JEE MAIN Continuity Previous Year Questions (PYQs) – Page 1 of 4

Let $f:\mathbb{R}\to\mathbb{R}$ be a function given by $ f(x)= \begin{cases} \dfrac{1-\cos 2x}{x^2}, & x<0,\\[6pt] \alpha, & x=0,\\[6pt] \dfrac{\beta\sqrt{\,1-\cos x\,}}{x}, & x>0, \end{cases} $ where $\alpha,\beta\in\mathbb{R}$. If $f$ is continuous at $x=0$, then $\alpha^2+\beta^2$ is equal to:

If the function $ f(x)= \begin{cases} \dfrac{7^{x}-9^{x}-8^{x}+1}{\sqrt{2}-\sqrt{1+\cos^{2}x}}, & x\neq0,\\[6pt] a\log_{e}2\log_{e}3, & x=0 \end{cases} $ is continuous at $x=0$, then the value of $a^{2}$ is equal to:

Let $f:\left( { - {\pi \over 4},{\pi \over 4}} \right) \to R$ be defined as $f(x) = \left\{ {\matrix{ {{{(1 + |\sin x|)}^{{{3a} \over {|\sin x|}}}}} & , & { - {\pi \over 4} < x < 0} \cr b & , & {x = 0} \cr {{e^{\cot 4x/\cot 2x}}} & , & {0 < x < {\pi \over 4}} \cr } } \right.$ If f is continuous at x = 0, then the value of 6a + b² is equal to :

Consider the function $ f(x)= \begin{cases} \dfrac{a\,(7x-12-x^{2})}{\,b\,\lfloor x^{2}-7x+12\rfloor\,}, & x<3,\\[6pt] \dfrac{\sin(x-3)}{2^{\,x-1}}, & x>3,\\[6pt] b, & x=3, \end{cases} $ where $\lfloor x\rfloor$ denotes the greatest integer $\le x$. If $S$ denotes the set of all ordered pairs $(a,b)$ such that $f(x)$ is continuous at $x=3$, then the number of elements in $S$ is:

Let $f:[-1,2]\to\mathbb{R}$ be given by $f(x)=2x^{2}+x+\lfloor x^{2}\rfloor-\lfloor x\rfloor$, where $\lfloor t\rfloor$ denotes the greatest integer $\le t$. The number of points where $f$ is not continuous is:

Let f(x) = $\left\{ {\matrix{ {{{\left( {x - 1} \right)}^{{1 \over {2 - x}}}},} & {x > 1,x \ne 2} \cr {k\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {,x = 2} \cr } } \right.$

Thevaue of k for which f s continuous at x = 2 is :

The function $f : \mathbb{R} \to \mathbb{R}$ defined by $$ f(x) = \lim_{n \to \infty} \frac{\cos(2 \pi x) - x^{2n} \sin(x-1)}{1 + x^{2n+1} - x^{2n}} $$ is continuous for all $x$ in :

Let $[x]$ denote the greatest integer function, and let $m$ and $n$ respectively be the numbers of the points where the function $f(x) = [x] + |x - 2|$, $-2 < x < 3$, is not continuous and not differentiable. Then $m + n$ is equal to:

$\text { If } f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0 & , x=0 \end{array}\right. \text {, then }$

Let \( f(x) = \begin{cases} \dfrac{ax^2 + 2ax + 3}{4x^2 + 4x - 3}, & x \ne -\frac{3}{2}, \frac{1}{2} \\ b, & x = -\frac{3}{2}, \frac{1}{2} \end{cases} \) be continuous at \( x = -\frac{3}{2} \). If \( f \circ f(x) = \frac{7}{5} \), then \( x \) is equal to: