Continuity and Differentiability JEE Main previous year questions with solutions

5 solved JEE Main questions on Continuity and Differentiability, free to read — no sign-in needed. The full chapter has 156 questions; sign in to attempt the remaining 151 in the exam simulator.

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1. Q1JEE Main 2026 (08 Apr, Shift 2)Continuity
   Let $f(x) = \begin{cases} \dfrac{1}{3}, & x \leq \pi/2 \\ \dfrac{b(1-\sin x)}{(\pi-2x)^2}, & x \gt \pi/2 \end{cases}$. If $f$ is continuous at $x=\pi/2$, then the value of $\displaystyle\int_{0}^{3b-6} |x^2+2x-3|\,dx$ is:
   1. A.$5$
   2. B.$2$
   3. C.$3$
   4. D.$4$
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   Answer: (D)

   Since $f(x)$ is continuous at $x = \pi/2$, the right-hand limit must equal the value of the function at $x = \pi/2$. $\lim_{x \to \pi/2^+} f(x) = f(\pi/2)$ $\lim_{x \to \pi/2^+} \dfrac{b(1-\sin x)}{(\pi-2x)^2} = \dfrac{1}{3}$ Let $x = \pi/2 + h$. As $x \to \pi/2^+$, $h \to 0^+$. $\lim_{h \to 0^+} \dfrac{b(1-\sin(\pi/2+h))}{(\pi-2(\pi/2+h))^2} = \dfrac{1}{3}$ $\lim_{h \to 0^+} \dfrac{b(1-\cos h)}{(-2h)^2} = \dfrac{1}{3}$ $\lim_{h \to 0^+} \dfrac{b(1-\cos h)}{4h^2} = \dfrac{1}{3}$ Using the standard limit $\lim_{h \to 0} \dfrac{1-\cos h}{h^2} = \dfrac{1}{2}$, we get: $\dfrac{b}{4} \times \dfrac{1}{2} = \dfrac{1}{3} \Rightarrow \dfrac{b}{8} = \dfrac{1}{3} \Rightarrow b = \dfrac{8}{3}$ The upper limit of the integral is $3b - 6 = 3\left(\dfrac{8}{3}\right) - 6 = 2$. The integral to evaluate is $I = \displaystyle\int_{0}^{2} |x^2+2x-3|\,dx$. Factoring the quadratic expression gives $x^2+2x-3 = (x+3)(x-1)$. For $x \in [0, 1]$, $(x+3)(x-1) \leq 0$, so $|x^2+2x-3| = 3 - 2x - x^2$. For $x \in [1, 2]$, $(x+3)(x-1) \geq 0$, so $|x^2+2x-3| = x^2+2x-3$. Splitting the integral at $x = 1$: $I = \displaystyle\int_{0}^{1} (3 - 2x - x^2)\,dx + \displaystyle\int_{1}^{2} (x^2+2x-3)\,dx$ Evaluating the first integral: $\displaystyle\int_{0}^{1} (3 - 2x - x^2)\,dx = \left[ 3x - x^2 - \dfrac{x^3}{3} \right]_0^1 = 3 - 1 - \dfrac{1}{3} = \dfrac{5}{3}$ Evaluating the second integral: $\displaystyle\int_{1}^{2} (x^2+2x-3)\,dx = \left[ \dfrac{x^3}{3} + x^2 - 3x \right]_1^2$ $= \left( \dfrac{8}{3} + 4 - 6 \right) - \left( \dfrac{1}{3} + 1 - 3 \right) = \left( \dfrac{8}{3} - 2 \right) - \left( \dfrac{1}{3} - 2 \right) = \dfrac{7}{3}$ Adding the two parts: $I = \dfrac{5}{3} + \dfrac{7}{3} = \dfrac{12}{3} = 4$ Answer: $4$
2. Q2JEE Main 2025 (04 Apr, Shift 1)Differentiability
   Let $m$ and $n$ be the number of points at which the function $f(\mathrm{x})=\max \left\{\mathrm{x}, \mathrm{x}^3, \mathrm{x}^5, \ldots ., \mathrm{x}^{21}\right\}, \mathrm{x} \in \mathbb{R}$, is not differentiable and not continuous, respectively. Then $\mathrm{m}+\mathrm{n}$ is equal to ________ .
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   Answer: 3

