(C) Since $f(x)$ is continuous on $R$,it must be continuous at $x=0$ and $x=2$. At $x=0$,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$. $\cos

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(C) Since $f(x)$ is continuous on $R$,it must be continuous at $x=0$ and $x=2$. At $x=0$,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$. $\cos(0) = 3(0) + \alpha \implies 1 = \alpha$. At $x=2$,$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$. $3(2) + \alpha = \beta(2) + 3$. Substituting $\alpha = 1$: $6 + 1 = 2\beta + 3 \implies 7 = 2\beta + 3 \implies 2\beta = 4 \implies \beta = 2$. Therefore,$\alpha^2 + \beta^2 = (1)^2 + (2)^2 = 1 + 4 = 5$.

Class 12 Mathematics Continuity and Differentiation Continuity

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If the function $f$ defined by $f(x) = \begin{cases} \cos x, & \text{if } x \leq 0 \\ 3x + \alpha, & \text{if } 0 < x < 2 \\ \beta x + 3, & \text{if } 2 \leq x \leq 4 \\ 11, & \text{if } x > 4 \end{cases}$ where $\alpha$ and $\beta$ are real constants,is continuous on $R$,then $\alpha^2 + \beta^2 =$

AP EAMCET 2018 All AP EAMCET

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$3$
B
$9$
C
$5$
D
$1$

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Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx + 1, & \text{if } x \le \pi \\ \cos x, & \text{if } x > \pi \end{cases}$ at $x = \pi$.

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Let $f(x) = \begin{cases} 0, & x < 0 \\ x^2, & x \ge 0 \end{cases}$,then for all values of $x$

MediumIIT 1984

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For $a, b > 0$,let $f(x) = \begin{cases} \frac{\tan((a+1)x) + b \tan x}{x}, & x < 0 \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b \sqrt{a} x \sqrt{x}}, & x > 0 \end{cases}$ be a continuous function at $x = 0$. Then $\frac{b}{a}$ is equal to

DifficultJEE MAIN 2024

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Let $\alpha \in R$ be such that the function $f(x) = \begin{cases} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x \neq 0 \\ \alpha, & x=0 \end{cases}$ is continuous at $x=0$,where $\{x\} = x - [x]$ and $[x]$ is the greatest integer less than or equal to $x$. Then:

DifficultJEE MAIN 2021

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Let $f, g: R \to R$ be two functions defined by $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right), & x \ne 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = x f(x)$.
Statement $I$: $f$ is a continuous function at $x = 0$.
Statement $II$: $g$ is a differentiable function at $x = 0$.

DifficultJEE MAIN 2014

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Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx + 1, & \text{if } x \le \pi \\ \cos x, & \text{if } x > \pi \end{cases}$ at $x = \pi$.

Let $f(x) = \begin{cases} 0, & x < 0 \\ x^2, & x \ge 0 \end{cases}$,then for all values of $x$

For $a, b > 0$,let $f(x) = \begin{cases} \frac{\tan((a+1)x) + b \tan x}{x}, & x < 0 \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b \sqrt{a} x \sqrt{x}}, & x > 0 \end{cases}$ be a continuous function at $x = 0$. Then $\frac{b}{a}$ is equal to

Let $\alpha \in R$ be such that the function $f(x) = \begin{cases} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x \neq 0 \\ \alpha, & x=0 \end{cases}$ is continuous at $x=0$,where $\{x\} = x - [x]$ and $[x]$ is the greatest integer less than or equal to $x$. Then:

Let $f, g: R \to R$ be two functions defined by $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right), & x \ne 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = x f(x)$.Statement $I$: $f$ is a continuous function at $x = 0$.Statement $II$: $g$ is a differentiable function at $x = 0$.

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Let $f, g: R \to R$ be two functions defined by $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right), & x \ne 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = x f(x)$.
Statement $I$: $f$ is a continuous function at $x = 0$.
Statement $II$: $g$ is a differentiable function at $x = 0$.