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) $(4,5)$

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
$$\mathbf { X } < a < \mathbf { Y } .$$

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) $- 1 \leqq a < 3$

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) $\frac { 3 } { a }$

The parabola $f(x)$ and the line $d$ are shown in the Cartesian coordinate plane above.
Accordingly, which of the following systems of inequalities has the shaded region as its solution set?
A) $\left.\begin{array}{l} y - x^{2} + 2x \leq 0 \\ y - x + 2 \geq 0 \end{array}\right\}$
B) $\left.\begin{array}{l} y - x^{2} + 2x \geq 0 \\ 2y - x + 2 \geq 0 \end{array}\right\}$
C) $\left.\begin{array}{l} y - x^{2} + 4x \leq 0 \\ 2y - x + 2 \leq 0 \end{array}\right\}$
D) $\left.\begin{array}{l} y + x^{2} - 4x \leq 0 \\ 2y - x + 4 \leq 0 \end{array}\right\}$
E) $\left.\begin{array}{l} y + x^{2} - 4x \leq 0 \\ 2y - x + 2 \geq 0 \end{array}\right\}$

$$\left. \begin{array} { l } x ( 3 - x ) > 0 \\ ( 2 x + 1 ) ( x - 2 ) < 0 \end{array} \right\}$$
If the solution set of the inequality system given above is the open interval $(\mathbf { a } , \mathbf { b })$, what is the difference $\mathbf { a - b }$?
A) - 2
B) 0
C) 1
D) $\frac { 1 } { 2 }$
E) $\frac { 3 } { 2 }$

In the Cartesian coordinate plane, the graphs of functions $f$, $g$ and $h$ whose domains consist of real numbers are given in the figure.
Accordingly, for $x \in [ - 2,2 ]$,
$$\begin{aligned} & f ( x ) \cdot g ( x ) > 0 \\ & g ( x ) \cdot h ( x ) < 0 \end{aligned}$$
the solution set of the system of inequalities is which of the following?
A) $( - 2 , - 1 )$ B) $( - 1,0 )$ C) $( 1,2 )$ D) $( - 2 , - 1 ) \cup ( 1,2 )$

Let $a$, $b$, $c$, and $d$ be real numbers such that
$$\begin{aligned} & a x ^ { 2 } + b x + 12 \geq 0 \\ & c x ^ { 2 } + d x + 24 \leq 0 \end{aligned}$$
To find the solution set of this system of inequalities, the following table is constructed and the solution set is found to be $[ - 2 , - 1 ] \cup [ 4,6 ]$.
What is the sum $a + b + c + d$?
A) 15
B) 16
C) 17
D) 18
E) 19

Let $a$ and $b$ be positive integers,
$$\begin{aligned} & (x - a)(x + 2a) < 0 \\ & (x - b)(x + 2b) > 0 \end{aligned}$$
Given this system of inequalities.
If $a + b = 8$ and the solution set of this system of inequalities contains 16 integers, what is the product $a \cdot b$?
A) 7 B) 10 C) 12 D) 15 E) 16