JEE Main 2020 Mathematics September 6 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 111
Date : 6
th
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 The region represented by {z = x + iy ? C : |z| – Re(z) ? 1} is also given by the
inequality:
{z = x + iy ? C : |z| – Re(z) ? 1}
(1) y
2
 ? 2
?
?
?
?
?
?
?
2
1
x
(2) y
2
 ? x + 
2
1
(3) y
2
 ? 2(x + 1) (4) y
2
 ? x + 1
Sol. 1
{z = x + iy ? C : |z| –Re(z) ? 1}
|z| = 
2 2
x y ?
Re(z) = x
|z| – Re(z) ? 1
? 
2 2
x y ? – x ? 1
? 
2 2
x y ? ? 1 + x
? ?x
2
 + y
2
 ? 1 + x
2
 + 2x
? ?y
2
 ? 2 
1
2
x
? ?
?
? ?
? ?
Q.2 The negation of the Boolean expression p ? (~p ? q) is equivalent to:
(1) p ? ~q (2) ~p ? ~q (3) ~p ? q (4) ~p ? ~q
Sol. 4
p ? (~p ? q)
(p ? ~ p) ? ?(p ? ?q)
t ? ?(p ? ?q)
p ? q
~ (p ? ? ?~ p ? ?q)) = ~ (p ? ?q)
= (~ p) ? ?(~ q)
Q.3 The general solution of the differential equation 
2 2 2 2
y x y x 1 ? ? ? + xy
dx
dy
 = 0 is:
(where C is a constant of integration)
(1) 
2
y 1 ? + 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
? ?
?
1 x 1
1 – x 1
2
2
 + C
(2) 
2
y 1 ? – 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
? ?
?
1 x 1
1 – x 1
2
2
 + C
(3) 
2
y 1 ? + 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
?
? ?
1 – x 1
1 x 1
2
2
 + C
(4) 
2
y 1 ? – 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
?
? ?
1 – x 1
1 x 1
2
2
 + C
Page 2


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 111
Date : 6
th
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 The region represented by {z = x + iy ? C : |z| – Re(z) ? 1} is also given by the
inequality:
{z = x + iy ? C : |z| – Re(z) ? 1}
(1) y
2
 ? 2
?
?
?
?
?
?
?
2
1
x
(2) y
2
 ? x + 
2
1
(3) y
2
 ? 2(x + 1) (4) y
2
 ? x + 1
Sol. 1
{z = x + iy ? C : |z| –Re(z) ? 1}
|z| = 
2 2
x y ?
Re(z) = x
|z| – Re(z) ? 1
? 
2 2
x y ? – x ? 1
? 
2 2
x y ? ? 1 + x
? ?x
2
 + y
2
 ? 1 + x
2
 + 2x
? ?y
2
 ? 2 
1
2
x
? ?
?
? ?
? ?
Q.2 The negation of the Boolean expression p ? (~p ? q) is equivalent to:
(1) p ? ~q (2) ~p ? ~q (3) ~p ? q (4) ~p ? ~q
Sol. 4
p ? (~p ? q)
(p ? ~ p) ? ?(p ? ?q)
t ? ?(p ? ?q)
p ? q
~ (p ? ? ?~ p ? ?q)) = ~ (p ? ?q)
= (~ p) ? ?(~ q)
Q.3 The general solution of the differential equation 
2 2 2 2
y x y x 1 ? ? ? + xy
dx
dy
 = 0 is:
(where C is a constant of integration)
(1) 
2
y 1 ? + 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
? ?
?
1 x 1
1 – x 1
2
2
 + C
(2) 
2
y 1 ? – 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
? ?
?
1 x 1
1 – x 1
2
2
 + C
(3) 
2
y 1 ? + 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
?
? ?
1 – x 1
1 x 1
2
2
 + C
(4) 
2
y 1 ? – 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
?
? ?
1 – x 1
1 x 1
2
2
 + C
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 112
Sol. 3
2 2 2 2
y x y x 1 ? ? ? + xy
dx
dy
 = 0
2 2
1 1 0
dy
( x )( y ) xy
dx
? ? ? ?
2
1 ( x )dx
x
?
= –
2
1
y
y ?
dy
Integrate the equation
2
1 x
x
?
?
dx = –
2
1
y
y ?
? dy
1 + x
2
 = t
2
1 + y
2
 = z
2
2xdx = 2tdt
dx = 
t
x
dt 2ydy = 2zdz
2
1
t.tdt zdx
–
z
t –
?
? ?
2
2
1 1
1
t –
dt
t –
?
?
= – z + c
2
1
1
1
dt
t –
?
? ?
dt = – z + c
t + 
1
2
ln 
1
1
t –
t
? ?
? ?
?
? ?
= – z + c
2
2
2
1 1 1
1
2
1 1
x –
x ln
x
? ?
?
? ?
? ?
? ?
? ?
? ?
= –
2
1 y c ? ?
2 2
1
1 1
2
y x ? ? ? ? ln 
2
2
1 1
1 1
x
x
? ?
? ?
? ?
? ?
? ?
? ?
+ c
Q.4 Let L
1
 be a tangent to the parabola y
2
 = 4(x + 1) and L
2
 be a tangent to the parabola
y
2
 = 8(x + 2) such that L
1
 and L
2
 intersect at right angles. Then L
1
 and L
2
 meet on the
straight line:
(1) x + 2y = 0 (2) x + 2 = 0 (3) 2x + 1 = 0 (4) x + 3 = 0
Sol. 4
Let t
1 
tangent of y
2
 = 4(x + 1)
L
1 
: t
1
y = (x + 1) + t
1
2
 .......(i)
and t
2
 tangent of y
2
 = 8 (x + 2)
L
2
 : t
2
y = (x + 2) + 2 t
2
2
L
1
 
?
L
2
Page 3


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 111
Date : 6
th
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 The region represented by {z = x + iy ? C : |z| – Re(z) ? 1} is also given by the
inequality:
{z = x + iy ? C : |z| – Re(z) ? 1}
(1) y
2
 ? 2
?
?
?
?
?
?
?
2
1
x
(2) y
2
 ? x + 
2
1
(3) y
2
 ? 2(x + 1) (4) y
2
 ? x + 1
Sol. 1
{z = x + iy ? C : |z| –Re(z) ? 1}
|z| = 
2 2
x y ?
Re(z) = x
|z| – Re(z) ? 1
? 
2 2
x y ? – x ? 1
? 
2 2
x y ? ? 1 + x
? ?x
2
 + y
2
 ? 1 + x
2
 + 2x
? ?y
2
 ? 2 
1
2
x
? ?
?
? ?
? ?
Q.2 The negation of the Boolean expression p ? (~p ? q) is equivalent to:
(1) p ? ~q (2) ~p ? ~q (3) ~p ? q (4) ~p ? ~q
Sol. 4
p ? (~p ? q)
(p ? ~ p) ? ?(p ? ?q)
t ? ?(p ? ?q)
p ? q
~ (p ? ? ?~ p ? ?q)) = ~ (p ? ?q)
= (~ p) ? ?(~ q)
Q.3 The general solution of the differential equation 
2 2 2 2
y x y x 1 ? ? ? + xy
dx
dy
 = 0 is:
(where C is a constant of integration)
(1) 
2
y 1 ? + 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
? ?
?
1 x 1
1 – x 1
2
2
 + C
(2) 
2
y 1 ? – 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
? ?
?
1 x 1
1 – x 1
2
2
 + C
(3) 
2
y 1 ? + 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
?
? ?
1 – x 1
1 x 1
2
2
 + C
(4) 
2
y 1 ? – 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
?
? ?
1 – x 1
1 x 1
2
2
 + C
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 112
Sol. 3
2 2 2 2
y x y x 1 ? ? ? + xy
dx
dy
 = 0
2 2
1 1 0
dy
( x )( y ) xy
dx
? ? ? ?
2
1 ( x )dx
x
?
= –
2
1
y
y ?
dy
Integrate the equation
2
1 x
x
?
?
dx = –
2
1
y
y ?
? dy
1 + x
2
 = t
2
1 + y
2
 = z
2
2xdx = 2tdt
dx = 
t
x
dt 2ydy = 2zdz
2
1
t.tdt zdx
–
z
t –
?
? ?
2
2
1 1
1
t –
dt
t –
?
?
= – z + c
2
1
1
1
dt
t –
?
? ?
dt = – z + c
t + 
1
2
ln 
1
1
t –
t
? ?
? ?
?
? ?
= – z + c
2
2
2
1 1 1
1
2
1 1
x –
x ln
x
? ?
?
? ?
? ?
? ?
? ?
? ?
= –
2
1 y c ? ?
2 2
1
1 1
2
y x ? ? ? ? ln 
2
2
1 1
1 1
x
x
? ?
? ?
? ?
? ?
? ?
? ?
+ c
Q.4 Let L
1
 be a tangent to the parabola y
2
 = 4(x + 1) and L
2
 be a tangent to the parabola
y
2
 = 8(x + 2) such that L
1
 and L
2
 intersect at right angles. Then L
1
 and L
2
 meet on the
straight line:
(1) x + 2y = 0 (2) x + 2 = 0 (3) 2x + 1 = 0 (4) x + 3 = 0
Sol. 4
Let t
1 
tangent of y
2
 = 4(x + 1)
L
1 
: t
1
y = (x + 1) + t
1
2
 .......(i)
and t
2
 tangent of y
2
 = 8 (x + 2)
L
2
 : t
2
y = (x + 2) + 2 t
2
2
L
1
 
?
L
2
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 113
1 2
1 1
.
t t
= –1
t
1
t
2
 = –1
t
2
(i) – t
1
 (ii)
t
1
t
2
y = t
2
 (x + 1) + t
2
. t
1
2
t
1
t
2
y = t
1
 (x + 2) + 2t
2
2
. t
1
–     –            –
_______________________
(t
2
–t
1
) x + (t
2
–2t
1
) + t
2
t
1
(t
1
–2t
2
) = 0
(t
2
 – t
1
) x + (t
2
 – 2t
1
) – (t
1
 – 2t
2
) = 0
(t
2
 – t
1
) x + 3t
2
 – 3t
1
 = 0
? x + 3 = 0
Q.5 The area (in sq. units) of the region A = {(x, y): |x| + |y| ? 1, 2y
2
 ? |x|}
(1) 
6
1
(2) 
6
5
(3) 
3
1
(4) 
6
7
Sol. 2
(–1/2, 1/2)
(1/2, 1/2)
(1/2, –1/2)
(–1/2,–1/2)
(0, 1)
(–1,0)
(1,0)
(0,–1)
Total area = 
1 2
0
4 1
2
/
x
( – x)– dx
? ?
? ?
? ? ? ?
? ?
? ?
? ? ? ?
?
= 4 
2 3 2
1 2 1
2 3 2 0
2
/
/ x x
x– –
/
? ?
? ?
? ?
? ?
= 4 
3 2
1 1 2 1
2 8 3 2
/
– –
? ?
? ?
? ?
? ?
? ? ? ?
? ?
=  
5 5
4
24 6
? ?
Q.6 The shortest distance between the lines 
0
1 – x
 = 
1 –
1 y ?
 = 
1
z
 and x + y + z + 1 = 0,
2x – y + z + 3 = 0 is:
(1) 1 (2) 
2
1
(3) 
3
1
(4) 
2
1
Page 4


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 111
Date : 6
th
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 The region represented by {z = x + iy ? C : |z| – Re(z) ? 1} is also given by the
inequality:
{z = x + iy ? C : |z| – Re(z) ? 1}
(1) y
2
 ? 2
?
?
?
?
?
?
?
2
1
x
(2) y
2
 ? x + 
2
1
(3) y
2
 ? 2(x + 1) (4) y
2
 ? x + 1
Sol. 1
{z = x + iy ? C : |z| –Re(z) ? 1}
|z| = 
2 2
x y ?
Re(z) = x
|z| – Re(z) ? 1
? 
2 2
x y ? – x ? 1
? 
2 2
x y ? ? 1 + x
? ?x
2
 + y
2
 ? 1 + x
2
 + 2x
? ?y
2
 ? 2 
1
2
x
? ?
?
? ?
? ?
