If $f(x) = \frac{(e^{2x} - 1) \sin x^{\circ}}{x^2}, x \neq 0$ is continuous at $x = 0$,then $f(0) =$

If the function given by $f(x) = \begin{cases} -2 \sin x & -\pi \leq x < -\pi/2 \\ a \sin x + b & -\pi/2 \leq x \leq \pi/2 \\ \cos x & \pi/2 < x \leq \pi \end{cases}$ is continuous in $[-\pi, \pi]$,then the value of $(3a + 2b)^3$ is

If $f(x) = \begin{cases} \frac{1-\cos Kx}{x \sin x}, & \text{if } x \neq 0 \\ \frac{1}{2}, & \text{if } x=0 \end{cases}$ is continuous at $x=0$,then the value of $K$ is

The function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$ is not defined at $x = 0$. The value which should be assigned to $f$ at $x = 0$ so that it is continuous at $x = 0$ is:

If $f(x)= \begin{cases}-2 \sin x & , \quad x \leqslant-\frac{\pi}{2} \\ a \sin x+b & , \quad \frac{-\pi}{2} < x < \frac{\pi}{2} \\ \cos x & , \quad x \geqslant \frac{\pi}{2}\end{cases}$ is continuous at $x=-\frac{\pi}{2}$ and $x=\frac{\pi}{2}$,then the value of $2a+b$ is

If $f(x) = \begin{cases} 1 + \cos x, & x \le 0 \\ a - x, & 0 < x < 2 \\ (x - b)^2, & x \ge 2 \end{cases}$ is continuous at $x=0$ and $x=2$, then find the value of $a^2+b^2$.