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 154 questions; sign in to attempt the remaining 149 in the exam simulator.

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 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: $3$
4. Q4JEE Main 2026 (04 Apr, Shift 2)Continuity
   Let $f(x)=\begin{cases} e^{x-1}, & x\lt 0 \\ x^2-5x+6, & x \geq 0 \end{cases}$ and $g(x)=f(|x|)+|f(x)|$. If the number of points where $g$ is not continuous and is not differentiable are $\alpha$ and $\beta$ respectively, then $\alpha+\beta$ is equal to ______
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   Answer: 4

   We are given the function: $f(x) = \begin{cases} e^{x-1}, & x \lt 0 \\ x^2 - 5x + 6, & x \geq 0 \end{cases}$ We need to analyze the continuity and differentiability of $g(x) = f(|x|) + |f(x)|$. For $x \lt 0$, $|x| = -x \gt 0$. Thus, $f(|x|) = (-x)^2 - 5(-x) + 6 = x^2 + 5x + 6$. Also, for $x \lt 0$, $f(x) = e^{x-1} \gt 0$, so $|f(x)| = e^{x-1}$. Therefore, for $x \lt 0$, $g(x) = x^2 + 5x + 6 + e^{x-1}$. For $x \geq 0$, $|x| = x$. Thus, $f(|x|) = f(x) = x^2 - 5x + 6$. Therefore, for $x \geq 0$, $g(x) = x^2 - 5x + 6 + |x^2 - 5x + 6|$. Let us check the continuity of $g(x)$ at $x = 0$: $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (x^2 + 5x + 6 + e^{x-1}) = 6 + \dfrac{1}{e}$ $\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (x^2 - 5x + 6 + |x^2 - 5x + 6|) = 6 + 6 = 12$ $g(0) = 12$ Since $\lim_{x \to 0^-} g(x) \neq \lim_{x \to 0^+} g(x)$, $g(x)$ is discontinuous at $x = 0$. For all other $x$, $g(x)$ is a sum of continuous functions and is therefore continuous. Thus, the number of points of discontinuity is $\alpha = 1$. Now, let us check the differentiability of $g(x)$. Since $g(x)$ is discontinuous at $x = 0$, it is not differentiable at $x = 0$. For $x \lt 0$, $g(x) = x^2 + 5x + 6 + e^{x-1}$, which is differentiable everywhere in its domain. For $x \gt 0$, we can rewrite $g(x)$ by analyzing the sign of $x^2 - 5x + 6 = (x-2)(x-3)$: $g(x) = \begin{cases} 2(x^2 - 5x + 6), & x \in (0, 2] \cup [3, \infty) \\ 0, & x \in (2, 3) \end{cases}$ Differentiating $g(x)$ for $x \gt 0, x \neq 2, 3$: $g'(x) = \begin{cases} 4x - 10, & x \in (0, 2) \cup (3, \infty) \\ 0, & x \in (2, 3) \end{cases}$ Checking differentiability at $x = 2$: $g'(2^-) = 4(2) - 10 = -2$ $g'(2^+) = 0$ Since $g'(2^-) \neq g'(2^+)$, $g(x)$ is not differentiable at $x = 2$. Checking differentiability at $x = 3$: $g'(3^-) = 0$ $g'(3^+) = 4(3) - 10 = 2$ Since $g'(3^-) \neq g'(3^+)$, $g(x)$ is not differentiable at $x = 3$. Thus, $g(x)$ is not differentiable at exactly three points: $x = 0, 2, 3$. So, the number of points of non-differentiability is $\beta = 3$. Finally, $\alpha + \beta = 1 + 3 = 4$. Answer: $4$
5. Q5JEE Main 2026 (02 Apr, Shift 2)Continuity
   The number of points in the interval $[2, 4]$, at which the function $f(x) = \left[x^2 - x - \dfrac{1}{2}\right]$, where $[\cdot]$ denotes the greatest integer function, is discontinuous, is _______.
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   Answer: 10

   Let $g(x) = x^2 - x - \dfrac{1}{2}$. Differentiating with respect to $x$, we get $g'(x) = 2x - 1$. For $x \in [2, 4]$, $g'(x) \gt 0$, which implies that $g(x)$ is strictly increasing in the interval $[2, 4]$. The values of $g(x)$ at the endpoints are: $g(2) = 2^2 - 2 - \dfrac{1}{2} = 1.5$ $g(4) = 4^2 - 4 - \dfrac{1}{2} = 11.5$ The function $f(x) = [g(x)]$ is discontinuous at all points where $g(x)$ is an integer, as $g(x)$ is strictly monotonic and crosses these integer values. The integers between $1.5$ and $11.5$ are $2, 3, 4, 5, 6, 7, 8, 9, 10, 11$. Since $g(x)$ is strictly increasing, it attains each of these $10$ integer values exactly once in the interval $(2, 4)$. At the endpoints $x = 2$ and $x = 4$, $g(x)$ is not an integer, so $f(x)$ is continuous from the right at $x = 2$ and continuous from the left at $x = 4$. Thus, the number of points of discontinuity in the interval $[2, 4]$ is $10$. Answer: $10$