Q.2 The negation of the Boolean expression p ? (~p ? q) is equivalent to:
(1) p ? ~q (2) ~p ? ~q (3) ~p ? q (4) ~p ? ~q
Sol. 4
p ? (~p ? q)
(p ? ~ p) ? ?(p ? ?q)
t ? ?(p ? ?q)
p ? q
~ (p ? ? ?~ p ? ?q)) = ~ (p ? ?q)
= (~ p) ? ?(~ q)
Q.3 The general solution of the differential equation 
2 2 2 2
y x y x 1 ? ? ? + xy
dx
dy
 = 0 is:
(where C is a constant of integration)
(1) 
2
y 1 ? + 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
? ?
?
1 x 1
1 – x 1
2
2
 + C
(2) 
2
y 1 ? – 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
? ?
?
1 x 1
1 – x 1
2
2
 + C
(3) 
2
y 1 ? + 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
?
? ?
1 – x 1
1 x 1
2
2
 + C
(4) 
2
y 1 ? – 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
?
? ?
1 – x 1
1 x 1
2
2
 + C
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 112
Sol. 3
2 2 2 2
y x y x 1 ? ? ? + xy
dx
dy
 = 0
2 2
1 1 0
dy
( x )( y ) xy
dx
? ? ? ?
2
1 ( x )dx
x
?
= –
2
1
y
y ?
dy
Integrate the equation
2
1 x
x
?
?
dx = –
2
1
y
y ?
? dy
1 + x
2
 = t
2
1 + y
2
 = z
2
2xdx = 2tdt
dx = 
t
x
dt 2ydy = 2zdz
2
1
t.tdt zdx
–
z
t –
?
? ?
2
2
1 1
1
t –
dt
t –
?
?
= – z + c
2
1
1
1
dt
t –
?
? ?
dt = – z + c
t + 
1
2
ln 
1
1
t –
t
? ?
? ?
?
? ?
= – z + c
2
2
2
1 1 1
1
2
1 1
x –
x ln
x
? ?
?
? ?
? ?
? ?
? ?
? ?
= –
2
1 y c ? ?
2 2
1
1 1
2
y x ? ? ? ? ln 
2
2
1 1
1 1
x
x
? ?
? ?
? ?
? ?
? ?
? ?
+ c
Q.4 Let L
1
 be a tangent to the parabola y
2
 = 4(x + 1) and L
2
 be a tangent to the parabola
y
2
 = 8(x + 2) such that L
1
 and L
2
 intersect at right angles. Then L
1
 and L
2
 meet on the
straight line:
(1) x + 2y = 0 (2) x + 2 = 0 (3) 2x + 1 = 0 (4) x + 3 = 0
Sol. 4
Let t
1 
tangent of y
2
 = 4(x + 1)
L
1 
: t
1
y = (x + 1) + t
1
2
 .......(i)
and t
2
 tangent of y
2
 = 8 (x + 2)
L
2
 : t
2
y = (x + 2) + 2 t
2
2
L
1
 
?
L
2
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 113
1 2
1 1
.
t t
= –1
t
1
t
2
 = –1
t
2
(i) – t
1
 (ii)
t
1
t
2
y = t
2
 (x + 1) + t
2
. t
1
2
t
1
t
2
y = t
1
 (x + 2) + 2t
2
2
. t
1
–     –            –
_______________________
(t
2
–t
1
) x + (t
2
–2t
1
) + t
2
t
1
(t
1
–2t
2
) = 0
(t
2
 – t
1
) x + (t
2
 – 2t
1
) – (t
1
 – 2t
2
) = 0
(t
2
 – t
1
) x + 3t
2
 – 3t
1
 = 0
? x + 3 = 0
Q.5 The area (in sq. units) of the region A = {(x, y): |x| + |y| ? 1, 2y
2
 ? |x|}
(1) 
6
1
(2) 
6
5
(3) 
3
1
(4) 
6
7
Sol. 2
(–1/2, 1/2)
(1/2, 1/2)
(1/2, –1/2)
(–1/2,–1/2)
(0, 1)
(–1,0)
(1,0)
(0,–1)
Total area = 
1 2
0
4 1
2
/
x
( – x)– dx
? ?
