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JEE Main Previous Year Questions (2026): Complex Numbers and Quadratic Equations
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Page 1
JEE Main Previous Year Questions
(2025): Complex Numbers and
Quadratic Equations
Q1: Let ?? , ?? be the roots of the equation ?? ?? - ???? - ?? = ?? with ???? ( ?? ) < ???? ( ?? ) . Let
?? ?? = ?? ?? - ?? ?? . If ?? ?? = -?? v?? ?? , ?? ?? = -?? v?? ?? , ?? ?? = ???? v?? ?? and ?? ?? = ???? v?? ?? , then ?? ?? +
?? ?? is equal to ____ .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 31
Solution:
We begin with the equations for the roots:
?? + ?? = a
???? = -b
Given:
P
6
= aP
5
+ bP
4
P
5
= aP
4
+ bP
3
Using the given values:
For P
6
:
45v7?? = a × 11v7?? + b( -3v7) ??
Simplifying, we obtain:
45 = 11a - 3 b ( Equation 1 )
For P
5
:
11v7?? = a( -3v7?? )+ b( -5v7?? )
Simplifying, we obtain:
11 = -3?? - 5?? (Equation 2)
Solving these linear equations, we find:
a = 3
b = -4
Now, we calculate ?? 4
+ ?? 4
using the relation:
?? 4
+ ?? 4
= v( ?? 4
- ?? 4
)
2
+ 4( ?? 4
?? 4
)
From b = -4, we know:
???? = -b = 4 ? ?? 4
?? 4
= ( ???? )
4
= 4
4
= 256
Substitute into the relation:
?? 4
+ ?? 4
= v( -63)+ 1024
= v961 = 31
Thus, ?? 4
+ ?? 4
is equal to 31 .
Page 2
JEE Main Previous Year Questions
(2025): Complex Numbers and
Quadratic Equations
Q1: Let ?? , ?? be the roots of the equation ?? ?? - ???? - ?? = ?? with ???? ( ?? ) < ???? ( ?? ) . Let
?? ?? = ?? ?? - ?? ?? . If ?? ?? = -?? v?? ?? , ?? ?? = -?? v?? ?? , ?? ?? = ???? v?? ?? and ?? ?? = ???? v?? ?? , then ?? ?? +
?? ?? is equal to ____ .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 31
Solution:
We begin with the equations for the roots:
?? + ?? = a
???? = -b
Given:
P
6
= aP
5
+ bP
4
P
5
= aP
4
+ bP
3
Using the given values:
For P
6
:
45v7?? = a × 11v7?? + b( -3v7) ??
Simplifying, we obtain:
45 = 11a - 3 b ( Equation 1 )
For P
5
:
11v7?? = a( -3v7?? )+ b( -5v7?? )
Simplifying, we obtain:
11 = -3?? - 5?? (Equation 2)
Solving these linear equations, we find:
a = 3
b = -4
Now, we calculate ?? 4
+ ?? 4
using the relation:
?? 4
+ ?? 4
= v( ?? 4
- ?? 4
)
2
+ 4( ?? 4
?? 4
)
From b = -4, we know:
???? = -b = 4 ? ?? 4
?? 4
= ( ???? )
4
= 4
4
= 256
Substitute into the relation:
?? 4
+ ?? 4
= v( -63)+ 1024
= v961 = 31
Thus, ?? 4
+ ?? 4
is equal to 31 .
Q2: Let integers ?? , ?? ? [-?? , ?? ] be such that ?? + ?? ? ?? . Then the number of all possible
ordered pairs ( ?? , ?? ) , for which
?? -?? ?? +?? | = ?? and |
?? + ?? ?? ?? ?? ?? ?? + ?? ?? ?? ?? ?? ?? ?? + ?? | = ?? , ?? ? ?? ,
where ?? and ?? ?? are the roots of ?? ?? + ?? + ?? = ?? , is equal to ____ -
JEE Main 2025 (Online) 29th January Evening Shift
Ans: 10
Solution:
?? , ?? ? ?? , -3 = ?? , ?? = 3, ?? + ?? ? 0
|?? - ?? | = |?? + ?? |
?? + 1 ?? ?? 2
?? ?? + ?? 2
1
?? 2
1 ?? + ?? = 1
?
?? ?? ?? ?? ?? + ?? 2
1
?? 2
1 ?? + ?? | = 1
?
1 1 1
?? ?? + ?? 2
1
?? 2
1 ?? + ?? | = 1
?
1 0
?? ?? + ?? 2
- ?? 0
?? 2
1 - ?? 2
1 - ?? ?? + ?? - ?? 2
| = 1
? ?? 3
= 1
? ?? = ?? , ?? 2
, 1
Now
|1 - a| = |1 + b|
? 10 pairs
Q3: Let ?? = {?? ? ?? : |?? - ?? - ?? | = ?? }, ?? = {?? ? ?? : ???? ( ?? - ???? ) = ?? } and ?? = ?? n ?? .
Then ?
?? ??? ?|?? |
?? is equal to ____ .
JEE Main 2025 (Online) 4th April Morning Shift
Ans: 22
Solution:
Page 3
JEE Main Previous Year Questions
(2025): Complex Numbers and
Quadratic Equations
Q1: Let ?? , ?? be the roots of the equation ?? ?? - ???? - ?? = ?? with ???? ( ?? ) < ???? ( ?? ) . Let
?? ?? = ?? ?? - ?? ?? . If ?? ?? = -?? v?? ?? , ?? ?? = -?? v?? ?? , ?? ?? = ???? v?? ?? and ?? ?? = ???? v?? ?? , then ?? ?? +
?? ?? is equal to ____ .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 31
Solution:
We begin with the equations for the roots:
?? + ?? = a
???? = -b
Given:
P
6
= aP
5
+ bP
4
P
5
= aP
4
+ bP
3
Using the given values:
For P
6
:
45v7?? = a × 11v7?? + b( -3v7) ??
Simplifying, we obtain:
45 = 11a - 3 b ( Equation 1 )
For P
5
:
11v7?? = a( -3v7?? )+ b( -5v7?? )
Simplifying, we obtain:
11 = -3?? - 5?? (Equation 2)
Solving these linear equations, we find:
a = 3
b = -4
Now, we calculate ?? 4
+ ?? 4
using the relation:
?? 4
+ ?? 4
= v( ?? 4
- ?? 4
)
2
+ 4( ?? 4
?? 4
)
From b = -4, we know:
???? = -b = 4 ? ?? 4
?? 4
= ( ???? )
4
= 4
4
= 256
Substitute into the relation:
?? 4
+ ?? 4
= v( -63)+ 1024
= v961 = 31
Thus, ?? 4
+ ?? 4
is equal to 31 .
Q2: Let integers ?? , ?? ? [-?? , ?? ] be such that ?? + ?? ? ?? . Then the number of all possible
ordered pairs ( ?? , ?? ) , for which
?? -?? ?? +?? | = ?? and |
?? + ?? ?? ?? ?? ?? ?? + ?? ?? ?? ?? ?? ?? ?? + ?? | = ?? , ?? ? ?? ,
where ?? and ?? ?? are the roots of ?? ?? + ?? + ?? = ?? , is equal to ____ -
JEE Main 2025 (Online) 29th January Evening Shift
Ans: 10
Solution:
?? , ?? ? ?? , -3 = ?? , ?? = 3, ?? + ?? ? 0
|?? - ?? | = |?? + ?? |
?? + 1 ?? ?? 2
?? ?? + ?? 2
1
?? 2
1 ?? + ?? = 1
?
?? ?? ?? ?? ?? + ?? 2
1
?? 2
1 ?? + ?? | = 1
?
1 1 1
?? ?? + ?? 2
1
?? 2
1 ?? + ?? | = 1
?
1 0
?? ?? + ?? 2
- ?? 0
?? 2
1 - ?? 2
1 - ?? ?? + ?? - ?? 2
| = 1
? ?? 3
= 1
? ?? = ?? , ?? 2
, 1
Now
|1 - a| = |1 + b|
? 10 pairs
Q3: Let ?? = {?? ? ?? : |?? - ?? - ?? | = ?? }, ?? = {?? ? ?? : ???? ( ?? - ???? ) = ?? } and ?? = ?? n ?? .
Then ?
?? ??? ?|?? |
?? is equal to ____ .
JEE Main 2025 (Online) 4th April Morning Shift
Ans: 22
Solution:
L e t ?? = ?? + ????
|?? - 2 - ?? | = 3 ? ( ?? - 2)
2
+ ( ?? - 1)
2
= 3
2
Re( ?? - ???? ) = Re( ?? + ???? - ???? + ?? ) = ?? + ?? ? ?? + ?? = 2
? ?? = {( ?? , ?? ) : ( ?? - 2)
2
+ ( ?? - 1)
2
= 3
2
, ?? , ?? ? ?? }
?? = {( ?? , ?? ) : ?? + ?? = 2}
? ?? - 2 = -?? ? ?? 2
+ ( ?? - 1)
2
= 3
2
? 2?? 2
- 2?? - 8 = 0 ? ?? 2
- ?? - 4 = 0
?? 1
+ ?? 2
= 1, ?? 1
?? 2
= -4
? ?? 1
2
+ ?? 2
2
= ( ?? 1
+ ?? 2
)
2
- 2?? 1
?? 2
= 9
? ?? 1
+ ?? 2
= 4( ?? 1
+ ?? 2
) = 3
?? 1
?? 2
= ( 2 - ?? 1
) ( 2 - ?? 2
) = 4 - 2( ?? 1
+ ?? 2
)+ ?? 1
?? 2
= -2
? ?? 1
2
+ ?? 2
2
= ( ?? 1
+ ?? 2
)
2
- 2?? 1
?? 2
= 13
? ?? = {( ?? 1
, ?? 1
) , ( ?? 2
, ?? 2
) }
? ?
?? ??? ?|?? |
2
= ( ?? 1
2
+ ?? 1
2
)+ ( ?? 2
2
+ ?? 2
2
)= 22
Q4: If ?? is a root of the equation ?? ?? + ?? + ?? = ?? and ?
?? =?? ?? ?(?? ?? +
?? ?? ?? )
?? = ???? , then ?? is
equal to ____ .
JEE Main 2025 (Online) 4th April Evening Shift
Ans: 11
Solution:
?? is root of equation 1 + ?? + ?? 2
= 0, ?? = ?? or ?? 2
(?? ?? +
1
?? ?? )
2
= ?? 2?? +
1
?? 2?? + 2 = ?? ?? +
1
?? ?? + 2
? ?? ?? +
1
?? ?? + 2 = {
4,3 divides ?? 1,3 does not divide ??
? ?
?? =1
?? ?(?? ?? +
1
?? ?? )
2
= 20
? ( 1 + 1 + 4)+ ( 1 + 1 + 4)+ ( 1 + 1 + 4)+ ( 1 + 1)
= 20
? ?? = 11
Q5: Let ?? ?? , ?? ?? and ?? ?? be three complex numbers on the circle |?? | = ?? with ?????? ( ?? ?? ) =
-?? ?? , ?????? ( ?? ?? ) = ?? and ?????? ( ?? ?? ) =
?? ?? . If |?? ?? ???
?? + ?? ?? ???
?? + ?? ?? ???
?? |
?? = ?? + ?? v?? , ?? , ?? ? ?? ,
then the value of ?? ?? + ?? ?? is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 41
B. 29
C. 24
D. 31
Page 4
JEE Main Previous Year Questions
(2025): Complex Numbers and
Quadratic Equations
Q1: Let ?? , ?? be the roots of the equation ?? ?? - ???? - ?? = ?? with ???? ( ?? ) < ???? ( ?? ) . Let
?? ?? = ?? ?? - ?? ?? . If ?? ?? = -?? v?? ?? , ?? ?? = -?? v?? ?? , ?? ?? = ???? v?? ?? and ?? ?? = ???? v?? ?? , then ?? ?? +
?? ?? is equal to ____ .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 31
Solution:
We begin with the equations for the roots:
?? + ?? = a
???? = -b
Given:
P
6
= aP
5
+ bP
4
P
5
= aP
4
+ bP
3
Using the given values:
For P
6
:
45v7?? = a × 11v7?? + b( -3v7) ??
Simplifying, we obtain:
45 = 11a - 3 b ( Equation 1 )
For P
5
:
11v7?? = a( -3v7?? )+ b( -5v7?? )
Simplifying, we obtain:
11 = -3?? - 5?? (Equation 2)
Solving these linear equations, we find:
a = 3
b = -4
Now, we calculate ?? 4
+ ?? 4
using the relation:
?? 4
+ ?? 4
= v( ?? 4
- ?? 4
)
2
+ 4( ?? 4
?? 4
)
From b = -4, we know:
???? = -b = 4 ? ?? 4
?? 4
= ( ???? )
4
= 4
4
= 256
Substitute into the relation:
?? 4
+ ?? 4
= v( -63)+ 1024
= v961 = 31
Thus, ?? 4
+ ?? 4
is equal to 31 .
Q2: Let integers ?? , ?? ? [-?? , ?? ] be such that ?? + ?? ? ?? . Then the number of all possible
ordered pairs ( ?? , ?? ) , for which
?? -?? ?? +?? | = ?? and |
?? + ?? ?? ?? ?? ?? ?? + ?? ?? ?? ?? ?? ?? ?? + ?? | = ?? , ?? ? ?? ,
where ?? and ?? ?? are the roots of ?? ?? + ?? + ?? = ?? , is equal to ____ -
JEE Main 2025 (Online) 29th January Evening Shift
Ans: 10
Solution:
?? , ?? ? ?? , -3 = ?? , ?? = 3, ?? + ?? ? 0
|?? - ?? | = |?? + ?? |
?? + 1 ?? ?? 2
?? ?? + ?? 2
1
?? 2
1 ?? + ?? = 1
?
?? ?? ?? ?? ?? + ?? 2
1
?? 2
1 ?? + ?? | = 1
?
1 1 1
?? ?? + ?? 2
1
?? 2
1 ?? + ?? | = 1
?
1 0
?? ?? + ?? 2
- ?? 0
?? 2
1 - ?? 2
1 - ?? ?? + ?? - ?? 2
| = 1
? ?? 3
= 1
? ?? = ?? , ?? 2
, 1
Now
|1 - a| = |1 + b|
? 10 pairs
Q3: Let ?? = {?? ? ?? : |?? - ?? - ?? | = ?? }, ?? = {?? ? ?? : ???? ( ?? - ???? ) = ?? } and ?? = ?? n ?? .
Then ?
?? ??? ?|?? |
?? is equal to ____ .
JEE Main 2025 (Online) 4th April Morning Shift
Ans: 22
Solution:
L e t ?? = ?? + ????
|?? - 2 - ?? | = 3 ? ( ?? - 2)
2
+ ( ?? - 1)
2
= 3
2
Re( ?? - ???? ) = Re( ?? + ???? - ???? + ?? ) = ?? + ?? ? ?? + ?? = 2
? ?? = {( ?? , ?? ) : ( ?? - 2)
2
+ ( ?? - 1)
2
= 3
2
, ?? , ?? ? ?? }
?? = {( ?? , ?? ) : ?? + ?? = 2}
? ?? - 2 = -?? ? ?? 2
+ ( ?? - 1)
2
= 3
2
? 2?? 2
- 2?? - 8 = 0 ? ?? 2
- ?? - 4 = 0
?? 1
+ ?? 2
= 1, ?? 1
?? 2
= -4
? ?? 1
2
+ ?? 2
2
= ( ?? 1
+ ?? 2
)
2
- 2?? 1
?? 2
= 9
? ?? 1
+ ?? 2
= 4( ?? 1
+ ?? 2
) = 3
?? 1
?? 2
= ( 2 - ?? 1
) ( 2 - ?? 2
) = 4 - 2( ?? 1
+ ?? 2
)+ ?? 1
?? 2
= -2
? ?? 1
2
+ ?? 2
2
= ( ?? 1
+ ?? 2
)
2
- 2?? 1
?? 2
= 13
? ?? = {( ?? 1
, ?? 1
) , ( ?? 2
, ?? 2
) }
? ?
