For a quadratic function $f ( x )$ with leading coefficient 1 and the function $$g ( x ) = \left\{ \begin{array} { c c } \frac { 1 } { \ln ( x + 1 ) } & ( x \neq 0 ) \\ 8 & ( x = 0 ) \end{array} \right.$$ when the function $f ( x ) g ( x )$ is continuous on the interval $( - 1 , \infty )$, what is the value of $f ( 3 )$? [3 points]
(1) 6
(2) 9
(3) 12
(4) 15
(5) 18

6. The domain of the function $f ( x ) = \sqrt { 4 - | x | } + \lg \frac { x ^ { 2 } - 5 x + 6 } { x - 3 }$ is
A. $ ( 2,3 )$
B. $ ( 2,4 ]$
C. $ ( 2,3 ) \cup ( 3,4 ]$
D. $ ( - 1,3 ) \cup ( 3,6 ]$

Let $f ( x ) = \begin{cases} 2 ^ { -x } & x \leq 0 \\ 1 & x > 0 \end{cases}$. Then the range of $x$ satisfying $f ( x + 1 ) < f ( 2 x )$ is
A. $( - \infty , - 1 ]$
B. $( 0 , + \infty )$
C. $( - 1,0 )$
D. $( - \infty , 0 )$

Let $E _ { 1 } = \left\{ x \in \mathbb { R } : x \neq 1 \right.$ and $\left. \frac { x } { x - 1 } > 0 \right\}$ and $E _ { 2 } = \left\{ x \in E _ { 1 } : \sin ^ { - 1 } \left( \log _ { e } \left( \frac { x } { x - 1 } \right) \right) \right.$ is a real number $\}$. (Here, the inverse trigonometric function $\sin ^ { - 1 } x$ assumes values in $\left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right]$.) Let $f : E _ { 1 } \rightarrow \mathbb { R }$ be the function defined by $f ( x ) = \log _ { e } \left( \frac { x } { x - 1 } \right)$ and $g : E _ { 2 } \rightarrow \mathbb { R }$ be the function defined by $g ( x ) = \sin ^ { - 1 } \left( \log _ { e } \left( \frac { x } { x - 1 } \right) \right)$.
LIST-I P. The range of $f$ is Q. The range of $g$ contains R. The domain of $f$ contains S. The domain of $g$ is
LIST-II

1. $\left( - \infty , \frac { 1 } { 1 - e } \right] \cup \left[ \frac { e } { e - 1 } , \infty \right)$
2. $( 0,1 )$
3. $\left[ - \frac { 1 } { 2 } , \frac { 1 } { 2 } \right]$
4. $( - \infty , 0 ) \cup ( 0 , \infty )$
5. $\left( - \infty , \frac { e } { e - 1 } \right]$
6. $( - \infty , 0 ) \cup \left( \frac { 1 } { 2 } , \frac { e } { e - 1 } \right]$

The correct option is:
(A) $\mathbf { P } \rightarrow \mathbf { 4 } ; \mathbf { Q } \rightarrow \mathbf { 2 } ; \mathbf { R } \rightarrow \mathbf { 1 } ; \mathbf { S } \rightarrow \mathbf { 1 }$
(B) $\mathbf { P } \rightarrow \mathbf { 3 } ; \mathbf { Q } \rightarrow \mathbf { 3 } ; \mathbf { R } \rightarrow \mathbf { 6 } ; \mathbf { S } \rightarrow \mathbf { 5 }$
(C) $\mathbf { P } \rightarrow \mathbf { 4 } ; \mathbf { Q } \rightarrow \mathbf { 2 } ; \mathbf { R } \rightarrow \mathbf { 1 } ; \mathbf { S } \rightarrow \mathbf { 6 }$
(D) $\mathrm { P } \rightarrow 4 ; \mathrm { Q } \rightarrow 3 ; \mathrm { R } \rightarrow 6 ; \mathrm { S } \rightarrow 5$

The largest interval lying in $\left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$ for which the function $\left[ f ( x ) = 4 ^ { - x ^ { 2 } } + \cos ^ { - 1 } \left( \frac { x } { 2 } - 1 \right) + \log ( \cos x ) \right]$ is defined, is
(1) $[ 0 , \pi ]$
(2) $\left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$
(3) $\left[ - \frac { \pi } { 4 } , \frac { \pi } { 2 } \right)$
(4) $\left[ 0 , \frac { \pi } { 2 } \right)$

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x) = \frac{x}{1+x^2}$, $x \in \mathbb{R}$. Then the range of $f$ is:
(1) $\mathbb{R} - [-1, 1]$
(2) $(-1, 1) - \{0\}$
(3) $\left[-\frac{1}{2}, \frac{1}{2}\right]$
(4) $\left(-\frac{1}{2}, \frac{1}{2}\right)$

The domain of the definition of the function $f ( x ) = \frac { 1 } { 4 - x ^ { 2 } } + \log _ { 10 } \left( x ^ { 3 } - x \right)$ is:
(1) $( - 1,0 ) \cup ( 1,2 ) \cup ( 2 , \infty )$
(2) $( 1,2 ) \cup ( 2 , \infty )$
(3) $( - 2 , - 1 ) \cup ( - 1,0 ) \cup ( 2 , \infty )$
(4) $( - 1,0 ) \cup ( 1,2 ) \cup ( 3 , \infty )$

Let $f:(1,3) \rightarrow R$ be a function defined by $f(x) = \frac{x[x]}{1 + x^{2}}$, where $[x]$ denotes the greatest integer $\leq x$. Then the range of $f$ is
(1) $\left(\frac{2}{5}, \frac{3}{5}\right] \cup \left(\frac{3}{4}, \frac{4}{5}\right)$
(2) $\left(\frac{2}{5}, \frac{1}{2}\right) \cup \left(\frac{3}{5}, \frac{4}{5}\right]$
(3) $\left(\frac{2}{5}, \frac{4}{5}\right]$
(4) $\left(\frac{3}{5}, \frac{4}{5}\right)$

The domain of the function $f(x) = \sin^{-1}\left(\frac{|x| + 5}{x^{2} + 1}\right)$ is $(-\infty, -a] \cup [a, \infty)$, then $a$ is equal to
(1) $\frac{\sqrt{17}}{2}$
(2) $\frac{\sqrt{17} - 1}{2}$
(3) $\frac{1 + \sqrt{17}}{2}$
(4) $\frac{\sqrt{17}}{2} + 1$

Let $f ( x ) = \sin ^ { - 1 } x$ and $g ( x ) = \frac { x ^ { 2 } - x - 2 } { 2 x ^ { 2 } - x - 6 }$. If $g ( 2 ) = \lim _ { x \rightarrow 2 } g ( x )$, then the domain of the function $f o g$ is
(1) $( - \infty , - 1 ] \cup [ 2 , \infty )$
(2) $( - \infty , - 2 ] \cup \left[ - \frac { 3 } { 2 } , \infty \right)$
(3) $( - \infty , - 2 ] \cup \left[ - \frac { 4 } { 3 } , \infty \right)$
(4) $( - \infty , - 2 ] \cup [ - 1 , \infty )$

Let $f : R \rightarrow R$ be defined as $f ( x ) = \begin{cases} 2 \sin \left( - \frac { \pi x } { 2 } \right) , & \text { if } x < - 1 \\ \left| a x ^ { 2 } + x + b \right| , & \text { if } - 1 \leq x \leq 1 \\ \sin ( \pi x ) , & \text { if } x > 1 \end{cases}$ If $f ( x )$ is continuous on $R$, then $a + b$ equals :
(1) 1
(2) 3
(3) - 3
(4) - 1

Let $\alpha \in R$ be such that the function $f ( x ) = \left\{ \begin{array} { l l } \frac { \cos ^ { - 1 } \left( 1 - \{ x \} ^ { 2 } \right) \sin ^ { - 1 } ( 1 - \{ x \} ) } { \{ x \} - \{ x \} ^ { 3 } } , & x \neq 0 \\ \alpha , & x = 0 \end{array} \right.$ is continuous at $x = 0$, where $\{ x \} = x - [ x ] , [ x ]$ is the greatest integer less than or equal to $x$. Then :
(1) $\alpha = \frac { \pi } { \sqrt { 2 } }$
(2) $\alpha = 0$
(3) no such $\alpha$ exists
(4) $\alpha = \frac { \pi } { 4 }$

Let $[ x ]$ denote the greatest integer $\leq x$, where $x \in R$. If the domain of the real valued function $f ( x ) = \sqrt { \frac { | [ x ] | - 2 } { | [ x ] | - 3 } }$ is $( - \infty , a ) \cup [ b , c ) \cup [ 4 , \infty ) , a < b < c$, then the value of $a + b + c$ is:
(1) 8
(2) 1
(3) $- 2$
(4) $- 3$

Let a function $f : R \rightarrow R$ be defined as, $f ( x ) = \begin{cases} \sin x - e ^ { x } & \text { if } x \leq 0 \\ a + [ - x ] & \text { if } 0 < x < 1 \\ 2 x - b & \text { if } x \geq 1 \end{cases}$
Where $[ x ]$ is the greatest integer less than or equal to $x$. If $f$ is continuous on $R$, then ( $a + b$ ) is equal to:
(1) 4
(2) 3
(3) 2
(4) 5

Let the functions $f : R \rightarrow R$ and $g : R \rightarrow R$ be defined as : $f ( x ) = \left\{ \begin{array} { l l } x + 2 , & x < 0 \\ x ^ { 2 } , & x \geq 0 \end{array} \right.$ and $g ( x ) = \begin{cases} x ^ { 3 } , & x < 1 \\ 3 x - 2 , & x \geq 1 \end{cases}$ Then, the number of points in $R$ where $( f \circ g ) ( x )$ is NOT differentiable is equal to :
(1) 3
(2) 1
(3) 0
(4) 2

The domain of $f ( x ) = \frac { \cos ^ { - 1 } \left( \frac { x ^ { 2 } - 5 x + 6 } { x ^ { 2 } - 9 } \right) } { \log \left( x ^ { 2 } - 3 x + 2 \right) }$ is
(1) $x \in \left[ \frac { - 1 } { 2 } , 1 \right) \cup ( 2 , \infty ) - \{ 3 \}$
(2) $x \in \left[ \frac { - 1 } { 2 } , 1 \right] \cup ( 2 , \infty ) - \{ 3 \}$
(3) $x \in \left( \frac { - 1 } { 2 } , 1 \right) \cup [ 2 , \infty ) - \{ 3 \}$
(4) $x \in \left[ \frac { - 1 } { 2 } , 1 \right) \cup [ 2 , \infty ) - \{ 3 \}$

The domain of the function $\cos ^ { - 1 } \left( \frac { 2 \sin ^ { - 1 } \left( \frac { 1 } { 4 x ^ { 2 } - 1 } \right) } { \pi } \right)$ is
(1) $\left( - \infty , - \frac { 1 } { \sqrt { 2 } } \right] \cup \left[ \frac { 1 } { \sqrt { 2 } } , \infty \right) \cup \{ 0 \}$
(2) $\left( - \infty , - \frac { 1 } { \sqrt { 2 } } \right] \cup \left[ \frac { 1 } { \sqrt { 2 } } , \infty \right)$
(3) $\left( - \infty , - \frac { 1 } { \sqrt { 2 } } \right) \cup \left( \frac { 1 } { 2 } , \infty \right) \cup \{ 0 \}$
(4) $R - \left\{ - \frac { 1 } { 2 } , \frac { 1 } { 2 } \right\}$

The domain of the function $f(x) = \sin^{-1}\left(\frac{x^2 - 3x + 2}{x^2 + 2x + 7}\right)$ is
(1) $[1, \infty)$
(2) $(-1, 2]$
(3) $[-1, \infty)$
(4) $(-\infty, 2]$

The domain of the function $f ( x ) = \sin ^ { - 1 } \left[ 2 x ^ { 2 } - 3 \right] + \log _ { 2 } \left( \log _ { \frac { 1 } { 2 } } \left( x ^ { 2 } - 5 x + 5 \right) \right)$, where $[ t ]$ is the greatest integer function, is
(1) $\left( - \sqrt { \frac { 5 } { 2 } } , \frac { 5 - \sqrt { 5 } } { 2 } \right)$
(2) $\left( \frac { 5 - \sqrt { 5 } } { 2 } , \frac { 5 + \sqrt { 5 } } { 2 } \right)$
(3) $\left( 1 , \frac { 5 - \sqrt { 5 } } { 2 } \right)$
(4) $\left[ 1 , \frac { 5 + \sqrt { } 5 } { 2 } \right)$

The function $f : R \rightarrow R$ defined by $f ( x ) = \lim _ { n \rightarrow \infty } \frac { \cos ( 2 \pi x ) - x ^ { 2 n } \sin ( x - 1 ) } { 1 + x ^ { 2 n + 1 } - x ^ { 2 n } }$ is continuous for all $x$ in
(1) $R - \{ - 1 \}$
(2) $R - \{ - 1,1 \}$
(3) $R - \{ 1 \}$
(4) $R - \{ 0 \}$

The range of $f(x) = 4 \sin ^ { - 1 } \left( \frac { x ^ { 2 } } { x ^ { 2 } + 1 } \right)$ is
(1) $[ 0,2 \pi ]$
(2) $[ 0 , \pi ]$
(3) $[ 0,2 \pi )$
(4) $[ 0 , \pi )$

If the domain of the function $f ( x ) = \log _ { e } \left( 4 x ^ { 2 } + 11 x + 6 \right) + \sin ^ { - 1 } ( 4 x + 3 ) + \cos ^ { - 1 } \left( \frac { 10 x + 6 } { 3 } \right)$ is $( \alpha , \beta ]$, then $36 | \alpha + \beta |$ is equal to
(1) 54
(2) 72
(3) 63
(4) 45

The domain of $f ( x ) = \frac { \log _ { ( x + 1 ) } ( x - 2 ) } { e ^ { 2 \log _ { e } x } - ( 2 x + 3 ) } , x \in R$ is
(1) $\mathbb { R } - \{ - 1,3 \}$
(2) $( 2 , \infty ) - \{ 3 \}$
(3) $( - 1 , \infty ) - \{ 3 \}$
(4) $\mathbb { R } - \{ 3 \}$

The range of the function $f(x) = \sqrt{3 - x} + \sqrt{2 + x}$ is
(1) $[\sqrt{5}, \sqrt{10}]$
(2) $[2\sqrt{2}, \sqrt{11}]$
(3) $[\sqrt{5}, \sqrt{13}]$
(4) $[\sqrt{2}, \sqrt{7}]$

If the domain of the function $f(x) = \frac{x}{1+\lfloor x \rfloor^2}$, where $\lfloor x \rfloor$ is greatest integer $\leq x$, is $[2,6)$, then its range is
(1) $\left\{\frac{5}{26}, \frac{2}{5}\right\} \cup \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$
(2) $\left[\frac{5}{26}, \frac{2}{5}\right]$
(3) $\left\{\frac{5}{37}, \frac{2}{5}\right\} \cup \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$
(4) $\left[\frac{5}{37}, \frac{2}{5}\right]$