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JEE Main Previous Year Questions (2025): Continuity and Differentiability
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JEE Main Previous Year Questions
(2025): Continuity and
Differentiability
Q1: Let the function,
?? (?? )={
-?? ???? ?? -?? , ?? <?? ?? ?? +?? ?? , ?? ???
be differentiable for all ?? ??? , where ?? >?? , ?? ??? . If the area of the region enclosed
by ?? =?? (?? ) and the line ?? =-???? is ?? +?? v?? ,?? ,?? ??? , then the value of ?? +?? is
____ .
JEE Main 2025 (Online) 22nd January Morning Shift
Ans: 34
Solution:
f(x) is continuous and differentiable
at ?? =1; LHL=RHL,LHD=RHD
-3?? -2=?? 2
+?? ,-6?? =?? ?? =2,1;?? =-12
?? (?? )={
-6?? 2
-2, ?? <1
4-12?? , ?? =1
Area =? ?
1
-v3
?(-6?? 2
-2+20)???? +? ?
2
1
?(4-12?? +20)?? ?? ]
=16+12v3+6=22+12v3
? ?? +?? =34
Page 2
JEE Main Previous Year Questions
(2025): Continuity and
Differentiability
Q1: Let the function,
?? (?? )={
-?? ???? ?? -?? , ?? <?? ?? ?? +?? ?? , ?? ???
be differentiable for all ?? ??? , where ?? >?? , ?? ??? . If the area of the region enclosed
by ?? =?? (?? ) and the line ?? =-???? is ?? +?? v?? ,?? ,?? ??? , then the value of ?? +?? is
____ .
JEE Main 2025 (Online) 22nd January Morning Shift
Ans: 34
Solution:
f(x) is continuous and differentiable
at ?? =1; LHL=RHL,LHD=RHD
-3?? -2=?? 2
+?? ,-6?? =?? ?? =2,1;?? =-12
?? (?? )={
-6?? 2
-2, ?? <1
4-12?? , ?? =1
Area =? ?
1
-v3
?(-6?? 2
-2+20)???? +? ?
2
1
?(4-12?? +20)?? ?? ]
=16+12v3+6=22+12v3
? ?? +?? =34
Q2: Let ?? (?? )={
?? ?? , ?? <?? ?????? {?? +?? +[?? ],?? +?? [?? ]}, ?? =?? =?? ?? , ?? >??
where [.] denotes greatest integer function. If ?? and ?? are the number of points,
where ?? is not continuous and is not differentiable, respectively, then ?? +?? equals
____ .
Ans: 5
Solution:
?? (?? )={
3?? ; ?? <0
min{1+?? ,?? } ; 0=?? <1
min{2+?? ,?? +2} ; 1=?? <2
?? (?? )={
3x ;?? <0
x ;
x+2 ;1=?? <1
5 ;?? >2
Not continuous at x?{1,2}??? =2
Not diff. at ?? ?{0,1,2}??? =3
?? +?? =5
Q3: Let ?? (?? )=?????? ?? ?8
??
?? =?? ?? ?(
?????? (?? /?? ?? +?? )+??????
?? (?? /?? ?? +?? )
?? -??????
?? (?? /?? ?? +?? )
) Then ?????? ?? ??? ?
?? ?? -?? ?? (?? )
(?? -?? (?? ))
is equal to
____ .
JEE Main 2025 (Online) 28th January Evening Shift
Ans: 1
Page 3
JEE Main Previous Year Questions
(2025): Continuity and
Differentiability
Q1: Let the function,
?? (?? )={
-?? ???? ?? -?? , ?? <?? ?? ?? +?? ?? , ?? ???
be differentiable for all ?? ??? , where ?? >?? , ?? ??? . If the area of the region enclosed
by ?? =?? (?? ) and the line ?? =-???? is ?? +?? v?? ,?? ,?? ??? , then the value of ?? +?? is
____ .
JEE Main 2025 (Online) 22nd January Morning Shift
Ans: 34
Solution:
f(x) is continuous and differentiable
at ?? =1; LHL=RHL,LHD=RHD
-3?? -2=?? 2
+?? ,-6?? =?? ?? =2,1;?? =-12
?? (?? )={
-6?? 2
-2, ?? <1
4-12?? , ?? =1
Area =? ?
1
-v3
?(-6?? 2
-2+20)???? +? ?
