Continuity and Differentiability JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Continuity and Differentiability, free to read — no sign-in needed. The full chapter has 22 questions; sign in to attempt the remaining 17 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   For a real number $\alpha$, let $[\alpha]$ denote the greatest integer less than or equal to $\alpha$. For a finite set $S$, let $|S|$ denote the number of elements in the set $S$. Consider the functions $f : (-3, 3) \rightarrow (-\infty, \infty)$ and $g : (-3, 3) \rightarrow (-\infty, \infty)$ defined by $f(x) = [x^3]\log_e(1 + \sin^2(\pi(x - [x])))$ and $g(x) = x^3 \sin^2(\pi \log_e(1 + x - [x]))$. Let $A = \{x \in (-3, 3) : f \text{ is discontinuous at } x\}$ and $B = \{x \in (-3, 3) : g \text{ is discontinuous at } x\}$. Then the value of $|A| + 2|B| - |A \cap B|$ is ___________.
   Show answer & solution

   Answer: 56

   For the function $f(x) = [x^3]\log_e(1 + \sin^2(\pi(x - [x])))$, we can simplify the argument of the sine function. Since $[x]$ is an integer, $\sin(\pi(x - [x])) = \sin(\pi x - \pi[x]) = \pm \sin(\pi x)$. Thus, $\sin^2(\pi(x - [x])) = \sin^2(\pi x)$. The function $f(x)$ can be rewritten as $f(x) = [x^3]\log_e(1 + \sin^2(\pi x))$. The term $\log_e(1 + \sin^2(\pi x))$ is continuous everywhere. The term $[x^3]$ has jump discontinuities where $x^3$ is an integer. For $x \in (-3, 3)$, $x^3 \in (-27, 27)$. The integer values of $x^3$ in this interval are $k \in \{-26, -25, \dots, 26\}$, which gives $53$ points of the form $x = k^{1/3}$. At these points, $[x^3]$ is discontinuous. For $f(x)$ to be continuous at $x = k^{1/3}$, the continuous factor must be zero: $\log_e(1 + \sin^2(\pi x)) = 0 \implies \sin^2(\pi x) = 0 \implies x$ is an integer. The integers in $(-3, 3)$ are $-2, -1, 0, 1, 2$. Their cubes are $-8, -1, 0, 1, 8$, which are $5$ values among the $53$ points. At these $5$ points, $f(x)$ is continuous. At the remaining $53 - 5 = 48$ points, $f(x)$ is discontinuous. Thus, $|A| = 48$. For the function $g(x) = x^3 \sin^2(\pi \log_e(1 + x - [x]))$, we can write $x - [x] = \{x\}$, which is the fractional part of $x$. The function $g(x) = x^3 \sin^2(\pi \log_e(1 + \{x\}))$ is continuous everywhere except possibly at the integers, where $\{x\}$ is discontinuous. The integers in $(-3, 3)$ are $c \in \{-2, -1, 0, 1, 2\}$. Let's check the continuity at $x = c$: Right-hand limit ($x \to c^+$): $\{x\} \to 0$, so $g(x) \to c^3 \sin^2(\pi \log_e 1) = 0$. Left-hand limit ($x \to c^-$): $\{x\} \to 1$, so $g(x) \to c^3 \sin^2(\pi \log_e 2)$. For $g(x)$ to be continuous at $x = c$, we must have $c^3 \sin^2(\pi \log_e 2) = 0$. Since $\log_e 2 \approx 0.693$, $\pi \log_e 2$ is not an integer multiple of $\pi$, meaning $\sin^2(\pi \log_e 2) \neq 0$. Therefore, we must have $c^3 = 0 \implies c = 0$. So, $g(x)$ is continuous at $x = 0$ and discontinuous at $x \in \{-2, -1, 1, 2\}$. Thus, $B = \{-2, -1, 1, 2\}$ and $|B| = 4$. Since $A$ contains no integers and $B$ contains only integers, $A \cap B = \emptyset$, which means $|A \cap B| = 0$. Finally, we calculate the required value: $|A| + 2|B| - |A \cap B| = 48 + 2(4) - 0 = 48 + 8 = 56$. Answer: $56$
2. Q2JEE Advanced Adv 2026 (Paper 1)
   Let $\mathbb{R}$ denote the set of all real numbers. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be an arbitrary function and let $g : \mathbb{R} \rightarrow \mathbb{R}$ be the function defined by $g(x) = x f(x)$, for all $x \in \mathbb{R}$. Then which of the following statements is (are) TRUE?
   1. A.The function $g$ is always continuous at $x = 0$
   2. B.If $f$ is continuous at $x = 0$, then $g$ is differentiable at $x = 0$
   3. C.If $g$ is differentiable at $x = 0$, then $f$ is continuous at $x = 0$
   4. D.If $g$ is differentiable at $x = 0$, then $\lim_{x \to 0} f(x)$ exists
   Show answer & solution

   Answer: B,D

   Given $g(x) = x f(x)$ for all $x \in \mathbb{R}$. For option A: Let $f(x) = \dfrac{1}{x}$ for $x \neq 0$ and $f(0) = 0$. Then $g(x) = 1$ for $x \neq 0$ and $g(0) = 0$. $\lim_{x \to 0} g(x) = 1 \neq g(0)$. Thus, $g$ is not necessarily continuous at $x = 0$. Option A is false. For option B: The derivative of $g$ at $x = 0$ is given by: $g'(0) = \lim_{x \to 0} \dfrac{g(x) - g(0)}{x - 0} = \lim_{x \to 0} \dfrac{x f(x) - 0}{x} = \lim_{x \to 0} f(x)$. If $f$ is continuous at $x = 0$, then $\lim_{x \to 0} f(x) = f(0)$. Since $f(0)$ is a finite real number, $g'(0)$ exists and $g$ is differentiable at $x = 0$. Option B is true. For option C: Let $f(x) = 0$ for $x \neq 0$ and $f(0) = 1$. Then $g(x) = 0$ for all $x \in \mathbb{R}$. Here, $g$ is differentiable at $x = 0$ with $g'(0) = 0$. However, $\lim_{x \to 0} f(x) = 0 \neq f(0)$, so $f$ is not continuous at $x = 0$. Option C is false. For option D: If $g$ is differentiable at $x = 0$, then $g'(0)$ exists. Since $g'(0) = \lim_{x \to 0} \dfrac{g(x) - g(0)}{x} = \lim_{x \to 0} f(x)$, the limit $\lim_{x \to 0} f(x)$ must exist. Option D is true. Answer: If $f$ is continuous at $x = 0$, then $g$ is differentiable at $x = 0$; If $g$ is differentiable at $x = 0$, then $\lim_{x \to 0} f(x)$ exists
3. Q3JEE Advanced Adv 2020 (Paper 2)
   Let $f:ℝ\rightarrow ℝ$ and $g:ℝ\rightarrow ℝ$ be functions satisfying $f\left(x+y\right)=f\left(x\right)+f\left(y\right)+f\left(x\right)f\left(y\right)\text{and}f\left(x\right)=xg\left(x\right)$ for all $x,y\in ℝ.$ If $\lim _{x\rightarrow 0}g\left(x\right)=1,$ then which of the following statements is/are TRUE?
   1. A.$f$ is differentiable at every $x\in ℝ$
   2. B.If $g\left(0\right)=1,$ then $g$ is differentiable at every$x\in ℝ$
   3. C.The derivative ${f}^{'}\left(1\right)$ is equal to $1$
   4. D.The derivative ${f}^{'}\left(0\right)$ is equal to $1$
   Show answer & solution

