Continuity and Differentiability — JEE Main PYQs

143 previous year questions on Continuity and Differentiability (Mathematics), newest first. Open any question for its complete step-by-step solution.

Let $f : R \to R$ be a function defined as
$f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 1 ) x + s i n 2 x}{2 x}, b \frac{x + b x ^{3} - x}{b x ^{5/2}} if x < 0, if x = 0, if x > 0$
If $f$ is continuous at $x = 0$ , then …
JEE Main 2021View solution →
Let $f : R \to R$ satisfy the equation $f (x + y) = f (x) \cdot f (y)$ for all $x, y \in R$ and $f (x) \neq = 0$ for any $x \in R$ . If the function $f$ is differentiable at $x = 0$ and $f^{'} (0) = 3$ , then $lim_{h \to 0} \frac{1}{h} (f (h) - 1)$ is equal to
JEE Main 2021View solution →
If $f (x) = {\frac{1}{∣ x ∣} a x^{2} + b; ∣ x ∣ \geq 1; ∣ x ∣ < 1$ is differentiable at every point of the domain, then the values of $a$ and $b$ are respectively ;
JEE Main 2021View solution →
Let $S$ be the set of points where the function, $f (x) = ∣2 - ∣ x - 3∣∣, x \in R$ , is not differentiable. Then $\sum_{x \in S} f (f (x))$ is equal to ____
JEE Main 2020View solution →
If the function $f$ defined on $(- \frac{1}{3}, \frac{1}{3})$ by $f (x) = {\frac{1}{x} lo g_{e} (\frac{1 + 3 x}{1 - 2 x}), when x \neq = 0 k, when x = 0$ is continuous, then k is equal to ____
JEE Main 2020View solution →
Let $f (x) = x cos^{- 1} (- sin ∣ x ∣), x \in [- \frac{π}{2}, \frac{π}{2}]$ , then which of the following is true?
JEE Main 2020View solution →
Let S be the set of all functions $\int : [0, 1] \to R$ , which are continuous on $[0, 1]$ and differentiable on $(0, 1)$ . Then for every $\int$ in $S$ , there exists $ac \in (0, 1)$ , depending on $f$ , such that:
JEE Main 2020View solution →
Let $[t]$ denote the greatest integer $\leq t$ and $lim_{x \to 0} x [\frac{4}{x}] = A$ . Then the function, $f (x) = [x^{2}] sin (π x)$ is discontinuous, when $x$ is equal to:
JEE Main 2020View solution →
If $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x}; b \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}; x < 0; x = 0 x > 0$ …
JEE Main 2020View solution →
Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{2}}$ . Let $S$ denote the set of all points in $R$ , where $f$ is not differentiable. Then:
JEE Main 2020View solution →
If the function $f (x) = {k_{1} (x - π)^{2} - 1 k_{2} cos x x \leq π x > π$ is twice differentiable, then the ordered pair $(k_{1}, k_{2})$ is equal to:
JEE Main 2020View solution →
Let $f (x) = x \cdot [\frac{x}{2}]$ , for $- 10 < x < 10$ , where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of $f$ is equal to ______
JEE Main 2020View solution →
Let $f : R \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ x^{5} sin (\frac{1}{x}) + 5 x^{2}, 0, x^{5} cos (\frac{1}{x}) + λ x^{2}, x < 0 x = 0 x > 0$

The value of $λ$ for which $f^{''} (0)$ exis …
JEE Main 2020View solution →
Suppose a differentiable function $f (x)$ satisfies the identity $f (x + y) = f (x) + f (y) + x y^{2} + x^{2} y$ , for all real $x$ and $y$ . If $lim_{x \to 0} \frac{f ( x )}{x} = 1$ then $f^{'} (3)$ is equal to ____ .
JEE Main 2020View solution →
The function $f (x) = {\frac{π}{4} + tan^{- 1} x, \frac{1}{2} (∣ x ∣ - 1), ∣ x ∣ \leq 1 ∣ x ∣ > 1$ is:
JEE Main 2020View solution →
Let $f : (0, \infty) \to (0, \infty)$ be a differentiable function such that $f (1) = e$ and $lim_{t \to x} \frac{t ^{2} f ^{2} ( x ) - x ^{2} f ^{2} ( t )}{t - x} = 0$ . If $f (x) = 1$ , then $x$ is equal to:
JEE Main 2020View solution →
Let $f : [- 1, 3] \to R$ be defined as
$f (x) = ⎩ ⎨ ⎧ ∣ x ∣ + [x], x + ∣ x ∣, x + [x], - 1 \leq x < 1 1 \leq x < 2 2 \leq x \leq 3$