2
1
?(4-12?? +20)?? ?? ]
=16+12v3+6=22+12v3
? ?? +?? =34
Q2: Let ?? (?? )={
?? ?? , ?? <?? ?????? {?? +?? +[?? ],?? +?? [?? ]}, ?? =?? =?? ?? , ?? >??
where [.] denotes greatest integer function. If ?? and ?? are the number of points,
where ?? is not continuous and is not differentiable, respectively, then ?? +?? equals
____ .
Ans: 5
Solution:
?? (?? )={
3?? ; ?? <0
min{1+?? ,?? } ; 0=?? <1
min{2+?? ,?? +2} ; 1=?? <2
?? (?? )={
3x ;?? <0
x ;
x+2 ;1=?? <1
5 ;?? >2
Not continuous at x?{1,2}??? =2
Not diff. at ?? ?{0,1,2}??? =3
?? +?? =5
Q3: Let ?? (?? )=?????? ?? ?8
??
?? =?? ?? ?(
?????? (?? /?? ?? +?? )+??????
?? (?? /?? ?? +?? )
?? -??????
?? (?? /?? ?? +?? )
) Then ?????? ?? ??? ?
?? ?? -?? ?? (?? )
(?? -?? (?? ))
is equal to
____ .
JEE Main 2025 (Online) 28th January Evening Shift
Ans: 1
Solution:
?? (?? )=lim
?? ?8
??
?? =0
?? ?(tan
?? 2
?? -tan
?? 2
?? +1
)=tan ??
lim
?? ?0
?(
?? ?? -?? tan ?? ?? -tan ?? )=lim
?? ?0
??? tan ?? (?? ?? -tan ?? -1)
(?? -tan ?? )
=1
Q4: Let [?? ] be the greatest integer less than or equal to ?? . Then the least value of ?? ?
?? for which ?????? ?? ??? +?(?? ([
?? ?? ]+[
?? ?? ]+?+[
?? ?? ])-?? ?? ([
?? ?? ?? ]+[
?? ?? ?? ?? ]+?+[
?? ?? ?? ?? ])=?? is
equal to ____ .
JEE Main 2025 (Online) 29th January Morning Shift
Ans: 24
Solution:
To find the least natural number ?? for which the following inequality holds:
lim
?? ?0
+?(?? ([
1
?? ]+[
2
?? ]+?+[
?? ?? ])-?? 2
([
1
?? 2
]+[
2
2
?? 2
]+?+[
9
2
?? 2
]))=1
we simplify the expression inside the limit.
As ?? ?0
+
,[
?? ?? ] approximates to
?? ?? . Thus, the problem becomes finding:
(1+2+?+?? )-(1
2
+2
2
+?+9
2
)=1
The sum of the first ?? natural numbers is given by:
?? (?? +1)
2
And the sum of the squares of the first 9 natural numbers is:
1
2
+2
2
+?+9
2
=
9·10·19
6
Thus, the inequality becomes:
?? (?? +1)
2
-
9·10·19
6
=1
Solving this, we rewrite:
?? (?? +1)=572
The least natural number ?? satisfying this condition is 24 .
Q5: If ?????? ?? ??? ?(
?????? ?? ?? )
?? ?? ?? =?? , then ???? ?????? ?? ?? is equal to ____
JEE Main 2025 (Online) 3rd April Evening Shift
Ans: 32
Solution:
Page 4
JEE Main Previous Year Questions
(2025): Continuity and
Differentiability
Q1: Let the function,
?? (?? )={
-?? ???? ?? -?? , ?? <?? ?? ?? +?? ?? , ?? ???
be differentiable for all ?? ??? , where ?? >?? , ?? ??? . If the area of the region enclosed
by ?? =?? (?? ) and the line ?? =-???? is ?? +?? v?? ,?? ,?? ??? , then the value of ?? +?? is
____ .
JEE Main 2025 (Online) 22nd January Morning Shift
Ans: 34
Solution:
f(x) is continuous and differentiable
at ?? =1; LHL=RHL,LHD=RHD
-3?? -2=?? 2
+?? ,-6?? =?? ?? =2,1;?? =-12
?? (?? )={
-6?? 2
-2, ?? <1
4-12?? , ?? =1
Area =? ?
1
-v3
?(-6?? 2
-2+20)???? +? ?
