If the function $f(x)$ is continuous on its domain $[-2, 2]$,where $f(x) = \begin{cases} \frac{\sin ax}{x} + 3, & -2 \leq x < 0 \\ x + 5, & 0 \leq x \leq 1 \\ \sqrt{x^2 + 8} - b, & 1 < x \leq 2 \end{cases}$,then $7a + b + 1$ is equal to:
- A$10$
- B$11$
- C$14$
- D$12$
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The function $f$ is defined by $f(x) = \begin{cases} 2x - 1, & \text{if } x > 2 \\ k, & \text{if } x = 2 \\ x^2 - 1, & \text{if } x < 2 \end{cases}$. If $f$ is continuous at $x = 2$,then the value of $k$ is equal to:
Easy
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If the function $f(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$ is continuous at $x = 0$,then $f(0)$ is equal to . . . . . . .
DifficultJEE MAIN 2025
View SolutionDefine $f: R \rightarrow R$ by $f(x) = [x] + \sqrt{x - [x]}$ for $x \in R$,where $[x]$ denotes the greatest integer function. Then the set of points at which $f$ is continuous is
The values of $A$ and $B$ such that the function $f(x) = \begin{cases} -2\sin x, & x \le -\frac{\pi}{2} \\ A\sin x + B, & -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & x \ge \frac{\pi}{2} \end{cases}$ is continuous everywhere are
Medium
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