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:

Let $f, g: R \rightarrow R$ be functions defined by $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = x f(x)$. Consider the following statements: $(i)$ $f(x)$ is continuous at $x = 0$ but not differentiable at $x = 0$. $(ii)$ $g(x)$ is differentiable at $x = 0$, but $g'(x)$ is not continuous at $x = 0$. Then, which one of the following is true?

If $f(x)$ defined as given below is continuous on $R$,then the value of $a+b$ is equal to: $f(x) = \begin{cases} \sin x, & x \leq 0 \\ x^2+a, & 0 < x < 1 \\ b x+3, & 1 \leq x \leq 3 \\ -3, & x > 3 \end{cases}$

If $f: [-2, 2] \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{\sqrt{1 + cx} - \sqrt{1 - cx}}{x}, & -2 \leq x < 0 \\ \frac{x + 3}{x + 1}, & 0 \leq x \leq 2 \end{cases}$ is continuous on $[-2, 2]$,then $c$ is equal to

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

If $f(x) = \frac{(27-2x)^{1/3}-3}{9-3(243+5x)^{1/5}}, x \neq 0$ is continuous at $x=0$, then the value of $f(0)$ is