JEE Exam > JEE Notes > Mathematics (Maths) Main & Advanced > Inequalities: JEE Main Previous Year Questions (2021-2026)
Inequalities: JEE Main Previous Year Questions (2021-2026)
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
JEE Main Previous Year Questions
(2025): Linear Inequalities
Q1: If the equation ?? ( ?? - ?? ) ?? ?? + ?? ( ?? - ?? ) ?? + ?? ( ?? - ?? ) = ?? has equal roots, where
?? + ?? = ???? and ?? =
????
?? , then ?? ?? + ?? ?? is equal to _ _ _ _
JEE Main 2025 (Online) 23rd January Morning Shift
Ans: 117
Solution:
To solve the given problem, we start with the quadratic equation:
?? ( ?? - ?? ) ?? 2
+ ?? ( ?? - ?? ) ?? + ?? ( ?? - ?? ) = 0
Given that the roots are equal (let's assume both roots are 1 ), we know the sum of the roots,
?? + ?? , is twice the value of one root, which leads us to:
?? + ?? = 2
Using the formula for the sum of roots for a quadratic equation, ?? + ?? = -
?? ( ?? - ?? )
?? ( ?? - ?? )
, we set this
equal to 2 :
-
?? ( ?? - ?? )
?? ( ?? - ?? )
= 2
Solving for this:
- ???? + ???? = 2 ???? - 2 ???? 2 ???? = ???? + ???? 2 ???? = ?? ( ?? + ?? )
Given that ?? + ?? = 15 and ?? =
36
5
, substitute these into the equation:
2 ???? = 15 ?? 2 ???? = 15 ×
36
5
= 108 ???? = 54
Now, using the equation ?? + ?? = 15 and ???? = 54, find ?? 2
+ ?? 2
:
?? 2
+ ?? 2
= ( ?? + ?? )
2
- 2 ???? = 15
2
- 2 × 54 ?? 2
+ ?? 2
= 225 - 108 = 117
Therefore, ?? 2
+ ?? 2
is equal to 117 .
Q2: If the set of all ?? ? ?? - { ?? }, for which the roots of the equation ( ?? - ?? ) ?? ?? +
?? ( ?? - ?? ) ?? + ?? = ?? are positive is ( - 8 , - ?? ] ? [ ?? , ?? ) , then ?? ?? + ?? + ?? is equal to
JEE Main 2025 (Online) 2nd April Evening Shift
Ans: 7
Solution:
?? ( ?? ) = ( 1 - ?? ) ?? 2
+ 2 ( ?? - 3 ) ?? + 9 , ?? ( 0 ) = 9 > 0
?? = 0 ? 4 ( ?? - 3 )
2
= 4 ( 1 - ?? ) · 9
? ?? ? ( - 8 , - 3 ] ? [ 0 , 8 ) ( ?? )
Page 2
JEE Main Previous Year Questions
(2025): Linear Inequalities
Q1: If the equation ?? ( ?? - ?? ) ?? ?? + ?? ( ?? - ?? ) ?? + ?? ( ?? - ?? ) = ?? has equal roots, where
?? + ?? = ???? and ?? =
????
?? , then ?? ?? + ?? ?? is equal to _ _ _ _
JEE Main 2025 (Online) 23rd January Morning Shift
Ans: 117
Solution:
To solve the given problem, we start with the quadratic equation:
?? ( ?? - ?? ) ?? 2
+ ?? ( ?? - ?? ) ?? + ?? ( ?? - ?? ) = 0
Given that the roots are equal (let's assume both roots are 1 ), we know the sum of the roots,
?? + ?? , is twice the value of one root, which leads us to:
?? + ?? = 2
Using the formula for the sum of roots for a quadratic equation, ?? + ?? = -
?? ( ?? - ?? )
?? ( ?? - ?? )
, we set this
equal to 2 :
-
?? ( ?? - ?? )
?? ( ?? - ?? )
= 2
Solving for this:
- ???? + ???? = 2 ???? - 2 ???? 2 ???? = ???? + ???? 2 ???? = ?? ( ?? + ?? )
Given that ?? + ?? = 15 and ?? =
36
5
, substitute these into the equation:
2 ???? = 15 ?? 2 ???? = 15 ×
36
5
= 108 ???? = 54
Now, using the equation ?? + ?? = 15 and ???? = 54, find ?? 2
+ ?? 2
:
?? 2
+ ?? 2
= ( ?? + ?? )
2
- 2 ???? = 15
2
- 2 × 54 ?? 2
+ ?? 2
= 225 - 108 = 117
Therefore, ?? 2
+ ?? 2
is equal to 117 .
Q2: If the set of all ?? ? ?? - { ?? }, for which the roots of the equation ( ?? - ?? ) ?? ?? +
?? ( ?? - ?? ) ?? + ?? = ?? are positive is ( - 8 , - ?? ] ? [ ?? , ?? ) , then ?? ?? + ?? + ?? is equal to
JEE Main 2025 (Online) 2nd April Evening Shift
Ans: 7
Solution:
?? ( ?? ) = ( 1 - ?? ) ?? 2
+ 2 ( ?? - 3 ) ?? + 9 , ?? ( 0 ) = 9 > 0
?? = 0 ? 4 ( ?? - 3 )
2
= 4 ( 1 - ?? ) · 9
? ?? ? ( - 8 , - 3 ] ? [ 0 , 8 ) ( ?? )
?? 1
+ ?? 2
=
- 2 ( ?? - 3 )
1 - ?? , ?? 1
?? 2
=
9
1 - ??
?? 1
+ ?? 2
> 0 ?
?? - 3
?? - 1
> 0 ? ?? ? ( - 8 , 1 ) ? ( 3 , 8 ) … ( ???? )
?? 1
?? 2
> 0 ? 1 - ?? > 0 ? ?? ? ( - 8 , 1 ) ( ?????? )
? Interaction of (i), (ii) and (iii)
?? ? ( - 8 , - 3 ] ? [ 0 , 1 )
? ?? = 3 , ?? = 0 , ?? = 1 ? 2 ?? + ?? + ?? = 7
Q3: Let ?? ?? and ?? ?? be the distinct roots of ?? ?? ?? + ( ?? ?? ?? ? ?? ) ?? - ?? = ?? , ?? ? ( ?? , ?? ?? ) . If ??
and ?? are the minimum and the maximum values of ?? ?? ?? + ?? ?? ?? , then ???? ( ?? + ?? )
equals :
JEE Main 2025 (Online) 22nd January Evening Shift
Options:
A. 27
B. 17
C. 25
D. 24
Ans: C
Solution:
To find the sum of the fourth powers of the roots ?? ?? and ?? ?? of the quadratic equation 2 ?? 2
+
( c os ? ?? ) ?? - 1 = 0, we start analyzing the expression ?? ?? 4
+ ?? ?? 4
.
Page 3
JEE Main Previous Year Questions
(2025): Linear Inequalities
Q1: If the equation ?? ( ?? - ?? ) ?? ?? + ?? ( ?? - ?? ) ?? + ?? ( ?? - ?? ) = ?? has equal roots, where
?? + ?? = ???? and ?? =
????
?? , then ?? ?? + ?? ?? is equal to _ _ _ _
JEE Main 2025 (Online) 23rd January Morning Shift
Ans: 117
Solution:
To solve the given problem, we start with the quadratic equation:
?? ( ?? - ?? ) ?? 2
+ ?? ( ?? - ?? ) ?? + ?? ( ?? - ?? ) = 0
Given that the roots are equal (let's assume both roots are 1 ), we know the sum of the roots,
?? + ?? , is twice the value of one root, which leads us to:
?? + ?? = 2
Using the formula for the sum of roots for a quadratic equation, ?? + ?? = -
?? ( ?? - ?? )
?? ( ?? - ?? )
, we set this
equal to 2 :
-
?? ( ?? - ?? )
?? ( ?? - ?? )
= 2
Solving for this:
- ???? + ???? = 2 ???? - 2 ???? 2 ???? = ???? + ???? 2 ???? = ?? ( ?? + ?? )
Given that ?? + ?? = 15 and ?? =
36
5
, substitute these into the equation:
2 ???? = 15 ?? 2 ???? = 15 ×
36
5
= 108 ???? = 54
Now, using the equation ?? + ?? = 15 and ???? = 54, find ?? 2
+ ?? 2
:
?? 2
+ ?? 2
= ( ?? + ?? )
2
- 2 ???? = 15
2
- 2 × 54 ?? 2
+ ?? 2
= 225 - 108 = 117
Therefore, ?? 2
+ ?? 2
is equal to 117 .
Q2: If the set of all ?? ? ?? - { ?? }, for which the roots of the equation ( ?? - ?? ) ?? ?? +
?? ( ?? - ?? ) ?? + ?? = ?? are positive is ( - 8 , - ?? ] ? [ ?? , ?? ) , then ?? ?? + ?? + ?? is equal to
JEE Main 2025 (Online) 2nd April Evening Shift
Ans: 7
Solution:
?? ( ?? ) = ( 1 - ?? ) ?? 2
+ 2 ( ?? - 3 ) ?? + 9 , ?? ( 0 ) = 9 > 0
?? = 0 ? 4 ( ?? - 3 )
2
= 4 ( 1 - ?? ) · 9
? ?? ? ( - 8 , - 3 ] ? [ 0 , 8 ) ( ?? )
?? 1
+ ?? 2
=
- 2 ( ?? - 3 )
1 - ?? , ?? 1
?? 2
=
9
1 - ??
?? 1
+ ?? 2
> 0 ?
?? - 3
?? - 1
> 0 ? ?? ? ( - 8 , 1 ) ? ( 3 , 8 ) … ( ???? )
?? 1
?? 2
> 0 ? 1 - ?? > 0 ? ?? ? ( - 8 , 1 ) ( ?????? )
? Interaction of (i), (ii) and (iii)
?? ? ( - 8 , - 3 ] ? [ 0 , 1 )
? ?? = 3 , ?? = 0 , ?? = 1 ? 2 ?? + ?? + ?? = 7
Q3: Let ?? ?? and ?? ?? be the distinct roots of ?? ?? ?? + ( ?? ?? ?? ? ?? ) ?? - ?? = ?? , ?? ? ( ?? , ?? ?? ) . If ??
and ?? are the minimum and the maximum values of ?? ?? ?? + ?? ?? ?? , then ???? ( ?? + ?? )
equals :
JEE Main 2025 (Online) 22nd January Evening Shift
Options:
A. 27
B. 17
C. 25
D. 24
Ans: C
Solution:
To find the sum of the fourth powers of the roots ?? ?? and ?? ?? of the quadratic equation 2 ?? 2
+
( c os ? ?? ) ?? - 1 = 0, we start analyzing the expression ?? ?? 4
+ ?? ?? 4
.
The equation can be rewritten with its roots using:
?? + ?? = -
c os ? ?? 2
, ???? = -
1
2
We need to calculate ?? 2
+ ?? 2
and ?? 2
?? 2
:
?? 2
+ ?? 2
= ( ?? + ?? )
2
- 2 ???? = ( -
c os ? ?? 2
)
2
- 2 ( -
1
2
) =
c os
2
? ?? 4
+ 1
?? 2
?? 2
= ( ???? )
2
= ( -
1
2
)
2
=
1
4
Substitute these into:
?? 4
+ ?? 4
= ( ?? 2
+ ?? 2
)
2
- 2 ?? 2
?? 2
= (
c o s
2
? ?? 4
+ 1 )
2
-
1
2
Maximize and minimize (
c os
2
? ?? 4
+ 1 )
2
:
Zero of c os ? ?? leads to:
(
0
2
4
+ 1 )
2
= 1
Max value c os
2
? ?? = 1 :
(
1
4
+ 1 )
2
= (
5
4
)
2
=
25
16
Substitute back:
Max:
25
16
-
1
2
=
25
16
-
8
16
=
17
16
Min: 1 -
1
2
=
1
2
Finally, compute 16 ( ?? + ?? ) :
16 (
17
16
+
1
2
) = 16 (
17
16
+
8
16
) = 16 ×
25
16
= 25
Q4: The product of all the rational roots of the equation ( ?? ?? - ?? ?? + ???? )
?? - ( ?? -
?? ) ( ?? - ?? ) = ?? , is equal to
JEE Main 2025 (Online) 24th January Morning Shift
Options:
A. 7
B. 21
C. 28
D. 14
Ans: D
Solution:
To solve the given equation, start by rewriting the expression:
( ?? 2
- 9 ?? + 11 )
2
- ( ?? - 4 ) ( ?? - 5 ) = 3
First, simplify the second part of the expression:
( ?? - 4 ) ( ?? - 5 ) = ?? 2
- 9 ?? + 20
Page 4
JEE Main Previous Year Questions
(2025): Linear Inequalities
Q1: If the equation ?? ( ?? - ?? ) ?? ?? + ?? ( ?? - ?? ) ?? + ?? ( ?? - ?? ) = ?? has equal roots, where
?? + ?? = ???? and ?? =
????
?? , then ?? ?? + ?? ?? is equal to _ _ _ _
JEE Main 2025 (Online) 23rd January Morning Shift
Ans: 117
Solution:
To solve the given problem, we start with the quadratic equation:
?? ( ?? - ?? ) ?? 2
+ ?? ( ?? - ?? ) ?? + ?? ( ?? - ?? ) = 0
Given that the roots are equal (let's assume both roots are 1 ), we know the sum of the roots,
?? + ?? , is twice the value of one root, which leads us to:
?? + ?? = 2
Using the formula for the sum of roots for a quadratic equation, ?? + ?? = -
?? ( ?? - ?? )
?? ( ?? - ?? )
, we set this
equal to 2 :
-
?? ( ?? - ?? )
?? ( ?? - ?? )
= 2
Solving for this:
- ???? + ???? = 2 ???? - 2 ???? 2 ???? = ???? + ???? 2 ???? = ?? ( ?? + ?? )
Given that ?? + ?? = 15 and ?? =
36
5
, substitute these into the equation:
2 ???? = 15 ?? 2 ???? = 15 ×
36
5
= 108 ???? = 54
Now, using the equation ?? + ?? = 15 and ???? = 54, find ?? 2
+ ?? 2
:
?? 2
+ ?? 2
= ( ?? + ?? )
2
- 2 ???? = 15
2
- 2 × 54 ?? 2
+ ?? 2
= 225 - 108 = 117
Therefore, ?? 2
+ ?? 2
is equal to 117 .
Q2: If the set of all ?? ? ?? - { ?? }, for which the roots of the equation ( ?? - ?? ) ?? ?? +
?? ( ?? - ?? ) ?? + ?? = ?? are positive is ( - 8 , - ?? ] ? [ ?? , ?? ) , then ?? ?? + ?? + ?? is equal to
JEE Main 2025 (Online) 2nd April Evening Shift
Ans: 7
Solution:
?? ( ?? ) = ( 1 - ?? ) ?? 2
+ 2 ( ?? - 3 ) ?? + 9 , ?? ( 0 ) = 9 > 0
?? = 0 ? 4 ( ?? - 3 )
2
= 4 ( 1 - ?? ) · 9
? ?? ? ( - 8 , - 3 ] ? [ 0 , 8 ) ( ?? )
?? 1
+ ?? 2
=
- 2 ( ?? - 3 )
1 - ?? , ?? 1
?? 2
=
9
1 - ??
?? 1
+ ?? 2
> 0 ?
?? - 3
?? - 1
> 0 ? ?? ? ( - 8 , 1 ) ? ( 3 , 8 ) … ( ???? )
?? 1
?? 2
> 0 ? 1 - ?? > 0 ? ?? ? ( - 8 , 1 ) ( ?????? )
? Interaction of (i), (ii) and (iii)
?? ? ( - 8 , - 3 ] ? [ 0 , 1 )
? ?? = 3 , ?? = 0 , ?? = 1 ? 2 ?? + ?? + ?? = 7
Q3: Let ?? ?? and ?? ?? be the distinct roots of ?? ?? ?? + ( ?? ?? ?? ? ?? ) ?? - ?? = ?? , ?? ? ( ?? , ?? ?? ) . If ??
and ?? are the minimum and the maximum values of ?? ?? ?? + ?? ?? ?? , then ???? ( ?? + ?? )
equals :
JEE Main 2025 (Online) 22nd January Evening Shift
Options:
A. 27
B. 17
C. 25
D. 24
Ans: C
Solution:
To find the sum of the fourth powers of the roots ?? ?? and ?? ?? of the quadratic equation 2 ?? 2
+
( c os ? ?? ) ?? - 1 = 0, we start analyzing the expression ?? ?? 4
+ ?? ?? 4
.
