee
x
[sin(log x) ? cos(log x)] C
2
+
(2)  
ee
x
[cos(log x) sin(log x)] C
2
++
(3) 
ee
x[cos(log x) sin(log x)] C ++
(4) 
ee
x[cos(log x) sin(log x)] C -+
Ans. (2)
Sol. I cos( n x)dx =
ò
l
I cos(ln x).x sin( n x) dx =+
ò
l
cos( n x)x [sin( n x).x cos( n x)dx] +-
ò
l ll
x
I [sin( n x) cos( n x)] C
2
= ++ ll
6. Let S
k
 = 
1 2 3 .... k
k
+ + ++
. If
222
1 2 10
5
S S .... S A
12
+++ =
, then A is equal to :
(1) 303 (2) 283 (3) 156 (4) 301
Ans. (1)
Sol.
K
K1
S
2
+
=
2
k
5
SA
12
S=
2
222 10
K1
K1 2 3 115
A
2 4 12
=
+ + +--+ æö
==
ç÷
èø
å
11 12 23 5
1A
63
´´
-=
5
505 A
3
=
,   A = 303
7. Let S = {1,2,3, ...., 100}. The number of non-
empty subsets A of S such that the product of
elements in A is even is :-
(1) 2
50
(2
50
?1) (2) 2
100
?1
(3) 2
50
?1 (4) 2
50
+1
Ans. (1)
Sol. S = {1,2,3------100}
= Total non empty subsets-subsets with product
of element is odd
= 2
100
?1?1[(2
50
?1)]
= 2
100
 ? 2
50
= 2
50
(2
50
?1)
8. If the sum of the deviations of 50 observations
from 30 is 50, then the mean of these
observation is :
(1) 50 (2) 51 (3) 30 (4) 31
Ans. (4)
Sol.
50
i
i1
(x 30) 50
=
-=
å
i
x 50 30 50 S =´=
i
x 50 50 30 S = ++
Mean = 
i
x 50 30 50
x 30 1 31
n 50
S ´+
= = = +=
9. If a variable line, 3x+4y?l=0 is such that the
two circles x
2
 + y
2
 ? 2x ? 2y + 1 = 0 and
x
2
+y
2
?18x?2y+78 = 0 are on its opposite sides,
then the set of all values of l is the interval :-
(1) [12, 21] (2) (2, 17)
(3) (23, 31) (4) [13, 23]
Ans. (1)
Sol. Centre of circles are opposite side of line
(3 + 4 ? l) (27 + 4 ? l) < 0
(l ? 7) (l ? 31) < 0
l Î (7, 31)
distance from S
1
34
1
5
+ -l
³ Þ lÎ(?¥, 2] È[(12,¥]
distance from S
2
E 3
JEE (Main) Examination–2019/Morning Session/12-01-2019
274
2
5
+ -l
³
 Þ lÎ (?¥, 21] È[41, ¥)
so lÎ [12, 21]
10. A ratio of the 5
th
 term from the beginning to
the 5
th
 term from the end in the binomial
expansion of 
10
1
3
1
3
1
2
2(3)
æö
+
ç÷
ç÷
èø
 is :
(1) 1 : 
1
3
4(16) (2) 1 : 
1
3
2(6)
(3) 
1
3
2(36) : 1 (4) 
1
3
4(36) : 1
Ans. (4)
Sol.
4
10 1/ 3 10 4
4 1/3
1/3 5
104 1
5
10 1/34
4 1/3
1
C(2)
T 2(3)
4.(36)
T
1
C (2)
2(3)
-
-
æö
ç÷
èø
==
æö
ç÷
èø
11. let C
1
 and C
2
 be the centres of the circles
x
2
+y
2
?2x?2y?2 = 0 and x
2
+y
2
?6x?6y+14 = 0
respectively. If P and Q are the points of
intersection of these circles, then the area
(in sq. units) of the quadrilateral PC
1
QC
2
 is :
(1) 8 (2) 6 (3) 9 (4) 4
Ans. (4)
Sol.
2
(1,1)
(3,3)
2
Area = 2 ×  
1
.42
2
=
12. In a random experiment, a fair die is rolled until
two fours are obtained in succession. The
probability that the experiment will end in the
fifth throw of the die is equal to :
(1) 
5
150
6
(2) 
5
175
6
(3) 
5
200
6
(4) 
5
225
6
Ans. (2)
Sol. 4 4
32
1
23 35
1 5 2C .5 175
66 66
æö
+=
ç÷
èø
13. If the straight line, 2x?3y+17 = 0 is
perpendicular to the line passing through the
points (7, 17) and (15, b), then b equals :-
(1) ?5 (2) 
35
?
3
(3) 
35
3
(4) 5
Ans. (4)
Sol.
172
1
83
-b
´ =-
-
b = 5
14. Let f and g be continuous functions on [0, a]
such that f(x) = f(a?x) and g(x)+g(a?x)=4,
then
a
0
f(x)g(x)dx
ò
 is equal to :-
(1) 
a
0
4 f(x)dx
ò (2) 
a
0
2 f(x)dx
ò
(3) 
a
0
?3 f(x)dx
ò (4) 
a
0
f(x)dx
ò
Ans. (2)
Sol.
a
0
I f(x)g(x)dx =
ò
a
0
I f(a x)g(a x)dx = --
ò
a
0
I f(x)(4 g(x)dx =-
ò
a
0
I 4 f(x)dx I =-
ò
Þ
a
0
I 2 f(x)dx =
ò
15. The maximum area (in sq. units) of a rectangle
having its base on the x-axis and its other two
vertices on the parabola, y = 12?x
2
 such that
the rectangle lies inside the parabola, is :-
(1) 
202
(2) 
183
(3) 32 (4) 36
E 4
JEE (Main) Examination–2019/Morning Session/12-01-2019
Ans. (3)
Sol. f(a) = 2a(12 ? a)
2
(a, 0)
(a, 12 – a )
2
(0, 0)
f'(a) = 2(12 ? 3a
2
)
maximum at a = 2
maximum area = f(2) = 32
16. The Boolean expression
((p q) (p ~q)) (~p ~q) Ù ÚÚ ÙÙ is equivalent to:
(1) p (~ q) Ù (2) p (~ q) Ú
(3) (~ p) (~ q) Ù (4) pq Ù
Ans. (3)
17.
( )
3
x
4
cot x tanx
lim
cosx
4
p
®
-
p
+
 is :
(1) 4 (2) 8
2
(3) 8 (4) 
42
Ans. (3)
Sol.
3
x /4
cot x tan x
lim
cosx
4
®p
-
p æö
+
ç÷
èø
4
x /4
(1 tan x)
lim
cos(x / 4)
®p
-
+p
2
x /4
(1 tan x)
2 lim
cos(x / 4)
®p
-
+p
22
2
x /4
cosx sinx1
R lim
cosx sinx
cosx
2
®p
-
-
x /4
4 2 lim (cosx sin x) 8
®p
+=
18. Considering only the principal values
of inverse functions, the set
{ }
11
A x0:tan(2x)tan(3x)
4
--
p
=³ +=
(1) is an empty set
(2) Contains more than two elements
(3) Contains two elements
(4) is a singleton
Ans. (4)
Sol. tan
?1
(2x) + tan
?1
(3x) = p/4
Þ
2
5x
1
1 6x
=
-
Þ 6x
2
 + 5x ? 1 = 0
x = ?1 or 
1
x
6
=
1
x x0
6
=> Q
19. An ordered pair(a,b) for which the system of
linear equations
(1+a)x + by+z = 2
ax+(1+b)y+z = 3
ax+by+2z = 2   has a unique solution is
(1) (1,?3) (2) (?3,1)
(3) (2, 4) (4) (?4, 2)
Ans. (3)
Sol. For unique solution
11
0 11
2
+ab
D¹ Þ a +b ¹0
ab
1 10
0 1 10
2
-
- ¹ Þ a +b ¹ -2
ab
20. The area (in sq. units) of the region bounded
by the parabola, y = x
2
 + 2 and the lines,
y = x + 1, x = 0 and x = 3, is :
(1) 
15
4
(2) 
15
2
(3) 
21
2
(4) 
17
4
Ans. (2)
Sol.
1
0 3
4
Req. area = 
3
2
0
1 15 15
(x 2)dx .5.3 9 6
2 22
+ - =+- =
ò
E 5
JEE (Main) Examination–2019/Morning Session/12-01-2019
21. If l be the ratio of the roots of the quadratic
equation in x, 3m
2
x
2
+m(m?4)x+2 = 0, then the
least value of m for which 
1
1 l+=
l
, is :
(1) 
23 -
(2) 
4 32 -
(3) 
22 -+
(4) 
4 23 -
Ans. (2)
Sol. 3m
2
x
2
 + m(m ? 4) x + 2 = 0
1
1 l+=
l
, 
1
ab
+=
ba
, a
2
 + b
2
 = ab
(a + b)
2
 = 3ab
2
22
m(m 4) 3(2)
3m 3m
- æö
-=
ç÷
èø
, 
2
2
(m4)6
9m 3m
-
=
(m ? 4)
2
 = 18, m = 4 18, 4 32 ±±
22. If the vertices of a hyperbola be at (?2, 0) and
(2, 0) and one of its foci be at (?3, 0), then
which one of the following points does not lie
on this hyperbola?
(1) ( ) 4, 15
(2) ( ) 6,2 10 -
(3) ( ) 6,52
(4) ( ) 2 6,5
Ans. (3)
Sol.
(–2,0)
(–a,0)
(–2,0)
(a,0)
(–ae,0)
ae = 3 
ae = 3, 
3
e
2
=
, 
2
9
b 41
4
æö
=-
ç÷
èø
, b
2
 = 5
22
xy
1
45
-=
23. If 
z
( R)
z
-a
aÎ
+a
 is a purely imaginary number
and |z| = 2, then a value of a is :
(1) 1 (2) 2 (3) 
2
(4) 
1
2
Ans. (2)
Sol.
zz
0
zz
-a -a
+=
+a +a
22
zzz z zzz z 0 + a-a -a + - a+ a-a =
|z|
2
 = a
2
,  
 
a = ±2
24. Let P(4, ?4) and Q(9, 6) be two points on the
parabola, y
2
= 4x and let X be any point on the
arc POQ of this parabola, where O is the vertex
of this parabola, such that the area of DPXQ is
maximum. Then this maximum area (in sq.
units) is :
(1) 
125
4
(2) 
125
2
(3) 
625
4
(4) 
75
2
Ans. (1)
Sol.
P
(4,–4) t=–2
X
(t , 2t)
2
Q(9,6)
y
2
 = 4x
2yy' = 4
1
y'2
t
==
, 
1
t
2
=
Area = 
1
11
4
1 125
9 61
24
4 41
=
-
25. the perpendicular distance from the origin to
the plane containing the two lines,
x2 y2 z5
3 57
+ -+
==
 and 
x 1 y4 z4
1 47
- -+
==
,
is:
(1) 
11
6
(2) 
6 11
(3) 11 (4) 
116
Ans. (1)
Sol.
i jk
3 57
1 47
?? ?
i(35 28) j(21.7) k(12 5) - - +-
?? ?
7i 14j 7k -+
?? ?
i 2jk -+
1(x + 2) ? 2(y ? 2) + 1 (z+15) = 0
x ? 2y + z + 11 = 0
11 11
4116
=
++