If $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ for $x \neq \pi$ is continuous at $x = \pi$,then $f(\pi) =$

If $f(x) = \begin{cases} 1+6x-3x^2, & x \leq 1 \\ x+\log_2(b^2+7), & x > 1 \end{cases}$ is continuous at all real $x$,then $b=$

Which of the following functions is discontinuous at $x = 0$?

Let $f$ be defined by $f(x) = \begin{cases} \frac{\tan x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$.
Statement-$1$: $x = 0$ is a point of local minima for $f$.
Statement-$2$: $f'(0) = 0$.

If $f(x)= \begin{cases} \frac{x-[x]}{x-2}, & x>2 \\ b, & x=2 \\ \frac{|x^2-x-2|}{a(2+x-x^2)}, & -1 < x \leq 2 \\ 2a-b, & x \leq -1 \end{cases}$ is continuous on $R$,then $\lim _{x \rightarrow 0} \frac{\sin ^2 ax+x \tan bx}{x^2}=$

Let $f(x) = \begin{cases} 0, & \text{if } -1 \leq x < 0 \\ 1, & \text{if } x = 0 \\ 2, & \text{if } 0 < x \leq 1 \end{cases}$ and let $F(x) = \int_{-1}^{x} f(t) \, dt, -1 \leq x \leq 1$. Then: