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 $$\mathbf{J} < x < \mathbf{KL}.$$

4. The complete set of values of $x$ for which $\left( x ^ { 2 } - 1 \right) ( x - 2 ) > 0$ is
A $x < - 1,1 < x < 2$
B $x < - 1 , x > 2$
C $- 1 < x < 2$
D $x < 1 , x > 2$
E $\quad - 1 < x < 1 , x > 2$

The set of solutions to the inequality $x ^ { 2 } + b x + c < 0$ is the interval $p < x < q$ where $b , c , p$ and $q$ are real constants with $c < 0$.
In terms of $p , q$ and $c$, what is the set of solutions to the inequality $x ^ { 2 } + b c x + c ^ { 3 } < 0$ ?
A $\frac { p } { c } < x < \frac { q } { c }$
B $\frac { q } { c } < x < \frac { p } { c }$
C $p c < x < q c$
D $q c < x < p c$
E $p c ^ { 2 } < x < q c ^ { 2 }$
F $q c ^ { 2 } < x < p c ^ { 2 }$

Consider the statement () about a real number $x$ : () There exists a real number $y$ such that $x - x y + y$ is negative.
For how many real values of $x$ is (*) true?
A no values of $x$
B exactly one value of $x$
C exactly two values of $x$
D all except exactly two values of $x$
E all except exactly one value of $x$ F all values of $x$

$$(2x-1)\left(4x^{2}-1\right)<0$$
Which of the following open intervals is the solution set of the inequality in real numbers?
A) $\left(-\infty, \frac{-1}{2}\right)$
B) $\left(\frac{-1}{2}, 0\right)$
C) $\left(\frac{-1}{2}, \frac{1}{2}\right)$
D) $\left(\frac{1}{4}, \frac{1}{2}\right)$
E) $\left(\frac{1}{2}, \infty\right)$

Let $f : \mathbf { R } \backslash \{ 0 \} \rightarrow \mathbf { R }$ with
$$f ( x ) = \frac { 2 } { x } - x + 1$$
For this function, which of the following is the set of all $x$ points such that $f ( x ) \in ( 0 , \infty )$?
A) $( - \infty , 0 )$
B) $( - 1 , \infty )$
C) $( 0,1 ) \cup ( 2 , \infty )$
D) $( - 2,0 ) \cup ( 2 , \infty )$
E) $( - \infty , - 1 ) \cup ( 0,2 )$

$\frac { 6 x + 1 } { ( x + 1 ) ^ { 2 } } > 1$\ Which of the following is the set of all real numbers that satisfy this inequality?\ A) $( - 1,4 )$\ B) $( - 1,6 )$\ C) $( 0,4 )$\ D) $( 0 , \infty )$\ E) $( 2 , \infty )$

Let a be a real number. Regarding the inequality $x + 1 \leq a$, the following are known.

• $\mathrm { x } = 0$ satisfies this inequality.
• $x = 4$ does not satisfy this inequality.

Accordingly, what is the widest interval expressing the values that the number a can take?
A) $( 0,4 ]$
B) $[ 0,4 )$
C) $[ 1,4 ]$
D) $( 1,5 ]$
E) $[ 1,5 )$