2
1
?(4-12?? +20)?? ?? ]
=16+12v3+6=22+12v3
? ?? +?? =34
Q2: Let ?? (?? )={
?? ?? , ?? <?? ?????? {?? +?? +[?? ],?? +?? [?? ]}, ?? =?? =?? ?? , ?? >??
where [.] denotes greatest integer function. If ?? and ?? are the number of points,
where ?? is not continuous and is not differentiable, respectively, then ?? +?? equals
____ .
Ans: 5
Solution:
?? (?? )={
3?? ; ?? <0
min{1+?? ,?? } ; 0=?? <1
min{2+?? ,?? +2} ; 1=?? <2
?? (?? )={
3x ;?? <0
x ;
x+2 ;1=?? <1
5 ;?? >2
Not continuous at x?{1,2}??? =2
Not diff. at ?? ?{0,1,2}??? =3
?? +?? =5
Q3: Let ?? (?? )=?????? ?? ?8
??
?? =?? ?? ?(
?????? (?? /?? ?? +?? )+??????
?? (?? /?? ?? +?? )
?? -??????
?? (?? /?? ?? +?? )
) Then ?????? ?? ??? ?
?? ?? -?? ?? (?? )
(?? -?? (?? ))
is equal to
____ .
JEE Main 2025 (Online) 28th January Evening Shift
Ans: 1
Solution:
?? (?? )=lim
?? ?8
??
?? =0
?? ?(tan
?? 2
?? -tan
?? 2
?? +1
)=tan ??
lim
?? ?0
?(
?? ?? -?? tan ?? ?? -tan ?? )=lim
?? ?0
??? tan ?? (?? ?? -tan ?? -1)
(?? -tan ?? )
=1
Q4: Let [?? ] be the greatest integer less than or equal to ?? . Then the least value of ?? ?
?? for which ?????? ?? ??? +?(?? ([
?? ?? ]+[
?? ?? ]+?+[
?? ?? ])-?? ?? ([
?? ?? ?? ]+[
?? ?? ?? ?? ]+?+[
?? ?? ?? ?? ])=?? is
equal to ____ .
JEE Main 2025 (Online) 29th January Morning Shift
Ans: 24
Solution:
To find the least natural number ?? for which the following inequality holds:
lim
?? ?0
+?(?? ([
1
?? ]+[
2
?? ]+?+[
?? ?? ])-?? 2
([
1
?? 2
]+[
2
2
?? 2
]+?+[
9
2
?? 2
]))=1
we simplify the expression inside the limit.
As ?? ?0
+
,[
?? ?? ] approximates to
?? ?? . Thus, the problem becomes finding:
(1+2+?+?? )-(1
2
+2
2
+?+9
2
)=1
The sum of the first ?? natural numbers is given by:
?? (?? +1)
2
And the sum of the squares of the first 9 natural numbers is:
1
2
+2
2
+?+9
2
=
9·10·19
6
Thus, the inequality becomes:
?? (?? +1)
2
-
9·10·19
6
=1
Solving this, we rewrite:
?? (?? +1)=572
The least natural number ?? satisfying this condition is 24 .
Q5: If ?????? ?? ??? ?(
?????? ?? ?? )
?? ?? ?? =?? , then ???? ?????? ?? ?? is equal to ____
JEE Main 2025 (Online) 3rd April Evening Shift
Ans: 32
Solution:
To solve the given limit problem, we start by analyzing the expression:
lim
?? ?0
?(
tan ?? ?? )
1
?? 2
=??
This limit exhibits the indeterminate form 1
8
. To handle this form, we use the transformation:
?? =?? lim
?? ?0
?(
tan ?? ?? -1)
1
?? 2
Expanding tan ?? using its Taylor series near ?? =0, we have:
tan ?? =?? +
?? 3
3
+
2
15
?? 5
+?
Substituting this expansion into the limit, we get:
tan ?? -?? ?? 3
=
(?? +
?? 3
3
+
2
15
?? 5
+?-?? )
?? 3
=
?? 3
3
+
2
15
?? 5
+?
?? 3
This simplifies to:
?? 3
3?? 3
=
1
3
Thus, the limit becomes:
?? =?? 1
3
log
?? ?? =
1
3
Finally, computing 96log
?? ?? :
96log
?? ?? =96·
1
3
=32
Q6: Let ?? and ?? be the number of points at which the function ?? (?? )=
?????? {?? ,?? ?? ,?? ?? ,…?? ????
},?? ?R, is not differentiable and not continuous, respectively.
Then ?? +?? is equal to ____ .
JEE Main 2025 (Online) 4th April Morning Shift
Ans: 3
Solution:
for ?? =1,?? 21
=?? 19
=?=?? .
