Let $f: R \rightarrow R$ be a differentiable function such that $f(0)=0$,$f(\frac{\pi}{2})=3$ and $f^{\prime}(0)=1$. If $g(x)=\int_x^{\pi / 2} [f^{\prime}(t) \operatorname{cosec} t - f(t) \operatorname{cosec} t \cot t] dt$ for $x \in (0, \frac{\pi}{2}]$,then $\lim _{x \rightarrow 0} g(x)=$

If $f(x) = \begin{cases} -x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \leq 0 \\ x-1, & 0 < x \leq 1 \\ \ln x, & x > 1 \end{cases}$,then which of the following statements are true?
$(A)$ $f(x)$ is continuous at $x=-\frac{\pi}{2}$
$(B)$ $f(x)$ is not differentiable at $x=0$
$(C)$ $f(x)$ is differentiable at $x=1$
$(D)$ $f(x)$ is differentiable at $x=-\frac{3}{2}$

Let $f:R \to R$ be a continuous function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$.
Statement-$1$: $f(c) = \frac{1}{3}$ for some $c \in R$.
Statement-$2$: $0 < f(x) < \frac{1}{2\sqrt{2}}$ for all $x \in R$.

If $f(x) = \begin{cases} \int_{0}^{x} (5 + |1-t|) \, dt, & x > 2 \\ 5x + 1, & x \leq 2 \end{cases}$,then:

If a function $f$ is defined by:
$\begin{cases} f(x) = x-1, & \text{when } -\infty < x < 1 \\ f(x) = 0, & \text{when } x=1 \\ f(x) = x^3-1, & \text{when } 1 < x < \infty \end{cases}$
then at $x=1$,$f$ is:

Let $f(x)$ be a differentiable function,$f^{\prime}(x) > f(x)$ and $f(0) = 0$. Then