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
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