The function $f(x) = \begin{cases} |x - 3| & x \geqslant 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4} & x < 1 \end{cases}$ is :

Let $f(x) = \lim_{n}$ ${\rightarrow \infty} \left( \frac{n^n(x+n)(x+\frac{n}{2}) \cdots (x+\frac{n}{n})}{n!(x^2+n^2)(x^2+\frac{n^2}{4}) \cdots (x^2+\frac{n^2}{n^2})} \right)^{\frac{x}{n}}$,for all $x > 0$. Then
$(A)$ $f(\frac{1}{2}) \geq f(1)$
$(B)$ $f(\frac{1}{3}) \leq f(\frac{2}{3})$
$(C)$ $f^{\prime}(2) \leq 0$
$(D)$ $\frac{f^{\prime}(3)}{f(3)} \geq \frac{f^{\prime}(2)}{f(2)}$

Let $f(x)=x^2+a x+b$,where $a, b \in R$. If $f(x)=0$ has all its roots imaginary,then the roots of $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0$ are

Let a function $f: R \rightarrow R$ be defined as :
$f(x)=\begin{cases} \int_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}+b x, & x \leq 4 \end{cases}$
where $b \in R$. If $f$ is continuous at $x=4$,then which of the following statements is $NOT$ true?

Consider the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \begin{cases} (2 - \sin(\frac{1}{x}))|x|, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then $f$ is

Let $f(x)$ be defined in $[-2, 2]$ by
$f(x) = \begin{cases} \max(4 - x^2, 1 + x^2), & -2 < x < 0 \\ \min(4 - x^2, 1 + x^2), & 0 < x < 2 \end{cases}$
Then $f(x)$: