JEE Mains 9 January 2019 Question Paper Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

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E
JEE (Main) Examination–2019/Morning Session/09-01-2019
1
1. The area (in sq. units) bounded by the
parabola y = x
2
 – 1, the tangent at the
point (2, 3) to it and the y-axis is :
(1) 
14
3
(2) 
56
3
(3) 
8
3
(4) 
32
3
Ans. (3)
Sol.
A(2,3)
(0,3)
Y-axis
C
0
B(0,–5)
y = (4x – 5)
X –axis
(0,–1)
Equation of tangent at (2,3) on
y = x
2
 – 1, is y = (4x – 5) ....(i)
\ Required shaded area
= ar (DABC) 
-
-+
ò
3
1
y 1 dy
( )() ( )
( )
3
3/2
1
12
.8.2 y1
23
-
= -+
168
8
33
=-=
 (square units)
2. The maximum volume (in cu. m) of the right
circular cone having slant height 3m is :
(1) 
33
 p (2) 6 p
(3) 
23
 p (4) 
4
3
 p
Ans. (3)
Sol.
q
h
r
l = 3(given)
\ h = 3 cos q
r = 3 sin q
Now,
( ) ( )
22
1
V r h 9sin . 3cos
33
p
=p= qq
MATHEMATICS
TEST PAPER OF JEE(MAIN) EXAMINATION – 2019
(Held On Wednesday 09
th
 JANUARY , 2019)    TIME : 9 : 30 AM To 12 : 30 PM
\
=
q
dV
0
d
Þ   
2
sin
3
q=
Also,
q=
ù
ú
q
û
2
2
2
sin
3
dV
d
 = negative
Þ Volume is maximum,
when
2
sin
3
q=
\ max
2
V sin
3
æö
q=
ç÷
ç÷
èø
 = 
23p
 (in cu. m)
3. For x
2
 ¹ np + 1, n Î N (the set of natural
numbers), the integral
( ) ( )
( ) ( )
22
22
2sinx1 sin2x1
x dx
2sinx1 sin2x1
---
-+-
ò
is equal to :
(where c is a constant of integration)
(1) 
2
e
x1
logsecc
2
æö -
+
ç÷
èø
(2) ( )
22
e
1
log sec x 1c
2
-+
(3) 
2
2
e
1 x1
logsecc
22
æö -
+
ç÷
èø
(4) 
2
e
1
log sec(x 1) c
2
-+
Ans. (1)
Sol. Put (x
2
 – 1) = 1
Þ 2xdx = dt
\
1 1 cost
I dt
2 1 cost
-
=
+
ò
1t
tan dt
22
æö
=
ç÷
èø
ò
t
lnsecc
2
=+
2
x1
Ilnsecc
2
- æö
=+
ç÷
èø