?? ??? ?|?? |
2
= ( ?? 1
2
+ ?? 1
2
)+ ( ?? 2
2
+ ?? 2
2
)= 22
Q4: If ?? is a root of the equation ?? ?? + ?? + ?? = ?? and ?
?? =?? ?? ?(?? ?? +
?? ?? ?? )
?? = ???? , then ?? is
equal to ____ .
JEE Main 2025 (Online) 4th April Evening Shift
Ans: 11
Solution:
?? is root of equation 1 + ?? + ?? 2
= 0, ?? = ?? or ?? 2
(?? ?? +
1
?? ?? )
2
= ?? 2?? +
1
?? 2?? + 2 = ?? ?? +
1
?? ?? + 2
? ?? ?? +
1
?? ?? + 2 = {
4,3 divides ?? 1,3 does not divide ??
? ?
?? =1
?? ?(?? ?? +
1
?? ?? )
2
= 20
? ( 1 + 1 + 4)+ ( 1 + 1 + 4)+ ( 1 + 1 + 4)+ ( 1 + 1)
= 20
? ?? = 11
Q5: Let ?? ?? , ?? ?? and ?? ?? be three complex numbers on the circle |?? | = ?? with ?????? ( ?? ?? ) =
-?? ?? , ?????? ( ?? ?? ) = ?? and ?????? ( ?? ?? ) =
?? ?? . If |?? ?? ???
?? + ?? ?? ???
?? + ?? ?? ???
?? |
?? = ?? + ?? v?? , ?? , ?? ? ?? ,
then the value of ?? ?? + ?? ?? is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 41
B. 29
C. 24
D. 31
Ans: B
Solution:
To solve the problem, we start with the given information about the complex numbers ?? 1
, ?? 2
,
and ?? 3
, which lie on the unit circle |?? | = 1. Their arguments are as follows:
arg ( ?? 1
) = -
?? 4
arg ( ?? 2
) = 0
arg ( ?? 3
) =
?? 4
Thus, the complex numbers can be represented as:
?? 1
= ?? -?? ?? 4
=
1
v2
-
?? v2
?? 2
= ?? ?? ·0
= 1
?? 3
= ?? ?? ?? 4
=
1
v2
+
?? v2
Next, calculate the conjugates needed:
?? ?
2
= 1
?? ?
3
=
1
v2
-
?? v2
?? ?
1
=
1
v2
+
?? v2
We need to evaluate:
?? 1
?? ?
2
+ ?? 2
?? ?
3
+ ?? 3
?? ?
1
Calculate each term separately:
?? 1
?? ?
2
= (
1
v2
-
?? v2
) · 1 =
1
v2
-
?? v2
?? 2
?? ?
3
= 1 · (
1
v2
-
?? v2
) =
1
v2
-
?? v2
?? 3
?? ?
1
= (
1
v2
+
?? v2
) · (
1
v2
+
?? v2
) =
( 1 + ?? )
2
2
=
1 + 2?? - 1
2
= ??
Sum the evaluated terms:
?? 1
?? ?
2
+ ?? 2
?? ?
3
+ ?? 3
?? ?
1
= (
1
v2
-
?? v2
) + (
1
v2
-
?? v2
) + ??
Simplify:
= v2 + ?? - v2?? = v2 + ?? ( 1 - v2)
Calculate the modulus squared:
|v2 + ?? ( 1 - v2) |
2
= ( v2)
2
+ ( 1 - v2)
2
= 2 + ( 1 - 2v2 + 2) = 5 - 2v2
Thus, the expression |?? 1
?? ?
2
+ ?? 2
?? ?
3
+ ?? 3
?? ?
1
|
2
simplifies as follows:
?? = 5
?? = -2
Finally, compute ?? 2
+ ?? 2
:
?? 2
+ ?? 2
= 5
2
+ ( -2)
2
= 25 + 4 = 29
Therefore, the value of ?? 2
+ ?? 2
is 29 .
