If the function $g ( x ) = \left\{ \begin{array} { c c } k \sqrt { x + 1 } , & 0 \leq x \leq 3 \\ m x + 2 , & 3 < x \leq 5 \end{array} \right.$ is differentiable, then the value of $k + m$ is
(1) 4
(2) 2
(3) $\frac { 16 } { 5 }$
(4) $\frac { 10 } { 3 }$

Let $S = \left\{ t \in R : f ( x ) = | x - \pi | \cdot \left( e ^ { | x | } - 1 \right) \sin | x |\right.$ is not differentiable at $\left. t \right\}$. Then, the set $S$ is equal to:
(1) $\{ 0 , \pi \}$
(2) $\phi$ (an empty set)
(3) $\{ 0 \}$
(4) $\{ \pi \}$

$\lim _ { x \rightarrow a } \frac { ( a + 2 x ) ^ { \frac { 1 } { 3 } } - ( 3 x ) ^ { \frac { 1 } { 3 } } } { ( 3 a + x ) ^ { \frac { 1 } { 3 } } - ( 4 x ) ^ { \frac { 1 } { 3 } } } ( a \neq 0 )$ is equal to:
(1) $\left( \frac { 2 } { 9 } \right) \left( \frac { 2 } { 3 } \right) ^ { \frac { 1 } { 3 } }$
(2) $\left( \frac { 2 } { 3 } \right) ^ { \frac { 4 } { 3 } }$
(3) $\left( \frac { 2 } { 9 } \right) ^ { \frac { 4 } { 3 } }$
(4) $\left( \frac { 2 } { 3 } \right) \left( \frac { 2 } { 9 } \right) ^ { \frac { 1 } { 3 } }$

If $\alpha$ is the positive root of the equation, $p ( x ) = x ^ { 2 } - x - 2 = 0$, then $\lim _ { x \rightarrow \alpha ^ { + } } \frac { \sqrt { 1 - \cos p ( x ) } } { x + \alpha - 4 }$ is equal to
(1) $\frac { 3 } { 2 }$
(2) $\frac { 3 } { \sqrt { 2 } }$
(3) $\frac { 1 } { \sqrt { 2 } }$
(4) $\frac { 1 } { 2 }$

$\lim_{x\rightarrow 0} \frac{x\left(e^{\left(\sqrt{1+x^2+x^4}-1\right)/x}-1\right)}{\sqrt{1+x^2+x^4}-1}$
(1) is equal to $\sqrt{e}$
(2) is equal to 1
(3) is equal to 0
(4) does not exist

$\lim _ { x \rightarrow 0 } \left( \tan \left( \frac { \pi } { 4 } + x \right) \right) ^ { 1 / x }$ is equal to
(1) $e$
(2) 2
(3) 1
(4) $e ^ { 2 }$

Let $f ( x )$ be a differentiable function at $x = a$ with $f ^ { \prime } ( a ) = 2$ and $f ( a ) = 4$. Then $\lim _ { x \rightarrow a } \frac { x f ( a ) - a f ( x ) } { x - a }$ equals:
(1) $a + 4$
(2) $2 a - 4$
(3) $4 - 2 a$
(4) $2 a + 4$

$f ( x ) = \left| \begin{array} { c c c } a & - 1 & 0 \\ a x & a & - 1 \\ a x ^ { 2 } & a x & a \end{array} \right| , a \in R$. Then the sum of the squares of all the values of $a$ for $2 f ^ { \prime } ( 10 ) - f ^ { \prime } ( 5 ) + 100 = 0$ is
(1) 117
(2) 106
(3) 125
(4) 136

Let $f: (-2, 2) \rightarrow \mathbb{R}$ be defined by $f(x) = \begin{cases} x\lfloor x\rfloor, & 0 \leq x < 2 \\ (x-1)\lfloor x\rfloor, & -2 < x < 0 \end{cases}$ where $\lfloor x \rfloor$ denotes the greatest integer function. If $m$ and $n$ respectively are the number of points in $(-2, 2)$ at which $y = f(x)$ is not continuous and not differentiable, then $m + n$ is equal to $\_\_\_\_$.

For a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$, $$f'(x) = 2x^{2} - 1, \quad f(2) = 4$$ Given this, what is the value of the limit $\displaystyle\lim_{x \rightarrow 2} \frac{f(x)-4}{x-2}$?
A) 3
B) 4
C) 5
D) 6
E) 7

$$\lim _ { x \rightarrow 1 } \frac { f ( x + 1 ) - 3 } { x - 1 } = 2$$
Given that, what is the value of the limit $\lim _ { x \rightarrow 2 } \frac { x \cdot f ( x ) - 6 } { x - 2 }$?
A) 5
B) 6
C) 7
D) 8
E) 9

$$\lim _ { x \rightarrow 0 } \frac { \sqrt { x + 5 } - \sqrt { 5 } } { x }$$
What is the value of this limit?
A) $\frac { \sqrt { 5 } } { 5 }$
B) $\frac { 2 \sqrt { 5 } } { 5 }$
C) $\frac { \sqrt { 5 } } { 10 }$
D) 0
E) $2 \sqrt { 5 }$