Question 178
O conjunto solução da inequação $2x - 3 > 7$ é
(A) $\{x \in \mathbb{R} \mid x < 5\}$ (B) $\{x \in \mathbb{R} \mid x > 5\}$ (C) $\{x \in \mathbb{R} \mid x < 2\}$ (D) $\{x \in \mathbb{R} \mid x > 2\}$ (E) $\{x \in \mathbb{R} \mid x > 10\}$

O conjunto solução da inequação $2x - 5 > 3$ no conjunto dos números reais é
(A) $\{x \in \mathbb{R} \mid x < 4\}$ (B) $\{x \in \mathbb{R} \mid x > 4\}$ (C) $\{x \in \mathbb{R} \mid x < -4\}$ (D) $\{x \in \mathbb{R} \mid x > -4\}$ (E) $\{x \in \mathbb{R} \mid x = 4\}$

QUESTION 148
The solution set of the inequality $2x - 5 > 3$ is
(A) $x > 1$
(B) $x > 2$
(C) $x > 3$
(D) $x > 4$
(E) $x > 5$

The solution set of the inequality $2x - 3 > 7$ is:
(A) $x > 2$
(B) $x > 3$
(C) $x > 4$
(D) $x > 5$
(E) $x > 6$

The product of all real roots of the irrational equation $x ^ { 2 } - 2 x + 2 \sqrt { x ^ { 2 } - 2 x } = 8$ is? [3 points]
(1) - 5
(2) - 4
(3) - 3
(4) - 2
(5) - 1

If $a > 1$, then the range of values of $x$ satisfying $\log_a(x^2 - 2) < \log_a x$ is
A. $(\sqrt{2}, +\infty)$
B. $(\sqrt{2}, 2)$
C. $(1, \sqrt{2})$
D. $(1, 2)$

The solution set of the inequality $\frac{x-4}{x-1} < 2$ is ( )
A. $\{x \mid -2 \leq x \leq 1\}$
B. $\{x \mid x < -2\}$
C. $\{x \mid -2 \leq x < 1\}$
D. $\{x \mid x > 1\}$

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 all real numbers $x$ such that $x ^ { 3 } ( x + 1 ) ( x - 2 ) \geq 0$ is:
(a) the interval $2 \leq x < \infty$
(b) the interval $0 \leq x < \infty$
(c) the interval $- 1 \leq x < \infty$
(d) none of the above.

The set of all real numbers $x$ such that $x ^ { 3 } ( x + 1 ) ( x - 2 ) \geq 0$ is:
(a) the interval $2 \leq x < \infty$
(b) the interval $0 \leq x < \infty$
(c) the interval $- 1 \leq x < \infty$
(d) none of the above.

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 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

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 $x > 2$.

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 $5,5 r , 5 r ^ { 2 }$ 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 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

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

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

$$(2x-1)\left(4x^{2}-1\right)<0$$
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) $[ 1,5 )$