The equation can be rewritten with its roots using:
?? + ?? = -
c os ? ?? 2
, ???? = -
1
2
We need to calculate ?? 2
+ ?? 2
and ?? 2
?? 2
:
?? 2
+ ?? 2
= ( ?? + ?? )
2
- 2 ???? = ( -
c os ? ?? 2
)
2
- 2 ( -
1
2
) =
c os
2
? ?? 4
+ 1
?? 2
?? 2
= ( ???? )
2
= ( -
1
2
)
2
=
1
4
Substitute these into:
?? 4
+ ?? 4
= ( ?? 2
+ ?? 2
)
2
- 2 ?? 2
?? 2
= (
c o s
2
? ?? 4
+ 1 )
2
-
1
2
Maximize and minimize (
c os
2
? ?? 4
+ 1 )
2
:
Zero of c os ? ?? leads to:
(
0
2
4
+ 1 )
2
= 1
Max value c os
2
? ?? = 1 :
(
1
4
+ 1 )
2
= (
5
4
)
2
=
25
16
Substitute back:
Max:
25
16
-
1
2
=
25
16
-
8
16
=
17
16
Min: 1 -
1
2
=
1
2
Finally, compute 16 ( ?? + ?? ) :
16 (
17
16
+
1
2
) = 16 (
17
16
+
8
16
) = 16 ×
25
16
= 25
Q4: The product of all the rational roots of the equation ( ?? ?? - ?? ?? + ???? )
?? - ( ?? -
?? ) ( ?? - ?? ) = ?? , is equal to
JEE Main 2025 (Online) 24th January Morning Shift
Options:
A. 7
B. 21
C. 28
D. 14
Ans: D
Solution:
To solve the given equation, start by rewriting the expression:
( ?? 2
- 9 ?? + 11 )
2
- ( ?? - 4 ) ( ?? - 5 ) = 3
First, simplify the second part of the expression:
( ?? - 4 ) ( ?? - 5 ) = ?? 2
- 9 ?? + 20
Now the equation becomes:
( ?? 2
- 9 ?? + 11 )
2
- ( ?? 2
- 9 ?? + 20 ) = 3
Introduce a substitution for simplification:
Let ?? = ?? 2
- 9 ??
Thus, the equation transforms to:
?? 2
+ 22 ?? + 121 - ?? - 20 - 3 = 0
Simplify further:
?? 2
+ 21 ?? + 98 = 0
Factor the quadratic:
( ?? + 14 ) ( ?? + 7 ) = 0
This gives:
?? = - 7 ? or ? ?? = - 14
Address each case where ?? = ?? 2
- 9 ?? :
?? 2
- 9 ?? = - 7
?? 2
- 9 ?? + 7 = 0
Solving this quadratic equation, we find the roots:
?? =
9 ± v 81 - 4 × 7
2
=
9 ± v 53
2
?? 2
- 9 ?? = - 14
?? 2
- 9 ?? + 14 = 0
Solving this quadratic equation:
?? =
9 ± v 81 - 4 × 14
2
=
9 ± v 25
2
?? =
9 ± 5
2
= 7 ? or ? ?? = 2
The rational roots from the second equation are 7 and 2 . Thus, the product of all the rational
roots is:
7 × 2 = 14
Q5: The number of real solution(s) of the equation
?? ?? + ?? ?? + ?? = ?? ???? { | ?? - ?? | , | ?? + ?? | } is :
JEE Main 2025 (Online) 24th January Evening Shift
Options:
A. 2
B. 3
C. 1
D. 0
Ans: A
Solution:
Page 5
JEE Main Previous Year Questions
(2025): Linear Inequalities
Q1: If the equation ?? ( ?? - ?? ) ?? ?? + ?? ( ?? - ?? ) ?? + ?? ( ?? - ?? ) = ?? has equal roots, where
?? + ?? = ???? and ?? =
????
?? , then ?? ?? + ?? ?? is equal to _ _ _ _
JEE Main 2025 (Online) 23rd January Morning Shift
Ans: 117
Solution:
To solve the given problem, we start with the quadratic equation:
?? ( ?? - ?? ) ?? 2
+ ?? ( ?? - ?? ) ?? + ?? ( ?? - ?? ) = 0
Given that the roots are equal (let's assume both roots are 1 ), we know the sum of the roots,
?? + ?? , is twice the value of one root, which leads us to:
?? + ?? = 2
Using the formula for the sum of roots for a quadratic equation, ?? + ?? = -
?? ( ?? - ?? )
?? ( ?? - ?? )
, we set this
equal to 2 :
-
?? ( ?? - ?? )
?? ( ?? - ?? )
= 2
Solving for this:
- ???? + ???? = 2 ???? - 2 ???? 2 ???? = ???? + ???? 2 ???? = ?? ( ?? + ?? )
Given that ?? + ?? = 15 and ?? =
36
5
, substitute these into the equation:
2 ???? = 15 ?? 2 ???? = 15 ×
36
5
= 108 ???? = 54
Now, using the equation ?? + ?? = 15 and ???? = 54, find ?? 2
+ ?? 2
:
?? 2
+ ?? 2
= ( ?? + ?? )
2
- 2 ???? = 15
2
- 2 × 54 ?? 2
+ ?? 2
= 225 - 108 = 117
Therefore, ?? 2
+ ?? 2
is equal to 117 .
Q2: If the set of all ?? ? ?? - { ?? }, for which the roots of the equation ( ?? - ?? ) ?? ?? +
?? ( ?? - ?? ) ?? + ?? = ?? are positive is ( - 8 , - ?? ] ? [ ?? , ?? ) , then ?? ?? + ?? + ?? is equal to
JEE Main 2025 (Online) 2nd April Evening Shift
Ans: 7
Solution:
?? ( ?? ) = ( 1 - ?? ) ?? 2
+ 2 ( ?? - 3 ) ?? + 9 , ?? ( 0 ) = 9 > 0
?? = 0 ? 4 ( ?? - 3 )
2
= 4 ( 1 - ?? ) · 9
? ?? ? ( - 8 , - 3 ] ? [ 0 , 8 ) ( ?? )
?? 1
+ ?? 2
=
- 2 ( ?? - 3 )
1 - ?? , ?? 1
?? 2
=
9
1 - ??
?? 1
+ ?? 2
> 0 ?
?? - 3
?? - 1
> 0 ? ?? ? ( - 8 , 1 ) ? ( 3 , 8 ) … ( ???? )
?? 1
?? 2
> 0 ? 1 - ?? > 0 ? ?? ? ( - 8 , 1 ) ( ?????? )
? Interaction of (i), (ii) and (iii)
?? ? ( - 8 , - 3 ] ? [ 0 , 1 )
? ?? = 3 , ?? = 0 , ?? = 1 ? 2 ?? + ?? + ?? = 7
Q3: Let ?? ?? and ?? ?? be the distinct roots of ?? ?? ?? + ( ?? ?? ?? ? ?? ) ?? - ?? = ?? , ?? ? ( ?? , ?? ?? ) . If ??
and ?? are the minimum and the maximum values of ?? ?? ?? + ?? ?? ?? , then ???? ( ?? + ?? )
equals :
JEE Main 2025 (Online) 22nd January Evening Shift
Options:
A. 27
B. 17
C. 25
D. 24
Ans: C
Solution:
To find the sum of the fourth powers of the roots ?? ?? and ?? ?? of the quadratic equation 2 ?? 2
+
( c os ? ?? ) ?? - 1 = 0, we start analyzing the expression ?? ?? 4
+ ?? ?? 4
.
The equation can be rewritten with its roots using:
?? + ?? = -
c os ? ?? 2
, ???? = -
1
2
We need to calculate ?? 2
+ ?? 2
and ?? 2
?? 2
:
?? 2
+ ?? 2
= ( ?? + ?? )
2
- 2 ???? = ( -
c os ? ?? 2
)
2
- 2 ( -
1
2
) =
c os
2
? ?? 4
+ 1
?? 2
?? 2
= ( ???? )
2
= ( -
1
2
)
2
=
1
4
Substitute these into:
?? 4
+ ?? 4
= ( ?? 2
+ ?? 2
)
2
- 2 ?? 2
?? 2
= (
c o s
2
? ?? 4
+ 1 )
2
-
1
2
Maximize and minimize (
c os
2
? ?? 4
+ 1 )
2
:
Zero of c os ? ?? leads to:
(
0
2
4
+ 1 )
2
= 1
Max value c os
2
? ?? = 1 :
(
1
4
+ 1 )
2
= (
5
4
)
2
=
25
16
Substitute back:
Max:
25
16
-
1
2
=
25
16
-
8
16
=
17
16
Min: 1 -
1
2
=
1
2
Finally, compute 16 ( ?? + ?? ) :
16 (
17
16
+
1
2
) = 16 (
17
16
+
8
16
) = 16 ×
25
16
= 25
Q4: The product of all the rational roots of the equation ( ?? ?? - ?? ?? + ???? )
?? - ( ?? -
?? ) ( ?? - ?? ) = ?? , is equal to
JEE Main 2025 (Online) 24th January Morning Shift
Options:
A. 7
B. 21
C. 28
D. 14
Ans: D
Solution:
To solve the given equation, start by rewriting the expression:
( ?? 2
- 9 ?? + 11 )
2
- ( ?? - 4 ) ( ?? - 5 ) = 3
First, simplify the second part of the expression:
( ?? - 4 ) ( ?? - 5 ) = ?? 2
- 9 ?? + 20
Now the equation becomes:
( ?? 2
- 9 ?? + 11 )
2
- ( ?? 2
- 9 ?? + 20 ) = 3
Introduce a substitution for simplification:
Let ?? = ?? 2
- 9 ??
Thus, the equation transforms to:
?? 2
+ 22 ?? + 121 - ?? - 20 - 3 = 0
Simplify further:
?? 2
+ 21 ?? + 98 = 0
Factor the quadratic:
( ?? + 14 ) ( ?? + 7 ) = 0
This gives:
?? = - 7 ? or ? ?? = - 14
Address each case where ?? = ?? 2
- 9 ?? :
?? 2
- 9 ?? = - 7
?? 2
- 9 ?? + 7 = 0
Solving this quadratic equation, we find the roots:
?? =
9 ± v 81 - 4 × 7
2
=
9 ± v 53
2
?? 2
- 9 ?? = - 14
?? 2
- 9 ?? + 14 = 0
Solving this quadratic equation:
?? =
9 ± v 81 - 4 × 14
2
=
9 ± v 25
2
?? =
9 ± 5
2
= 7 ? or ? ?? = 2
The rational roots from the second equation are 7 and 2 . Thus, the product of all the rational
roots is:
7 × 2 = 14
Q5: The number of real solution(s) of the equation
?? ?? + ?? ?? + ?? = ?? ???? { | ?? - ?? | , | ?? + ?? | } is :
JEE Main 2025 (Online) 24th January Evening Shift
Options:
A. 2
B. 3
C. 1
D. 0
Ans: A
Solution:
Only 2 solutions.
Q6: The sum, of the squares of all the roots of the equation ?? ?? + | ?? ?? - ?? | - ?? = ?? , is
JEE Main 2025 (Online) 28th January Morning Shift
Options:
A. 6 ( 2 - v 2 )
B. 3 ( 3 - v 2 )
C. 3 ( 2 - v 2 )
D. 6 ( 3 - v 2 )
Ans: A
Solution:
To find the sum of the squares of all the roots of the equation ?? 2
+ | 2 ?? - 3 | - 4 = 0 :
Case I: ?? =
3
2
For ?? =
3
2
, the expression | 2 ?? - 3 | becomes 2 ?? - 3. Thus, the equation becomes:
?? 2
+ 2 ?? - 3 - 4 = 0
Simplifying gives:
?? 2
+ 2 ?? - 7 = 0
Solving this quadratic equation, we find:
?? = 2 v 2 - 1
Case II: ?? <
3
2
For ?? <
3
2
, the expression | 2 ?? - 3 | becomes - ( 2 ?? - 3 ) = - 2 ?? + 3. The equation therefore
becomes:
?? 2
+ 3 - 2 ?? - 4 = 0
Simplifying gives:
Read MoreFAQs on Inequalities: JEE Main Previous Year Questions (2021-2026)
| 1. How do I identify which inequality technique to use for JEE Main problems? |  |
Ans. Different JEE Main inequality problems require distinct approaches: AM-GM inequality works best for product-to-sum conversions, Cauchy-Schwarz handles squared terms and weighted sums, and algebraic manipulation suits direct comparisons. Recognising whether variables are positive, whether constraints exist, and the inequality's structure determines the optimal method. Practice with previous year questions (2021-2026) to develop pattern recognition skills quickly.
| 2. What's the difference between solving strict and non-strict inequalities in JEE exams? | |
Ans. Strict inequalities (< or >) exclude boundary values, affecting solution intervals and final answers significantly. Non-strict inequalities (≤ or ≥) include endpoints, often changing whether equality cases count as valid solutions. In JEE Main problems, boundary conditions frequently appear in answer choices, making this distinction critical for scoring marks accurately and avoiding common mistakes.
| 3. Why do AM-GM and Cauchy-Schwarz inequality problems keep appearing in JEE Main? | |
Ans. AM-GM and Cauchy-Schwarz are foundational tools for proving inequalities efficiently in competitive exams, appearing repeatedly across JEE Main papers (2021-2026). They simplify complex algebraic expressions, reduce calculation time, and test conceptual understanding rather than mechanical solving. Mastering these classical inequalities unlocks multiple problem types and boosts overall exam performance significantly.
| 4. How do I avoid sign errors when working with inequalities involving negative numbers? | |
Ans. Multiplying or dividing both sides by negative numbers reverses the inequality sign-a critical rule students frequently forget. Always isolate negative terms carefully, verify direction changes at each step, and test boundary values by substitution. Refer to flashcards and mind maps on EduRev covering inequality properties to reinforce this concept through visual reinforcement and practice examples.
| 5. Which inequality topics appear most frequently in JEE Main previous year papers between 2021 and 2026? | |
Ans. Quadratic inequalities, modulus-based inequalities, and rational inequality problems dominate JEE Main papers from 2021-2026. Problems combining inequalities with quadratic functions, absolute values, and fraction analysis appear consistently across question sets. Analysing previous year solutions reveals these patterns, helping students prioritise high-yield topics and allocate study time efficiently for maximum marks potential.
About this Document
4.71/5 Rating
Apr 25, 2026 Last updated
Related Exams
Document Description: Inequalities: JEE Main Previous Year Questions (2021-2026) for JEE 2026 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The notes and questions for Inequalities: JEE Main Previous Year Questions (2021-2026) have been prepared according to the JEE exam syllabus. Information about Inequalities: JEE Main Previous Year Questions (2021-2026) covers topics like and Inequalities: JEE Main Previous Year Questions (2021-2026) Example, for JEE 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Inequalities: JEE Main Previous Year Questions (2021-2026).
Introduction of Inequalities: JEE Main Previous Year Questions (2021-2026) in English is available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & Inequalities: JEE Main Previous Year Questions (2021-2026) in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: Inequalities: JEE Main Previous Year Questions (2021-2026)
Description
Inequalities: JEE Main Previous Year Questions (2021 of Maths is important for the exam. Get access to all the questions with solutions asked in past years of JEE exam. Download it from EduRev.
Information about Inequalities: JEE Main Previous Year Questions (2021-2026)
In this doc you can find the meaning of Inequalities: JEE Main Previous Year Questions (2021-2026) defined & explained in the simplest way possible. Besides explaining types of Inequalities: JEE Main Previous Year Questions (2021-2026) theory, EduRev gives you an ample number of questions to practice Inequalities: JEE Main Previous Year Questions (2021-2026) tests, examples and also practice JEE tests
Related Searches
Free, video lectures, Viva Questions, ppt, Inequalities: JEE Main Previous Year Questions (2021-2026), Sample Paper, Semester Notes, mock tests for examination, Previous Year Questions with Solutions, Objective type Questions, shortcuts and tricks, Important questions, pdf , Inequalities: JEE Main Previous Year Questions (2021-2026), past year papers, Extra Questions, Summary, Exam, study material, MCQs, practice quizzes, Inequalities: JEE Main Previous Year Questions (2021-2026);