Best Study Material for JEE Exam
JEE Exam > JEE Notes > JEE Main & Advanced Previous Year Papers > JEE Main 2020 January 8 Question Paper Shift 2
JEE Main 2020 January 8 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
8
th
January 2020 (Shift 2), Mathematics Page | 1
Date: 8
th
January 2020 (Shift 2)
Time: 2:30 P.M. to 5:30 P.M.
Subject: Mathematics
1. Solution set of 3
?? (3
?? - 1) + 2 = |3
?? - 1| + |3
?? - 2| contains
a. exactly one element b. at least four elements
c. two elements d. infinite elements
Answer: (?? )
Solution:
3
?? (3
?? - 1) + 2 = |3
?? - 1| + |3
?? - 2|
Let 3
?? = ??
?? (?? - 1) + 2 = |?? - 1| + |?? - 2|
? ?? 2
- ?? + 2 = |?? - 1| + |?? - 2|
We plot ?? 2
- ?? + 2 and |?? - 1| + |?? - 2|
As 3
?? is always positive, therefore only positive values of ?? will be the solution.
Therefore, we have only one solution.
2. Which of the following is a tautology?
a. ~(?? ? ~?? ) ? (?? ? ?? ) b. (~?? ? ?? ) ? (?? ? ?? )
c. ~(?? ? ~?? ) ? (?? ? ?? ) d. ~(?? ? ~?? ) ? (?? ? ?? )
Answer: (?? )
Solution:
Page 2
8
th
January 2020 (Shift 2), Mathematics Page | 1
Date: 8
th
January 2020 (Shift 2)
Time: 2:30 P.M. to 5:30 P.M.
Subject: Mathematics
1. Solution set of 3
?? (3
?? - 1) + 2 = |3
?? - 1| + |3
?? - 2| contains
a. exactly one element b. at least four elements
c. two elements d. infinite elements
Answer: (?? )
Solution:
3
?? (3
?? - 1) + 2 = |3
?? - 1| + |3
?? - 2|
Let 3
?? = ??
?? (?? - 1) + 2 = |?? - 1| + |?? - 2|
? ?? 2
- ?? + 2 = |?? - 1| + |?? - 2|
We plot ?? 2
- ?? + 2 and |?? - 1| + |?? - 2|
As 3
?? is always positive, therefore only positive values of ?? will be the solution.
Therefore, we have only one solution.
2. Which of the following is a tautology?
a. ~(?? ? ~?? ) ? (?? ? ?? ) b. (~?? ? ?? ) ? (?? ? ?? )
c. ~(?? ? ~?? ) ? (?? ? ?? ) d. ~(?? ? ~?? ) ? (?? ? ?? )
Answer: (?? )
Solution:
8
th
January 2020 (Shift 2), Mathematics Page | 2
~(?? ? ~?? ) ? (?? ? ?? )
= (?? ? ~?? ) ? (?? ? ?? )
= (?? ? ?? ) ? (?? ? ~?? )
= ?? ? ??
= ??
3. If a hyperbola has vertices (±6, 0) and ?? (10, 16) lies on it, then the equation of normal
at ?? is
a. 2?? + 5?? = 10 b. 2?? + 5?? = 100
c. 2?? - 5?? = 100 d. 5?? + 2?? = 100
Answer: (?? )
Solution:
Vertex of hyperbola is (±?? , 0) = (±6, 0) ? ?? = 6
Let the equation of hyperbola be
?? 2
?? 2
-
?? 2
?? 2
= 1
?
?? 2
36
-
?? 2
?? 2
= 1
As ?? (10, 16) lies on the hyperbola.
100
36
-
256
?? 2
= 1
?
64
36
=
256
?? 2
? ?? 2
= 144
Equation of hyperbola becomes
?? 2
36
-
?? 2
144
= 1
Equation of normal is
?? 2
?? ?? 1
+
?? 2
?? ?? 1
= ?? 2
+ ?? 2
?
