JEE Mains 10 January 2019 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


E 1
JEE (Main) Examination–2019/Evening Session/10-01-2019
1. Let z = 
55
3 i 3i
2 2 22
æ öæö
+ +-
ç ÷ç÷
ç ÷ç÷
è øèø
. If R(z) and I[z]
respectively denote the real and imaginary parts
of z, then :
(1) R(z) > 0 and I(z) > 0
(2) R(z) < 0 and I(z) > 0
(3) R(z) = –3
(4) I(z) = 0
Ans. (4)
Sol. z = 
55
3 i 3i
22
æ öæö
+-
+
ç ÷ç÷
ç ÷ç÷
è øèø
z = 
( ) ( )
55
i /6 i /6
ee
p -p
+
  = 
i5 /6 i5 /6
ee
p -p
+
  = cos
5sin5 55
i cos isin
6 6 66
p p -p -p æö æö
+ ++
ç÷ ç÷
èø èø
  = 2 cos
5
6
p
 < 0
I(z) = 0  and Re(z) < 0
Option (4)
2. Let a
1
,a
2
,a
3
, ...., a
10
 be in G.P. with a
i
 > 0 for
i = 1,2,...., 10 and S be the set of pairs (r,k),
r kÎN (the set of natural numbers) for which
r k r k rk
e 1 2 e2 3 e3 4
r k rk rk
e 4 5 e5 6 e6 7
rk r k rk
e 78 e 8 9 e 910
logaa logaa logaa
logaa logaa logaa 0
logaa logaa logaa
=
Then the number of elements in S, is :
(1) Infinitely many (2) 4
(3) 10 (4) 2
Ans. (1)
Sol. Apply
C
3
 ® C
3
 – C
2
C
2
 ® C
2
 – C
1
We get   D = 0
Option (1)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Thursday 10
th
 JANUARY , 2019)    TIME : 2 : 30 PM To 5 : 30 PM
3. The positive value of l for which the
co-efficient of x
2
 in the expression
10
2
2
xx
x
l æö
+
ç÷
èø
 is 720, is :
(1) 
5
(2) 4
(3) 
22
(4) 3
Ans. (2)
Sol.
()
r
10r
2 10
r 2
x Cx
x
- æö
l æö
ç÷
ç÷
ç÷
èø
èø
10r
2 10 r 2r
2
r
x C (x) ( ) (x)
-
-
éù
l
êú
ëû
10 5r
210r
2
r
x Cx
-
éù
l
êú
ëû
\  r = 2
Hence, 
10
C
2
 l
2
 = 720
l
2
 = 16
l = ±4
Option (2)
4. The value of 
2 3 10 10
cos cos ..... cos sin
2 2 22
p p pp
× ×××
is :
(1) 
256
1
(2) 
1
2
(3) 
1
512
(4) 
1
1024
Ans. (3)
Sol.
10 10 2
2sin cos cos
2 22
p pp
¼¼
9
1
sin
22
p
= 
1
512
Option (3)
Page 2


E 1
JEE (Main) Examination–2019/Evening Session/10-01-2019
1. Let z = 
55
3 i 3i
2 2 22
æ öæö
+ +-
ç ÷ç÷
ç ÷ç÷
è øèø
. If R(z) and I[z]
respectively denote the real and imaginary parts
of z, then :
(1) R(z) > 0 and I(z) > 0
(2) R(z) < 0 and I(z) > 0
(3) R(z) = –3
(4) I(z) = 0
Ans. (4)
Sol. z = 
55
3 i 3i
22
æ öæö
+-
+
ç ÷ç÷
ç ÷ç÷
è øèø
z = 
( ) ( )
55
i /6 i /6
ee
p -p
+
  = 
i5 /6 i5 /6
ee
p -p
+
  = cos
5sin5 55
i cos isin
6 6 66
p p -p -p æö æö
+ ++
ç÷ ç÷
èø èø
  = 2 cos
5
6
p
 < 0
I(z) = 0  and Re(z) < 0
Option (4)
2. Let a
1
,a
2
,a
3
, ...., a
10
 be in G.P. with a
i
 > 0 for
i = 1,2,...., 10 and S be the set of pairs (r,k),
r kÎN (the set of natural numbers) for which
r k r k rk
e 1 2 e2 3 e3 4
r k rk rk
e 4 5 e5 6 e6 7
rk r k rk
e 78 e 8 9 e 910
logaa logaa logaa
logaa logaa logaa 0
logaa logaa logaa
=
Then the number of elements in S, is :
(1) Infinitely many (2) 4
(3) 10 (4) 2
Ans. (1)
Sol. Apply
C
3
 ® C
3
 – C
2
C
2
 ® C
2
 – C
1
We get   D = 0
Option (1)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Thursday 10
th
 JANUARY , 2019)    TIME : 2 : 30 PM To 5 : 30 PM
3. The positive value of l for which the
co-efficient of x
2
 in the expression
10
2
2
xx
x
l æö
+
ç÷
èø
 is 720, is :
(1) 
5
(2) 4
(3) 
22
(4) 3
Ans. (2)
Sol.
()
r
10r
2 10
r 2
x Cx
x
- æö
l æö
ç÷
ç÷
ç÷
èø
èø
10r
2 10 r 2r
2
r
x C (x) ( ) (x)
-
-
éù
l
êú
ëû
10 5r
210r
2
r
x Cx
-
éù
l
êú
ëû
\  r = 2
Hence, 
10
C
2
 l
2
 = 720
l
2
 = 16
l = ±4
Option (2)
4. The value of 
2 3 10 10
cos cos ..... cos sin
2 2 22
p p pp
× ×××
is :
(1) 
256
1
(2) 
1
2
(3) 
1
512
(4) 
1
1024
Ans. (3)
Sol.
10 10 2
2sin cos cos
2 22
p pp
¼¼
9
1
sin
22
p
= 
1
512
Option (3)
E 2
JEE (Main) Examination–2019/Evening Session/10-01-2019
5. The value of 
/2
/2
dx
,
[x] [sinx] 4
p
-p
++
ò
where [t]
denotes the greatest integer less than or equal
to t, is :
(1) ( )
1
75
12
p+
(2) ( )
3
43
10
p-
(3) ( )
1
75
12
p-
(4) ( )
3
43
20
p-
Ans. (4)
Sol. I = 
2
2
dx
[x] [sin x] 4
p
-p
++
ò
 
10
1
2
dx dx
2 14 1 14
-
-p -
=+
- -+ - -+
òò
        
1
2
01
dx dx
004 1 0 4
p
++
+ + ++
òò
1 01 2
1 01
2
dx dx dx dx
1 245
p
-
-p -
+ ++
ò ò òò
1 11
1 (01)1
2 2 4 52
pp æ ö æö
-+ + +++-
ç ÷ ç÷
è ø èø
1 11
1
2 4 5 2 10
pp
-+ + - ++
20 10 5 4 6
20 10
-+ +-p
+
93
205
-p
+
Option (4)
6. If the probability of hitting a target by a shooter,
in any shot, is 1/3, then the minimum number
of independent shots at the target required by
him so that the probability of hitting the target
at least once is greater than 
5
6
, is :
(1) 6 (2) 5
(3) 4 (4) 3
Ans. (2)
Sol.
0n
n
0
1 25
1C
3 36
æ öæö
->
ç ÷ç÷
è øèø
n
12
63
æö
>
ç÷
èø
 Þ  0.1666 > 
n
2
3
æö
ç÷
èø
n
min
 = 5  Þ  Option (2)
7. If mean and standard deviation of
5 observations x
1
, x
2
, x
3
,x
4
,x
5
 are 10 and 3,
respectively, then the variance of 6
observations x
1
,x
2
, ....,x
5
 and –50 is equal to :
(1) 582.5 (2) 507.5
(3) 586.5 (4) 509.5
Ans. (2)
Sol.
5
i
i1
x 10 x 50
=
= Þ=
å
5
2
i
2 i1
x
S.D. (x)8
5
=
= -=
å
Þ
5
2
i
i1
(x)
=
å
= 109
variance =
5
22
i 5
i i1
i1
(x ) ( 50)
x 50
66
=
=
+-
æ -ö
-
ç÷
èø
å
å
       = 507.5
Option (2)
8. The length of the chord of the parabola x
2
 = 4y
having equationx 2y 4 20 - += is :
(1) 
2 11
(2) 
32
(3) 
63
(4) 
82
Ans. (3)
Page 3


E 1
JEE (Main) Examination–2019/Evening Session/10-01-2019
1. Let z = 
55
3 i 3i
2 2 22
æ öæö
+ +-
ç ÷ç÷
ç ÷ç÷
è øèø
. If R(z) and I[z]
respectively denote the real and imaginary parts
of z, then :
(1) R(z) > 0 and I(z) > 0
(2) R(z) < 0 and I(z) > 0
(3) R(z) = –3
(4) I(z) = 0
Ans. (4)
Sol. z = 
55
3 i 3i
22
æ öæö
+-
+
ç ÷ç÷
ç ÷ç÷
è øèø
z = 
( ) ( )
55
i /6 i /6
ee
p -p
+
  = 
i5 /6 i5 /6
ee
p -p
+
  = cos
5sin5 55
i cos isin
6 6 66
p p -p -p æö æö
+ ++
ç÷ ç÷
èø èø
  = 2 cos
5
6
p
 < 0
I(z) = 0  and Re(z) < 0
Option (4)
2. Let a
1
,a
2
,a
3
, ...., a
10
 be in G.P. with a
i
 > 0 for
i = 1,2,...., 10 and S be the set of pairs (r,k),
r kÎN (the set of natural numbers) for which
r k r k rk
e 1 2 e2 3 e3 4
r k rk rk
e 4 5 e5 6 e6 7
rk r k rk
e 78 e 8 9 e 910
logaa logaa logaa
logaa logaa logaa 0
logaa logaa logaa
=
Then the number of elements in S, is :
(1) Infinitely many (2) 4
(3) 10 (4) 2
Ans. (1)
Sol. Apply
C
3
 ® C
3
 – C
2
C
2
 ® C
2
 – C
1
We get   D = 0
Option (1)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Thursday 10
th
 JANUARY , 2019)    TIME : 2 : 30 PM To 5 : 30 PM
3. The positive value of l for which the
co-efficient of x
2
 in the expression
10
2
2
xx
x
l æö
+
ç÷
èø
 is 720, is :
(1) 
5
(2) 4
(3) 
22
(4) 3
Ans. (2)
Sol.
()
r
10r
2 10
r 2
x Cx
x
- æö
l æö
ç÷
ç÷
ç÷
èø
èø
10r
2 10 r 2r
2
r
x C (x) ( ) (x)
-
-
éù
l
êú
ëû
10 5r
210r
2
r
x Cx
-
éù
l
êú
ëû
\  r = 2
Hence, 
10
C
2
 l
2
 = 720
l
2
 = 16
l = ±4
Option (2)
4. The value of 
2 3 10 10
cos cos ..... cos sin
2 2 22
p p pp
× ×××
is :
(1) 
256
1
(2) 
1
2
(3) 
1
512
(4) 
1
1024
Ans. (3)
Sol.
10 10 2
2sin cos cos
2 22
p pp
¼¼
9
1
sin
22
p
= 
1
512
Option (3)
E 2
JEE (Main) Examination–2019/Evening Session/10-01-2019
5. The value of 
/2
/2
dx
,
[x] [sinx] 4
p
-p
++
ò
where [t]
denotes the greatest integer less than or equal
to t, is :
(1) ( )
1
75
12
p+
(2) ( )
3
43
10
p-
(3) ( )
1
75
12
p-
(4) ( )
3
43
20
p-
Ans. (4)
Sol. I = 
2
2
dx
[x] [sin x] 4
p
-p
++
ò
 
10
1
2
dx dx
2 14 1 14
-
-p -
=+
- -+ - -+
òò
        
1
2
01
dx dx
004 1 0 4
p
++
+ + ++
òò
1 01 2
1 01
2
dx dx dx dx
1 245
p
-
-p -
+ ++
ò ò òò
1 11
1 (01)1
2 2 4 52
pp æ ö æö
-+ + +++-
ç ÷ ç÷
è ø èø
1 11
1
2 4 5 2 10
pp
-+ + - ++
20 10 5 4 6
20 10
-+ +-p
+
93
205
-p
+
Option (4)
6. If the probability of hitting a target by a shooter,
in any shot, is 1/3, then the minimum number
of independent shots at the target required by
him so that the probability of hitting the target
at least once is greater than 
5
6
, is :
(1) 6 (2) 5
(3) 4 (4) 3
Ans. (2)
Sol.
0n
n
0
1 25
1C
3 36
æ öæö
->
ç ÷ç÷
è øèø
n
12
63
æö
>
ç÷
èø
 Þ  0.1666 > 
n
2
3
æö
ç÷
èø
n
min
 = 5  Þ  Option (2)
7. If mean and standard deviation of
5 observations x
1
, x
2
, x
3
,x
4
,x
5
 are 10 and 3,
respectively, then the variance of 6
observations x
1
,x
2
, ....,x
5
 and –50 is equal to :
(1) 582.5 (2) 507.5
(3) 586.5 (4) 509.5
Ans. (2)
Sol.
5
i
i1
x 10 x 50
=
= Þ=
å
5
2
i
2 i1
x
S.D. (x)8
5
=
= -=
å
Þ
5
2
i
i1
(x)
=
å
= 109
variance =
5
22
i 5
i i1
i1
(x ) ( 50)
x 50
66
=
=
+-
æ -ö
-
ç÷
èø
å
å
       = 507.5
Option (2)
8. The length of the chord of the parabola x
2
 = 4y
having equationx 2y 4 20 - += is :
(1) 
2 11
(2) 
32
(3) 
63
(4) 
82
Ans. (3)
E 3
JEE (Main) Examination–2019/Evening Session/10-01-2019
Sol. x
2
 = 4y
x 2y 4 20 - +=
Solving together we get
2
x 42
x4
2
æö
+
=
ç÷
ç÷
èø
2
2x 4x 162 ++
2
2x 4x 16 20 --=
x
1
 + x
2
 = 2 2 ;     x
1
x
2
 = 
162
2
-
= –16
Similarly,
( )
2
2y 4 2 4y -=
2y
2
 + 32 – 16y = 4y
2y
2
 – 20y + 32 = 0 
y + y = 10
12
y y = 16
12
B(x, y)
22
A
(x, y)
11
l
AB
 = 
22
2 1 21
(x x) (y y) - +-
= 
( )
2
2
2 2 64 (10) 4(16) + +-
8 64 100 64 = + +-
108 6 3 ==
Option (3)
9. Let 
2
2 b1
A b b 1b
1 b2
éù
êú
=+
êú
ëû
 where b > 0. Then the
minimum value of 
det(A)
b
 is :
(1)
3
(2) 
3 -
(3) 23 - (4)23
Ans. (4)
Sol. A = 
2
2 b1
b b 1b
1 b2
éù
êú
+
êú
êú
ëû
 (b > 0)
|A| = 2(2b
2
 + 2 – b
2
) – b(2b – b) + 1 (b
2
 – b
2
 – 1)
|A| = 2(b
2
 + 2) – b
2
 – 1
|A| = b
2
 + 3
|A|3
b
bb
=+
  Þ  
3
b
b
3
2
+
³
3
b 23
b
+³
Option (4)
10. The tangent to the curve,
2
x
y xe = passing
through the point (1,e) also passes through the
point :
(1) 
4
,2e
3
æö
ç÷
èø
(2) (2,3e)
(3) 
5
,2e
3
æö
ç÷
èø
(4) (3,6e)
Ans. (1)
Sol.
2
x
y xe =
( )
22
xx
(1, e)
(1, e)
dy
e·e ·2x e
dx
=+
= 2 · e + e = 3e
T : y – e = 3e (x – 1)
y = 3ex – 3e + e
y = (3e)x – 2e
4
, 2e
3
æö
ç÷
èø
  lies on it
Option (1)
11. The number of values of qÎ(0,p) for which  the
system of linear equations
x + 3y + 7z = 0
–x + 4y + 7z = 0
(sin 3q)x + (cos 2q) y + 2z = 0
has a non-trivial solution, is :
(1) One (2) Three
(3) Four (4) Two
Ans. (4)
Page 4


