Limits Continuity and Differentiability JEE Main PYQs (2020–2026)

Solved Limits Continuity and Differentiability PYQs

3 recent questions with answers. The full set of 136 is free to practise.

1. JEE Main 2026MediumContinuity
   Q1

   The number of points in the interval $[2,4]$, at which the function $f(x)=\left[x^2-x-\frac{1}{2}\right]$, where $[\cdot ]$ denotes the greatest integer function, is discontinuous, is ________.

   Numerical answer type

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   Correct answer: 10

   For the greatest integer function $[\cdot ]$, $f(x)=\left[x^2-x-\frac{1}{2}\right]$ is discontinuous at those points where $x^2-x-\frac{1}{2}$ takes an integer value.
   
   Let $g(x)=x^2-x-\frac{1}{2}$ on $[2,4]$. $g(2)=4-2-\frac{1}{2}=\frac{3}{2}$. $g(4)=16-4-\frac{1}{2}=\frac{23}{2}=11.5$.
   
   $g^{\prime }(x)=2x-1>0$ for $x\in [2,4]$, so $g$ is strictly increasing and takes every value from $\frac{3}{2}$ to $\frac{23}{2}$ exactly once.
   
   Integers in $[1.5,11.5]$: $2,3,4,5,6,7,8,9,10,11\to 10$ integers.
   
   Each integer value corresponds to exactly one $x$, so $f(x)$ is discontinuous at exactly 10 points.
   
   Answer: 10
2. JEE Main 2026EasyEvaluation of Limits
   Q2

   Let $f:\mathbb{R}\to (0,\infty )$ be a twice differentiable function such that $f(3)=18$, $f^{\prime }(3)=0$ and $f^{\prime \prime }(3)=4$. Then ${\lim }_{x\to 1}{\log }_e\left({\left(\frac{f(2+x)}{f(3)}\right)}^{\frac{18}{(x-1{)}^2}}\right)$ is equal to :

   • A9
   • B18
   • C1
   • D2

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   Correct answer: (D)

   $\ln \left({\lim }_{x\to 1}{\left(\frac{f(x+2)}{f(3)}\right)}^{\frac{18}{(x-1{)}^2}}\right)$ (form $1^{\infty }$)
   
   $$=ln\left(e^{{\lim }_{x\to 1}\frac{18}{(x-1{)}^2}\left(\frac{f(x+2)}{f(3)}-1\right)}\right)$$
   
   $=ln\left(e^{{\lim }_{x\to 1}\frac{f^{\prime }(x+2)}{2(x-1)}}\right)$ (L'Hospital, using $f(3)=18$)
   
   $=ln\left(e^{{\lim }_{x\to 1}\frac{f^{\prime \prime }(x+2)}{2}}\right)$ (L'Hospital again)
   
   $=\frac{f^{\prime \prime }(3)}{2}=\frac{4}{2}=2$.
   
   Answer: D (2)
3. JEE Main 2026HardDifferentiability
   Q3

   Let $\alpha ,\beta \in \mathbb{R}$ be such that the function $f(x)=\{\begin{array}{cc}2\alpha \left(x^2-2\right)+2\beta x,& x<1\\ (\alpha +3)x+(\alpha -\beta ),& x\ge 1\end{array}$ be differentiable at all $x\in \mathbb{R}$. Then $34(\alpha +\beta )$ is equal to

   • A48
   • B84
   • C36
   • D24

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   Correct answer: (A)

   $f$ is differentiable at all $x\in \mathbb{R}$, so $f$ is continuous and differentiable at $x=1$.
   
   1. Continuity at $x=1$: ${\lim }_{x\to 1^{-}}\left[2\alpha \left(x^2-2\right)+2\beta x\right]={\lim }_{x\to 1^{+}}\left[(\alpha +3)x+(\alpha -\beta )\right]$
   
   $$2\alpha (1-2)+2\beta =\alpha +3+\alpha -\beta \Rightarrow -2\alpha +2\beta =2\alpha -\beta +3\Rightarrow 3\beta -4\alpha =3\dots (1)$$
   
   2. Differentiability at $x=1$: $f^{\prime }(x)=2\alpha (2x)+2\beta$ for $x<1$; $\alpha +3$ for $x>1$.
   
   $$f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)\Rightarrow 4\alpha +2\beta =\alpha +3\Rightarrow 3\alpha +2\beta =3\dots (2)$$
   
   Solving $(1)\times 2-(2)\times 3$: $-8\alpha -9\alpha =-3\Rightarrow \alpha =\frac{3}{17}$.
   
   In $(1)$: $3\beta -\frac{12}{17}=3\Rightarrow 3\beta =\frac{63}{17}\Rightarrow \beta =\frac{21}{17}$.
   
   $34(\alpha +\beta )=34\left(\frac{3}{17}+\frac{21}{17}\right)=6+42=48$.
   
   Answer: A

Topic-wise question count

Where JEE Main has asked from Limits Continuity and Differentiability, most frequent first.

• Differentiability44
• Continuity43
• Evaluation of Limits25
• Methods of Differentiation17
• Mean Value Theorems7

Year-wise Limits Continuity and Differentiability questions in JEE Main

Number of questions from this chapter in each year's paper(s).