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:

Let a function $f: R \rightarrow R$ be defined as
$f(x) = \begin{cases} \sin x - e^x & \text{if } x \leq 0 \\ a + [-x] & \text{if } 0 < x < 1 \\ 2x - b & \text{if } x \geq 1 \end{cases}$
where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous on $R$,then $(a + b)$ is equal to:

The function $f : R \rightarrow R$ defined by $f(x) = \lim_{n \rightarrow \infty} \frac{\cos(2 \pi x) - x^{2n} \sin(x-1)}{1 + x^{2n+1} - x^{2n}}$ is continuous for all $x$ in.

$f(x) = \begin{cases} 3x - 8 & \text{if } x \leq 5 \\ 2k & \text{if } x > 5 \end{cases}$ is continuous,find $k$.

Let $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\ 2x \cot x+b, & \frac{\pi}{4} \leq x < \frac{\pi}{2} \\ a \cos 2x-b \sin x, & \frac{\pi}{2} \leq x \leq \pi \end{cases}$. If $f(x)$ is continuous for $0 \leq x \leq \pi$,then:

The function $f(x) = \begin{cases} x, & \text{if } 0 \le x \le 1 \\ 1, & \text{if } 1 < x \le 2 \end{cases}$ is