If $f$ is defined by $f(x) = \begin{cases} x & \text{for } 0 \leq x < 1 \\ 2-x & \text{for } x \geq 1 \end{cases}$,then at $x=1$,$f(x)$ is

Let $f: R \rightarrow R$ be a function defined by $f(x) = \begin{cases} \max_{t \leq x} \{t^3 - 3t\} & x \leq 2 \\ x^2 + 2x - 6 & 2 < x < 3 \\ [x-3] + 9 & 3 \leq x \leq 5 \\ 2x + 1 & x > 5 \end{cases}$ where $[t]$ is the greatest integer less than or equal to $t$. Let $m$ be the number of points where $f$ is not differentiable and $I = \int_{-2}^{2} f(x) dx$. Then the ordered pair $(m, I)$ is equal to:

Consider the functions $f_{1}(x) = x$ and $f_{2}(x) = 2 + \ln x$ for $x > 0$. The graphs of these functions intersect:

Let $f$ and $g$ be real-valued functions defined on the interval $(-1, 1)$ such that $g^{\prime \prime}(x)$ is continuous,$g(0) \neq 0$,$g^{\prime}(0) = 0$,$g^{\prime \prime}(0) \neq 0$,and $f(x) = g(x) \sin x$.
$STATEMENT-1$: $\lim_{x \rightarrow 0} [g(x) \cot x - g(0) \operatorname{cosec} x] = f^{\prime \prime}(0)$.
$STATEMENT-2$: $f^{\prime}(0) = g(0)$.

$A$ function $f(x) = \begin{cases} 1 + x, & x \le 2 \\ 5 - x, & x > 2 \end{cases}$ is

Let $y = f(x) = \begin{cases} e^{-\frac{1}{x^2}}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$. Then which of the following can best represent the graph of $y = f(x)$?