36?? 10
+
144?? 16
= 180
?
?? 50
+
?? 20
= 1
? 2?? + 5?? = 100
4. If ?? = ???? + ?? is a tangent to the circle (?? - 3)
2
+ ?? 2
= 1 and also perpendicular to the
tangent to the circle ?? 2
+ ?? 2
= 1 at (
1
v 2
,
1
v 2
), then
a. ?? 2
- 6?? - 7 = 0 b. ?? 2
- 6?? + 7 = 0
c. ?? 2
+ 6?? - 7 = 0 d. ?? 2
+ 6?? + 7 = 0
Answer: (?? )
Page 3
8
th
January 2020 (Shift 2), Mathematics Page | 1
Date: 8
th
January 2020 (Shift 2)
Time: 2:30 P.M. to 5:30 P.M.
Subject: Mathematics
1. Solution set of 3
?? (3
?? - 1) + 2 = |3
?? - 1| + |3
?? - 2| contains
a. exactly one element b. at least four elements
c. two elements d. infinite elements
Answer: (?? )
Solution:
3
?? (3
?? - 1) + 2 = |3
?? - 1| + |3
?? - 2|
Let 3
?? = ??
?? (?? - 1) + 2 = |?? - 1| + |?? - 2|
? ?? 2
- ?? + 2 = |?? - 1| + |?? - 2|
We plot ?? 2
- ?? + 2 and |?? - 1| + |?? - 2|
As 3
?? is always positive, therefore only positive values of ?? will be the solution.
Therefore, we have only one solution.
2. Which of the following is a tautology?
a. ~(?? ? ~?? ) ? (?? ? ?? ) b. (~?? ? ?? ) ? (?? ? ?? )
c. ~(?? ? ~?? ) ? (?? ? ?? ) d. ~(?? ? ~?? ) ? (?? ? ?? )
Answer: (?? )
Solution:
8
th
January 2020 (Shift 2), Mathematics Page | 2
~(?? ? ~?? ) ? (?? ? ?? )
= (?? ? ~?? ) ? (?? ? ?? )
= (?? ? ?? ) ? (?? ? ~?? )
= ?? ? ??
= ??
3. If a hyperbola has vertices (±6, 0) and ?? (10, 16) lies on it, then the equation of normal
at ?? is
a. 2?? + 5?? = 10 b. 2?? + 5?? = 100
c. 2?? - 5?? = 100 d. 5?? + 2?? = 100
Answer: (?? )
Solution:
Vertex of hyperbola is (±?? , 0) = (±6, 0) ? ?? = 6
Let the equation of hyperbola be
?? 2
?? 2
-
?? 2
?? 2
= 1
?
?? 2
36
-
?? 2
?? 2
= 1
As ?? (10, 16) lies on the hyperbola.
100
36
-
256
?? 2
= 1
?
64
36
=
256
?? 2
? ?? 2
= 144
Equation of hyperbola becomes
?? 2
36
-
?? 2
144
= 1
Equation of normal is
?? 2
?? ?? 1
+
?? 2
?? ?? 1
= ?? 2
+ ?? 2
?
36?? 10
+
144?? 16
= 180
?
?? 50
+
?? 20
= 1
? 2?? + 5?? = 100
4. If ?? = ???? + ?? is a tangent to the circle (?? - 3)
2
+ ?? 2
= 1 and also perpendicular to the
tangent to the circle ?? 2
+ ?? 2
= 1 at (
1
v 2
,
1
v 2
), then
a. ?? 2
- 6?? - 7 = 0 b. ?? 2
- 6?? + 7 = 0
c. ?? 2
+ 6?? - 7 = 0 d. ?? 2
+ 6?? + 7 = 0
Answer: (?? )
8
th
January 2020 (Shift 2), Mathematics Page | 3
Solution:
For circle, ?? 2
+ ?? 2
= 1
2?? + 2?? ?? '
= 0 ? ?? '
= -
?? ??
Slope of tangent to ?? 2
+ ?? 2
= 1 at (
1
v 2
,
1
v 2
) = -1
?Slope of tangent to (?? - 3)
2
+ ?? 2
= 1 is 1 ? ?? = 1
Tangent to (?? - 3)
2
+ ?? 2
= 1 is ?? = ?? + ??
Perpendicular distance of tangent ?? = ?? + ?? from centre (3, 0) is equal to radius = 1
|
3 + ?? v 2
| = 1
? ?? + 3 = ±v 2
? ?? 2
+ 6?? + 9 = 2
? ?? 2
+ 6?? + 7 = 0
5. If ?? ? = ?? ^ - 2?? ^ + ?? ^
, ?? ? ?
= ?? ^ - ?? ^ + ?? ^
and ?? ? is non-zero vector and ?? ? ?
× ?? ? ?? = ?? ? ?
× ?? ? ? ??, ?? ? ? ??. ?? ? ?? = 0
then ?? ? ?
. ?? ? is equal to
a.
1
2
b. -
1
3
c. -
1
2
d.
1
3
Answer: (?? )
Solution:
?? ? ? ?? = ?? ^ - 2?? ^ + ?? ^
?? ? ?
= ?? ^ - ?? ^ + ?? ^
?? ? ?
× ?? ? ?? = ?? ? ?
× ?? ? ? ??
? ?? ? ? ?? × (??
? ? ??
× ?? ? ??) = ?? ? ? ?? × (??
? ? ??
× ?? ? ? ??)
? (?? ? ? ??. ?? ? ??)??
? ? ??
- (?? ? ? ??. ??
? ? ??
)?? ? ?? = (?? ? ? ??. ?? ? ? ??)??
? ? ??
- (?? ? ? ??. ??
? ? ??
)?? ? ? ??
? -(?? ? ? ??. ??
? ? ??
)?? ? ?? = (?? ? ? ??. ?? ? ? ??)??
? ? ??
- (?? ? ? ??. ??
? ? ??
)?? ? ? ??
? -4?? ? ?? = 6( ?? ^ - ?? ^ + ?? ^
) - 4(?? ^ - 2?? ^ + ?? ^
)
? ?? ? ?? = -
1
2
( ?? ^ + ?? ^ + ?? ^
)
? ??
? ? ??
. ?? ? ?? = -
1
2
.
Page 4
8
th
January 2020 (Shift 2), Mathematics Page | 1
Date: 8
th
January 2020 (Shift 2)
Time: 2:30 P.M. to 5:30 P.M.
Subject: Mathematics
1. Solution set of 3
?? (3
?? - 1) + 2 = |3
?? - 1| + |3
?? - 2| contains
a. exactly one element b. at least four elements
c. two elements d. infinite elements
Answer: (?? )
Solution:
3
?? (3
?? - 1) + 2 = |3
?? - 1| + |3
?? - 2|
Let 3
?? = ??
?? (?? - 1) + 2 = |?? - 1| + |?? - 2|
? ?? 2
- ?? + 2 = |?? - 1| + |?? - 2|
We plot ?? 2
- ?? + 2 and |?? - 1| + |?? - 2|
As 3
?? is always positive, therefore only positive values of ?? will be the solution.
Therefore, we have only one solution.
2. Which of the following is a tautology?
a. ~(?? ? ~?? ) ? (?? ? ?? ) b. (~?? ? ?? ) ? (?? ? ?? )
c. ~(?? ? ~?? ) ? (?? ? ?? ) d. ~(?? ? ~?? ) ? (?? ? ?? )
Answer: (?? )
Solution:
8
th
January 2020 (Shift 2), Mathematics Page | 2
~(?? ? ~?? ) ? (?? ? ?? )
= (?? ? ~?? ) ? (?? ? ?? )
= (?? ? ?? ) ? (?? ? ~?? )
= ?? ? ??
= ??
3. If a hyperbola has vertices (±6, 0) and ?? (10, 16) lies on it, then the equation of normal
at ?? is
a. 2?? + 5?? = 10 b. 2?? + 5?? = 100
c. 2?? - 5?? = 100 d. 5?? + 2?? = 100
Answer: (?? )
Solution:
Vertex of hyperbola is (±?? , 0) = (±6, 0) ? ?? = 6
Let the equation of hyperbola be
?? 2
?? 2
-
?? 2
?? 2
= 1
?
?? 2
36
-
?? 2
?? 2
= 1
As ?? (10, 16) lies on the hyperbola.
100
36
-
256
?? 2
= 1
?
64
36
=
256
?? 2
? ?? 2
= 144
Equation of hyperbola becomes
?? 2
36
-
?? 2
144
= 1
Equation of normal is
?? 2
?? ?? 1
+
?? 2
?? ?? 1
= ?? 2
+ ?? 2
?
36?? 10
+
144?? 16
= 180
?
?? 50
+
?? 20
= 1
? 2?? + 5?? = 100
4. If ?? = ???? + ?? is a tangent to the circle (?? - 3)
2
+ ?? 2
= 1 and also perpendicular to the
tangent to the circle ?? 2
+ ?? 2
= 1 at (
1
v 2
,
1
v 2
), then
a. ?? 2
- 6?? - 7 = 0 b. ?? 2
- 6?? + 7 = 0
c. ?? 2
+ 6?? - 7 = 0 d. ?? 2
+ 6?? + 7 = 0
Answer: (?? )
8
th
January 2020 (Shift 2), Mathematics Page | 3
Solution:
For circle, ?? 2
+ ?? 2
= 1
2?? + 2?? ?? '
= 0 ? ?? '
= -
?? ??
Slope of tangent to ?? 2
+ ?? 2
= 1 at (
1
v 2
,
1
v 2
) = -1
?Slope of tangent to (?? - 3)
2
+ ?? 2
= 1 is 1 ? ?? = 1
Tangent to (?? - 3)
2
+ ?? 2
= 1 is ?? = ?? + ??
Perpendicular distance of tangent ?? = ?? + ?? from centre (3, 0) is equal to radius = 1
|
3 + ?? v 2
| = 1
? ?? + 3 = ±v 2
? ?? 2
+ 6?? + 9 = 2
? ?? 2
+ 6?? + 7 = 0
5. If ?? ? = ?? ^ - 2?? ^ + ?? ^
, ?? ? ?
= ?? ^ - ?? ^ + ?? ^
and ?? ? is non-zero vector and ?? ? ?
× ?? ? ?? = ?? ? ?
× ?? ? ? ??, ?? ? ? ??. ?? ? ?? = 0
then ?? ? ?
. ?? ? is equal to
a.
1
2
b. -
1
3
c. -
1
2
d.
1
3
Answer: (?? )
Solution:
?? ? ? ?? = ?? ^ - 2?? ^ + ?? ^
?? ? ?
= ?? ^ - ?? ^ + ?? ^
?? ? ?
× ?? ? ?? = ?? ? ?
× ?? ? ? ??
? ?? ? ? ?? × (??
? ? ??
× ?? ? ??) = ?? ? ? ?? × (??
? ? ??
× ?? ? ? ??)
? (?? ? ? ??. ?? ? ??)??
? ? ??
- (?? ? ? ??. ??
? ? ??
)?? ? ?? = (?? ? ? ??. ?? ? ? ??)??
? ? ??
- (?? ? ? ??. ??
? ? ??
)?? ? ? ??
? -(?? ? ? ??. ??
? ? ??
)?? ? ?? = (?? ? ? ??. ?? ? ? ??)??
? ? ??
- (?? ? ? ??. ??
? ? ??
)?? ? ? ??
? -4?? ? ?? = 6( ?? ^ - ?? ^ + ?? ^
) - 4(?? ^ - 2?? ^ + ?? ^
)
? ?? ? ?? = -
1
2
( ?? ^ + ?? ^ + ?? ^
)
? ??
? ? ??
. ?? ? ?? = -
1
2
.
8
th
January 2020 (Shift 2), Mathematics Page | 4
6. If the coefficient of ?? 4
and ?? 2
in the expansion of (?? + v ?? 2
- 1 )
6
+ (?? - v?? 2
- 1 )
6
is
?? and ?? ,then ?? - ?? is equal to
a. 48 b. -60
c. 60 d. -132
Answer: (?? )
Solution:
(?? +
v
?? 2
- 1 )
6
+ (?? -
v
?? 2
- 1 )
6
= 2[ ?? 0
?? 6
6
+ ?? 2
?? 4
(?? 2
- 1) + ?? 4
?? 2
(?? 2
- 1)
2
+ ?? 6
(?? 2
- 1)
3
6
6
6
]
= 2[32?? 6
- 48?? 4
+ 18?? 2
- 1]
? ?? = -96, ?? = 36
? ?? - ?? = -132
7. Differential equation of ?? 2
= 4?? (?? + ?? ), where ?? is a parameter, is
a. ?? (
????
????
)
2
= 2?? (
????
????
) + ??
b. ?? (
????
????
)
2
= 2?? (
????
????
) + ?? 2
c. ?? (
????
????
)
2
= ?? (
????
????
) + ?? 2
d. ?? (
????
????
)
2
= ?? (
????
????
) + 2?? 2
Answer: (?? )
Solution:
?? 2
= 4?? (?? + ?? ) … (1)
Differentiating both the sides w.r.t. ?? , we get
? 2?? = 4?? ?? '
? ?? =
?? 2?? '
Putting the value of ?? in (1), we get
? ?? 2
=
2?? ?? '
(?? +
?? 2?? '
)
? ?? 2
=
2????
?? '
+
?? 2
?? '2
? ?? ?? '2
= 2?? ?? '
+ ??
Page 5
8
th
January 2020 (Shift 2), Mathematics Page | 1
Date: 8
th
January 2020 (Shift 2)
Time: 2:30 P.M. to 5:30 P.M.
Subject: Mathematics
1. Solution set of 3
?? (3
?? - 1) + 2 = |3
?? - 1| + |3
?? - 2| contains
a. exactly one element b. at least four elements
c. two elements d. infinite elements
Answer: (?? )
Solution:
3
?? (3
?? - 1) + 2 = |3
?? - 1| + |3
?? - 2|
Let 3
?? = ??
?? (?? - 1) + 2 = |?? - 1| + |?? - 2|
? ?? 2
- ?? + 2 = |?? - 1| + |?? - 2|
We plot ?? 2
- ?? + 2 and |?? - 1| + |?? - 2|
As 3
?? is always positive, therefore only positive values of ?? will be the solution.
Therefore, we have only one solution.
2. Which of the following is a tautology?
a. ~(?? ? ~?? ) ? (?? ? ?? ) b. (~?? ? ?? ) ? (?? ? ?? )
c. ~(?? ? ~?? ) ? (?? ? ?? ) d. ~(?? ? ~?? ) ? (?? ? ?? )
Answer: (?? )
Solution:
8
th
January 2020 (Shift 2), Mathematics Page | 2
~(?? ? ~?? ) ? (?? ? ?? )
= (?? ? ~?? ) ? (?? ? ?? )
= (?? ? ?? ) ? (?? ? ~?? )
= ?? ? ??
= ??
3. If a hyperbola has vertices (±6, 0) and ?? (10, 16) lies on it, then the equation of normal
at ?? is
a. 2?? + 5?? = 10 b. 2?? + 5?? = 100
c. 2?? - 5?? = 100 d. 5?? + 2?? = 100
Answer: (?? )
Solution:
Vertex of hyperbola is (±?? , 0) = (±6, 0) ? ?? = 6
Let the equation of hyperbola be
?? 2
?? 2
-
?? 2
?? 2
= 1
?
?? 2
36
-
?? 2
?? 2
= 1
As ?? (10, 16) lies on the hyperbola.
100
36
-
256
?? 2
= 1
?
64
36
=
256
?? 2
? ?? 2
= 144
Equation of hyperbola becomes
?? 2
36
-
?? 2
144
= 1
Equation of normal is
?? 2
?? ?? 1
+
?? 2
?? ?? 1
= ?? 2
+ ?? 2
?
36?? 10
+
144?? 16
= 180
?
?? 50
+
?? 20
= 1
? 2?? + 5?? = 100
4. If ?? = ???? + ?? is a tangent to the circle (?? - 3)
2
+ ?? 2
= 1 and also perpendicular to the
tangent to the circle ?? 2
+ ?? 2
= 1 at (
1
v 2
,
1
v 2
), then
a. ?? 2
- 6?? - 7 = 0 b. ?? 2
- 6?? + 7 = 0
c. ?? 2
+ 6?? - 7 = 0 d. ?? 2
+ 6?? + 7 = 0
Answer: (?? )
8
th
January 2020 (Shift 2), Mathematics Page | 3
Solution:
For circle, ?? 2
+ ?? 2
= 1
2?? + 2?? ?? '
= 0 ? ?? '
= -
?? ??
Slope of tangent to ?? 2
+ ?? 2
= 1 at (
1
v 2
,
1
v 2
) = -1
?Slope of tangent to (?? - 3)
2
+ ?? 2
= 1 is 1 ? ?? = 1
Tangent to (?? - 3)
2
+ ?? 2
= 1 is ?? = ?? + ??
Perpendicular distance of tangent ?? = ?? + ?? from centre (3, 0) is equal to radius = 1
|
3 + ?? v 2
| = 1
? ?? + 3 = ±v 2
? ?? 2
+ 6?? + 9 = 2
? ?? 2
+ 6?? + 7 = 0
5. If ?? ? = ?? ^ - 2?? ^ + ?? ^
, ?? ? ?
= ?? ^ - ?? ^ + ?? ^
and ?? ? is non-zero vector and ?? ? ?
× ?? ? ?? = ?? ? ?
× ?? ? ? ??, ?? ? ? ??. ?? ? ?? = 0
then ?? ? ?
. ?? ? is equal to
a.
1
2
b. -
1
3
c. -
1
2
d.
1
3
Answer: (?? )
Solution:
?? ? ? ?? = ?? ^ - 2?? ^ + ?? ^
?? ? ?
= ?? ^ - ?? ^ + ?? ^
?? ? ?
× ?? ? ?? = ?? ? ?
× ?? ? ? ??
? ?? ? ? ?? × (??
? ? ??
× ?? ? ??) = ?? ? ? ?? × (??
? ? ??
× ?? ? ? ??)
? (?? ? ? ??. ?? ? ??)??
? ? ??
- (?? ? ? ??. ??
? ? ??
)?? ? ?? = (?? ? ? ??. ?? ? ? ??)??
? ? ??
- (?? ? ? ??. ??
? ? ??
)?? ? ? ??
? -(?? ? ? ??. ??
? ? ??
)?? ? ?? = (?? ? ? ??. ?? ? ? ??)??
? ? ??
- (?? ? ? ??. ??
? ? ??
)?? ? ? ??
? -4?? ? ?? = 6( ?? ^ - ?? ^ + ?? ^
) - 4(?? ^ - 2?? ^ + ?? ^
)
? ?? ? ?? = -
1
2
( ?? ^ + ?? ^ + ?? ^
)
? ??
? ? ??
. ?? ? ?? = -
1
2
.
8
th
January 2020 (Shift 2), Mathematics Page | 4
6. If the coefficient of ?? 4
and ?? 2
in the expansion of (?? + v ?? 2
- 1 )
6
+ (?? - v?? 2
- 1 )
6
is
?? and ?? ,then ?? - ?? is equal to
a. 48 b. -60
c. 60 d. -132
Answer: (?? )
Solution:
(?? +
v
?? 2
- 1 )
6
+ (?? -
v
?? 2
- 1 )
6
= 2[ ?? 0
?? 6
6
+ ?? 2
?? 4
(?? 2
- 1) + ?? 4
?? 2
(?? 2
- 1)
2
+ ?? 6
(?? 2
- 1)
3
6
6
6
]
= 2[32?? 6
- 48?? 4
+ 18?? 2
- 1]
? ?? = -96, ?? = 36
? ?? - ?? = -132
7. Differential equation of ?? 2
= 4?? (?? + ?? ), where ?? is a parameter, is
a. ?? (
????
????
)
2
= 2?? (
????
????
) + ??
b. ?? (
????
????
)
2
= 2?? (
????
????
) + ?? 2
c. ?? (
????
????
)
2
= ?? (
????
????
) + ?? 2
d. ?? (
????
????
)
2
= ?? (
????
????
) + 2?? 2
Answer: (?? )
Solution:
?? 2
= 4?? (?? + ?? ) … (1)
Differentiating both the sides w.r.t. ?? , we get
? 2?? = 4?? ?? '
? ?? =
?? 2?? '
Putting the value of ?? in (1), we get
? ?? 2
=
2?? ?? '
(?? +
?? 2?? '
)
? ?? 2
=
2????
?? '
+
?? 2
?? '2
? ?? ?? '2
= 2?? ?? '
+ ??
8
th
January 2020 (Shift 2), Mathematics Page | 5
? ?? (
????
????
)
2
= 2?? (
????
????
) +??
8. Image of point (1, 2, 3) w.r.t a plane is (-
7
3
, -
4
3
, -
1
3
) then which of the following points
lie on this plane
a. (1, 1, -1) b. (-1, -1, 1)
c. (-1, 1, -1) d. (-1, -1, -1)
Answer: (?? )
Solution:
Image of point ?? (1, 2, 3) w.r.t. a plane ???? + ???? + ???? + ?? = 0 is ?? (-
7
3
, -
4
3
, -
1
3
)
Direction ratios of ???? : -
10
3
, -
10
3
, -
10
3
= 1, 1, 1
Direction ratios of normal to plane is 1, 1, 1
Mid-point of ???? lies on the plane
? The mid-point of ???? = (-
2
3
,
1
3
,
4
3
)
? Equation of plane is ?? +
2
3
+ ?? -
1
3
+ ?? -
4
3
= 0
? ?? + ?? + ?? = 1
(1, 1, -1) satisfies the equation of the plane.
9. lim
?? ?0
? ?? sin 10?? ????
?? 0
?? is equal to
a. 10 b. 0
c. 1 d. 5
Answer: (?? )
Solution:
lim
?? ?0
? ?? sin 10?? ????
?? 0
??
Applying L’Hospital’s Rule:
=lim
?? ?0
?? sin 10?? 1
= 0
10. Let ?? be the set of points (?? , ?? ) such that (?? 2
= ?? = -2?? + 3 ). Then area bounded by
points in ?? is
Read More
|
264 docs|1 tests
|
FAQs on JEE Main 2020 January 8 Question Paper Shift 2 - JEE Main & Advanced Previous Year Papers
| 1. What is the JEE Main 2020 exam? |  |
| 2. When was the JEE Main 2020 January 8 exam held? | |
Ans. The JEE Main 2020 January 8 exam was held in two shifts - Shift 1 and Shift 2.
| 3. What is the difficulty level of the JEE Main 2020 January 8 Question Paper Shift 2? | |
Ans. The difficulty level of the JEE Main 2020 January 8 Question Paper Shift 2 was moderate to difficult, with questions covering a wide range of topics from the syllabus.
