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                                               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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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?
Ans. JEE Main 2020 is a national level engineering entrance exam conducted by the National Testing Agency (NTA) for admission to various undergraduate engineering programs in India.
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.
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