For two constants $a , b$, if $\lim _ { x \rightarrow 3 } \frac { \sqrt { x + a } - b } { x - 3 } = \frac { 1 } { 4 }$, what is the value of $a + b$? [2 points]
(1) 3
(2) 5
(3) 7
(4) 9
(5) 11

120. Suppose $f(x) = \cos^2(2x) + ax^2 + b$, $\displaystyle\lim_{x \to 0^+} \frac{f(x)}{x} = 0$, and $\displaystyle\lim_{x \to 0^-} \frac{f'(x)}{x} = 2$. What is $a + b$?
(1) $8$ (2) $6$ (3) $4$ (4) $-8$
121. The tangent lines to the curve $f(x) = |\sin(2x)| + 1$ at the point $x = 0$ with length 4 are drawn. If $A$ and $B$ are respectively the second and fourth intersection points of the tangent lines with the $x$-axis, what is the length of segment $AB$?
(1) zero (2) $\dfrac{2\sqrt{7}}{3}$ (3) $\dfrac{4\sqrt{7}}{3}$ (4) $2\sqrt{2}$

16. $\mathrm { f } ( \mathrm { x } ) = \begin{cases} \mathrm { b } \sin ^ { - 1 } \left( \frac { \mathrm { x } + \mathrm { c } } { 2 } \right) , & - \frac { 1 } { 2 } < \mathrm { x } < 0 \\ \frac { 1 } { 2 } , & \mathrm { x } = 0 \\ \frac { \mathrm { e } ^ { \frac { \mathrm { a } } { 2 } \mathrm { x } } - 1 } { \mathrm { x } } , & 0 < \mathrm { x } < \frac { 1 } { 2 } \end{cases}$
If $\mathrm { f } ( \mathrm { x } )$ is differentiable at $\mathrm { x } = 0$ and $| \mathrm { c } | < \frac { 1 } { 2 }$ then find the value of ' a ' and prove that $64 \mathrm {~b} ^ { 2 } = \left( 4 - \mathrm { c } ^ { 2 } \right)$. Sol. $f \left( 0 ^ { + } \right) = f \left( 0 ^ { - } \right) = f ( 0 )$ Here $\mathrm { f } \left( 0 ^ { + } \right) = \lim _ { \mathrm { x } \rightarrow \infty } \frac { \mathrm { e } ^ { \frac { \mathrm { ax } } { 2 } } - 1 } { \mathrm { x } } = \lim _ { \mathrm { x } \rightarrow \infty } \frac { \mathrm { e } ^ { \frac { \mathrm { ax } } { 2 } } - 1 } { \frac { \mathrm { ax } } { 2 } } \cdot \frac { \mathrm { a } } { 2 } = \frac { \mathrm { a } } { 2 }$. $\Rightarrow \mathrm { b } \sin ^ { - 1 } \frac { \mathrm { c } } { 2 } = \frac { \mathrm { a } } { 2 } = \frac { 1 } { 2 } \Rightarrow \mathrm { a } = 1$. $\mathrm { L } \mathrm { f } ^ { \prime } \left( 0 _ { - } \right) = \lim _ { \mathrm { h } \rightarrow 0 ^ { - } } \frac { \mathrm { b } \sin ^ { - 1 } \frac { ( \mathrm {~h} + \mathrm { c } ) } { 2 } - \frac { 1 } { 2 } } { \mathrm {~h} } = \frac { \mathrm { b } / 2 } { \sqrt { 1 - \frac { \mathrm { c } ^ { 2 } } { 4 } } }$ $R f ^ { \prime } \left( 0 _ { + } \right) = \lim _ { h \rightarrow 0 ^ { + } } \frac { \frac { e ^ { h / 2 } - 1 } { h } - \frac { 1 } { 2 } } { h } = \frac { 1 } { 8 }$ Now $\mathrm { L } ^ { \prime } \left( 0 _ { - } \right) = \mathrm { Rf } ^ { \prime } \left( 0 _ { + } \right) \Rightarrow \frac { \frac { \mathrm { b } } { 2 } } { \sqrt { 1 - \frac { \mathrm { c } ^ { 2 } } { 4 } } } = \frac { 1 } { 8 }$ $4 b = \sqrt { 1 - \frac { c ^ { 2 } } { 4 } } \Rightarrow 16 b ^ { 2 } = \frac { 4 - c ^ { 2 } } { 4 } \Rightarrow 64 b ^ { 2 } = 4 - c ^ { 2 }$.

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) $\frac { 10 } { 3 }$

If $\lim _ { x \rightarrow 0 } \frac { a e ^ { x } - b \cos x + c e ^ { - x } } { x \sin x } = 2$, then $a + b + c$ is equal to

If the function $f ( x ) = \begin{cases} \frac { 1 } { x } \log _ { \mathrm { e } } \left( \frac { 1 + \frac { x } { a } } { 1 - \frac { x } { b } } \right) & , x < 0 \\ k & , x = 0 \\ \frac { \cos ^ { 2 } x - \sin ^ { 2 } x - 1 } { \sqrt { x ^ { 2 } + 1 } - 1 } & , x > 0 \end{cases}$ is continuous at $x = 0$, then $\frac { 1 } { a } + \frac { 1 } { b } + \frac { 4 } { k }$ is equal to:
(1) 4
(2) 5
(3) $- 4$
(4) $- 5$

Let $g(x)$ be a linear function and $f(x) = \begin{cases} g(x), & x \leq 0 \\ \frac { 1 + x } { 2 + x } , & x > 0 \end{cases}$, is continuous at $x = 0$. If $f'(1) = f(-1)$, then the value of $g(3)$ is
(1) $\frac { 1 } { 3 } \log _ { e } \frac { 4 } { e^{1/3} }$
(2) $\frac { 1 } { 3 } \log _ { e } \frac { 4 } { 9 } + 1$
(3) $\log _ { e } \frac { 4 } { 9 } - 1$
(4) $\log _ { e } \frac { 4 } { 9 e ^ { 1/3 } }$

If the function $f ( x ) = \left\{ \begin{array} { l } \frac { 2 } { x } \left\{ \sin \left( k _ { 1 } + 1 \right) x + \sin \left( k _ { 2 } - 1 \right) x \right\} , \quad x < 0 \\ 4 , \quad x = 0 \end{array} \quad \right.$ is continuous at $\mathrm { x } = 0$, then $\mathrm { k } _ { 1 } ^ { 2 } + \mathrm { k } _ { 2 } ^ { 2 }$ is equal to
(1) 20
(2) 5
(3) 8
(4) 10

Q72. If the function $f ( x ) = \frac { \sin 3 x + \alpha \sin x - \beta \cos 3 x } { x ^ { 3 } } , x \in \mathbf { R }$, is continuous at $x = 0$, then $f ( 0 )$ is equal to :
(1) 2
(2) - 2
(3) 4
(4) - 4

Q72. $\quad$ For $\mathrm { a } , \mathrm { b } > 0$, let $f ( x ) = \left\{ \begin{array} { c l } \frac { \tan ( ( \mathrm { a } + 1 ) x ) + \mathrm { b } \tan x } { x } , & x < 0 \\ 3 , & x = 0 \\ \frac { \sqrt { \mathrm { a } x + \mathrm { b } ^ { 2 } x ^ { 2 } } - \sqrt { \mathrm { a } x } } { \mathrm {~b} \sqrt { \mathrm { a } } \sqrt { x } } , & x > 0 \end{array} \right.$ be a continous function at $x = 0$. Then $\frac { \mathrm { b } } { \mathrm { a } }$ is equal to :
(1) 6
(2) 4
(3) 5
(4) 8

$f ( x ) = \frac { e ^ { x } \left( e ^ { \tan x - x } - 1 \right) + \log ( \sec x + \tan x ) - x } { \tan x - x }$ If $\mathrm { f } ( \mathrm { x } )$ is continuous at $\mathrm { x } = 0$, then find $\mathrm { f } ( 0 )$