where $[t]$ denotes the greatest integer less than or equal to $t$ . Then, $f$ is discontinuous at:
JEE Main 2019View solution →
Let $f (x) = 15 - ∣ x - 10∣; x \in R$ . then the set of all values of $x$ , at which the function, $g (x) = f (f (x))$ is not differentiable, is:
JEE Main 2019View solution →
If the function f defined on $(\frac{π}{6}, \frac{π}{3})$ by $f (x) = {\frac{2 c o s x - 1}{c o t x - 1}, k, x \neq = \frac{π}{4} x = \frac{π}{4}$ is co …
JEE Main 2019View solution →
Let $f (x) = {max {∣ x ∣, x^{2}}, 8 - 2∣ x ∣, ∣ x ∣ \leq 2 2 < ∣ x ∣ \leq 4$ . Let S be the set of points in the interval $(- 4, 4)$ at which $f$ is not differentiable. Then $S$ :
JEE Main 2019View solution →
Let K be the set of all real values of x where the function $f (x) = sin ∣ x ∣ - ∣ x ∣ + 2 (x - π) cos ∣ x ∣$ is not differentiable. Then the set $K$ is equal to:
JEE Main 2019View solution →
Let $f (x) = {- 1, x^{2} - 1, - 2 \leq x < 0 0, \leq x \leq 2$ and $g (x) = ∣ f (x) ∣ + f (∣ x ∣)$ , Then, in the interval ( $- 2, 2)$ , $g$ is:
JEE Main 2019View solution →
Let S be the set of all points in $(- π, π)$ at which the function, $f (x) = min {sin x, cos x}$ is non-differentiable. Then S is a subset of which of the following?
JEE Main 2019View solution →
If the function $f (x) = {a ∣ π - x ∣ + 1, x \leq 5 b ∣ π - x ∣ + 3, x > 5$ is continuous at $x = 5$ , then the value of $a - b$ is
JEE Main 2019View solution →
Let $f$ be a differentiable function $R$ to $R$ such that $∣ f (x) - f (y) ∣ \leq 2∣ x - y ∣^{\frac{3}{2}}$ , for all $x, y \in R$ . If $f (0) = 1$ then $\int_{0}^{1} f^{2} (x) d x$ is equal to
JEE Main 2019View solution →
Let $f : R \to R$ be a function defined as $f (x) = ⎩ ⎨ ⎧ 5, a + b x, b + 5 x, 30, if if if if x \leq 1 1 < x < 3 3 \leq x < 5 x \geq 5$ . Then $f$ is:
JEE Main 2019View solution →
If $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( p + 1 ) x + s i n x}{x} q \frac{x + x ^{2} - x}{x ^{3/2}}, x < 0, x = 0, x > 0$

Is continuous at $x = 0$ …
JEE Main 2019View solution →
Let $f : R \to R$ be differentiable at $c \in R$ and $f (c) = 0$ . If $g (x) = ∣ f (x) ∣$ , then at $x = c, g$ is:
JEE Main 2019View solution →
Let $S = {t \in R : f (x) = ∣ x - π ∣ \cdot (e^{∣ x ∣} - 1) sin ∣ x ∣$ is not differentiable at t $}$ . Then the set S is equal to :
JEE Main 2018View solution →
For $x \in R, f (x) = ∣ lo g 2 - sin x ∣$ and $g (x) = f (f (x))$ , then:
JEE Main 2016View solution →
If the function. $g (x) = {k x + 1, m x + 2, 0 \leq x \leq 3 3 < x \leq 5$ is differentiable, then the value of $k + m$ is :
JEE Main 2015View solution →
If $f : R \to R$ is a function defined by $f (x) = [x] cos (\frac{2 x - 1}{2}) π$ , where $[x]$ denotes the greatest integer function, then f is
JEE Main 2012View solution →
Consider the function $f (x) = ∣ x - 2∣ + ∣ x - 5∣, x \in R$ .

Statement 1: $f^{'} (4) = 0$

Statement 2: $f$ is continuous in [2, 5], differentiable in (2,5) and $f (2) = f (5)$ .
JEE Main 2012View solution →
The value of $p$ and $q$ for which the function $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( p + 1 ) x + s i n x}{x} q \frac{x + x ^{2} - x}{x ^{3/2}}, x < 0, x = 0, x > 0$
is continuous for all $x$ in $R$ , is
JEE Main 2011View solution →
Let $f : R \to R$ be a continuous function defined by $f (x) = \frac{1}{e ^{x} + 2 e ^{- x}}$ .

Statement-1: $f (c) = \frac{1}{3}$ , for some $c \in R$ .
Statement-2: $0 < f (x) \leq \frac{1}{2 2}$ , for all $x \in R$
JEE Main 2010View solution →
Let $f (x) = x ∣ x ∣$ and $g (x) = sin x$ .

Statement-1: $g \circ f$ is differentiable at $x = 0$ and its derivative is continuous at that point.
Statement-2: $g \circ f$ is twice differentiable at $x = 0$ .
JEE Main 2009View solution →
Let $f (x) = {(x - 1) sin (\frac{1}{x - 1}), 0, if x \neq = 1 if x = 1$ . Then which one of the following is true?
JEE Main 2008View solution →
Let $f : R \to R$ be a function defined by $f (x) = Min {x + 1, ∣ x ∣ + 1}$ . Then which of the following is true?
JEE Main 2007View solution →
The function $f : R \sim {0} \to R$ 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
JEE Main 2007View solution →
The set of points where $f (x) = \frac{x}{1 + ∣ x ∣}$ is differentiable is
JEE Main 2006View solution →
Suppose $f (x)$ is differentiable at $x = 1$ and $lim_{h \to 0} \frac{1}{h} f (1 + h) = 5$ , then $f^{'} (1)$ equals
JEE Main 2005View solution →
If $f$ is a real-valued differentiable function satisfying $∣ f (x) - f (y) ∣ \leq (x - y)^{2}, x, y \in R$ and $f (0) = 0$ , then $f (1)$ equals
JEE Main 2005View solution →
Let $f (x) = \frac{1 - t a n x}{4 x - π}, x \neq = \frac{π}{4}, x \in [0, \frac{π}{2}]$ . If $f (x)$ is continuous in $[0, \frac{π}{2}]$ , then $f (\frac{π}{4})$ is
JEE Main 2004View solution →