   Answer: A,B,D

   Given $f\left(x+y\right)=f\left(x\right)+f\left(y\right)+f\left(x\right)f\left(y\right)$ Put $x=y=0$ in given relation. $\Rightarrow f\left(0\right)=f\left(0\right)+f\left(0\right)+{f}^{2}\left(0\right)$ $\Rightarrow f\left(0\right)=0$ or $-1$ $∵f\left(x+y\right)=f\left(x\right)+f\left(y\right)+f\left(x\right)\cdot f\left(y\right)$ $\Rightarrow \dfrac{f\left(x+y\right)-f\left(x\right)}{y}=\dfrac{f\left(y\right)\left(1+f\left(x\right)\right)}{y}$ $\Rightarrow \lim _{y\rightarrow 0}\left(\dfrac{f(x+y)-f(x)}{y}\right)=\lim _{y\rightarrow 0}\left(1+f\left(x\right)\right)\cdot \dfrac{f\left(y\right)}{y}$ $∵\lim _{x\rightarrow 0}g\left(x\right)=\lim _{x\rightarrow 0}\dfrac{f\left(x\right)}{x}=1$ $\Rightarrow {f}^{'}\left(x\right)=1+f\left(x\right)$ $\Rightarrow {f}^{'}\left(0\right)=1+f\left(0\right)$ $\Rightarrow {f}^{'}\left(0\right)=1+0$ $\Rightarrow {f}^{'}\left(0\right)=1$ Again $\dfrac{{f}^{'}\left(x\right)}{1+f\left(x\right)}=1\Rightarrow \int \dfrac{{f}^{'}\left(x\right)dx}{1+f\left(x\right)}dx=\int dx$ $\Rightarrow ln\left(1+f\left(x\right)\right)=x+C$ $\Rightarrow ln\left[1+f\left(x\right)\right]=x\left(∵C=0\right)$ $\Rightarrow 1+f\left(x\right)={e}^{x}$ $\Rightarrow f\left(x\right)={e}^{x}-1\Rightarrow {f}^{'}\left(x\right)={e}^{x}$ $\Rightarrow {f}^{'}\left(1\right)=e$ Also, $f\left(x\right)$ is differentiable for every $x\in R$. $g\left(x\right)=\dfrac{f\left(x\right)}{x}=\dfrac{{e}^{x}-1}{x}{y}^{'}\left({0}^{+}\right)=\lim _{h\rightarrow 0}\dfrac{g\left(0+h\right)-g\left(0\right)}{h}$ If $g\left(0\right)=1$ then ${g}^{'}\left({0}^{+}\right)=\lim _{h\rightarrow 0}\dfrac{\dfrac{{e}^{h}-1}{h}-1}{h}=\lim _{h\rightarrow 0}\dfrac{{e}^{h}-1-h}{{h}^{2}}=\dfrac{1}{2}$ ${g}^{'}\left({0}^{-}\right)=\lim _{h\rightarrow 0}\dfrac{g\left(0-h\right)-g\left(0\right)}{-h}=\lim _{h\rightarrow 0}\dfrac{{e}^{-h}-1}{\dfrac{-h}{-h}}$ $\lim _{h\rightarrow 0}\dfrac{{e}^{-h}-1+h}{{h}^{2}}=\dfrac{1}{2}$ $g\left(x\right)$ is differentiable for every $x\in R$.
4. Q4JEE Advanced Adv 2020 (Paper 1)
   Let the function $f:R\rightarrow R$ be defined by $f\left(x\right)={x}^{3}-{x}^{2}+\left(x-1\right)\sin x$ and let $g:R\rightarrow R$ be an arbitrary function. Let $fg:R\rightarrow R$ be the product function defined by $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$. Then which of the following statements is/are TRUE?
   1. A.If $g$ is continuous at $x=1$, then $fg$ is differentiable at $x=1$
   2. B.If $fg$ is differentiable at $x=1$, then $g$ is continuous at $x=1$
   3. C.If $g$ is differentiable at $x=1$, then $fg$ is differentiable at $x=1$
   4. D.If $fg$ is differentiable at $x=1$, then $g$ is differentiable at $x=1$
   Show answer & solution

   Answer: A,C

   Differentiability of $fg$ at $x=1$ Left-hand derivative : ${\left(fg\right)}^{'}\left({1}^{-}\right)=\lim _{h\rightarrow 0}\dfrac{fg\left(1-h\right)-fg\left(1\right)}{-h}$ $=\lim _{h\rightarrow 0}\dfrac{\left\{{\left(1-h\right)}^{3}-{\left(1-h\right)}^{2}-h\sin \left(1-h\right)\right\}g\left(1-h\right)-0}{-h}$ $=\lim _{h\rightarrow 0}\left\{{\left(1-h\right)}^{2}+\sin \left(1-h\right)\right\}g\left(1-h\right).......\left(i\right)$ Right-hand derivative : ${\left(fg\right)}^{'}\left({1}^{+}\right)=\lim _{h\rightarrow 0}\dfrac{fg\left(1+h\right)-fg\left(1\right)}{h}$ $=\lim _{h\rightarrow 0}\dfrac{\left\{{\left(1+h\right)}^{3}-{\left(1+h\right)}^{2}+h\sin \left(1+h\right)\right\}g\left(1+h\right)-0}{h}$ $=\lim _{h\rightarrow 0}\left\{{\left(1+h\right)}^{2}+\sin \left(1+h\right)\right\}g\left(1+h\right).......\left(ii\right)$ If $g$ is continuous at $x=1$, then $\lim _{h\rightarrow 0}g\left(1+h\right)=\lim _{h\rightarrow 0}g\left(1-h\right)=g\left(1\right).........\left(iii\right)$ From equations $\left(i\right),\left(ii\right)\&\left(iii\right)$, we get $\lim _{h\rightarrow 0}{\left(fg\right)}^{'}\left({1}^{+}\right)=\lim _{h\rightarrow 0}{\left(fg\right)}^{'}\left({1}^{-}\right)=\left(1+\sin 1\right)g\left(1\right)$ $∴fg$ is differentiable at $x=1$. So, option $\left(A\right)$ is correct. Now, from equations $\left(i\right)\&\left(ii\right)$, we can say that for $fg$ to be differentiable, we need only $g\left(1+h\right)=g\left(1-h\right)$. But for $g$ to be continuous, we need $g\left(1+h\right)=g\left(1-h\right)=g\left(1\right)$. So, option $\left(B\right)$ is incorrect. Now, if $g$ is differentiable at $x=1$ and $f$ is already differentiable at $x=1$ as ${f}^{'}\left({1}^{-}\right)={f}^{'}\left({1}^{+}\right)=1+\sin 1$, so product of two differentiable functions is also differentiable. $∴fg$ is differentiable at $x=1$. So, option $\left(C\right)$ is correct. Now, from option $\left(B\right)$, if $fg$ is differentiable at $x=1$, we cannot guarantee $g$ to be continuous at $x=1$. So, we also cannot guarantee $g$ to be differentiable at $x=1$. So, option $\left(D\right)$ is incorrect.
5. Q5JEE Advanced Adv 2017 (Paper 1)
   Let $\left[x\right]$ be the greatest integer less than or equals to $x$. Then, at which of the following point(s) the function $f\left(x\right)=x\cos ⁡\left(\pi \left(x+\left[x\right]\right)\right)$ is discontinuous?
   1. A.$x=1$
   2. B.$x=-1$
   3. C.$x=0$
   4. D.$x=2$
   Show answer & solution

   Answer: A,B,D

   $f\left(x\right)=x\cos ⁡\left(\pi x+\left[x\right]\pi \right)$ $\Rightarrow f\left(x\right)={\left(-1\right)}^{\left[x\right]}x\cos \left(\pi x\right)$ $[x]$ is discontinuous at all integers and $x\cos \pi x$ is continuous everywhere. At $x=0$, $f(x)=0$, so it's continuous. At other integral points, $f(x)$ is discontinuous.