The set of all real numbers $x$ satisfying the inequality $x ^ { 3 } ( x + 1 ) ( x - 2 ) \geq 0$ is
(A) the interval $[ 2 , \infty )$
(B) the interval $[ 0 , \infty )$
(C) the interval $[ - 1 , \infty )$
(D) none of the above

The set of values of $m$ for which $m x ^ { 2 } - 6 m x + 5 m + 1 > 0$ for all real $x$ is
(A) $m < \frac { 1 } { 4 }$
(B) $m \geq 0$
(C) $0 \leq m \leq \frac { 1 } { 4 }$
(D) $0 \leq m < \frac { 1 } { 4 }$

Let $\alpha$ denote a real number. The range of values of $| \alpha - 4 |$ such that $| \alpha - 1 | + | \alpha + 3 | \leq 8$ is
(A) $( 0,7 )$
(B) $( 1,8 )$
(C) $[ 1,9 ]$
(D) $[ 2,5 ]$.

For a real number $x$, $$x ^ { 3 } - 7 x + 6 > 0$$ if and only if
(A) $x > 2$.
(B) $- 3 < x < 1$.
(C) $x < - 3$ or $1 < x < 2$.
(D) $- 3 < x < 1$ or $x > 2$.

Let $P = \{(x, y) : x + 1 \geqslant y,\, x \geqslant -1,\, y \geqslant 2x\}$. Then the minimum value of $(x + y)$ where $(x, y)$ varies over the set $P$ is
(A) $-1$
(B) $-3$
(C) $3$
(D) $0$

The real number $x$ satisfies $$\frac{|x|^2 - |x| - 2}{2|x| - |x|^2 - 2} > 2$$ if and only if $x$ belongs to
(A) $(-2, -1) \cup (1, 2)$
(B) $(-2/3, 0) \cup (0, 2/3)$
(C) $(-1, -2/3) \cup (2/3, 1)$
(D) $(-1, 0) \cup (0, 1)$

8. If $\mathrm { a } , \mathrm { b } , \mathrm { c } , \mathrm { d }$ are positive real numbers such that $\mathrm { a } + \mathrm { b } + \mathrm { c } + \mathrm { d } = 2$, then $\mathrm { M } = ( \mathrm { a } + \mathrm { b } ) ( \mathrm { c } +$ d) satisfies the relation :
(A) $0 \leq M \leq 1$
(B) $1 \leq \mathrm { M } \leq 2$
(C) $2 \leq M \leq 3$
(D) $3 \leq M \leq 4$

8. The set of all real numbers $x$ for which $x ^ { 2 } - | x + 2 | + x > 0$ is
(A) $( - \infty , - 2 ) \cup ( 2 , \infty )$
(B) $( - \infty , - \sqrt { } 2 ) \cup ( \sqrt { } 2 , \infty )$
(C) $( - \infty , - 1 ) \cup ( 1 , \infty )$
(D) $( \sqrt { } 2 , \infty )$

19. If $a \hat { I } ( 0 , \Pi / 2 )$ then $\sqrt { } \left( x ^ { 2 } + x \right)$ is always greater than or equal to :
(a) $2 \tan a$
(b) 1
(c) 2
(d) $\quad \sec ^ { 2 } a$

The least value of $\alpha \in \mathbb{R}$ for which $4\alpha x^2 + \frac{1}{x} \geq 1$, for all $x > 0$, is
(A) $\frac{1}{64}$
(B) $\frac{1}{32}$
(C) $\frac{1}{27}$
(D) $\frac{1}{25}$

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 least integral value $\alpha$ of $x$ such that $\frac { x - 5 } { x ^ { 2 } + 5 x - 14 } > 0$, satisfies :
(1) $\alpha ^ { 2 } + 3 \alpha - 4 = 0$
(2) $\alpha ^ { 2 } - 5 \alpha + 4 = 0$
(3) $\alpha ^ { 2 } - 7 \alpha + 6 = 0$
(4) $\alpha ^ { 2 } + 5 \alpha - 6 = 0$

If $x$ is a solution of the equation $\sqrt { 2 x + 1 } - \sqrt { 2 x - 1 } = 1 , \left( x \geq \frac { 1 } { 2 } \right)$, then $\sqrt { 4 x ^ { 2 } - 1 }$ is equal to :
(1) $\frac { 3 } { 4 }$
(2) $\frac { 1 } { 2 }$
(3) $2 \sqrt { 2 }$
(4) 2

Let $S = \{ x \in R : x \geq 0 \& 2 | \sqrt { x } - 3 | + \sqrt { x } ( \sqrt { x } - 6 ) + 6 = 0 \}$. Then $S$ :
(1) Contains exactly four elements
(2) Is an empty set
(3) Contains exactly one element
(4) Contains exactly two elements

If both the roots of the quadratic equation $x^2 - mx + 4 = 0$ are real and distinct and they lie in the interval $(1,5)$, then $m$ lies in the interval: Note: In the actual JEE paper interval was $[1,5]$
(1) $(-5,-4)$
(2) $(3,4)$
(3) $(5,6)$
(4) $(4,5)$

If $5,5 r , 5 r ^ { 2 }$ are the lengths of the sides of a triangle, then $r$ can not be equal to:
(1) $\frac { 3 } { 4 }$
(2) $\frac { 3 } { 2 }$
(3) $\frac { 5 } { 4 }$
(4) $\frac { 7 } { 4 }$

If $A = \{ x \in R : | x | < 2 \}$ and $B = \{ x \in R : | x - 2 | \geq 3 \}$; then
(1) $A \cap B = ( - 2 , - 1 )$
(2) $B - A = R - ( - 2,5 )$
(3) $A \cup B = R - ( 2,5 )$
(4) $A - B = [ - 1,2 )$

If the equation $\cos ^ { 4 } \theta + \sin ^ { 4 } \theta + \lambda = 0$ has real solutions for $\theta$ then $\lambda$ lies in interval
(1) $\left( - \frac { 5 } { 4 } , - 1 \right)$
(2) $\left[ - 1 , - \frac { 1 } { 2 } \right]$
(3) $\left( - \frac { 1 } { 2 } , - \frac { 1 } { 4 } \right]$
(4) $\left[ - \frac { 3 } { 2 } , - \frac { 5 } { 4 } \right]$

Let a complex number $z , | z | \neq 1$, satisfy $\log _ { \frac { 1 } { \sqrt { 2 } } } \left( \frac { | z | + 11 } { ( | z | - 1 ) ^ { 2 } } \right) \leq 2$. Then, the largest value of $| z |$ is equal to
(1) 8
(2) 7
(3) 6
(4) 5

Let $A = \{x \in R : |x + 1| < 2\}$ and $B = \{x \in R : |x - 1| \geq 2\}$. Then which one of the following statements is NOT true?
(1) $A - B = (-1,1)$
(2) $B - A = R - (-3,1)$
(3) $A \cap B = (-3,-1]$
(4) $A \cup B = R - [1,3)$

Let $S = \left\{ x \in [ - 6,3 ] - \{ - 2,2 \} : \frac { | x + 3 | - 1 } { | x | - 2 } \geq 0 \right\}$ and $T = \left\{ x \in Z : x ^ { 2 } - 7 | x | + 9 \leq 0 \right\}$. Then the number of elements in $S \cap T$ is
(1) 7
(2) 5
(3) 4
(4) 3

The number of real roots of the equation $\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}$, is:
(1) 0
(2) 1
(3) 3
(4) 2

The number of elements in the set $\{n \in \mathbb{Z} : n^2 - 10n + 19 < 6\}$ is $\_\_\_\_$.

Let $S$ be the set of positive integral values of $a$ for which $\frac { a x ^ { 2 } + 2 a + 1 x + 9 a + 4 } { x ^ { 2 } - 8 x + 32 } < 0 , \quad \forall x \in \mathbb { R }$. Then, the number of elements in $S$ is:
(1) 1
(2) 0
(3) $\infty$
(4) 3

Let $R$ be the interior region between the lines $3 x - y + 1 = 0$ and $x + 2 y - 5 = 0$ containing the origin. The set of all values of $a$, for which the points $\mathrm { a } ^ { 2 } , \mathrm { a } + 1$ lie in R , is :
(1) $( - 3 , - 1 ) \cup - \frac { 1 } { 3 } , 1$
(2) $( - 3,0 ) \cup \frac { 1 } { 3 } , 1$
(3) $( - 3,0 ) \cup \frac { 2 } { 3 } , 1$
(4) $( - 3 , - 1 ) \cup \frac { 1 } { 3 } , 1$