E 1
JEE (Main) Examination–2019/Evening Session/10-01-2019
1. Let z = 
55
3 i 3i
2 2 22
æ öæö
+ +-
ç ÷ç÷
ç ÷ç÷
è øèø
. If R(z) and I[z]
respectively denote the real and imaginary parts
of z, then :
(1) R(z) > 0 and I(z) > 0
(2) R(z) < 0 and I(z) > 0
(3) R(z) = –3
(4) I(z) = 0
Ans. (4)
Sol. z = 
55
3 i 3i
22
æ öæö
+-
+
ç ÷ç÷
ç ÷ç÷
è øèø
z = 
( ) ( )
55
i /6 i /6
ee
p -p
+
  = 
i5 /6 i5 /6
ee
p -p
+
  = cos
5sin5 55
i cos isin
6 6 66
p p -p -p æö æö
+ ++
ç÷ ç÷
èø èø
  = 2 cos
5
6
p
 < 0
I(z) = 0  and Re(z) < 0
Option (4)
2. Let a
1
,a
2
,a
3
, ...., a
10
 be in G.P. with a
i
 > 0 for
i = 1,2,...., 10 and S be the set of pairs (r,k),
r kÎN (the set of natural numbers) for which
r k r k rk
e 1 2 e2 3 e3 4
r k rk rk
e 4 5 e5 6 e6 7
rk r k rk
e 78 e 8 9 e 910
logaa logaa logaa
logaa logaa logaa 0
logaa logaa logaa
=
Then the number of elements in S, is :
(1) Infinitely many (2) 4
(3) 10 (4) 2
Ans. (1)
Sol. Apply
C
3
 ® C
3
 – C
2
C
2
 ® C
2
 – C
1
We get   D = 0
Option (1)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Thursday 10
th
 JANUARY , 2019)    TIME : 2 : 30 PM To 5 : 30 PM
3. The positive value of l for which the
co-efficient of x
2
 in the expression
10
2
2
xx
x
l æö
+
ç÷
èø
 is 720, is :
(1) 
5
(2) 4
(3) 
22
(4) 3
Ans. (2)
Sol.
()
r
10r
2 10
r 2
x Cx
x
- æö
l æö
ç÷
ç÷
ç÷
èø
èø
10r
2 10 r 2r
2
r
x C (x) ( ) (x)
-
-
éù
l
êú
ëû
10 5r
210r
2
r
x Cx
-
éù
l
êú
ëû
\  r = 2
Hence, 
10
C
2
 l
2
 = 720
l
2
 = 16
l = ±4
Option (2)
4. The value of 
2 3 10 10
cos cos ..... cos sin
2 2 22
p p pp
× ×××
is :
(1) 
256
1
(2) 
1
2
(3) 
1
512
(4) 
1
1024
Ans. (3)
Sol.
10 10 2
2sin cos cos
2 22
p pp
¼¼
9
1
sin
22
p
= 
1
512
Option (3)
E 2
JEE (Main) Examination–2019/Evening Session/10-01-2019
5. The value of 
/2
/2
dx
,
[x] [sinx] 4
p
-p
++
ò
where [t]
denotes the greatest integer less than or equal
to t, is :
(1) ( )
1
75
12
p+
(2) ( )
3
43
10
p-
(3) ( )
1
75
12
p-
(4) ( )
3
43
20
p-
Ans. (4)
Sol. I = 
2
2
dx
[x] [sin x] 4
p
-p
++
ò
 
10
1
2
dx dx
2 14 1 14
-
-p -
=+
- -+ - -+
òò
        
1
2
01
dx dx
004 1 0 4
p
++
+ + ++
òò
1 01 2
1 01
2
dx dx dx dx
1 245
p
-
-p -
+ ++
ò ò òò
1 11
1 (01)1
2 2 4 52
pp æ ö æö
-+ + +++-
ç ÷ ç÷
è ø èø
1 11
1
2 4 5 2 10
pp
-+ + - ++
20 10 5 4 6
20 10
-+ +-p
+
93
205
-p
+
Option (4)
6. If the probability of hitting a target by a shooter,
in any shot, is 1/3, then the minimum number
of independent shots at the target required by
him so that the probability of hitting the target
at least once is greater than 
5
6
, is :
(1) 6 (2) 5
(3) 4 (4) 3
Ans. (2)
Sol.
0n
n
0
1 25
1C
3 36
æ öæö
->
ç ÷ç÷
è øèø
n
12
63
æö
>
ç÷
èø
 Þ  0.1666 > 
n
2
3
æö
ç÷
èø
n
min
 = 5  Þ  Option (2)
7. If mean and standard deviation of
5 observations x
1
, x
2
, x
3
,x
4
,x
5
 are 10 and 3,
respectively, then the variance of 6
observations x
1
,x
2
, ....,x
5
 and –50 is equal to :
(1) 582.5 (2) 507.5
(3) 586.5 (4) 509.5
Ans. (2)
Sol.
5
i
i1
x 10 x 50
=
= Þ=
å
5
2
i
2 i1
x
S.D. (x)8
5
=
= -=
å
Þ
5
2
i
i1
(x)
=
å
= 109
variance =
5
22
i 5
i i1
i1
(x ) ( 50)
x 50
66
=
=
+-
æ -ö
-
ç÷
èø
å
å
       = 507.5
Option (2)
8. The length of the chord of the parabola x
2
 = 4y
having equationx 2y 4 20 - += is :
(1) 
2 11
(2) 
32
(3) 
63
(4) 
82
Ans. (3)
E 3
JEE (Main) Examination–2019/Evening Session/10-01-2019
Sol. x
2
 = 4y
x 2y 4 20 - +=
Solving together we get
2
x 42
x4
2
æö
+
=
ç÷
ç÷
èø
2
2x 4x 162 ++
2
2x 4x 16 20 --=
x
1
 + x
2
 = 2 2 ;     x
1
x
2
 = 
162
2
-
= –16
Similarly,
( )
2
2y 4 2 4y -=
2y
2
 + 32 – 16y = 4y
2y
2
 – 20y + 32 = 0 
y + y = 10
12
y y = 16
12
B(x, y)
22
A
(x, y)
11
l
AB
 = 
22
2 1 21
(x x) (y y) - +-
= 
( )
2
2
2 2 64 (10) 4(16) + +-
8 64 100 64 = + +-
108 6 3 ==
Option (3)
9. Let 
2
2 b1
A b b 1b
1 b2
éù
êú
=+
êú
ëû
 where b > 0. Then the
minimum value of 
det(A)
b
 is :
(1)
3
(2) 
3 -
(3) 23 - (4)23
Ans. (4)
Sol. A = 
2
2 b1
b b 1b
1 b2
éù
êú
+
êú
êú
ëû
 (b > 0)
|A| = 2(2b
2
 + 2 – b
2
) – b(2b – b) + 1 (b
2
 – b
2
 – 1)
|A| = 2(b
2
 + 2) – b
2
 – 1
|A| = b
2
 + 3
|A|3
b
bb
=+
  Þ  
3
b
b
3
2
+
³
3
b 23
b
+³
Option (4)
10. The tangent to the curve,
2
x
y xe = passing
through the point (1,e) also passes through the
point :
(1) 
4
,2e
3
æö
ç÷
èø
(2) (2,3e)
(3) 
5
,2e
3
æö
ç÷
èø
(4) (3,6e)
Ans. (1)
Sol.
2
x
y xe =
( )
22
xx
(1, e)
(1, e)
dy
e·e ·2x e
dx
=+
= 2 · e + e = 3e
T : y – e = 3e (x – 1)
y = 3ex – 3e + e
y = (3e)x – 2e
4
, 2e
3
æö
ç÷
èø
  lies on it
Option (1)
11. The number of values of qÎ(0,p) for which  the
system of linear equations
x + 3y + 7z = 0
–x + 4y + 7z = 0
(sin 3q)x + (cos 2q) y + 2z = 0
has a non-trivial solution, is :
(1) One (2) Three
(3) Four (4) Two
Ans. (4)
E 4
JEE (Main) Examination–2019/Evening Session/10-01-2019
Sol.
1 37
1 4 70
sin3 cos22
-=
qq
(8 – 7 cos 2q) – 3(–2 – 7 sin 3q)
+ 7 (– cos 2q – 4 sin 3q) = 0
14 – 7 cos 2q + 21 sin 3q – 7 cos 2q
   – 28 sin 3q = 0
14 – 7 sin 3q – 14 cos 2q = 0
14 – 7 (3 sin q – 4 sin
3
 q) – 14 (1 – 2 sin
2
 q) = 0
–21 sin q + 28 sin
3
 q + 28 sin
2
 q = 0
7 sin q [–3 + 4 sin
2
 q + 4 sin q] = 0
sin q,
4 sin
2
 q + 6 sin q – 2 sin q – 3 = 0
2 sin q(2 sin q + 3) – 1 (2 sin q + 3) = 0
sin q = 
3
2
-
;  sin q = 
1
2
Hence, 2 solutions in (0, p)
Option (4)
12. If () ()
x1
22
0x
f t dt x t f t dt =+
òò
, then f'(1/2) is :
(1) 
6
25
(2) 
24
25
(3) 
18
25
(4) 
4
5
Ans. (2)
Sol.
x1
22
0x
f(t)dt x t f(t)dt =+
òò
            