?? (?? )={
?? ?? <-1
?? 21
-1=?? =0
?? 0<?? <1
?? 21
?? =1
Page 5
JEE Main Previous Year Questions
(2025): Continuity and
Differentiability
Q1: Let the function,
?? (?? )={
-?? ???? ?? -?? , ?? <?? ?? ?? +?? ?? , ?? ???
be differentiable for all ?? ??? , where ?? >?? , ?? ??? . If the area of the region enclosed
by ?? =?? (?? ) and the line ?? =-???? is ?? +?? v?? ,?? ,?? ??? , then the value of ?? +?? is
____ .
JEE Main 2025 (Online) 22nd January Morning Shift
Ans: 34
Solution:
f(x) is continuous and differentiable
at ?? =1; LHL=RHL,LHD=RHD
-3?? -2=?? 2
+?? ,-6?? =?? ?? =2,1;?? =-12
?? (?? )={
-6?? 2
-2, ?? <1
4-12?? , ?? =1
Area =? ?
1
-v3
?(-6?? 2
-2+20)???? +? ?
2
1
?(4-12?? +20)?? ?? ]
=16+12v3+6=22+12v3
? ?? +?? =34
Q2: Let ?? (?? )={
?? ?? , ?? <?? ?????? {?? +?? +[?? ],?? +?? [?? ]}, ?? =?? =?? ?? , ?? >??
where [.] denotes greatest integer function. If ?? and ?? are the number of points,
where ?? is not continuous and is not differentiable, respectively, then ?? +?? equals
____ .
Ans: 5
Solution:
?? (?? )={
3?? ; ?? <0
min{1+?? ,?? } ; 0=?? <1
min{2+?? ,?? +2} ; 1=?? <2
?? (?? )={
3x ;?? <0
x ;
x+2 ;1=?? <1
5 ;?? >2
Not continuous at x?{1,2}??? =2
Not diff. at ?? ?{0,1,2}??? =3
?? +?? =5
Q3: Let ?? (?? )=?????? ?? ?8
??
?? =?? ?? ?(
?????? (?? /?? ?? +?? )+??????
?? (?? /?? ?? +?? )
?? -??????
?? (?? /?? ?? +?? )
) Then ?????? ?? ??? ?
?? ?? -?? ?? (?? )
(?? -?? (?? ))
is equal to
____ .
JEE Main 2025 (Online) 28th January Evening Shift
Ans: 1
Solution:
?? (?? )=lim
?? ?8
??
?? =0
?? ?(tan
?? 2
?? -tan
?? 2
?? +1
)=tan ??
lim
?? ?0
?(
?? ?? -?? tan ?? ?? -tan ?? )=lim
?? ?0
??? tan ?? (?? ?? -tan ?? -1)
(?? -tan ?? )
=1
Q4: Let [?? ] be the greatest integer less than or equal to ?? . Then the least value of ?? ?
?? for which ?????? ?? ??? +?(?? ([
?? ?? ]+[
?? ?? ]+?+[
?? ?? ])-?? ?? ([
?? ?? ?? ]+[
?? ?? ?? ?? ]+?+[
?? ?? ?? ?? ])=?? is
equal to ____ .
JEE Main 2025 (Online) 29th January Morning Shift
Ans: 24
Solution:
To find the least natural number ?? for which the following inequality holds:
lim
?? ?0
+?(?? ([
1
?? ]+[
2
?? ]+?+[
?? ?? ])-?? 2
([
1
?? 2
]+[
2
2
?? 2
]+?+[
9
2
?? 2
]))=1
we simplify the expression inside the limit.
As ?? ?0
+
,[
?? ?? ] approximates to
?? ?? . Thus, the problem becomes finding:
(1+2+?+?? )-(1
2
+2
2
+?+9
2
)=1
The sum of the first ?? natural numbers is given by:
?? (?? +1)
2
And the sum of the squares of the first 9 natural numbers is:
1
2
+2
2
+?+9
2
=
9·10·19
6
Thus, the inequality becomes:
?? (?? +1)
2
-
9·10·19
6
=1
Solving this, we rewrite:
?? (?? +1)=572
The least natural number ?? satisfying this condition is 24 .
Q5: If ?????? ?? ??? ?(
?????? ?? ?? )
?? ?? ?? =?? , then ???? ?????? ?? ?? is equal to ____
JEE Main 2025 (Online) 3rd April Evening Shift
Ans: 32
Solution:
To solve the given limit problem, we start by analyzing the expression:
lim
?? ?0
?(
tan ?? ?? )
1
?? 2
=??
This limit exhibits the indeterminate form 1
8
. To handle this form, we use the transformation:
?? =?? lim
?? ?0
?(
tan ?? ?? -1)
1
?? 2
Expanding tan ?? using its Taylor series near ?? =0, we have:
tan ?? =?? +
?? 3
3
+
2
15
?? 5
+?
Substituting this expansion into the limit, we get:
tan ?? -?? ?? 3
=
(?? +
?? 3
3
+
2
15
?? 5
+?-?? )
?? 3
=
?? 3
3
+
2
15
?? 5
+?
?? 3
This simplifies to:
?? 3
3?? 3
=
1
3
Thus, the limit becomes:
?? =?? 1
3
log
?? ?? =
1
3
Finally, computing 96log
?? ?? :
96log
?? ?? =96·
1
3
=32
Q6: Let ?? and ?? be the number of points at which the function ?? (?? )=
?????? {?? ,?? ?? ,?? ?? ,…?? ????
},?? ?R, is not differentiable and not continuous, respectively.
Then ?? +?? is equal to ____ .
JEE Main 2025 (Online) 4th April Morning Shift
Ans: 3
Solution:
for ?? =1,?? 21
=?? 19
=?=?? .
?? (?? )={
?? ?? <-1
?? 21
-1=?? =0
?? 0<?? <1
?? 21
?? =1
Clearly, ?? (?? ) is continuous everywhere.
??? =0{1 ;?? <-1
?? '
(?? )={
21?? 20
;-1=?? =0
1 ;0<?? <1
21·?? 20
;?? =1
??? =3
??? +?? =3
Q7: The number of points of discontinuity of the function ?? (?? )=[
?? ?? ?? ]-[v?? ],?? ?
[?? ,?? ], where [·] denotes the greatest integer function, is ____ .
Solution:
To determine the points of discontinuity of the function ?? (?? )=[
?? 2
2
]-[v?? ], where [.] denotes
the greatest integer function, we need to identify possible values of ?? where discontinuities
might occur within the interval [0,4].
Discontinuity Analysis
For the term [
?? 2
2
] :
The probable values of ?? that could cause discontinuities are the roots or specific values where
the integer part changes between consecutive integers. The transitions happen when:
=1,2,3,4,5,6,7,8
??? =v2,2,v6,2v2,v10,2v3,v14,4
For the term [v?? ] :
The values of ?? where [v?? ] changes are straightforward. They occur at:
?? =1,2
Discontinuity Check
Read MoreFAQs on JEE Main Previous Year Questions (2025): Continuity and Differentiability
| 1. What is the definition of continuity in mathematics? |  |
Ans. Continuity in mathematics refers to a function that is continuous at a point if the following three conditions are met: the function is defined at that point, the limit of the function as it approaches that point exists, and the limit equals the function's value at that point. Essentially, a continuous function has no breaks, jumps, or holes in its graph.
| 2. How can we determine if a function is differentiable? | |
Ans. A function is differentiable at a point if it has a defined derivative at that point. This means that the function must be continuous at that point, and the limit of the difference quotient must exist as the interval approaches zero. In simpler terms, the graph of the function must have a tangent at that point, without any sharp corners or vertical tangents.
| 3. What are the conditions for a function to be continuous over an interval? | |
Ans. For a function to be continuous over an interval, it must be continuous at every point within that interval. This means that for each point in the interval, the function must be defined, the limit must exist, and the limit must equal the function's value. This applies to both closed intervals (including endpoints) and open intervals (excluding endpoints).
| 4. Can a function be continuous but not differentiable? | |
Ans. Yes, a function can be continuous but not differentiable. A classic example is the absolute value function, f(x) = |x|. It is continuous everywhere but not differentiable at x = 0 because there is a sharp corner at that point where the slope changes abruptly. Thus, while continuity is a requirement for differentiability, it is not a sufficient condition.
| 5. What is the significance of the Intermediate Value Theorem in relation to continuous functions? | |
Ans. The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b], and k is any value between f(a) and f(b), there exists at least one c in the interval (a, b) such that f(c) = k. This theorem is significant as it guarantees that continuous functions take on every value between their endpoints, which is crucial for understanding the behavior of functions in calculus and solving equations.
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