Limits and Derivatives: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Limits, Continuity and Differentiability  
 
(January 2026) 
Q1:  
(I) f (x) is continuous at x = 1. 
(II) f (x) is continuous at x = -1. 
Then, 
A: Neither (I) nor (II) is True 
B: Only (II) is True 
C: Only (I) is True 
D: Both (I) and (II) are True 
Answer: A 
Explanation: 
The function is de?ned as: 
 
Let n = 2/?. As ? ? 0, n ? 8 (speci?cally, let's consider ? approaching 0 such that the power 
grows). The behavior of the function depends on the value of x because of the term xn. 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Limits, Continuity and Differentiability  
 
(January 2026) 
Q1:  
(I) f (x) is continuous at x = 1. 
(II) f (x) is continuous at x = -1. 
Then, 
A: Neither (I) nor (II) is True 
B: Only (II) is True 
C: Only (I) is True 
D: Both (I) and (II) are True 
Answer: A 
Explanation: 
The function is de?ned as: 
 
Let n = 2/?. As ? ? 0, n ? 8 (speci?cally, let's consider ? approaching 0 such that the power 
grows). The behavior of the function depends on the value of x because of the term xn. 
 
Analyze the Statements 
Statement (I) : f(x) is discontinuous at x = 1. 
Value at x = 1  : f(1) = -1.
 
Conclusion : Statement (I) is False. 
Statement (II): f(x) is continuous at x = -1. 
Value at x = -1 : f(-1) = 1 - sin 2.
 
Since the limits do not match, the function is discontinuous at X = -1. 
Conclusion : Statement (II) is False. 
 
Q2: The value of 
 
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JEE Main Previous Year Questions (2021-2026): 
Limits, Continuity and Differentiability  
 
(January 2026) 
Q1:  
(I) f (x) is continuous at x = 1. 
(II) f (x) is continuous at x = -1. 
Then, 
A: Neither (I) nor (II) is True 
B: Only (II) is True 
C: Only (I) is True 
D: Both (I) and (II) are True 
Answer: A 
Explanation: 
The function is de?ned as: 
 
Let n = 2/?. As ? ? 0, n ? 8 (speci?cally, let's consider ? approaching 0 such that the power 
grows). The behavior of the function depends on the value of x because of the term xn. 
 
Analyze the Statements 
Statement (I) : f(x) is discontinuous at x = 1. 
Value at x = 1  : f(1) = -1.
 
Conclusion : Statement (I) is False. 
Statement (II): f(x) is continuous at x = -1. 
Value at x = -1 : f(-1) = 1 - sin 2.
 
Since the limits do not match, the function is discontinuous at X = -1. 
Conclusion : Statement (II) is False. 
 
Q2: The value of 
 
is equal to 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
 
Now understanding dominating term 
If A ? 1 ? ln A ˜ A - 1 
as x ? 0 ? sec(ex) ? 1 ? ln sec(ex) ˜ sec(ex) - 1 
so generalized x ? 0 ? ln sec(e?x) ˜ sec(e?x) - 1 
If A ? 0 ? e? - 1 ˜ A 
so in denominator as x ? 0 ? 2(cos x - 1) = 0 
so, e
(2(cos x-1)) - 1 ˜ 2(cos x - 1)
 
Substituting this dominating term in equation (1)
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JEE Main Previous Year Questions (2021-2026): 
Limits, Continuity and Differentiability  
 
(January 2026) 
Q1:  
(I) f (x) is continuous at x = 1. 
(II) f (x) is continuous at x = -1. 
Then, 
A: Neither (I) nor (II) is True 
B: Only (II) is True 
C: Only (I) is True 
D: Both (I) and (II) are True 
Answer: A 
Explanation: 
The function is de?ned as: 
 
Let n = 2/?. As ? ? 0, n ? 8 (speci?cally, let's consider ? approaching 0 such that the power 
grows). The behavior of the function depends on the value of x because of the term xn. 
 
Analyze the Statements 
Statement (I) : f(x) is discontinuous at x = 1. 
Value at x = 1  : f(1) = -1.
 
Conclusion : Statement (I) is False. 
Statement (II): f(x) is continuous at x = -1. 
Value at x = -1 : f(-1) = 1 - sin 2.
 
Since the limits do not match, the function is discontinuous at X = -1. 
Conclusion : Statement (II) is False. 
 
Q2: The value of 
 
is equal to 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
 
Now understanding dominating term 
If A ? 1 ? ln A ˜ A - 1 
as x ? 0 ? sec(ex) ? 1 ? ln sec(ex) ˜ sec(ex) - 1 
so generalized x ? 0 ? ln sec(e?x) ˜ sec(e?x) - 1 
If A ? 0 ? e? - 1 ˜ A 
so in denominator as x ? 0 ? 2(cos x - 1) = 0 
so, e
(2(cos x-1)) - 1 ˜ 2(cos x - 1)
 
Substituting this dominating term in equation (1)
 
 
e² + e4 + ··· + e²° is a G.P . whose ?rst term is e² common ratio is e² so sum of G.P . is a(rn - 1)/(r - 
1) where a is ?rst term and r is common ratio. 
 
 
Q3: Let y = y(x) be a differentiable function in the interval (0, 8) such that y(1) = 2 
  = 3 for each x > 0. Then 2y(2) is equal to : 
A: 27 
B: 18 
C: 23 
D: 12 
Answer: C 
Explanation: 
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JEE Main Previous Year Questions (2021-2026): 
Limits, Continuity and Differentiability  
 
(January 2026) 
Q1:  
(I) f (x) is continuous at x = 1. 
(II) f (x) is continuous at x = -1. 
Then, 
A: Neither (I) nor (II) is True 
B: Only (II) is True 
C: Only (I) is True 
D: Both (I) and (II) are True 
Answer: A 
Explanation: 
The function is de?ned as: 
 
Let n = 2/?. As ? ? 0, n ? 8 (speci?cally, let's consider ? approaching 0 such that the power 
grows). The behavior of the function depends on the value of x because of the term xn. 
 
Analyze the Statements 
Statement (I) : f(x) is discontinuous at x = 1. 
Value at x = 1  : f(1) = -1.
 
Conclusion : Statement (I) is False. 
Statement (II): f(x) is continuous at x = -1. 
Value at x = -1 : f(-1) = 1 - sin 2.
 
Since the limits do not match, the function is discontinuous at X = -1. 
Conclusion : Statement (II) is False. 
 
Q2: The value of 
 
is equal to 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
 
Now understanding dominating term 
If A ? 1 ? ln A ˜ A - 1 
as x ? 0 ? sec(ex) ? 1 ? ln sec(ex) ˜ sec(ex) - 1 
so generalized x ? 0 ? ln sec(e?x) ˜ sec(e?x) - 1 
If A ? 0 ? e? - 1 ˜ A 
so in denominator as x ? 0 ? 2(cos x - 1) = 0 
so, e
(2(cos x-1)) - 1 ˜ 2(cos x - 1)
 
Substituting this dominating term in equation (1)
 
 
e² + e4 + ··· + e²° is a G.P . whose ?rst term is e² common ratio is e² so sum of G.P . is a(rn - 1)/(r - 
1) where a is ?rst term and r is common ratio. 
 
 
Q3: Let y = y(x) be a differentiable function in the interval (0, 8) such that y(1) = 2 
  = 3 for each x > 0. Then 2y(2) is equal to : 
A: 27 
B: 18 
C: 23 
D: 12 
Answer: C 
Explanation: 
 
Use 
 
Equation (1) is differential equation can be written by replacing