If a function defined by $f(x) = \frac{(3^x - 1)^2}{\sin x \log(1 + x)}$,$x \neq 0$,is continuous at $x = 0$,then $f(0) =$

Let $f(x) = \begin{cases} x^{3}-x^{2}+10x-7, & x \leq 1 \\ -2x+\log_{2}(b^{2}-4), & x > 1 \end{cases}$. Then the set of all values of $b$,for which $f(x)$ has a maximum value at $x=1$,is:

If the function $f(x)=\begin{cases} \frac{1-\cos x}{x^{2}}, & \text{for } x \neq 0 \\ k, & \text{for } x=0 \end{cases}$ is continuous at $x=0$,then the value of $k$ is

The function $f(t) = \frac{1}{t^2 + t - 2}$,where $t = \frac{1}{x - 1}$,is discontinuous at

Let $f(x) = \begin{cases} \operatorname{sgn}(x^2 - 3x + 2) & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}$,then the number of points where $f(x)$ is continuous is (where $\operatorname{sgn}(x)$ denotes the signum function of $x$).

Let $f(x)$ be a real-valued function. If $f^{\prime}(x)$ is a constant for all $x \in R$,$f(0)=2$,and $f^{\prime}(0)=1$,then