   $f(x)=\left\{\begin{array}{cc} x, & x \lt -1 \\ x^{21}, & -1 \leq x \lt 0 \\ x, & 0 \leq x \lt 1 \\ x^{21}, & x \geq 1 \end{array}\right.$ $f(x)$ is continuous everywhere. $\begin{aligned} & \therefore \mathrm{n}=0 \\ & \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{cc} 1, & \mathrm{x} \lt -1 \\ 21 \mathrm{x}^{20}, & -1 \leq \mathrm{x} \lt 0 \\ 1, & 0 \lt \mathrm{x} \lt 1 \\ 21 \mathrm{x}^{20}, & \mathrm{x} \geq 1 \end{array}\right. \end{aligned}$ $\therefore \mathrm{f}(\mathrm{x})$ is non-differentiable at $\mathrm{x}=-1,0,1$ $\begin{aligned} & \therefore \mathrm{m}=3 \\ & \mathrm{~m}+\mathrm{n}=3 \end{aligned}$
3. Q3JEE Main 2007Exponential and logarithmic limits
   The function $f:R \sim\{0\} \rightarrow R$ given by $f(x)=\frac{1}{x}-\frac{2}{e^{2 x}-1}$ can be made continuous at $x=0$ by defining $f(0)$ as
   1. A.$2$
   2. B.$-1$
   3. C.$0$
   4. D.$1$
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   Answer: (D)

   $\lim _{x \rightarrow 0} \frac{1}{x}-\frac{2}{e^{2 x}-1}$ $\lim _{x \rightarrow 0} \frac{e^{2 x}-1-2 x}{x\left(e^{2 x}-1\right)}$ $\lim _{x \rightarrow 0} \frac{2 e^{2 x}-2}{\left(e^{2 x}-1\right)+2 x e^{2 x}}$ $\lim _{x \rightarrow 0} \frac{4 e^{2 x}}{4 e^{2 x}+4 x e^{2 x}}=1$.
4. Q4JEE Main 2004Trigonometric and Inverse Trigonometric limits
   Let $f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. If $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $f\left(\frac{\pi}{4}\right)$ is
   1. A.1
   2. B.$\frac{1}{2}$
   3. C.$-\frac{1}{2}$
   4. D.$-1$
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   Answer: (C)

   $f(x)=\frac{1-\tan x}{4 x-\pi} \Rightarrow \lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}=-\frac{1}{2}$
5. Q5JEE Main 2026 (06 Apr, Shift 2)Continuity
   Let $f(x) = \begin{cases} x^3 + 8 ; & x \lt 0 \\ x^2 - 4 ; & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} (x-8)^{1/3} ; & x \lt 0 \\ (x+4)^{1/2} ; & x \geq 0 \end{cases}$. Then the number of points, where the function $g \circ f$ is discontinuous, is __________.
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   Answer: 3

   The possible points of discontinuity for the composite function $g(f(x))$ are the points where $f(x)$ is discontinuous and the points where $f(x)$ is equal to a point of discontinuity of $g(x)$. First, we find the points of discontinuity of $f(x)$ and $g(x)$. For $f(x)$, the only possible point of discontinuity is at $x = 0$. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x^3 + 8) = 8$ $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 - 4) = -4$ Since $\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$, $f(x)$ is discontinuous at $x = 0$. For $g(x)$, the only possible point of discontinuity is at $x = 0$. $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (x - 8)^{1/3} = -2$ $\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (x + 4)^{1/2} = 2$ Since $\lim_{x \to 0^-} g(x) \neq \lim_{x \to 0^+} g(x)$, $g(x)$ is discontinuous at $x = 0$. Next, we find the points where $f(x)$ equals the point of discontinuity of $g(x)$, which is $f(x) = 0$. For $x \lt 0$, $x^3 + 8 = 0 \Rightarrow x = -2$. For $x \geq 0$, $x^2 - 4 = 0 \Rightarrow x = 2$. Thus, the possible points of discontinuity for $g(f(x))$ are $x = -2$, $x = 0$, and $x = 2$. We check the continuity at each of these points. At $x = -2$: As $x \to -2^-$, $f(x) \to 0^-$, so $\lim_{x \to -2^-} g(f(x)) = \lim_{y \to 0^-} g(y) = -2$ As $x \to -2^+$, $f(x) \to 0^+$, so $\lim_{x \to -2^+} g(f(x)) = \lim_{y \to 0^+} g(y) = 2$ Since the left-hand limit and right-hand limit are not equal, $g(f(x))$ is discontinuous at $x = -2$. At $x = 0$: $\lim_{x \to 0^-} g(f(x)) = g(8) = (8 + 4)^{1/2} = 2\sqrt{3}$ $\lim_{x \to 0^+} g(f(x)) = g(-4) = (-4 - 8)^{1/3} = (-12)^{1/3}$ Since the limits are not equal, $g(f(x))$ is discontinuous at $x = 0$. At $x = 2$: As $x \to 2^-$, $f(x) \to 0^-$, so $\lim_{x \to 2^-} g(f(x)) = \lim_{y \to 0^-} g(y) = -2$ As $x \to 2^+$, $f(x) \to 0^+$, so $\lim_{x \to 2^+} g(f(x)) = \lim_{y \to 0^+} g(y) = 2$ Since the limits are not equal, $g(f(x))$ is discontinuous at $x = 2$. Therefore, there are $3$ points of discontinuity. Answer: $3$