(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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If a real-valued function $f(x) = \begin{cases} \frac{2x^2+(k+2)x+9}{3x^2-7x-6} & , \text{for } x \neq 3 \\ l & , \text{for } x=3 \end{cases}$ is continuous at $x=3$ and $l$ is a finite value,then $l-k=$

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If a real-valued function $f(x) = \begin{cases} \frac{2x^2+(k+2)x+9}{3x^2-7x-6} & , \text{for } x \neq 3 \\ l & , \text{for } x=3 \end{cases}$ is continuous at $x=3$ and $l$ is a finite value,then $l-k=$

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$f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & x < 4 \\ a+b, & x=4 \\ \frac{x-4}{|x-4|} + b, & x > 4 \end{cases}$
If $f(x)$ given above is continuous at $x=4$,then find the values of '$a$' and '$b$'.