Continuity & Differentiability16 questions3 PYQ

Continuity & Differentiability — JEE Maths Practice Questions & Solutions

16 questions on Continuity & Differentiability with full step-by-step solutions, including past-year (PYQ) problems. Free to practice.

hardPYQ · JEE Main 2026
Let f(x)={ex−1,x<0x2−5x+6,x≥0f(x)=\begin{cases}e^{x-1}, & x<0\\ x^2-5x+6, & x\ge0\end{cases} and g(x)=f(∣x∣)+∣f(x)∣g(x)=f(|x|)+|f(x)|. If the number of points where gg is not continuous and is not differentiable are α\alpha and β\beta respectively, then α+β\alpha+\beta is equal to
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hardPYQ · JEE Main 2026
The number of points, at which the function f(x)=max⁡{6x, 2+3x2}+∣x−1∣cos⁡∣x2−14∣f(x)=\max\{6x,\ 2+3x^2\}+|x-1|\cos\left|x^2-\dfrac14\right|, x∈(−π,π)x\in(-\pi,\pi), is not differentiable, is
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mediumPYQ · JEE Main 2026
The number of points in the interval [2,4][2,4], at which the function f(x)=[x2−x−12]f(x)=\left[x^2-x-\dfrac12\right], where [⋅][\cdot] denotes the greatest integer function, is discontinuous, is
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hard
Let
f(x)={(15−3b){x}−(b2−4b−5)sgn⁡(x+1),−π2<x<0k([x]+[−x]),0≤x≤π(a+2cos⁡x)(1+tan⁡x)ln⁡(1+π2−2πx+x2),π<x<3π2f(x) = \begin{cases} \left(15-3b\right)\{x\}-\left(b^2-4b-5\right)\operatorname{sgn}(x+1), & -\dfrac{\pi}{2} < x < 0 \\[2mm] k\left(\left[x\right]+\left[-x\right]\right), & 0 \le x \le \pi \\[2mm] \dfrac{\left(a+2\cos x\right)\left(1+\tan x\right)}{\ln\left(1+\pi^2-2\pi x+x^2\right)}, & \pi < x < \dfrac{3\pi}{2} \end{cases}
 where [y][y], {y}\{y\} and sgn⁡(y)\operatorname{sgn}(y) denote greatest integer function, fractional part function and signum function of yy respectively.
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hard
Consider a function defined as
f(x)={sin⁡(π2(∣x∣−{x})),−1≤x<1,{x}8(x−2)2+2+2∣4x2−16x+15∣,1≤x≤3,f\left(x\right) = \begin{cases} \sin\left(\dfrac{\pi}2\left(\left|x\right|-\left\{x\right\}\right)\right), & -1 \le x<1,\\[2mm] \left\{x\right\}\sqrt{8\left(x-2\right)^2+2+2\left|4x^2-16x+15\right|}, & 1 \le x \le 3, \end{cases}
 (where {x}\left\{x\right\} represents the fractional part function)
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hard
Let
f(x)={a+(x−b)2for ∣x−b∣≤kc+∣x−b∣for ∣x−b∣>kf(x) = \begin{cases} a+(x-b)^{2} & \text{for } |x-b| \le k \\ c+|x-b| & \text{for } |x-b| > k \end{cases}
 then the positive value of kk so that f(x)f(x) becomes differentiable at x=b−kx = b-k and x=b+kx = b+k is
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medium
f:R→Rf:\mathbb R\to\mathbb R satisfies the equation (f(x))3−(f(x))2−x2f(x)+x2=0\left(f(x)\right)^3-\left(f(x)\right)^2-x^2f(x)+x^2 = 0. The maximum value of f(x)f(x) is 11 and the minimum value is 00. Match List-I with List-II.
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medium
Let f(x)f\left(x\right) be a real valued function defined by f(x)=x2−2∣x∣f\left(x\right) = x^2-2\left|x\right| and
g(x)={min⁡{f(t):−2≤t≤x},x∈[−2,0),max⁡{f(t):0≤t≤x},x∈[0,3].g\left(x\right) = \begin{cases} \min\left\{f\left(t\right) : -2 \le t \le x\right\}, & x \in \left[-2,0\right),\\[1mm] \max\left\{f\left(t\right) : 0 \le t \le x\right\}, & x \in \left[0,3\right]. \end{cases}
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medium
If a twice differentiable function satisfies the relation
f(x2y)=x2f(y)+y f(x2)∀ x,y>0f\left(x^2y\right) = x^2f(y)+y\,f\left(x^2\right) \qquad \forall\ x,y>0
 and f(1)=1f(1) = 1, then the value of f′′(17)f''\left(\dfrac17\right) is
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medium
Let
f(x)={lim⁡n→∞(pxn∑r=1n[r2−e−x+r−1]r(r+1))+λ,x>0q,x=0lim⁡n→∞∑r=1n{r2+r+ex−1}r(r+1),x<0f(x) = \begin{cases} \displaystyle\lim_{n\to\infty}\left(\frac{px}{n}\sum_{r=1}^{n}\frac{\left[r^2-e^{-x}+r-1\right]}{r(r+1)}\right)+\lambda, & x>0 \\[3mm] q, & x=0 \\[3mm] \displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\frac{\left\{r^2+r+e^{x}-1\right\}}{r(r+1)}, & x<0 \end{cases}
 is differentiable in R\mathbb{R} ([ ⋅ ][\,\cdot\,] is G.I.F. and { ⋅ }\{\,\cdot\,\} is F.P. of xx). Then
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medium
Let
f(x)={lim⁡n→∞(pxn∑r=1n[r2−e−x+r−1]r(r+1))+λ,x>0q,x=0lim⁡n→∞∑r=1n{r2+r+ex−1}r(r+1),x<0f(x) = \begin{cases} \displaystyle\lim_{n\to\infty}\left(\frac{px}{n}\sum_{r=1}^{n}\frac{\left[r^2-e^{-x}+r-1\right]}{r(r+1)}\right)+\lambda, & x>0 \\[3mm] q, & x=0 \\[3mm] \displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\frac{\left\{r^2+r+e^{x}-1\right\}}{r(r+1)}, & x<0 \end{cases}
 is differentiable in R\mathbb{R} ([ ⋅ ][\,\cdot\,] is G.I.F. and { ⋅ }\{\,\cdot\,\} is F.P. of xx). Then
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medium
If
f(x)={xmcos⁡(x−1x)x>00x=0f(x) = \begin{cases} x^{m}\cos\left(x-\dfrac1x\right) & x > 0 \\ 0 & x = 0 \end{cases}
 then
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medium
Let [x][x] be the greatest integer function. Then
f(x)=sin⁡(14π[x])[x]f(x) = \frac{\sin\left(\dfrac14\pi[x]\right)}{[x]}
 is
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medium
Let f(x)f(x) be a non-zero differentiable function for all x∈Rx \in \mathbb{R} satisfying
f(x+y)=f(x−y)∀ x,y∈R.f(x+y) = f(x-y) \quad \forall\ x, y \in \mathbb{R} .
 Then
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medium
If f(x)=1x2−17x+66f(x) = \dfrac{1}{x^{2}-17x+66}, then f(2x−2)f\left(\dfrac{2}{x-2}\right) is discontinuous at
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easy
Number of values of x∈[0,π]x \in \left[0,\pi\right] where f(x)=[4sin⁡x−7]f(x) = \left[4\sin x-7\right] is non-derivable is ([ ⋅ ][\,\cdot\,] is G.I.F.)
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