If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text{if } x \neq 0 \\ a, & \text{if } x=0 \end{cases}$, then the value of $a$ so that $f$ is continuous at $x=0$ is

If $f(x) = \begin{cases} \frac{5}{2} - x, & x < 2 \\ 1, & x = 2 \\ x - \frac{3}{2}, & x > 2 \end{cases}$,then:

Check the points where the constant function $f(x)=k$ is continuous.

If $f(x) = \begin{cases} \frac{a}{2}(x - |x|), & \text{for } x < 0 \\ 0, & \text{for } x = 0 \\ bx^2 \sin \left( \frac{1}{x} \right), & \text{for } x > 0 \end{cases}$ is continuous at $x = 0$,then

Let $f(x) = \begin{cases} |x|+3, & \text{if } x \leq -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x+2, & \text{if } x \geq 3 \end{cases}$. Determine the continuity of $f(x)$ at $x = -3$ and $x = 3$.

The values of $a$ and $b$,so that the function $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b, & \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous for $0 \leq x \leq \pi$,are respectively given by