| 4. How many questions were asked in the JEE Main 2020 January 8 Question Paper Shift 2? | |
Ans. The JEE Main 2020 January 8 Question Paper Shift 2 consisted of a total of 75 questions, with 25 questions each from Physics, Chemistry, and Mathematics.
| 5. What is the significance of the JEE Main 2020 exam for engineering aspirants? | |
Ans. JEE Main 2020 is a crucial exam for engineering aspirants as the scores obtained in this exam are used for admission to various top engineering colleges in India.
Related Exams
About this Document
Apr 03, 2025
Last updated
Document Description: JEE Main 2020 January 8 Question Paper Shift 2 for JEE 2025 is part of JEE Main & Advanced Previous Year Papers preparation.
The notes and questions for JEE Main 2020 January 8 Question Paper Shift 2 have been prepared according to the JEE exam syllabus. Information about JEE Main 2020 January 8 Question Paper Shift 2 covers topics
like and JEE Main 2020 January 8 Question Paper Shift 2 Example, for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main 2020 January 8 Question Paper Shift 2.
Introduction of JEE Main 2020 January 8 Question Paper Shift 2 in English is available as part of our JEE Main & Advanced Previous Year Papers
for JEE & JEE Main 2020 January 8 Question Paper Shift 2 in Hindi for JEE Main & Advanced Previous Year Papers course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: JEE Main 2020 January 8 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers
Description
Full syllabus notes, lecture & questions for JEE Main 2020 January 8 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers - JEE | Plus excerises question with solution to help you revise complete syllabus for JEE Main & Advanced Previous Year Papers | Best notes, free PDF download
Information about JEE Main 2020 January 8 Question Paper Shift 2
In this doc you can find the meaning of JEE Main 2020 January 8 Question Paper Shift 2 defined & explained in the simplest way possible. Besides explaining types of
JEE Main 2020 January 8 Question Paper Shift 2 theory, EduRev gives you an ample number of questions to practice JEE Main 2020 January 8 Question Paper Shift 2 tests, examples and also practice JEE
tests
Related Searches
Viva Questions
,MCQs
,shortcuts and tricks
,video lectures
,JEE Main 2020 January 8 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers
,Extra Questions
,mock tests for examination
,Sample Paper
,Previous Year Questions with Solutions
,Free
,JEE Main 2020 January 8 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers
,Exam
,Semester Notes
,Summary
,ppt
,Important questions
,Objective type Questions
,JEE Main 2020 January 8 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers
,past year papers
,study material
,practice quizzes
;
Additional Information about JEE Main 2020 January 8 Question Paper Shift 2 for JEE Preparation
JEE Main 2020 January 8 Question Paper Shift 2 Free PDF Download
The JEE Main 2020 January 8 Question Paper Shift 2 is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the JEE Main 2020 January 8 Question Paper Shift 2 now and kickstart your journey towards success in the JEE exam.
Importance of JEE Main 2020 January 8 Question Paper Shift 2
The importance of JEE Main 2020 January 8 Question Paper Shift 2 cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
JEE Main 2020 January 8 Question Paper Shift 2 Notes
JEE Main 2020 January 8 Question Paper Shift 2 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to JEE Main 2020 January 8 Question Paper Shift 2.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, JEE Main 2020 January 8 Question Paper Shift 2 Notes on EduRev are your ultimate resource for success.
JEE Main 2020 January 8 Question Paper Shift 2 JEE Questions
The "JEE Main 2020 January 8 Question Paper Shift 2 JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study JEE Main 2020 January 8 Question Paper Shift 2 on the App
Students of JEE can study JEE Main 2020 January 8 Question Paper Shift 2 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the JEE Main 2020 January 8 Question Paper Shift 2,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of JEE Main 2020 January 8 Question Paper Shift 2 is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily