JEE Main 2020 Mathematics September 2 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 1
Date : 2
nd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 A line parallel to the straight line 2x-y=0 is tangent to the hyperbola 
2 2
x y
1
4 2
? ? at the
point (x
1
,y
1
). Then 
2 2
1 1
x 5y ? is equal to :
(1) 6 (2) 10 (3) 8 (4) 5
Sol. 1
T : 
1
xx
4
– 
1
yy
2
=1 ....(1)
t : 2x — y = 0 is parallel  to T
? T : 2x — y = ? .......(2)
Now compare (1) & (2)
1
x
4
2
=
1
y
2
1
=
1
?
x
1
=8/ ? & y
1
 = 2/ ?
(x
1
,y
1
) lies on hyperbola ?  
2
64
4 ?
–
2
4
2 ?
=1
? 14 = ?
2
Now
2
1
x +
2
1
5y
=
?
2
64
+5
?
2
4
= 
84
14
= 6 Ans.
Q.2 The domain of the function ? ?
1
2
| x | 5
f x sin
x 1
?
? ? ?
?
? ?
?
? ?
 is ( , a] [a, ). ? ? ? ? ? Then a is equal to :
(1) 
17 1
2
?
(2) 
17
2
(3)  
1 17
2
?
(4) 
17
1
2
?
Sol. 3
–1 < 
2
| x | 5
x 1
?
?
< 1
–x
2
-1 < |x|+5 < x
2
+1
case - I
–x
2
-1 < |x|+5
this inequality is always right 
?
 x ? R
Page 2


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 1
Date : 2
nd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 A line parallel to the straight line 2x-y=0 is tangent to the hyperbola 
2 2
x y
1
4 2
? ? at the
point (x
1
,y
1
). Then 
2 2
1 1
x 5y ? is equal to :
(1) 6 (2) 10 (3) 8 (4) 5
Sol. 1
T : 
1
xx
4
– 
1
yy
2
=1 ....(1)
t : 2x — y = 0 is parallel  to T
? T : 2x — y = ? .......(2)
Now compare (1) & (2)
1
x
4
2
=
1
y
2
1
=
1
?
x
1
=8/ ? & y
1
 = 2/ ?
(x
1
,y
1
) lies on hyperbola ?  
2
64
4 ?
–
2
4
2 ?
=1
? 14 = ?
2
Now
2
1
x +
2
1
5y
=
?
2
64
+5
?
2
4
= 
84
14
= 6 Ans.
Q.2 The domain of the function ? ?
1
2
| x | 5
f x sin
x 1
?
? ? ?
?
? ?
?
? ?
 is ( , a] [a, ). ? ? ? ? ? Then a is equal to :
(1) 
17 1
2
?
(2) 
17
2
(3)  
1 17
2
?
(4) 
17
1
2
?
Sol. 3
–1 < 
2
| x | 5
x 1
?
?
< 1
–x
2
-1 < |x|+5 < x
2
+1
case - I
–x
2
-1 < |x|+5
this inequality is always right 
?
 x ? R
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 2
case - II
|x|+5 < x
2
+1
x
2 
– |x| > 4
|x|
2
–|x|–4 >0
1 17
| x |
2
? ? ? ?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
1 17
| x |
2
? ? ? ?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
> 0
|x| <
1 17
2
?
 (Not possible)
or
|x| > 
1 17
2
?
1 17
x ,
2
? ?
? ?
? ? ? ? ?
?
?
? ?
?
1 17
,
2
? ?
?
? ? ?
?
?
? ?
a = 
1 17
2
?
Q.3 If a function f(x) defined by 
? ?
x x
2
2
ae be , 1 x 1
f x cx , 1 x 3
ax 2cx ,3 x 4
?
? ? ? ? ?
?
? ? ?
?
?
? ? ?
?
 be continuous for some a,
b,c
?
R and  f’(0)+f’(2) =e, then the value of a is :
(1)  
2
1
e 3e 13 ? ?
(2)
2
e
e 3e 13 ? ?
(3) 
2
e
e 3e 13 ? ?
(4) 
2
e
e 3e 13 ? ?
Sol. 4
f(x) is continuous
at x=1 ?
at x=3 ? 9c = 9a + 6c ? c=3a
Now f 
’
(0) +f
 ‘
(2) = e
? a – b + 4c = e
? a – e (3a–ae) + 4.3a = e
? a – 3ae + ae
2
 + 12a = e
? 13a – 3ae + ae
2
=e
? 2
e
a
13 3e e
?
? ?
Q.4 The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie
in :
(1) ? ? , 9 3, ? ? ? ? ? ? ?
? ?
(2) ? 3, ? ? ?
?
(3) ? , 9 ? ? ?
?
(4) ? ? , 3 9, ? ? ? ? ? ? ?
? ?
Page 3


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 1
Date : 2
nd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 A line parallel to the straight line 2x-y=0 is tangent to the hyperbola 
2 2
x y
1
4 2
? ? at the
point (x
1
,y
1
). Then 
2 2
1 1
x 5y ? is equal to :
(1) 6 (2) 10 (3) 8 (4) 5
Sol. 1
T : 
1
xx
4
– 
1
yy
2
=1 ....(1)
t : 2x — y = 0 is parallel  to T
? T : 2x — y = ? .......(2)
Now compare (1) & (2)
1
x
4
2
=
1
y
2
1
=
1
?
x
1
=8/ ? & y
1
 = 2/ ?
(x
1
,y
1
) lies on hyperbola ?  
2
64
4 ?
–
2
4
2 ?
=1
? 14 = ?
2
Now
2
1
x +
2
1
5y
=
?
2
64
+5
?
2
4
= 
84
14
= 6 Ans.
Q.2 The domain of the function ? ?
1
2
| x | 5
f x sin
x 1
?
? ? ?
?
? ?
?
? ?
 is ( , a] [a, ). ? ? ? ? ? Then a is equal to :
(1) 
17 1
2
?
(2) 
17
2
(3)  
1 17
2
?
(4) 
17
1
2
?
Sol. 3
–1 < 
2
| x | 5
x 1
?
?
< 1
–x
2
-1 < |x|+5 < x
2
+1
case - I
–x
2
-1 < |x|+5
this inequality is always right 
?
 x ? R
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 2
case - II
|x|+5 < x
2
+1
x
2 
– |x| > 4
|x|
2
–|x|–4 >0
1 17
| x |
2
? ? ? ?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
1 17
| x |
2
? ? ? ?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
> 0
|x| <
1 17
2
?
 (Not possible)
or
|x| > 
1 17
2
?
1 17
x ,
2
? ?
? ?
? ? ? ? ?
?
?
? ?
?
1 17
,
2
? ?
?
? ? ?
?
?
? ?
a = 
1 17
2
?
Q.3 If a function f(x) defined by 
? ?
x x
2
2
ae be , 1 x 1
f x cx , 1 x 3
ax 2cx ,3 x 4
?
? ? ? ? ?
?
? ? ?
?
?
? ? ?
?
 be continuous for some a,
b,c
?
R and  f’(0)+f’(2) =e, then the value of a is :
(1)  
2
1
e 3e 13 ? ?
(2)
2
e
e 3e 13 ? ?
(3) 
2
e
e 3e 13 ? ?
(4) 
2
e
e 3e 13 ? ?
Sol. 4
f(x) is continuous
at x=1 ?
at x=3 ? 9c = 9a + 6c ? c=3a
Now f 
’
(0) +f
 ‘
(2) = e
? a – b + 4c = e
? a – e (3a–ae) + 4.3a = e
? a – 3ae + ae
2
 + 12a = e
? 13a – 3ae + ae
2
=e
? 2
e
a
13 3e e
?
? ?
Q.4 The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie
in :
(1) ? ? , 9 3, ? ? ? ? ? ? ?
? ?
(2) ? 3, ? ? ?
?
(3) ? , 9 ? ? ?
?
(4) ? ? , 3 9, ? ? ? ? ? ? ?
? ?
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 3
Sol. 4
a
.a.ar 27 a 3
r
? ? ?
a
r
+a+ar =S
1
r
+1+r =
S
3
r+
1
r
=
S
3
-1
r+
1
r
>2 or r +
1
2
r
? ?
S
3
3
?
 or 
S
1
3
? ?
S>9 or S<–3
S ? ? , 3 ? ? ? ?
?
? ? 9, ? ?
?
Q.5 If 
 
? ? ? ?
2 2
R x,y :x,y Z,x 3y 8 ? ? ? ? is a relation on the set of integers Z, then the domain
of R
-1
 is :
(1) ? ? 1,0,1 ? (2) ? ? 2, 1,1,2 ? ? (3)  ? ? 0,1 (4) 
? ? 2, 1,0,1,2 ? ?
Sol. 1
3y
2 
< 8 –x
2
R : {(0,1), (0,–1), (1,0),(–1,0), (1,1), (1,-1)
(-1,1), (-1,-1), (2,0), (-2,0),(-2,0),(2,1),(2,-1), (-2,1),(-2,-1)}
? R : {-2,-1,0,1,2} ?{-1,0,-1}
Hence R
-1
 : {-1,0,1} ?{-2,-1,0,1,2}
Q.6 The value of 
3
2 2
1 sin icos
9 9
2 2
1 sin icos
9 9
? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
 is :
(1) 
? ?
1
1 i 3
2
? ? (2) 
? ?
1
1 i 3
2
? (3) 
? ?
1
3 i
2
? ? (4) 
? ?
1
3 i
2
?
Sol. 3
3
2 2
1 sin icos
9 9
2 2
1 sin icos
9 9
? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
= 
3
2 2
1 cos isin
2 9 2 9
2 2
1 cos isin
2 9 2 9
? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
Page 4


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 1
Date : 2
nd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 A line parallel to the straight line 2x-y=0 is tangent to the hyperbola 
2 2
x y
1
4 2
? ? at the
point (x
1
,y
1
). Then 
2 2
1 1
x 5y ? is equal to :
(1) 6 (2) 10 (3) 8 (4) 5
Sol. 1
T : 
1
xx
4
– 
1
yy
2
=1 ....(1)
t : 2x — y = 0 is parallel  to T
? T : 2x — y = ? .......(2)
Now compare (1) & (2)
1
x
4
2
=
1
y
2
1
=
1
?
x
1
=8/ ? & y
1
 = 2/ ?
(x
1
,y
1
) lies on hyperbola ?  
2
64
4 ?
–
2
4
2 ?
=1
? 14 = ?
2
Now
2
1
x +
2
1
5y
=
?
2
64
+5
?
2
4
= 
84
14
= 6 Ans.
Q.2 The domain of the function ? ?
1
2
| x | 5
f x sin
x 1
?
? ? ?
?
? ?
?
? ?
 is ( , a] [a, ). ? ? ? ? ? Then a is equal to :
(1) 
17 1
2
?
(2) 
17
2
(3)  
1 17
2
?
(4) 
17
1
2
?
Sol. 3
–1 < 
2
| x | 5
x 1
?
?
< 1
–x
2
-1 < |x|+5 < x
2
+1
case - I
–x
2
-1 < |x|+5
this inequality is always right 
?
 x ? R
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 2
case - II
|x|+5 < x
2
+1
x
2 
– |x| > 4
|x|
2
–|x|–4 >0
1 17
| x |
2
? ? ? ?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
1 17
| x |
2
? ? ? ?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
> 0
|x| <
1 17
2
?
 (Not possible)
or
|x| > 
1 17
2
?
1 17
x ,
2
? ?
? ?
? ? ? ? ?
?
?
? ?
?
1 17
,
2
? ?
?
? ? ?
?
?
? ?
a = 
1 17
2
?
Q.3 If a function f(x) defined by 
? ?
x x
2
2
ae be , 1 x 1
f x cx , 1 x 3
ax 2cx ,3 x 4
?
? ? ? ? ?
?
? ? ?
?
?
? ? ?
?
 be continuous for some a,
b,c
?
R and  f’(0)+f’(2) =e, then the value of a is :
(1)  
2
1
e 3e 13 ? ?
(2)
2
e
e 3e 13 ? ?
(3) 
2
e
e 3e 13 ? ?
(4) 
2
e
e 3e 13 ? ?
Sol. 4
f(x) is continuous
at x=1 ?
at x=3 ? 9c = 9a + 6c ? c=3a
Now f 
’
(0) +f
 ‘
(2) = e
? a – b + 4c = e
? a – e (3a–ae) + 4.3a = e
? a – 3ae + ae
2
 + 12a = e
? 13a – 3ae + ae
2
=e
? 2
e
a
13 3e e
?
? ?
Q.4 The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie
in :
(1) ? ? , 9 3, ? ? ? ? ? ? ?
? ?
(2) ? 3, ? ? ?
?
(3) ? , 9 ? ? ?
?
(4) ? ? , 3 9, ? ? ? ? ? ? ?
? ?
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 3
Sol. 4
a
.a.ar 27 a 3
r
? ? ?
a
r
+a+ar =S
1
r
+1+r =
S
3
r+
1
r
=
S
3
-1
r+
1
r
>2 or r +
1
2
r
? ?
S
3
3
?
 or 
S
1
3
? ?
S>9 or S<–3
S ? ? , 3 ? ? ? ?
?
? ? 9, ? ?
?
Q.5 If 
 
? ? ? ?
2 2
R x,y :x,y Z,x 3y 8 ? ? ? ? is a relation on the set of integers Z, then the domain
of R
-1
 is :
(1) ? ? 1,0,1 ? (2) ? ? 2, 1,1,2 ? ? (3)  ? ? 0,1 (4) 
? ? 2, 1,0,1,2 ? ?
Sol. 1
3y
2 
< 8 –x
2
R : {(0,1), (0,–1), (1,0),(–1,0), (1,1), (1,-1)
(-1,1), (-1,-1), (2,0), (-2,0),(-2,0),(2,1),(2,-1), (-2,1),(-2,-1)}
? R : {-2,-1,0,1,2} ?{-1,0,-1}
Hence R
-1
 : {-1,0,1} ?{-2,-1,0,1,2}
Q.6 The value of 
3
2 2
1 sin icos
9 9
2 2
1 sin icos
9 9
? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
 is :
(1) 
? ?
1
1 i 3
2
? ? (2) 
? ?
1
1 i 3
2
? (3) 
? ?
1
3 i
2
? ? (4) 
? ?
1
3 i
2
?
Sol. 3
3
2 2
1 sin icos
9 9
2 2
1 sin icos
9 9
? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
= 
3
2 2
1 cos isin
2 9 2 9
2 2
1 cos isin
2 9 2 9
? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 4
= 
3
5 5
1 cos isin
18 18
5 5
1 cos isin
18 18
? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
= 
3
5 5 5
2cos cos isin
36 36 36
5 5 5
2cos cos isin
36 36 36
? ? ? ? ? ? ?
?
? ? ? ?
? ?
? ?
? ? ? ? ? ? ?
?
? ? ? ?
? ? ? ?
= 
3
5
cis
36
5
cis
36
? ? ? ? ?
? ? ? ?
? ?
? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
3
5
i
36
5
–i
36
e
e
? ? ?
? ?
? ?
? ? ?
? ?
? ?
? ?
? ?
?
? ?
? ?
? ?
3
5 5
i i
36 36
e
? ? ? ? ? ?
?
? ? ? ?
? ? ? ?
? ?
? ? ?
? ?
? ?
3
5
2i
36
e
? ? ?
? ?
? ?
? ?
? ? ?
? ?
? ?
30
i
36
e
? ? ?
? ?
? ?
?
5
i
6
e
? ? ?
? ?
? ?
?
= cis 
5
6
? ? ?
? ?
? ?
=
3 i
2 2
? ?
Q.7 Let P(h,k) be a point on the curve y=x
2
+7x+2, nearest to the line, y=3x-3. Then the
equation of the normal to the curve at P is:
(1) x+3y-62=0 (2) x-3y-11=0 (3) x-3y+22=0 (4) x+3y+26=0
Page 5


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 1
Date : 2
nd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Maths
Q.1 A line parallel to the straight line 2x-y=0 is tangent to the hyperbola 
2 2
x y
1
4 2
? ? at the
point (x
1
,y
1
). Then 
2 2
1 1
x 5y ? is equal to :
(1) 6 (2) 10 (3) 8 (4) 5
Sol. 1
T : 
1
xx
4
– 
1
yy
2
=1 ....(1)
t : 2x — y = 0 is parallel  to T
? T : 2x — y = ? .......(2)
Now compare (1) & (2)
1
x
4
2
=
1
y
2
1
=
1
?
x
1
=8/ ? & y
1
 = 2/ ?
(x
1
,y
1
) lies on hyperbola ?  
2
64
4 ?
–
2
4
2 ?
=1
? 14 = ?
2
Now
2
1
x +
2
1
5y
=
?
2
64
+5
?
2
4
= 
84
14
= 6 Ans.
Q.2 The domain of the function ? ?
1
2
| x | 5
f x sin
x 1
?
? ? ?
?
? ?
?
? ?
 is ( , a] [a, ). ? ? ? ? ? Then a is equal to :
(1) 
17 1
2
?
(2) 
17
2
(3)  
1 17
2
?
(4) 
17
1
2
?
Sol. 3
–1 < 
2
| x | 5
x 1
?
?
< 1
–x
2
-1 < |x|+5 < x
2
+1
case - I
–x
2
-1 < |x|+5
this inequality is always right 
?
 x ? R
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 2
case - II
|x|+5 < x
2
+1
x
2 
– |x| > 4
|x|
2
–|x|–4 >0
1 17
| x |
2
? ? ? ?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
1 17
| x |
2
? ? ? ?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
> 0
|x| <
1 17
2
?
 (Not possible)
or
|x| > 
1 17
2
?
1 17
x ,
2
? ?
? ?
? ? ? ? ?
?
?
? ?
?
1 17
,
2
? ?
?
? ? ?
?
?
? ?
a = 
1 17
2
?
Q.3 If a function f(x) defined by 
? ?
x x
2
2
ae be , 1 x 1
f x cx , 1 x 3
ax 2cx ,3 x 4
?
? ? ? ? ?
?
? ? ?
?
?
? ? ?
?
 be continuous for some a,
b,c
?
R and  f’(0)+f’(2) =e, then the value of a is :
(1)  
2
1
e 3e 13 ? ?
(2)
2
e
e 3e 13 ? ?
(3) 
2
e
e 3e 13 ? ?
(4) 
2
e
e 3e 13 ? ?
Sol. 4
f(x) is continuous
at x=1 ?
at x=3 ? 9c = 9a + 6c ? c=3a
Now f 
’
(0) +f
 ‘
(2) = e
? a – b + 4c = e
? a – e (3a–ae) + 4.3a = e
? a – 3ae + ae
2
 + 12a = e
? 13a – 3ae + ae
2
=e
? 2
e
a
13 3e e
?
? ?
Q.4 The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie
in :
(1) ? ? , 9 3, ? ? ? ? ? ? ?
? ?
(2) ? 3, ? ? ?
?
(3) ? , 9 ? ? ?
?
(4) ? ? , 3 9, ? ? ? ? ? ? ?
? ?
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 3
Sol. 4
a
.a.ar 27 a 3
r
? ? ?
a
r
+a+ar =S
1
r
+1+r =
S
3
r+
1
r
=
S
3
-1
r+
1
r
>2 or r +
1
2
r
? ?
S
3
3
?
 or 
S
1
3
? ?
S>9 or S<–3
S ? ? , 3 ? ? ? ?
?
? ? 9, ? ?
?
Q.5 If 
 
? ? ? ?
2 2
R x,y :x,y Z,x 3y 8 ? ? ? ? is a relation on the set of integers Z, then the domain
of R
-1
 is :
(1) ? ? 1,0,1 ? (2) ? ? 2, 1,1,2 ? ? (3)  ? ? 0,1 (4) 
? ? 2, 1,0,1,2 ? ?
Sol. 1
3y
2 
< 8 –x
2
R : {(0,1), (0,–1), (1,0),(–1,0), (1,1), (1,-1)
(-1,1), (-1,-1), (2,0), (-2,0),(-2,0),(2,1),(2,-1), (-2,1),(-2,-1)}
? R : {-2,-1,0,1,2} ?{-1,0,-1}
Hence R
-1
 : {-1,0,1} ?{-2,-1,0,1,2}
Q.6 The value of 
3
2 2
1 sin icos
9 9
2 2
1 sin icos
9 9
? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
 is :
(1) 
? ?
1
1 i 3
2
? ? (2) 
? ?
1
1 i 3
2
? (3) 
? ?
1
3 i
2
? ? (4) 
? ?
1
3 i
2
?
Sol. 3
3
2 2
1 sin icos
9 9
2 2
1 sin icos
9 9
? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
= 
3
2 2
1 cos isin
2 9 2 9
2 2
1 cos isin
2 9 2 9
? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 4
= 
3
5 5
1 cos isin
18 18
5 5
1 cos isin
18 18
? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
= 
3
5 5 5
2cos cos isin
36 36 36
5 5 5
2cos cos isin
36 36 36
? ? ? ? ? ? ?
?
? ? ? ?
? ?
? ?
? ? ? ? ? ? ?
?
? ? ? ?
? ? ? ?
= 
3
5
cis
36
5
cis
36
? ? ? ? ?
? ? ? ?
? ?
? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
3
5
i
36
5
–i
36
e
e
? ? ?
? ?
? ?
? ? ?
? ?
? ?
? ?
? ?
?
? ?
? ?
? ?
3
5 5
i i
36 36
e
? ? ? ? ? ?
?
? ? ? ?
? ? ? ?
? ?
? ? ?
? ?
? ?
3
5
2i
36
e
? ? ?
? ?
? ?
? ?
? ? ?
? ?
? ?
30
i
36
e
? ? ?
? ?
? ?
?
5
i
6
e
? ? ?
? ?
? ?
?
= cis 
5
6
? ? ?
? ?
? ?
=
3 i
2 2
? ?
Q.7 Let P(h,k) be a point on the curve y=x
2
+7x+2, nearest to the line, y=3x-3. Then the
equation of the normal to the curve at P is:
(1) x+3y-62=0 (2) x-3y-11=0 (3) x-3y+22=0 (4) x+3y+26=0
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Maths     Page | 5
Sol. 4
C : y = x
2
+ 7x+2
Let P : (h, k) lies on
C
L
P(h, k)
Curve k = h
2
+7h+2          .......(1)
Now for shortest distance
Slope of tangent line at point P = slope of line L
? ?
L
at P h,k
dy
m
dx
?
? ?
? ?
2
at P h ,k
d
x 7x 2 3
dx
? ? ?
? ?
? ? at P h ,k
2x 7 3 ? ?
2h + 7 = 3
h =-2
from equation (1)
k=-8
P : (-2,-8)
equation of normal to the curve is perpendicular to L : 3x – y = 3
N : x + 3y = ?
?
Pass (-2,-8)
? = -26
N : x + 3y + 26 = 0
Q.8 Let A be a 2×2 real matrix with entries from ? ? 0,1 and | A | 0 ? . Consider the following
two statements:
(P) If  A ? I
2
, then | A | 1 ? ?
(Q) If |A|=1, then tr(A) =2,
where I
2
 denotes 2×2 identity matrix and tr(A) denotes the sum of the diagonal entries
of A. Then:
(1) Both (P) and (Q) are false (2) (P) is true and (Q) is false
(3) Both (P) and (Q) are true (4) (P) is false and (Q) is true
Sol. 4
P : A = 
1 1
0 1
? ?
? ?
? ?
 ? I
2
 & |A| ? 0 & |A| = 1(false)
Q : |A| = 
1 1
0 1
? ?
? ?
? ?
 = 1 then Tr(A) = 2  (true)