If $f(x) = \begin{cases} \frac{8}{x^3} - 6x, & x \le 1 \\ \sqrt{x} + 1, & x > 1 \end{cases}$, then at $x = 1$, $f$ is:

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 $f: R \rightarrow R$ be defined as $f(x)=\begin{cases} \frac{a-b \cos 2 x}{x^2} & ; x<0 \\ x^2+c x+2 & ; 0 \leq x \leq 1 \\ 2 x+1 & ; x>1 \end{cases}$. If $f$ is continuous everywhere in $R$ and $m$ is the number of points where $f$ is $NOT$ differentiable,then $m+a+b+c$ equals:

Let $R$ be the set of all real numbers and $f(x) = \sin^{10} x (\cos^8 x + \cos^4 x + \cos^2 x + 1)$ for $x \in R$. Let $S = \{\lambda \in R : \text{there exists a point } c \in (0, 2\pi) \text{ with } f'(c) = \lambda f(c)\}$. Then,

Let $f: R \rightarrow (0, \infty)$ and $g: R \rightarrow R$ be twice differentiable functions such that $f^{\prime \prime}$ and $g^{\prime \prime}$ are continuous functions on $R$. Suppose $f^{\prime}(2) = g(2) = 0$,$f^{\prime \prime}(2) \neq 0$ and $g^{\prime}(2) \neq 0$. If $\lim_{x \rightarrow 2} \frac{f(x) g(x)}{f^{\prime}(x) g^{\prime}(x)} = 1$,then:

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