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 Page 1


JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 1
Date : 3
rd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 The value of (2.
1
P
0
-3
.2
P
1
+4
.3
P
2
-..... up to 51
th
 term) +(1!–2!+3!-...... up to 51
th
 term) is
equal to:
(1) 1-51(51)! (2) 1+(52)! (3) 1 (4) 1+ (51)!
Sol. 2
2. 
1
P
0
 = 2
3. 
2
P
1
 = 3
4. 
3
P
2
 = 4
( 2 - 3 + 4 - 5 + ........ 52) + ( 1 - 2 + 3 - 4 ...... + 5 1)
= 52 + 1
Q.2 Let P be a point on the parabola, y
2
=12x and N be the foot of the perpendicular drawn
from P on the axis of the parabola. A line is now drawn through the mid-point M of PN,
parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is  
4
3
,
then:
(1) PN=4 (2) MQ=
1
3
(3)  PN=3 (4) MQ=
1
4
Sol. 4
P(3t , 6t)
2
P(3t , 0)
2
N
Q
M
Q (h, 3t) lie on
Parabola
9t
2
 = 12 h
h = 
2
3
4
t
Q = 
2
3
3
4
t
, t
? ?
? ?
? ?
Page 2


JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 1
Date : 3
rd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 The value of (2.
1
P
0
-3
.2
P
1
+4
.3
P
2
-..... up to 51
th
 term) +(1!–2!+3!-...... up to 51
th
 term) is
equal to:
(1) 1-51(51)! (2) 1+(52)! (3) 1 (4) 1+ (51)!
Sol. 2
2. 
1
P
0
 = 2
3. 
2
P
1
 = 3
4. 
3
P
2
 = 4
( 2 - 3 + 4 - 5 + ........ 52) + ( 1 - 2 + 3 - 4 ...... + 5 1)
= 52 + 1
Q.2 Let P be a point on the parabola, y
2
=12x and N be the foot of the perpendicular drawn
from P on the axis of the parabola. A line is now drawn through the mid-point M of PN,
parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is  
4
3
,
then:
(1) PN=4 (2) MQ=
1
3
(3)  PN=3 (4) MQ=
1
4
Sol. 4
P(3t , 6t)
2
P(3t , 0)
2
N
Q
M
Q (h, 3t) lie on
Parabola
9t
2
 = 12 h
h = 
2
3
4
t
Q = 
2
3
3
4
t
, t
? ?
? ?
? ?
JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 2
Equation of NQ
y = 
2
2
3
3
3
4
t
t
t
? ?
?
? ?
? ?
(x - 3t
2
)
y = ? ?
2
2
4
3
3
t
x t
t
?
?
put x = 0
y = ? ?
2
4
3 4
3
t t
t
?
? ?
4t = 
4
3
 
?
  t = 
1
3
PN = 6t = 
1
6 2
3
. ?
M = 
1 1
1 1
3 12
, ,Q ,
? ? ? ?
? ? ? ?
? ? ? ?
MQ = 
1 1 1
3 12 4
? ?
Q.3 If ?=
x 2 2x 3 3x 4
2x 3 3x 4 4x 5
3x 5 5x 8 10x 17
? ? ?
? ? ?
? ? ?
= Ax
3
+Bx
2
+Cx+D, then B+C is equal to:
(1) 1 (2)-1 (3) -3 (4) 9
Sol. 3
? 
2 2 3 3 4
2 3 3 4 4 5
3 5 5 8 10 17
x x x
x x x
x x x
? ? ?
? ? ?
? ? ?
 = Ax
3
+Bx
2
+Cx+D
R
2
 ? R
2
 - R
1
 , R
3
 ? R
3
 - R
2
? 
? ? ? ?
3 2
2 2 3 3 4
1 1 1
2 2 2 6 2
? ? ?
? ? ? ?
x x x
x – x – x – Ax Bx Cx D
x – x – x –
 
? ? ? ?
3 2
2 2 3 3 4
1 2 1 1 1
1 2 6
? ? ?
? ? ? ? ?
x x x
x – x – Ax Bx Cx D
Page 3


JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 1
Date : 3
rd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 The value of (2.
1
P
0
-3
.2
P
1
+4
.3
P
2
-..... up to 51
th
 term) +(1!–2!+3!-...... up to 51
th
 term) is
equal to:
(1) 1-51(51)! (2) 1+(52)! (3) 1 (4) 1+ (51)!
Sol. 2
2. 
1
P
0
 = 2
3. 
2
P
1
 = 3
4. 
3
P
2
 = 4
( 2 - 3 + 4 - 5 + ........ 52) + ( 1 - 2 + 3 - 4 ...... + 5 1)
= 52 + 1
Q.2 Let P be a point on the parabola, y
2
=12x and N be the foot of the perpendicular drawn
from P on the axis of the parabola. A line is now drawn through the mid-point M of PN,
parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is  
4
3
,
then:
(1) PN=4 (2) MQ=
1
3
(3)  PN=3 (4) MQ=
1
4
Sol. 4
P(3t , 6t)
2
P(3t , 0)
2
N
Q
M
Q (h, 3t) lie on
Parabola
9t
2
 = 12 h
h = 
2
3
4
t
Q = 
2
3
3
4
t
, t
? ?
? ?
? ?
JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 2
Equation of NQ
y = 
2
2
3
3
3
4
t
t
t
? ?
?
? ?
? ?
(x - 3t
2
)
y = ? ?
2
2
4
3
3
t
x t
t
?
?
put x = 0
y = ? ?
2
4
3 4
3
t t
t
?
? ?
4t = 
4
3
 
?
  t = 
1
3
PN = 6t = 
1
6 2
3
. ?
M = 
1 1
1 1
3 12
, ,Q ,
? ? ? ?
? ? ? ?
? ? ? ?
MQ = 
1 1 1
3 12 4
? ?
Q.3 If ?=
x 2 2x 3 3x 4
2x 3 3x 4 4x 5
3x 5 5x 8 10x 17
? ? ?
? ? ?
? ? ?
= Ax
3
+Bx
2
+Cx+D, then B+C is equal to:
(1) 1 (2)-1 (3) -3 (4) 9
Sol. 3
? 
2 2 3 3 4
2 3 3 4 4 5
3 5 5 8 10 17
x x x
x x x
x x x
? ? ?
? ? ?
? ? ?
 = Ax
3
+Bx
2
+Cx+D
R
2
 ? R
2
 - R
1
 , R
3
 ? R
3
 - R
2
? 
? ? ? ?
3 2
2 2 3 3 4
1 1 1
2 2 2 6 2
? ? ?
? ? ? ?
x x x
x – x – x – Ax Bx Cx D
x – x – x –
 
? ? ? ?
3 2
2 2 3 3 4
1 2 1 1 1
1 2 6
? ? ?
? ? ? ? ?
x x x
x – x – Ax Bx Cx D
JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 3
   
? (x – 1)(x – 2){(x – 2)(6 – 2) – (2x – 3)(6 – 1) + (3x – 4)(2 – 1)} = Ax
3
 + Bx
2
 + Cx + D
? (x – 1) (x – 2){4x – 8 – 10x + 15 + 3x – 4} = Ax
3
 + Bx
2
 + Cx + D
? (x – 1)(x – 2)(–3x + 3) = Ax
3
 + Bx
2
 + Cx + D
? –3(x – 1)
2
 (x – 2) = Ax
3
 + Bx
2
 + Cx + D
? ?–3x
3
 + 12x
2
 – 15x + 6 =  Ax
3
 + Bx
2
 + Cx + D
comparing both side
A = –3, B = 12, C = –15
B + C = 12 – 15 = –3
Q.4 The foot of the perpendicular drawn from the point (4,2,3) to the line joining the points
(1,-2,3) and (1,1,0) lies on the plane:
(1) x-y-2z=1 (2) x-2y+z=1 (3)2x+y-z=1 (4) x+2y-z=1
Sol. 3
? ? ? ?
1 2 3 0 3 3 r , , , , ? ? ? ? ?
?
M
P
(4,2,3)
(0, 3, - 3) b =
?
?
? ? ? ? ?
PM b
0 ?
? ? ? ? ?
PM. b
(-3, 3 ? - 4, -3 ?). (0, 3, -3) = 0
? 0 + 9 ? - 12 + 9 ? = 0
  
? 
12 2
18 3
? ? ?
m = (1, 0, 1) are on 2x + y - z = 1
Q.5 If y
2
+log
e
(cos
2
x)=y, x , ,
2 2
? ? ? ?
? ?
? ?
? ?
then
(1) |y‘(0)|+|y“(0)|=1 (2) y“(0)=0
(3) |y‘(0)|+|y’’(0)|=3 (4) |y“(0)|=2
Sol. 4
2yy’ + 2 (-tanx) = y’ ....(1)
diff. w.r.t.x
2yy” + 2(y’)
2
 - 2 sec
2
 x = y” ....(2)
Put x = 0 in given equation we get  y = 0, 1
from (1) x = 0, y = 0 ? y’(0) = 0
x = 0, y = 1, ? y’(0) = 0
from (2) x = 0, y = 0, y’(0) = 0 ? y”(0) = -2
x = 0, y = 1, y’(0) = 0 ? y”(0) = 2
|y”(0)| = 2
Page 4


JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 1
Date : 3
rd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 The value of (2.
1
P
0
-3
.2
P
1
+4
.3
P
2
-..... up to 51
th
 term) +(1!–2!+3!-...... up to 51
th
 term) is
equal to:
(1) 1-51(51)! (2) 1+(52)! (3) 1 (4) 1+ (51)!
Sol. 2
2. 
1
P
0
 = 2
3. 
2
P
1
 = 3
4. 
3
P
2
 = 4
( 2 - 3 + 4 - 5 + ........ 52) + ( 1 - 2 + 3 - 4 ...... + 5 1)
= 52 + 1
Q.2 Let P be a point on the parabola, y
2
=12x and N be the foot of the perpendicular drawn
from P on the axis of the parabola. A line is now drawn through the mid-point M of PN,
parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is  
4
3
,
then:
(1) PN=4 (2) MQ=
1
3
(3)  PN=3 (4) MQ=
1
4
Sol. 4
P(3t , 6t)
2
P(3t , 0)
2
N
Q
M
Q (h, 3t) lie on
Parabola
9t
2
 = 12 h
h = 
2
3
4
t
Q = 
2
3
3
4
t
, t
? ?
? ?
? ?
JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 2
Equation of NQ
y = 
2
2
3
3
3
4
t
t
t
? ?
?
? ?
? ?
(x - 3t
2
)
y = ? ?
2
2
4
3
3
t
x t
t
?
?
put x = 0
y = ? ?
2
4
3 4
3
t t
t
?
? ?
4t = 
4
3
 
?
  t = 
1
3
PN = 6t = 
1
6 2
3
. ?
M = 
1 1
1 1
3 12
, ,Q ,
? ? ? ?
? ? ? ?
? ? ? ?
MQ = 
1 1 1
3 12 4
? ?
Q.3 If ?=
x 2 2x 3 3x 4
2x 3 3x 4 4x 5
3x 5 5x 8 10x 17
? ? ?
? ? ?
? ? ?
= Ax
3
+Bx
2
+Cx+D, then B+C is equal to:
(1) 1 (2)-1 (3) -3 (4) 9
Sol. 3
? 
2 2 3 3 4
2 3 3 4 4 5
3 5 5 8 10 17
x x x
x x x
x x x
? ? ?
? ? ?
? ? ?
 = Ax
3
+Bx
2
+Cx+D
R
2
 ? R
2
 - R
1
 , R
3
 ? R
3
 - R
2
? 
? ? ? ?
3 2
2 2 3 3 4
1 1 1
2 2 2 6 2
? ? ?
? ? ? ?
x x x
x – x – x – Ax Bx Cx D
x – x – x –
 
? ? ? ?
3 2
2 2 3 3 4
1 2 1 1 1
1 2 6
? ? ?
? ? ? ? ?
x x x
x – x – Ax Bx Cx D
JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 3
   
? (x – 1)(x – 2){(x – 2)(6 – 2) – (2x – 3)(6 – 1) + (3x – 4)(2 – 1)} = Ax
3
 + Bx
2
 + Cx + D
? (x – 1) (x – 2){4x – 8 – 10x + 15 + 3x – 4} = Ax
3
 + Bx
2
 + Cx + D
? (x – 1)(x – 2)(–3x + 3) = Ax
3
 + Bx
2
 + Cx + D
? –3(x – 1)
2
 (x – 2) = Ax
3
 + Bx
2
 + Cx + D
? ?–3x
3
 + 12x
2
 – 15x + 6 =  Ax
3
 + Bx
2
 + Cx + D
comparing both side
A = –3, B = 12, C = –15
B + C = 12 – 15 = –3
Q.4 The foot of the perpendicular drawn from the point (4,2,3) to the line joining the points
(1,-2,3) and (1,1,0) lies on the plane:
(1) x-y-2z=1 (2) x-2y+z=1 (3)2x+y-z=1 (4) x+2y-z=1
Sol. 3
? ? ? ?
1 2 3 0 3 3 r , , , , ? ? ? ? ?
?
M
P
(4,2,3)
(0, 3, - 3) b =
?
?
? ? ? ? ?
PM b
0 ?
? ? ? ? ?
PM. b
(-3, 3 ? - 4, -3 ?). (0, 3, -3) = 0
? 0 + 9 ? - 12 + 9 ? = 0
  
? 
12 2
18 3
? ? ?
m = (1, 0, 1) are on 2x + y - z = 1
Q.5 If y
2
+log
e
(cos
2
x)=y, x , ,
2 2
? ? ? ?
? ?
? ?
? ?
then
(1) |y‘(0)|+|y“(0)|=1 (2) y“(0)=0
(3) |y‘(0)|+|y’’(0)|=3 (4) |y“(0)|=2
Sol. 4
2yy’ + 2 (-tanx) = y’ ....(1)
diff. w.r.t.x
2yy” + 2(y’)
2
 - 2 sec
2
 x = y” ....(2)
Put x = 0 in given equation we get  y = 0, 1
from (1) x = 0, y = 0 ? y’(0) = 0
x = 0, y = 1, ? y’(0) = 0
from (2) x = 0, y = 0, y’(0) = 0 ? y”(0) = -2
x = 0, y = 1, y’(0) = 0 ? y”(0) = 2
|y”(0)| = 2
JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 4
Q.6
? ? ?
? ?
? ? ? ?
? ?
? ?
1 1 1
4 5 16
2 sin sin sin
5 13 65
is equal to:
(1) 
5
4
?
(2) 
3
2
?
(3) 
7
4
?
(4) 
2
?
Sol. 2
1 1 1
4 5 16
2
3 12 63
? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
tan tan tan
1 1
4 5
16
3 12
2
4 5
63
1
3 12
? ?
? ?
?
? ?
? ?
? ? ? ?
? ?
? ?
? ?
? ?
?
? ?
? ?
tan tan
.
 
1 1
48 15 16
2
36 20 63
? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
?
? ? ? ?
tan tan
 
1 1
63 63
2
16 16
? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
tan cot
?
 
?? ? ? ? 
2
?
3
2
?
?
Q.7 A hyperbola having the transverse axis of length 
2
 has the same foci as that of the
ellipse 3x
2
+4y
2
=12, then this hyperbola does not pass through which of the following
points ?
(1) 
3 1
,
2
2
? ?
? ?
? ?
? ?
(2) 
1
1,
2
? ?
?
? ?
? ?
(3) 
1
,0
2
? ?
? ?
? ?
(4) 
3
,1
2
? ?
? ? ?
? ?
? ?
Sol. 1
2 2
1
4 3
x y
? ?
? ?
2 2 2
1 1 1
1 b a e ? ?
3 = 4(1 - e
1
2
)
e
1
 = 
1
2
focus = (± a
1
e
1
, 0)
= (±1, 0)
Page 5


JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 1
Date : 3
rd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 The value of (2.
1
P
0
-3
.2
P
1
+4
.3
P
2
-..... up to 51
th
 term) +(1!–2!+3!-...... up to 51
th
 term) is
equal to:
(1) 1-51(51)! (2) 1+(52)! (3) 1 (4) 1+ (51)!
Sol. 2
2. 
1
P
0
 = 2
3. 
2
P
1
 = 3
4. 
3
P
2
 = 4
( 2 - 3 + 4 - 5 + ........ 52) + ( 1 - 2 + 3 - 4 ...... + 5 1)
= 52 + 1
Q.2 Let P be a point on the parabola, y
2
=12x and N be the foot of the perpendicular drawn
from P on the axis of the parabola. A line is now drawn through the mid-point M of PN,
parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is  
4
3
,
then:
(1) PN=4 (2) MQ=
1
3
(3)  PN=3 (4) MQ=
1
4
Sol. 4
P(3t , 6t)
2
P(3t , 0)
2
N
Q
M
Q (h, 3t) lie on
Parabola
9t
2
 = 12 h
h = 
2
3
4
t
Q = 
2
3
3
4
t
, t
? ?
? ?
? ?
JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 2
Equation of NQ
y = 
2
2
3
3
3
4
t
t
t
? ?
?
? ?
? ?
(x - 3t
2
)
y = ? ?
2
2
4
3
3
t
x t
t
?
?
put x = 0
y = ? ?
2
4
3 4
3
t t
t
?
? ?
4t = 
4
3
 
?
  t = 
1
3
PN = 6t = 
1
6 2
3
. ?
M = 
1 1
1 1
3 12
, ,Q ,
? ? ? ?
? ? ? ?
? ? ? ?
MQ = 
1 1 1
3 12 4
? ?
Q.3 If ?=
x 2 2x 3 3x 4
2x 3 3x 4 4x 5
3x 5 5x 8 10x 17
? ? ?
? ? ?
? ? ?
= Ax
3
+Bx
2
+Cx+D, then B+C is equal to:
(1) 1 (2)-1 (3) -3 (4) 9
Sol. 3
? 
2 2 3 3 4
2 3 3 4 4 5
3 5 5 8 10 17
x x x
x x x
x x x
? ? ?
? ? ?
? ? ?
 = Ax
3
+Bx
2
+Cx+D
R
2
 ? R
2
 - R
1
 , R
3
 ? R
3
 - R
2
? 
? ? ? ?
3 2
2 2 3 3 4
1 1 1
2 2 2 6 2
? ? ?
? ? ? ?
x x x
x – x – x – Ax Bx Cx D
x – x – x –
 
? ? ? ?
3 2
2 2 3 3 4
1 2 1 1 1
1 2 6
? ? ?
? ? ? ? ?
x x x
x – x – Ax Bx Cx D
JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 3
   
? (x – 1)(x – 2){(x – 2)(6 – 2) – (2x – 3)(6 – 1) + (3x – 4)(2 – 1)} = Ax
3
 + Bx
2
 + Cx + D
? (x – 1) (x – 2){4x – 8 – 10x + 15 + 3x – 4} = Ax
3
 + Bx
2
 + Cx + D
? (x – 1)(x – 2)(–3x + 3) = Ax
3
 + Bx
2
 + Cx + D
? –3(x – 1)
2
 (x – 2) = Ax
3
 + Bx
2
 + Cx + D
? ?–3x
3
 + 12x
2
 – 15x + 6 =  Ax
3
 + Bx
2
 + Cx + D
comparing both side
A = –3, B = 12, C = –15
B + C = 12 – 15 = –3
Q.4 The foot of the perpendicular drawn from the point (4,2,3) to the line joining the points
(1,-2,3) and (1,1,0) lies on the plane:
(1) x-y-2z=1 (2) x-2y+z=1 (3)2x+y-z=1 (4) x+2y-z=1
Sol. 3
? ? ? ?
1 2 3 0 3 3 r , , , , ? ? ? ? ?
?
M
P
(4,2,3)
(0, 3, - 3) b =
?
?
? ? ? ? ?
PM b
0 ?
? ? ? ? ?
PM. b
(-3, 3 ? - 4, -3 ?). (0, 3, -3) = 0
? 0 + 9 ? - 12 + 9 ? = 0
  
? 
12 2
18 3
? ? ?
m = (1, 0, 1) are on 2x + y - z = 1
Q.5 If y
2
+log
e
(cos
2
x)=y, x , ,
2 2
? ? ? ?
? ?
? ?
? ?
then
(1) |y‘(0)|+|y“(0)|=1 (2) y“(0)=0
(3) |y‘(0)|+|y’’(0)|=3 (4) |y“(0)|=2
Sol. 4
2yy’ + 2 (-tanx) = y’ ....(1)
diff. w.r.t.x
2yy” + 2(y’)
2
 - 2 sec
2
 x = y” ....(2)
Put x = 0 in given equation we get  y = 0, 1
from (1) x = 0, y = 0 ? y’(0) = 0
x = 0, y = 1, ? y’(0) = 0
from (2) x = 0, y = 0, y’(0) = 0 ? y”(0) = -2
x = 0, y = 1, y’(0) = 0 ? y”(0) = 2
|y”(0)| = 2
JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 4
Q.6
? ? ?
? ?
? ? ? ?
? ?
? ?
1 1 1
4 5 16
2 sin sin sin
5 13 65
is equal to:
(1) 
5
4
?
(2) 
3
2
?
(3) 
7
4
?
(4) 
2
?
Sol. 2
1 1 1
4 5 16
2
3 12 63
? ? ?
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
tan tan tan
1 1
4 5
16
3 12
2
4 5
63
1
3 12
? ?
? ?
?
? ?
? ?
? ? ? ?
? ?
? ?
? ?
? ?
?
? ?
? ?
tan tan
.
 
1 1
48 15 16
2
36 20 63
? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
?
? ? ? ?
tan tan
 
1 1
63 63
2
16 16
? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
tan cot
?
 
?? ? ? ? 
2
?
3
2
?
?
Q.7 A hyperbola having the transverse axis of length 
2
 has the same foci as that of the
ellipse 3x
2
+4y
2
=12, then this hyperbola does not pass through which of the following
points ?
(1) 
3 1
,
2
2
? ?
? ?
? ?
? ?
(2) 
1
1,
2
? ?
?
? ?
? ?
(3) 
1
,0
2
? ?
? ?
? ?
(4) 
3
,1
2
? ?
? ? ?
? ?
? ?
Sol. 1
2 2
1
4 3
x y
? ?
? ?
2 2 2
1 1 1
1 b a e ? ?
3 = 4(1 - e
1
2
)
e
1
 = 
1
2
focus = (± a
1
e
1
, 0)
= (±1, 0)
JEE Main 2020 Paper
         3
rd
 September 2020 | (Shift-1), Maths     Page | 5
Length of transverse axis 2a
2
 = 
2
 ? a
2
 = 
1
2
a
2
e
2
 = 1
= e
2
 = 
2
b
2
2
 = a
2
2
 (e
2
2 
- 1)
b
2
2
 = ? ?
1 1
2 1
2 2
? ?
equation  of Hyperbola
x
2
 - y
2
 = 
1
2
Q.8 For the frequency distribution:
Variate(x): x
1
x
2
x
3
....x
15
Frequency(f): f
1
f
2
f
3
.....f
15
where 0<x
1
<x
2
<x
3
<....<x
15 
?
 
10 and 
15
i
i 1
f 0,
?
?
? the standard deviation cannot be:
(1) 1 (2) 4 (3) 6 (4) 2
Sol. 3
2 2
1
4
(M m) ? ? ?
(M = upper bound of value of any random variable,
 m =  Lower bound of value of any random variable)
2 2
1
10 0
4
( ) ? ? ?
2
25 ? ?
 - 5 <  ? < 5
? ? 6
Q.9 A die is thrown two times and the sum of the scores appearing on the die is observed to
be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast
once is:
(1) 
1
3
(2) 
1
4
(3) 
1
8
(4) 
1
9
Sol. 4
Total Possibilities = (1, 3), (3, 1), (2, 2),
(2, 6), (6, 2) (4, 4)
(3, 5), (5, 3) (6, 6)
? ? ? ? ? ?
? ?
? ?
? ? ? ?
?
? ?
?
? ?
? ?
A B 4,4 ,n A B 1
P A B
B 1
Required probability P = =
A P A 9
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