Q6: Let the curve ?? ( ?? + ?? )+ ???( ?? - ?? ) = ?? , ?? ? ?? , divide the region |?? - ?? | = ?? into
two parts of areas ?? and ?? . Then |?? - ?? | equals :
Page 5
JEE Main Previous Year Questions
(2025): Complex Numbers and
Quadratic Equations
Q1: Let ?? , ?? be the roots of the equation ?? ?? - ???? - ?? = ?? with ???? ( ?? ) < ???? ( ?? ) . Let
?? ?? = ?? ?? - ?? ?? . If ?? ?? = -?? v?? ?? , ?? ?? = -?? v?? ?? , ?? ?? = ???? v?? ?? and ?? ?? = ???? v?? ?? , then ?? ?? +
?? ?? is equal to ____ .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 31
Solution:
We begin with the equations for the roots:
?? + ?? = a
???? = -b
Given:
P
6
= aP
5
+ bP
4
P
5
= aP
4
+ bP
3
Using the given values:
For P
6
:
45v7?? = a × 11v7?? + b( -3v7) ??
Simplifying, we obtain:
45 = 11a - 3 b ( Equation 1 )
For P
5
:
11v7?? = a( -3v7?? )+ b( -5v7?? )
Simplifying, we obtain:
11 = -3?? - 5?? (Equation 2)
Solving these linear equations, we find:
a = 3
b = -4
Now, we calculate ?? 4
+ ?? 4
using the relation:
?? 4
+ ?? 4
= v( ?? 4
- ?? 4
)
2
+ 4( ?? 4
?? 4
)
From b = -4, we know:
???? = -b = 4 ? ?? 4
?? 4
= ( ???? )
4
= 4
4
= 256
Substitute into the relation:
?? 4
+ ?? 4
= v( -63)+ 1024
= v961 = 31
Thus, ?? 4
+ ?? 4
is equal to 31 .
Q2: Let integers ?? , ?? ? [-?? , ?? ] be such that ?? + ?? ? ?? . Then the number of all possible
ordered pairs ( ?? , ?? ) , for which
?? -?? ?? +?? | = ?? and |
?? + ?? ?? ?? ?? ?? ?? + ?? ?? ?? ?? ?? ?? ?? + ?? | = ?? , ?? ? ?? ,
where ?? and ?? ?? are the roots of ?? ?? + ?? + ?? = ?? , is equal to ____ -
JEE Main 2025 (Online) 29th January Evening Shift
Ans: 10
Solution:
?? , ?? ? ?? , -3 = ?? , ?? = 3, ?? + ?? ? 0
|?? - ?? | = |?? + ?? |
?? + 1 ?? ?? 2
?? ?? + ?? 2
1
?? 2
1 ?? + ?? = 1
?
?? ?? ?? ?? ?? + ?? 2
1
?? 2
1 ?? + ?? | = 1
?
1 1 1
?? ?? + ?? 2
1
?? 2
1 ?? + ?? | = 1
?
1 0
?? ?? + ?? 2
- ?? 0
?? 2
1 - ?? 2
1 - ?? ?? + ?? - ?? 2
| = 1
? ?? 3
= 1
? ?? = ?? , ?? 2
, 1
Now
|1 - a| = |1 + b|
? 10 pairs
Q3: Let ?? = {?? ? ?? : |?? - ?? - ?? | = ?? }, ?? = {?? ? ?? : ???? ( ?? - ???? ) = ?? } and ?? = ?? n ?? .
Then ?
?? ??? ?|?? |
?? is equal to ____ .
JEE Main 2025 (Online) 4th April Morning Shift
Ans: 22
Solution:
L e t ?? = ?? + ????
|?? - 2 - ?? | = 3 ? ( ?? - 2)
2
+ ( ?? - 1)
2
= 3
2
Re( ?? - ???? ) = Re( ?? + ???? - ???? + ?? ) = ?? + ?? ? ?? + ?? = 2
? ?? = {( ?? , ?? ) : ( ?? - 2)
2
+ ( ?? - 1)
2
= 3
2
, ?? , ?? ? ?? }
?? = {( ?? , ?? ) : ?? + ?? = 2}
? ?? - 2 = -?? ? ?? 2
+ ( ?? - 1)
2
= 3
2
? 2?? 2
- 2?? - 8 = 0 ? ?? 2
- ?? - 4 = 0
?? 1
+ ?? 2
= 1, ?? 1
?? 2
= -4
? ?? 1
2
+ ?? 2
2
= ( ?? 1
+ ?? 2
)
2
- 2?? 1
?? 2
= 9
? ?? 1
+ ?? 2
= 4( ?? 1
+ ?? 2
) = 3
?? 1
?? 2
= ( 2 - ?? 1
) ( 2 - ?? 2
) = 4 - 2( ?? 1
+ ?? 2
)+ ?? 1
?? 2
= -2
? ?? 1
2
+ ?? 2
2
= ( ?? 1
+ ?? 2
)
2
- 2?? 1
?? 2
= 13
? ?? = {( ?? 1
, ?? 1
) , ( ?? 2
, ?? 2
) }
? ?
?? ??? ?|?? |
2
= ( ?? 1
2
+ ?? 1
2
)+ ( ?? 2
2
+ ?? 2
2
)= 22
Q4: If ?? is a root of the equation ?? ?? + ?? + ?? = ?? and ?
?? =?? ?? ?(?? ?? +
?? ?? ?? )
?? = ???? , then ?? is
equal to ____ .
JEE Main 2025 (Online) 4th April Evening Shift
Ans: 11
Solution:
?? is root of equation 1 + ?? + ?? 2
= 0, ?? = ?? or ?? 2
(?? ?? +
1
?? ?? )
2
= ?? 2?? +
1
?? 2?? + 2 = ?? ?? +
1
?? ?? + 2
? ?? ?? +
1
?? ?? + 2 = {
4,3 divides ?? 1,3 does not divide ??
? ?
?? =1
?? ?(?? ?? +
1
?? ?? )
2
= 20
? ( 1 + 1 + 4)+ ( 1 + 1 + 4)+ ( 1 + 1 + 4)+ ( 1 + 1)
= 20
? ?? = 11
Q5: Let ?? ?? , ?? ?? and ?? ?? be three complex numbers on the circle |?? | = ?? with ?????? ( ?? ?? ) =
-?? ?? , ?????? ( ?? ?? ) = ?? and ?????? ( ?? ?? ) =
?? ?? . If |?? ?? ???
?? + ?? ?? ???
?? + ?? ?? ???
?? |
?? = ?? + ?? v?? , ?? , ?? ? ?? ,
then the value of ?? ?? + ?? ?? is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 41
B. 29
C. 24
D. 31
Ans: B
Solution:
To solve the problem, we start with the given information about the complex numbers ?? 1
, ?? 2
,
and ?? 3
, which lie on the unit circle |?? | = 1. Their arguments are as follows:
arg ( ?? 1
) = -
?? 4
arg ( ?? 2
) = 0
arg ( ?? 3
) =
?? 4
Thus, the complex numbers can be represented as:
?? 1
= ?? -?? ?? 4
=
1
v2
-
?? v2
?? 2
= ?? ?? ·0
= 1
?? 3
= ?? ?? ?? 4
=
1
v2
+
?? v2
Next, calculate the conjugates needed:
?? ?
2
= 1
?? ?
3
=
1
v2
-
?? v2
?? ?
1
=
1
v2
+
?? v2
We need to evaluate:
?? 1
?? ?
2
+ ?? 2
?? ?
3
+ ?? 3
?? ?
1
Calculate each term separately:
?? 1
?? ?
2
= (
1
v2
-
?? v2
) · 1 =
1
v2
-
?? v2
?? 2
?? ?
3
= 1 · (
1
v2
-
?? v2
) =
1
v2
-
?? v2
?? 3
?? ?
1
= (
1
v2
+
?? v2
) · (
1
v2
+
?? v2
) =
( 1 + ?? )
2
2
=
1 + 2?? - 1
2
= ??
Sum the evaluated terms:
?? 1
?? ?
2
+ ?? 2
?? ?
3
+ ?? 3
?? ?
1
= (
1
v2
-
?? v2
) + (
1
v2
-
?? v2
) + ??
Simplify:
= v2 + ?? - v2?? = v2 + ?? ( 1 - v2)
Calculate the modulus squared:
|v2 + ?? ( 1 - v2) |
2
= ( v2)
2
+ ( 1 - v2)
2
= 2 + ( 1 - 2v2 + 2) = 5 - 2v2
Thus, the expression |?? 1
?? ?
2
+ ?? 2
?? ?
3
+ ?? 3
?? ?
1
|
2
simplifies as follows:
?? = 5
?? = -2
Finally, compute ?? 2
+ ?? 2
:
?? 2
+ ?? 2
= 5
2
+ ( -2)
2
= 25 + 4 = 29
Therefore, the value of ?? 2
+ ?? 2
is 29 .
Q6: Let the curve ?? ( ?? + ?? )+ ???( ?? - ?? ) = ?? , ?? ? ?? , divide the region |?? - ?? | = ?? into
two parts of areas ?? and ?? . Then |?? - ?? | equals :
JEE Main 2025 (Online) 22nd January Evening Shift
Options:
A. 1 +
?? 3
B. 1 +
?? 6
C. 1 +
?? 2
D. 1 +
?? 4
Ans: C
Solution:
Let ?? = ?? + ????
( ?? + ???? ) ( 1 + ?? )+ ( ?? - ???? ) ( 1 - ?? ) = 4
?? + ???? + ???? - ?? + ?? - ???? - ???? - ?? = 4
2?? - 2?? = 4
?? - ?? = 2
|?? - 3| = 1
( ?? - 3)
2
+ ?? 2
= 1
Area of shaded region =
?? ·1
2
4
-
1
2
· 1 · 1 =
?? 4
-
1
2
Area of unshaded region inside the circle
=
3
4
?? · 1
2
+
1
2
· 1 · 1 =
3?? 4
+
1
2
? difference of area = (
3?? 4
+
1
2
) - (
?? 4
-
1
2
)
=
?? 2
+ 1
Q7: Let |
???-?? ?? ???+?? | =
?? ?? , ?? ? ?? , be the equation of a circle with center at ?? . If the area of the
triangle, whose vertices are at the points ( ?? , ?? ) , ?? and ( ?? , ?? ) is 11 square units, then
?? ?? equals:
Read MoreFAQs on JEE Main Previous Year Questions (2026): Complex Numbers and Quadratic Equations
| 1. What are complex numbers and how are they represented? |  |
Ans. Complex numbers are numbers that have a real part and an imaginary part. They are represented in the form a + bi, where a is the real part, b is the imaginary part, and i is the imaginary unit, defined as the square root of -1. Complex numbers can also be represented in polar form as r(cos θ + i sin θ), where r is the magnitude and θ is the argument of the complex number.
| 2. How do you solve quadratic equations involving complex numbers? | |
Ans. Quadratic equations can be solved using the quadratic formula x = (-b ± √(b² - 4ac)) / 2a. If the discriminant (b² - 4ac) is negative, the solutions will involve complex numbers. In such cases, the solutions can be expressed as x = (-b ± i√(|b² - 4ac|)) / 2a, where i represents the imaginary unit.
| 3. What is the significance of the discriminant in quadratic equations? | |
Ans. The discriminant of a quadratic equation, given by D = b² - 4ac, determines the nature of the roots. If D > 0, the equation has two distinct real roots. If D = 0, there is exactly one real root (a repeated root). If D < 0, the roots are complex and conjugate pairs. Understanding the discriminant helps in predicting the type of solutions without actually solving the equation.
| 4. Can complex numbers be added and multiplied like real numbers? | |
Ans. Yes, complex numbers can be added and multiplied similarly to real numbers. To add two complex numbers, you combine their real parts and their imaginary parts separately: (a + bi) + (c + di) = (a + c) + (b + d)i. For multiplication, you use the distributive property: (a + bi)(c + di) = ac + adi + bci + bdi². Since i² = -1, this simplifies to (ac - bd) + (ad + bc)i.
| 5. What are the applications of complex numbers in real-world scenarios? | |
Ans. Complex numbers have various applications in fields such as engineering, physics, and applied mathematics. They are used in electrical engineering for analyzing AC circuits, in control theory for system stability, and in fluid dynamics. Additionally, complex numbers are essential in signal processing and in the study of wave phenomena, where they help in simplifying calculations involving oscillations and rotations.
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