1
f'
2
æö
ç÷
èø
 = ?
Differentiate w.r.t. 'x'
f(x) = 2x + 0 – x
2
 f(x)
f(x) =
2
2x
1x +
  Þ  f'(x) = 
2
22
(1 x )2 2x(2x)
(1 x)
+-
+
f'(x) = 
22
22
2x4x2
(1 x)
-+
+
2
13
22
1 48 24 42
f'
25
2 50 25
1
1
16
4
æö æö
-
ç÷ ç÷
æö
èø èø
= = ==
ç÷
èøæö
+
ç÷
èø
Option (2)
13. Let f : (–1,1)®R be a function defined by
f(x) = max
{ }
2
|x|, 1 x. - -- If K be the set of
all points at which f is not differentiable, then
K has exactly :
(1) Three elements (2) One element
(3) Five elements (4) Two elements
Ans. (1)
Sol. f : (–1, 1) ® R
f(x) = max
{ }
2
|x|, 1x - --
(–1, 0) (1, 0)
O
Non-derivable at 3 points in (–1, 1)
Option (1)
14. Let 
()
22
2
yx
S x,y R : 1,
1r 1r
ìü
= Î -=
íý
+-
îþ
 where
r ¹ ±1. Then S represents :
(1) A hyperbola whose eccentricity is
2
,
r1 +
where  0 < r < 1.
(2) An ellipse whose eccentricity is
1
,
r1 +
where r > 1
(3) A hyperbola whose eccentricity is
2
,
1r -
when 0 < r < 1.
(4) An ellipse whose eccentricity is
2
,
r1 +
when r > 1
Page 5


E 1
JEE (Main) Examination–2019/Evening Session/10-01-2019
1. Let z = 
55
3 i 3i
2 2 22
æ öæö
+ +-
ç ÷ç÷
ç ÷ç÷
è øèø
. If R(z) and I[z]
respectively denote the real and imaginary parts
of z, then :
(1) R(z) > 0 and I(z) > 0
(2) R(z) < 0 and I(z) > 0
(3) R(z) = –3
(4) I(z) = 0
Ans. (4)
Sol. z = 
55
3 i 3i
22
æ öæö
+-
+
ç ÷ç÷
ç ÷ç÷
è øèø
z = 
( ) ( )
55
i /6 i /6
ee
p -p
+
  = 
i5 /6 i5 /6
ee
p -p
+
  = cos
5sin5 55
i cos isin
6 6 66
p p -p -p æö æö
+ ++
ç÷ ç÷
èø èø
  = 2 cos
5
6
p
 < 0
I(z) = 0  and Re(z) < 0
Option (4)
2. Let a
1
,a
2
,a
3
, ...., a
10
 be in G.P. with a
i
 > 0 for
i = 1,2,...., 10 and S be the set of pairs (r,k),
r kÎN (the set of natural numbers) for which
r k r k rk
e 1 2 e2 3 e3 4
r k rk rk
e 4 5 e5 6 e6 7
rk r k rk
e 78 e 8 9 e 910
logaa logaa logaa
logaa logaa logaa 0
logaa logaa logaa
=
Then the number of elements in S, is :
(1) Infinitely many (2) 4
(3) 10 (4) 2
Ans. (1)
Sol. Apply
C
3
 ® C
3
 – C
2
C
2
 ® C
2
 – C
1
We get   D = 0
Option (1)
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Thursday 10
th
 JANUARY , 2019)    TIME : 2 : 30 PM To 5 : 30 PM
3. The positive value of l for which the
co-efficient of x
2
 in the expression
10
2
2
xx
x
l æö
+
ç÷
èø
 is 720, is :
(1) 
5
(2) 4
(3) 
22
(4) 3
Ans. (2)
Sol.
()
r
10r
2 10
r 2
x Cx
x
- æö
l æö
ç÷
ç÷
ç÷
èø
èø
10r
2 10 r 2r
2
r
x C (x) ( ) (x)
-
-
éù
l
êú
ëû
10 5r
210r
2
r
x Cx
-
éù
l
êú
ëû
\  r = 2
Hence, 
10
C
2
 l
2
 = 720
l
2
 = 16
l = ±4
Option (2)
4. The value of 
2 3 10 10
cos cos ..... cos sin
2 2 22
p p pp
× ×××
is :
(1) 
256
1
(2) 
1
2
(3) 
1
512
(4) 
1
1024
Ans. (3)
Sol.
10 10 2
2sin cos cos
2 22
p pp
¼¼
9
1
sin
22
p
= 
1
512
Option (3)
E 2
JEE (Main) Examination–2019/Evening Session/10-01-2019
5. The value of 
/2
/2
dx
,
[x] [sinx] 4
p
-p
++
ò
where [t]
denotes the greatest integer less than or equal
to t, is :
(1) ( )
1
75
12
p+
(2) ( )
3
43
10
p-
(3) ( )
1
75
12
p-
(4) ( )
3
43
20
p-
Ans. (4)
Sol. I = 
2
2
dx
[x] [sin x] 4
p
-p
++
ò
 
10
1
2
dx dx
2 14 1 14
-
-p -
=+
- -+ - -+
òò
        
1
2
01
dx dx
004 1 0 4
p
++
+ + ++
òò
1 01 2
1 01
2
dx dx dx dx
1 245
p
-
-p -
+ ++
ò ò òò
1 11
1 (01)1
2 2 4 52
pp æ ö æö
-+ + +++-
ç ÷ ç÷
è ø èø
1 11
1
2 4 5 2 10
pp
-+ + - ++
20 10 5 4 6
20 10
-+ +-p
+
93
205
-p
+
Option (4)
6. If the probability of hitting a target by a shooter,
in any shot, is 1/3, then the minimum number
of independent shots at the target required by
him so that the probability of hitting the target
at least once is greater than 
5
6
, is :
(1) 6 (2) 5
(3) 4 (4) 3
Ans. (2)
Sol.
0n
n
0
1 25
1C
3 36
æ öæö
->
ç ÷ç÷
è øèø
n
12
63
æö
>
ç÷
èø
 Þ  0.1666 > 
n
2
3
æö
ç÷
èø
n
min
 = 5  Þ  Option (2)
7. If mean and standard deviation of
5 observations x
1
, x
2
, x
3
,x
4
,x
5
 are 10 and 3,
respectively, then the variance of 6
observations x
1
,x
2
, ....,x
5
 and –50 is equal to :
(1) 582.5 (2) 507.5
(3) 586.5 (4) 509.5
Ans. (2)
Sol.
5
i
i1
x 10 x 50
=
= Þ=
å
5
2
i
2 i1
x
S.D. (x)8
5
=
= -=
å
Þ
5
2
i
i1
(x)
=
å
= 109
variance =
5
22
i 5
i i1
i1
(x ) ( 50)
x 50
66
=
=
+-
æ -ö
-
ç÷
èø
å
å
       = 507.5
Option (2)
8. The length of the chord of the parabola x
2
 = 4y
having equationx 2y 4 20 - += is :
(1) 
2 11
(2) 
32
(3) 
63
(4) 
82
Ans. (3)
E 3
JEE (Main) Examination–2019/Evening Session/10-01-2019
Sol. x
2
 = 4y
x 2y 4 20 - +=
Solving together we get
2
x 42
x4
2
æö
+
=
ç÷
ç÷
èø
2
2x 4x 162 ++
2
2x 4x 16 20 --=
x
1
 + x
2
 = 2 2 ;     x
1
x
2
 = 
162
2
-
= –16
Similarly,
( )
2
2y 4 2 4y -=
2y
2
 + 32 – 16y = 4y
2y
2
 – 20y + 32 = 0 
y + y = 10
12
y y = 16
12
B(x, y)
22
A
(x, y)
11
l
AB
 = 
22
2 1 21
(x x) (y y) - +-
= 
( )
2
2
2 2 64 (10) 4(16) + +-
8 64 100 64 = + +-
108 6 3 ==
Option (3)
9. Let 
2
2 b1
A b b 1b
1 b2
éù
êú
=+
êú
ëû
 where b > 0. Then the
minimum value of 
det(A)
b
 is :
(1)
3
(2) 
3 -
(3) 23 - (4)23
Ans. (4)
Sol. A = 
2
2 b1
b b 1b
1 b2
éù
êú
+
êú
êú
ëû
 (b > 0)
|A| = 2(2b
2
 + 2 – b
2
) – b(2b – b) + 1 (b
2
 – b
2
 – 1)
|A| = 2(b
2
 + 2) – b
2
 – 1
|A| = b
2
 + 3
|A|3
b
bb
=+
  Þ  
3
b
b
3
2
+
³
3
b 23
b
+³
Option (4)
10. The tangent to the curve,
2
x
y xe = passing
through the point (1,e) also passes through the
point :
(1) 
4
,2e
3
æö
ç÷
èø
(2) (2,3e)
(3) 
5
,2e
3
æö
ç÷
èø
(4) (3,6e)
Ans. (1)
Sol.
2
x
y xe =
( )
22
xx
(1, e)
(1, e)
dy
e·e ·2x e
dx
=+
= 2 · e + e = 3e
T : y – e = 3e (x – 1)
y = 3ex – 3e + e
y = (3e)x – 2e
4
, 2e
3
æö
ç÷
èø
  lies on it
Option (1)
11. The number of values of qÎ(0,p) for which  the
system of linear equations
x + 3y + 7z = 0
–x + 4y + 7z = 0
(sin 3q)x + (cos 2q) y + 2z = 0
has a non-trivial solution, is :
(1) One (2) Three
(3) Four (4) Two
Ans. (4)
E 4
JEE (Main) Examination–2019/Evening Session/10-01-2019
Sol.
1 37
1 4 70
sin3 cos22
-=
qq
(8 – 7 cos 2q) – 3(–2 – 7 sin 3q)
+ 7 (– cos 2q – 4 sin 3q) = 0
14 – 7 cos 2q + 21 sin 3q – 7 cos 2q
   – 28 sin 3q = 0
14 – 7 sin 3q – 14 cos 2q = 0
14 – 7 (3 sin q – 4 sin
3
 q) – 14 (1 – 2 sin
2
 q) = 0
–21 sin q + 28 sin
3
 q + 28 sin
2
 q = 0
7 sin q [–3 + 4 sin
2
 q + 4 sin q] = 0
sin q,
4 sin
2
 q + 6 sin q – 2 sin q – 3 = 0
2 sin q(2 sin q + 3) – 1 (2 sin q + 3) = 0
sin q = 
3
2
-
;  sin q = 
1
2
Hence, 2 solutions in (0, p)
Option (4)
12. If () ()
x1
22
0x
f t dt x t f t dt =+
òò
, then f'(1/2) is :
(1) 
6
25
(2) 
24
25
(3) 
18
25
(4) 
4
5
Ans. (2)
Sol.
x1
22
0x
f(t)dt x t f(t)dt =+
òò
            
1
f'
2
æö
ç÷
èø
 = ?
Differentiate w.r.t. 'x'
f(x) = 2x + 0 – x
2
 f(x)
f(x) =
2
2x
1x +
  Þ  f'(x) = 
2
22
(1 x )2 2x(2x)
(1 x)
+-
+
f'(x) = 
22
22
2x4x2
(1 x)
-+
+
2
13
22
1 48 24 42
f'
25
2 50 25
1
1
16
4
æö æö
-
ç÷ ç÷
æö
èø èø
= = ==
ç÷
èøæö
+
ç÷
èø
Option (2)
13. Let f : (–1,1)®R be a function defined by
f(x) = max
{ }
2
|x|, 1 x. - -- If K be the set of
all points at which f is not differentiable, then
K has exactly :
(1) Three elements (2) One element
(3) Five elements (4) Two elements
Ans. (1)
Sol. f : (–1, 1) ® R
f(x) = max
{ }
2
|x|, 1x - --
(–1, 0) (1, 0)
O
Non-derivable at 3 points in (–1, 1)
Option (1)
14. Let 
()
22
2
yx
S x,y R : 1,
1r 1r
ìü
= Î -=
íý
+-
îþ
 where
r ¹ ±1. Then S represents :
(1) A hyperbola whose eccentricity is
2
,
r1 +
where  0 < r < 1.
(2) An ellipse whose eccentricity is
1
,
r1 +
where r > 1
(3) A hyperbola whose eccentricity is
2
,
1r -
when 0 < r < 1.
(4) An ellipse whose eccentricity is
2
,
r1 +
when r > 1
E 5
JEE (Main) Examination–2019/Evening Session/10-01-2019
Ans. (4)
Sol.
22
yx
1
1 r 1r
-=
+-
for r > 1,    
22
yx
1
1 r r1
+=
+-
r1
e1
r1
- æö
=-
ç÷
+
èø
(r1)(r1)
(r 1)
+ --
=
+
22
r1 r1
==
++
Option (4)
15. If 
{ } ( )
25
50 50 r 50
r 25 r 25
r0
C C KC
-
-
=
×=
å
, then K is
equal to :
(1) 2
25 
– 1 (2) (25)
2
(3) 2
25
(4) 2
24
Ans. (3)
Sol.
25
50 50r
r 25r
r0
C·C
-
-
=
å
=
25
r0
50! (50 r)!
r! (50 r)! (25)! (25 r)!
=
-
´
--
å
= 
25
r0
50! 25!
25! 25! (25 r)! (r!)
=
´
-
å
= 
25
50 25
25r
r0
CC
=
å
= 
( )
25 50
25
2C
\  K = 2
25
Option (3)
16. Let N be the set of natural numbers and two
functions f and g be defined as f,g : N®N
such that : 
()
n1
if nisodd
2
fn
n
if n is even
2
+ æ
ç
=
ç
ç
è
and g(n) = n–(–1)
n
. The fog is :
(1) Both one-one and onto
(2) One-one but not onto
(3) Neither one-one nor onto
(4) onto but not one-one
Ans. (4)
Sol. f(x) = 
n1
  n is odd
2
n / 2 n is even
+ ì
ï
í
ï
î
g(x) = n – (–1)
n
 
n 1 ;  n is odd
n 1 ; n is even
+ ì
í
-
î
f(g(n)) =
n
; n is even
2
n1
; n is odd
2
ì
ï
ï
í
+
ï
ï
î
\  many one but onto
Option (4)
17. The values of l such that sum of the squares
of the roots of the quadratic equation,
x
2
 + (3 – l) x + 2 = l has the least value is :
(1) 2 (2) 
4
9
(3) 
15
8
(4) 1
Ans. (1)
Sol. a + b = l – 3
ab = 2 – l
a
2
 + b
2
 = (a + b)
2
 – 2ab = (l – 3)
2
 – 2(2 – l)
     = l
2
 + 9 – 6l – 4 + 2l
     = l
2
 – 4l + 5
     = (l – 2)
2
 + 1
\  l = 2
Option (1)
18. Two vertices of a triangle are (0,2) and (4,3).
If its orthocentre is at the origin, then its third
vertex lies in which quadrant ?
(1) Fourth
(2) Second
(3) Third
(4) First
Ans. (2)