? ?
? ? ? ?
? ?
? ?
? ? ? ?
?
= 4 
2 3 2
1 2 1
2 3 2 0
2
/
/ x x
x– –
/
? ?
? ?
? ?
? ?
= 4 
3 2
1 1 2 1
2 8 3 2
/
– –
? ?
? ?
? ?
? ?
? ? ? ?
? ?
=  
5 5
4
24 6
? ?
Q.6 The shortest distance between the lines 
0
1 – x
 = 
1 –
1 y ?
 = 
1
z
 and x + y + z + 1 = 0,
2x – y + z + 3 = 0 is:
(1) 1 (2) 
2
1
(3) 
3
1
(4) 
2
1
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 114
Sol. 3
Plane through line of intersection is
x + y + z + 1 + ? (2x –y + z + 3) = 0
It should be parallel to given line
0(1 + 2 ?) - 1(1 - ?) + 1(1 + ?) = 0 ? ? = 0
Plane: x + y + z + 1 = 0
Shortest distance of (1, –1, 0) from this plane
= 
2 2 2
1 1 0 1 1
3
1 1 1
| – | ? ?
?
? ?
Q.7 Let a, b, c, d and p be any non zero distinct real numbers such that
(a
2
 + b
2
 + c
2
)p
2
 – 2(ab + bc + cd)p + (b
2
 + c
2
 + d
2
) = 0. Then:
(1) a, c, p are in G.P. (2) a, b, c, d are in G.P.
(3) a, b, c, d are in A.P. (4) a, c, p are in A.P.
Sol. 2
(a
2
 + b
2
 + c
2
)p
2
 - 2 (ab + bc + cd)p + (b
2
 + c
2
 + d
2
) = 0
(a
2
p
2
 - 2abp + b
2
] + [b
2
p
2
 - 2bcp + c
2
] + [ c
2
p
2
 - 2cdp + d
2
] = 0
(ap - b)
2
 + (bp - c)
2
 + (cp - d)
2
 = 0
ap = b
b c d
p
a b c
? ? ?
bp = c
cp = d a, b, c, d are in G.P.
Q.8 Two families with three members each and one family with four members are to be seated
in a row. In how many ways can they be seated so that the same family members are not
separated?
(1) 2! 3! 4! (2) (3!)
3
?(4!) (3) 3! (4!)
3
(4) (3!)
2
?(4!)
Sol. 2
Total numbers in three familes = 3 + 3 + 4 = 10
so total arrangement = 10!
Family 1
3
Family 2
3
Family 3
4
Favourable cases
Arrangement of 3 Families
Interval Arrangement of families memebers
3! 3! 3! 4! ? ? ? ?
= (3!)
3
?(4!)
Q.9 The values of ? and µ for which the system of linear equations
x + y + z = 2
x + 2y + 3z = 5
x + 3y + ?z = µ
has infinitely many solutions are, respectively:
(1) 6 and 8 (2) 5 and 8 (3) 5 and 7 (4) 4 and 9
Page 5


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 111
Date : 6
th
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 The region represented by {z = x + iy ? C : |z| – Re(z) ? 1} is also given by the
inequality:
{z = x + iy ? C : |z| – Re(z) ? 1}
(1) y
2
 ? 2
?
?
?
?
?
?
?
2
1
x
(2) y
2
 ? x + 
2
1
(3) y
2
 ? 2(x + 1) (4) y
2
 ? x + 1
Sol. 1
{z = x + iy ? C : |z| –Re(z) ? 1}
|z| = 
2 2
x y ?
Re(z) = x
|z| – Re(z) ? 1
? 
2 2
x y ? – x ? 1
? 
2 2
x y ? ? 1 + x
? ?x
2
 + y
2
 ? 1 + x
2
 + 2x
? ?y
2
 ? 2 
1
2
x
? ?
?
? ?
? ?
Q.2 The negation of the Boolean expression p ? (~p ? q) is equivalent to:
(1) p ? ~q (2) ~p ? ~q (3) ~p ? q (4) ~p ? ~q
Sol. 4
p ? (~p ? q)
(p ? ~ p) ? ?(p ? ?q)
t ? ?(p ? ?q)
p ? q
~ (p ? ? ?~ p ? ?q)) = ~ (p ? ?q)
= (~ p) ? ?(~ q)
Q.3 The general solution of the differential equation 
2 2 2 2
y x y x 1 ? ? ? + xy
dx
dy
 = 0 is:
(where C is a constant of integration)
(1) 
2
y 1 ? + 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
? ?
?
1 x 1
1 – x 1
2
2
 + C
(2) 
2
y 1 ? – 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
? ?
?
1 x 1
1 – x 1
2
2
 + C
(3) 
2
y 1 ? + 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
?
? ?
1 – x 1
1 x 1
2
2
 + C
(4) 
2
y 1 ? – 
2
x 1 ?
 = 
2
1
log
e
?
?
?
?
?
?
?
?
?
? ?
1 – x 1
1 x 1
2
2
 + C
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 112
Sol. 3
2 2 2 2
y x y x 1 ? ? ? + xy
dx
dy
 = 0
2 2
1 1 0
dy
( x )( y ) xy
dx
? ? ? ?
2
1 ( x )dx
x
?
= –
2
1
y
y ?
dy
Integrate the equation
2
1 x
x
?
?
dx = –
2
1
y
y ?
? dy
1 + x
2
 = t
2
1 + y
2
 = z
2
2xdx = 2tdt
dx = 
t
x
dt 2ydy = 2zdz
2
1
t.tdt zdx
–
z
t –
?
? ?
2
2
1 1
1
t –
dt
t –
?
?
= – z + c
2
1
1
1
dt
t –
?
? ?
dt = – z + c
t + 
1
2
ln 
1
1
t –
t
? ?
? ?
?
? ?
= – z + c
2
2
2
1 1 1
1
2
1 1
x –
x ln
x
? ?
?
? ?
? ?
? ?
? ?
? ?
= –
2
1 y c ? ?
2 2
1
1 1
2
y x ? ? ? ? ln 
2
2
1 1
1 1
x
x
? ?
? ?
? ?
? ?
? ?
? ?
+ c
Q.4 Let L
1
 be a tangent to the parabola y
2
 = 4(x + 1) and L
2
 be a tangent to the parabola
y
2
 = 8(x + 2) such that L
1
 and L
2
 intersect at right angles. Then L
1
 and L
2
 meet on the
straight line:
(1) x + 2y = 0 (2) x + 2 = 0 (3) 2x + 1 = 0 (4) x + 3 = 0
Sol. 4
Let t
1 
tangent of y
2
 = 4(x + 1)
L
1 
: t
1
y = (x + 1) + t
1
2
 .......(i)
and t
2
 tangent of y
2
 = 8 (x + 2)
L
2
 : t
2
y = (x + 2) + 2 t
2
2
L
1
 
?
L
2
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 113
1 2
1 1
.
t t
= –1
t
1
t
2
 = –1
t
2
(i) – t
1
 (ii)
t
1
t
2
y = t
2
 (x + 1) + t
2
. t
1
2
t
1
t
2
y = t
1
 (x + 2) + 2t
2
2
. t
1
–     –            –
_______________________
(t
2
–t
1
) x + (t
2
–2t
1
) + t
2
t
1
(t
1
–2t
2
) = 0
(t
2
 – t
1
) x + (t
2
 – 2t
1
) – (t
1
 – 2t
2
) = 0
(t
2
 – t
1
) x + 3t
2
 – 3t
1
 = 0
? x + 3 = 0
Q.5 The area (in sq. units) of the region A = {(x, y): |x| + |y| ? 1, 2y
2
 ? |x|}
(1) 
6
1
(2) 
6
5
(3) 
3
1
(4) 
6
7
Sol. 2
(–1/2, 1/2)
(1/2, 1/2)
(1/2, –1/2)
(–1/2,–1/2)
(0, 1)
(–1,0)
(1,0)
(0,–1)
Total area = 
1 2
0
4 1
2
/
x
( – x)– dx
? ?
? ?
? ? ? ?
? ?
? ?
? ? ? ?
?
= 4 
2 3 2
1 2 1
2 3 2 0
2
/
/ x x
x– –
/
? ?
? ?
? ?
? ?
= 4 
3 2
1 1 2 1
2 8 3 2
/
– –
? ?
? ?
? ?
? ?
? ? ? ?
? ?
=  
5 5
4
24 6
? ?
Q.6 The shortest distance between the lines 
0
1 – x
 = 
1 –
1 y ?
 = 
1
z
 and x + y + z + 1 = 0,
2x – y + z + 3 = 0 is:
(1) 1 (2) 
2
1
(3) 
3
1
(4) 
2
1
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 114
Sol. 3
Plane through line of intersection is
x + y + z + 1 + ? (2x –y + z + 3) = 0
It should be parallel to given line
0(1 + 2 ?) - 1(1 - ?) + 1(1 + ?) = 0 ? ? = 0
Plane: x + y + z + 1 = 0
Shortest distance of (1, –1, 0) from this plane
= 
2 2 2
1 1 0 1 1
3
1 1 1
| – | ? ?
?
? ?
Q.7 Let a, b, c, d and p be any non zero distinct real numbers such that
(a
2
 + b
2
 + c
2
)p
2
 – 2(ab + bc + cd)p + (b
2
 + c
2
 + d
2
) = 0. Then:
(1) a, c, p are in G.P. (2) a, b, c, d are in G.P.
(3) a, b, c, d are in A.P. (4) a, c, p are in A.P.
Sol. 2
(a
2
 + b
2
 + c
2
)p
2
 - 2 (ab + bc + cd)p + (b
2
 + c
2
 + d
2
) = 0
(a
2
p
2
 - 2abp + b
2
] + [b
2
p
2
 - 2bcp + c
2
] + [ c
2
p
2
 - 2cdp + d
2
] = 0
(ap - b)
2
 + (bp - c)
2
 + (cp - d)
2
 = 0
ap = b
b c d
p
a b c
? ? ?
bp = c
cp = d a, b, c, d are in G.P.
Q.8 Two families with three members each and one family with four members are to be seated
in a row. In how many ways can they be seated so that the same family members are not
separated?
(1) 2! 3! 4! (2) (3!)
3
?(4!) (3) 3! (4!)
3
(4) (3!)
2
?(4!)
Sol. 2
Total numbers in three familes = 3 + 3 + 4 = 10
so total arrangement = 10!
Family 1
3
Family 2
3
Family 3
4
Favourable cases
Arrangement of 3 Families
Interval Arrangement of families memebers
3! 3! 3! 4! ? ? ? ?
= (3!)
3
?(4!)
Q.9 The values of ? and µ for which the system of linear equations
x + y + z = 2
x + 2y + 3z = 5
x + 3y + ?z = µ
has infinitely many solutions are, respectively:
(1) 6 and 8 (2) 5 and 8 (3) 5 and 7 (4) 4 and 9
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Maths     Page | 115
Sol. 2
x + y + z = 2
x + 2y + 3z = 5
x + 3y + ?z = µ
has infinitely many solutions
? = 
1 1 1
1 2 3
1 3 ?
= 0
R
2
 ? R
2
 – R
1
R
3
 ? R
3
 – R
1
1 1 1
0 1 2
0 2 1 – ?
= 0
( ? –1–4) = 0
? ? ? = 5
?
3
 = 
1 1 2
1 2 5
1 3 µ
= 0
R
2
 ? R
2
 – R
1
R
3
 ? R
3
 – R
1
1 1 2
0 1 3
0 2 2 µ–
= 0
(µ – 2–6) = 0
? µ = 8
? = 5, µ = 8
Q.10 Let m and M be respectively the minimum and maximum values of
x 2 sin 1 x sin x cos
x 2 sin x sin x cos 1
x 2 sin x sin 1 x cos
2 2
2 2
2 2
?
?
?
Then the ordered pair (m, M) is equal to:
(1) (–3, –1) (2) (–4, –1) (3) (1, 3) (4) (–3, 3)
Sol. 1
x 2 sin 1 x sin x cos
x 2 sin x sin x cos 1
x 2 sin x sin 1 x cos
2 2
2 2
2 2
?
?
?