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) $\frac { 7 } { 4 }$

Let $S$ be the set of all real roots of the equation, $3^{x}\left(3^{x} - 1\right) + 2 = \left|3^{x} - 1\right| + \left|3^{x} - 2\right|$, then
(1) contains exactly two elements.
(2) is a singleton.
(3) is an empty set.
(4) contains at least four elements.

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) $A - B = [ - 1,2 )$

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

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) $( - 3 , - 1 ) \cup \frac { 1 } { 3 } , 1$

Let $\mathrm { A } = \{ ( x , y ) \in \mathbf { R } \times \mathbf { R } : | x + y | \geqslant 3 \}$ and $\mathrm { B } = \{ ( x , y ) \in \mathbf { R } \times \mathbf { R } : | x | + | y | \leq 3 \}$. If $\mathrm { C } = \{ ( x , y ) \in \mathrm { A } \cap \mathrm { B } : x = 0$ or $y = 0 \}$, then $\sum _ { ( x , y ) \in \mathrm { C } } | x + y |$ is :
(1) 15
(2) 24
(3) 18
(4) 12

The number of real solution(s) of the equation $x^{2} + 3x + 2 = \min\{|x - 3|, |x + 2|\}$ is:
(1) 1
(2) 0
(3) 2
(4) 3

Consider the quadratic function of $x$
$$x ^ { 2 } + ( 4 a - 6 ) x + 2 a + b + 5 = 0 . \tag{1}$$
We are to find the conditions under which one solution is $-1$ and the other solution satisfies the inequality
$$| x + 2 a | < a + 1 . \tag{2}$$
(1) The condition for equation (1) to have the solution $-1$ is
$$b = \mathbf { L } a - \mathbf { M N } .$$
Denote the other solution by $\alpha$. When $\alpha$ is expressed in terms of $a$, we have
$$\alpha = \mathbf { O P } a + \mathbf { Q } .$$
(2) When $a > \mathbf{RS}$, the inequality (2) has a solution, and its solution is
$$\mathbf { TU } a - \mathbf { VV } < x < - a + \mathbf { W } .$$
Hence the conditions are: that $a$ and $b$ satisfy (3) and that $a$ satisfies
$$\mathbf { X } < a < \mathbf { Y } .$$

Let $x$ and $y$ be real numbers which satisfy
$$3x^2 + 2xy + 3y^2 = 32. \tag{1}$$
Then we are to find the ranges of the values of $x + y$ and $xy$.
First, we set
$$a = x + y. \tag{2}$$
By eliminating $y$ from (1) and (2), we obtain the quadratic equation in $x$
$$\mathbf{A}x^2 - \mathbf{B}ax + \mathbf{C}a^2 - 32 = 0.$$
Since $x$ is a real number, we have
$$\mathbf{DE} \leqq a \leqq \mathbf{F}.$$
Next, we set
$$b = xy. \tag{4}$$
From (1), (2) and (4) we obtain
$$b = \frac{\mathbf{G}}{\mathbf{H}}a^2 - \mathbf{I}.$$
Hence from (3) and (5) we have
$$\mathbf{JK} \leqq b \leqq \mathbf{L}.$$

Consider the following equation in $x$
$$|ax - 11| = 4x - 10, \tag{1}$$
where $a$ is a constant.
(1) Equation (1) can be rewritten without using the absolute value symbol as
$$\begin{aligned} & \text{when } ax \geqq 11, \quad \text{then } (a - \mathbf{N})x = \mathbf{O}; \\ & \text{when } ax < 11, \quad \text{then } (a + \mathbf{P})x = \mathbf{QR}. \end{aligned}$$
(2) When $a = \sqrt{7}$, the solution of equation (1) is
$$x = \frac{\mathrm{S}\left(\frac{\mathrm{T}}{\mathrm{V}} - \sqrt{\mathrm{U}}\right)}{\mathrm{V}}.$$
(3) Let $a$ be a positive integer. When equation (1) has a positive integral solution, we have $a = \mathbf{W}$, and that solution $x = \mathbf{X}$.

Let $a , b , c$ and $d$ be real numbers satisfying $a < b < c < d$. Suppose that the two subsets of real numbers
$$A = \{ x \mid a \leqq x \leqq c \} , \quad B = \{ x \mid b \leqq x \leqq d \}$$
satisfy
$$A \cap B = \left\{ x \mid x ^ { 2 } - 4 x + 3 \leqq 0 \right\} .$$
Then, answer the questions for cases (1) and (2).
(1) Let the union of $A$ and $B$ be
$$A \cup B = \left\{ x \mid x ^ { 2 } - 5 x - 24 \leqq 0 \right\} .$$
Then the values of $a , b , c$ and $d$ are
$$a = \mathbf { \text { NO } } , \quad b = \mathbf { P } , \quad c = \mathbf { Q } , \quad d = \mathbf { Q } .$$
(2) Let the intersection of $A$ and the complement $\bar { B }$ of $B$ be
$$A \cap \bar { B } = \left\{ x \mid x ^ { 2 } + 5 x - 6 \leqq 0 \text { and } x \neq 1 \right\} ,$$
and let the intersection of the complement $\bar { A }$ of $A$ and $B$ be
$$\bar { A } \cap B = \left\{ x \mid x ^ { 2 } - 9 x + 18 \leqq 0 \text { and } x \neq 3 \right\} .$$
Then the values of $a , b , c$ and $d$ are
$$a = \mathbf { S T } , \quad b = \mathbf { U } , \quad c = \mathbf { V } , \quad d = \mathbf { W } .$$

Let $a , b , c$ and $d$ be real numbers satisfying $a < b < c < d$. Suppose that the two subsets of real numbers
$$A = \{ x \mid a \leqq x \leqq c \} , \quad B = \{ x \mid b \leqq x \leqq d \}$$
satisfy
$$A \cap B = \left\{ x \mid x ^ { 2 } - 4 x + 3 \leqq 0 \right\} .$$
Then, answer the questions for cases (1) and (2).
(1) Let the union of $A$ and $B$ be
$$A \cup B = \left\{ x \mid x ^ { 2 } - 5 x - 24 \leqq 0 \right\} .$$
Then the values of $a , b , c$ and $d$ are
$$a = \mathbf { \text { NO } } , \quad b = \mathbf { P } , \quad c = \mathbf { Q } , \quad d = \mathbf { Q } .$$
(2) Let the intersection of $A$ and the complement $\bar { B }$ of $B$ be
$$A \cap \bar { B } = \left\{ x \mid x ^ { 2 } + 5 x - 6 \leqq 0 \text { and } x \neq 1 \right\} ,$$
and let the intersection of the complement $\bar { A }$ of $A$ and $B$ be
$$\bar { A } \cap B = \left\{ x \mid x ^ { 2 } - 9 x + 18 \leqq 0 \text { and } x \neq 3 \right\} .$$
Then the values of $a , b , c$ and $d$ are
$$a = \mathbf { S T } , \quad b = \mathbf { U } , \quad c = \mathbf { V } , \quad d = \mathbf { W } .$$

Let $a$ be a constant. Consider the quadratic inequality
$$x ^ { 2 } - 2 ( a + 2 ) x + 25 > 0 . \tag{1}$$
The left-hand side of inequality (1) can be transformed into
$$( x - a - \mathbf { A } ) ^ { 2 } - a ^ { 2 } - \mathbf { B } a + \mathbf { C D } .$$
Hence, we have the following results.
(1) The condition under which inequality (1) holds for all real numbers $x$ is
$$\mathbf { E F } < a < \mathbf { G } .$$
(2) The condition under which inequality (1) holds for all real numbers $x$ satisfying $x \geqq - 1$ is
$$\mathbf { H I J } < a < \mathbf { K } .$$

Suppose that an integer $x$ and a real number $y$ satisfy both the equation
$$2 ( y + 1 ) = x ( 8 - x ) \tag{1}$$
and the inequality
$$5 x - 4 y + 1 \leqq 0 . \tag{2}$$
We are to find $M$, the maximum value of $y$, and $m$, the minimum value of $y$.
First of all, let us transform (1) into
$$y = - \frac { 1 } { \mathbf { P } } ( x - \mathbf { Q } ) ^ { 2 } + \mathbf{R} .$$
Also, from (1) and (2) we obtain the inequality in $x$
$$2 x ^ { 2 } - \mathbf { S T } x + \mathbf { U V } \leqq 0 . \tag{3}$$
Thus when $x$ is an integer satisfying (3) if we consider the range of values which $y$ can take, we see that $y$ is maximized at $x = \square \mathbf{ V }$ and is minimized at $x = \square \mathbf{ W }$, and hence that
$$M = \mathbf { X } , \quad m = \frac { \mathbf { Y } } { \mathbf{Z} } .$$

For each of $\mathbf{A} \sim \mathbf{D}$ in the following questions, choose the correct answer from among (0) $\sim$ (5) below each question.
Consider the three quadratic inequalities
$$\begin{aligned} x ^ { 2 } + 3 x - 18 & < 0 \tag{1}\\ x ^ { 2 } - 2 x - 8 & > 0 \tag{2}\\ x ^ { 2 } + a x + b & < 0 . \tag{3} \end{aligned}$$
(1) The range of $x$ which satisfies both of the inequalities (1) and (2) is $\mathbf { A }$. Also, the range of $x$ which satisfies neither inequality (1) nor (2) is $\mathbf{B}$. (0) $3 \leqq x \leqq 4$
(1) $- 6 \leqq x \leqq - 2$
(2) $3 < x < 4$
(3) $2 < x < 6$
(4) $- 6 < x < - 2$
(5) $- 4 \leqq x \leqq - 3$
(2) The range of $x$ that satisfies at least one of the inequalities (1) and (3) will be $- 6 < x < 7$, if and only if $a$ and $b$ satisfy the equation $\square \mathbf{C}$, and $a$ satisfies the inequality $\square \mathbf{D}$. (0) $b = 6 a - 36$
(1) $b = 7 a - 49$
(2) $b = - 7 a - 49$
(3) $- 10 < a \leqq - 3$
(4) $- 10 < a \leqq - 1$
(5) $- 1 \leqq a < 3$

We have a triangle which has sides of the lengths 15, 19 and 23. We make it into an obtuse triangle by shortening each of its sides by $x$. What is the range of values that $x$ can take?
First, since $15 - x$, $19 - x$ and $23 - x$ can be the lengths of the sides of a triangle, it follows that $$x < \mathbf{AB}.$$
In addition, such a triangle is an obtuse triangle only when $x$ satisfies $$x^2 - \mathbf{CD}x + \mathbf{EF} < 0.$$
By solving this quadratic inequality, we have $$\mathbf{G} < x < \overline{\mathbf{HI}}.$$
Hence, the range of $x$ is $$\mathbf{J} < x < \mathbf{KL}.$$

For each of $\mathbf { A } \sim \mathbf { M }$ in the following statements, choose the correct answer from among (0) ~ (9) at the bottom of this page.
We are to solve the following simultaneous inequalities
$$\left\{ \begin{aligned} x ^ { 2 } - 2 x < 3 & \cdots \cdots \cdots (1)\\ a x ^ { 2 } - a x - x + 1 > 0 , & \cdots \cdots \cdots (2) \end{aligned} \right.$$
where $0 < a < 1$.
When we solve (1), we have
$$\mathbf { A } < x < \mathbf { B } .$$
Next, when we transform (2), we have
$$( a x - \mathbf { C } ) ( x - \mathbf { D } ) > 0 .$$
Hence, noting that $0 < a < 1$, we see that the solution of (2) is
$$x < \mathbf { E } \text { or } \mathbf { F } < x .$$
Thus, when $0 < a \leqq \mathbf { G }$, the solution of the simultaneous inequalities is
$$\mathbf { H } < x < \mathbf { I }$$
and when $\mathbf { G } < a < 1$, the solution is
$$\mathbf { J } < x < \mathbf { K } \text { or } \mathbf { L } < x < \mathbf { M }$$
where $\mathbf { K } < \mathbf { M }$.
(0) 0
(1) 1
(2) 2
(3) 3
(4) $- 1$
(5) $\frac { 1 } { 2 }$ (6) $\frac { 1 } { 3 }$ (7) $\frac { 1 } { a }$ (8) $\frac { 2 } { a }$ (9) $\frac { 3 } { a }$

Let us consider the real numbers $x , y , t$ and $u$ satisfying the following four conditions:
$$\begin{aligned} & y \geqq | x | \\ & x + y = t \\ & x ^ { 2 } + y ^ { 2 } = 12 \\ & x ^ { 3 } + y ^ { 3 } = u \end{aligned}$$
We are to find the ranges of values which $t$ and $u$ can take.
(1) From (1) and (3), we see that the point $( x , y )$ is located on the arc which is a quadrant of the circle having its center at the origin and the radius $\mathbf { A }$. Moreover, the coordinates of the end points of this arc are
$$( \sqrt { \mathbf { C } } , \sqrt { \mathbf { CD } } ) \text { and } ( - \sqrt { \mathbf { C } } , \sqrt { \mathbf { D } } ) .$$
From this and (2), we also see that the range of values which $t$ can take is
$$\mathbf { E } \leqq t \leqq \mathbf { F } . \mathbf { G } .$$
(2) Next, from (2) and (3), we have
$$x y = \frac { \mathbf { H } } { \mathbf { I } } \left( t ^ { 2 } - \mathbf { J K } \right)$$
and further, using (4) we also have
$$u = \frac { \mathbf { L } } { \mathbf{L}} \left( \mathbf{NO} \, t - t ^ { 3 } \right)$$
Hence, since
$$\frac { d u } { d t } = \frac { \mathbf { P } } { \mathbf { Q } } \left( \mathbf { RS } - t ^ { 2 } \right)$$
the range of values which $u$ can take under the condition (5) is
$$\mathbf { TL } \leqq u \leqq \mathbf { UV } \sqrt { \mathbf{ W } } .$$

Consider two squares as in the figure to the right. Let the coordinates of their vertexes be
$$\begin{array} { l l } \mathrm { A } ( 2 t , 0 ) , \quad \mathrm { B } ( 0,2 t ) , & \mathrm { C } ( - 2 t , 0 ) , \quad \mathrm { D } ( 0 , - 2 t ) , \\ \mathrm { P } \left( 4 - t ^ { 2 } , 4 - t ^ { 2 } \right) , & \mathrm { Q } \left( - 4 + t ^ { 2 } , 4 - t ^ { 2 } \right) , \\ \mathrm { R } \left( - 4 + t ^ { 2 } , - 4 + t ^ { 2 } \right) , & \mathrm { S } \left( 4 - t ^ { 2 } , - 4 + t ^ { 2 } \right) , \end{array}$$
where $0 < t < 2$. Denote the areas of the two squares ABCD and PQRS by $S _ { 1 }$ and $S _ { 2 }$, respectively.
Then we have
$$S _ { 1 } = \mathbf { M } t ^ { 2 } \text { and } S _ { 2 } = \mathbf { N } \left( t ^ { 2 } - \mathbf { O } \right) ^ { 2 } .$$
(1) $S _ { 1 } + S _ { 2 }$ is minimized at $t = \sqrt { \mathbf { P } }$, and the minimum value is $\mathbf { Q } \mathbf { R }$.
(2) For $\mathbf { W }$ and $\mathbf { X }$ below, choose the correct answer from among (0) $\sim$ (9), and for the other $\square$, enter the correct numbers.
We are to find the range of $t$ such that $S _ { 1 } < S _ { 2 }$. If $S _ { 1 } < S _ { 2 }$, then $t$ satisfies the inequality
$$t ^ { 4 } - \mathbf { ST } t ^ { 2 } + \mathbf { UV } > 0 .$$
From the above inequality, a condition on $t ^ { 2 }$ is $\mathbf { W }$. Hence, $S _ { 1 } < S _ { 2 }$ if and only if $t$ satisfies $\mathbf { X }$.
(0) $t ^ { 2 } < 4$ or $6 < t ^ { 2 }$ (1) $4 < t ^ { 2 } < 6$ (2) $t ^ { 2 } < 2$ or $8 < t ^ { 2 }$ (3) $2 < t ^ { 2 } < 8$ (4) $t ^ { 2 } \neq 4$ (5) $0 < t < 2$ (6) $0 < t < \sqrt { 2 }$ (7) $\sqrt { 2 } < t < 2$ (8) $2 < t < \sqrt { 6 }$ (9) $t \neq 2$