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

2. The solution set of the inequality $\frac { 2 - x } { x + 4 } > 0$ is $\_\_\_\_$.

1. The solution set of the inequality $\frac { 2 - x } { x + 4 } > 0$ is $\_\_\_\_$ $( - 4,2 )$.
Analysis: This examines the method for solving fractional inequalities. $\frac { 2 - x } { x + 4 } > 0$ is equivalent to $( x - 2 ) ( x + 4 ) < 0$, so $- 4 < x < 2$

6. The solution to the inequality $\frac{1}{x} < 1$ is $\_\_\_\_$

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. $(1, 2)$

23. Solution: (1) When $a = 1$, $f ( x ) = | x - 1 | x + | x - 2 | ( x - 1 )$ .
When $x < 1$, $f ( x ) = - 2 ( x - 1 ) ^ { 2 } < 0$ ; when $x \geq 1$, $f ( x ) \geq 0$ .
Therefore, the solution set of the inequality $f ( x ) < 0$ is $( - \infty , 1 )$ .
(2) Since $f ( a ) = 0$, we have $a \geq 1$ .
When $a \geq 1 , x \in ( - \infty , 1 )$, $f ( x ) = ( a - x ) x + ( 2 - x ) ( x - a ) = 2 ( a - x ) ( x - 1 ) < 0$ .
Therefore, the range of values for $a$ is $[ 1 , + \infty )$ .

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. $\{x \mid x > 1\}$

104- Suppose the solution set of the inequality $\dfrac{((m^2-1)x^2 - 4mx + 4)(x - 3\sqrt{x} + 2)}{3x - 3} > 0$, for $x > \dfrac{3}{2}$, is $[2, 4]$. What is the value of $m$?
(1) $-2$ (2) zero (3) $1$ (4) $2$

107-- If $\dfrac{1}{a^2+1} + \dfrac{1}{a^2-1} = 2$, then $\left(\dfrac{1}{a^2 - \sqrt{a^2+1}} + \dfrac{1}{a^2 + \sqrt{a^2+1}}\right)^{150}$ equals what?
(1) $2$ (2) $-2$ (3) $1$ (4) $-1$
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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 $x > 2$.

39. Match the following
(i) Two rays in the first quadrant $x + y = | a |$ and ax $- y = 1$ intersects each other in the interval $a \in \left( a _ { 0 } , \infty \right)$, the value of $a _ { 0 }$ is
(A) 2
(ii) Point ( $\alpha , \beta , \gamma$ ) lies on the plane $\mathrm { x } + \mathrm { y } + \mathrm { z } = 2$. Let $\overrightarrow { \mathrm { a } } = \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } + \gamma \hat { \mathrm { k } } , \hat { \mathrm { k } } \times ( \hat { \mathrm { k } } \times \overrightarrow { \mathrm { a } } ) = 0$, then $\gamma =$.
(B) $4 / 3$
(iii) $\left| \int _ { 0 } ^ { 1 } \left( 1 - y ^ { 2 } \right) d y \right| + \left| \int _ { 1 } ^ { 0 } \left( y ^ { 2 } - 1 \right) d y \right|$
(C) $\left| \int _ { 0 } ^ { 1 } \sqrt { 1 - \mathrm { x } } \mathrm { dx } \right| + \left| \int _ { - 1 } ^ { 0 } \sqrt { 1 + \mathrm { x } } \mathrm { dx } \right|$
(iv) If $\sin \mathrm { A } \sin \mathrm { B } \sin \mathrm { C } + \cos \mathrm { A } \cos \mathrm { B } = 1$, then the value of $\sin \mathrm { C } =$
(D) 1
Sol. (i) Solving the two equations of ray i.e. $x + y = | a |$ and $a x - y = 1$ we get $x = \frac { | a | + 1 } { a + 1 } > 0$ and $y = \frac { | a | - 1 } { a + 1 } > 0$ when $\mathrm { a } + 1 > 0$; we get $\mathrm { a } > 1 \quad \therefore \mathrm { a } _ { 0 } = 1$.
(ii) We have $\overrightarrow { \mathrm { a } } = \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } + \gamma \hat { \mathrm { k } } \Rightarrow \overrightarrow { \mathrm { a } } \cdot \hat { \mathrm { k } } = \gamma$
Now; $\hat { \mathrm { k } } \times ( \hat { \mathrm { k } } \times \hat { \mathrm { a } } ) = ( \hat { \mathrm { k } } \cdot \overrightarrow { \mathrm { a } } ) \hat { \mathrm { k } } - ( \hat { \mathrm { k } } \cdot \hat { \mathrm { k } } ) \overrightarrow { \mathrm { a } }$ $= \gamma \hat { \mathrm { k } } - ( \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } + \gamma \hat { \mathrm { k } } )$ $= \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } = \overrightarrow { 0 } \quad \Rightarrow \alpha = \beta = 0$ As $\alpha + \beta + \gamma = 2 \Rightarrow \gamma = 2$.
(iii) $\quad \left| \int _ { 0 } ^ { 1 } \left( 1 - y ^ { 2 } \right) d y \right| + \left| \int _ { 1 } ^ { 0 } \left( y ^ { 2 } - 1 \right) d y \right|$ $= 2 \int _ { 0 } ^ { 1 } \left( 1 - y ^ { 2 } \right) d y = \frac { 4 } { 3 }$ $\left| \int _ { 0 } ^ { 1 } \sqrt { 1 - \mathrm { x } } \mathrm { dx } \right| + \left| \int _ { - 1 } ^ { 0 } \sqrt { 1 + \mathrm { x } } \mathrm { dx } \right| = 2 \int _ { 0 } ^ { 1 } \sqrt { 1 - \mathrm { x } } \mathrm { dx }$ $= 2 \int _ { 0 } ^ { 1 } \sqrt { \mathrm { x } } \mathrm { d } x = \left. 2 \cdot \frac { 2 } { 3 } \cdot \mathrm { x } ^ { 3 / 2 } \right| _ { 0 } ^ { 1 } = \frac { 4 } { 3 }$.
(iv) $\quad \sin \mathrm { A } \sin \mathrm { B } \sin \mathrm { C } + \cos \mathrm { A } \cos \mathrm { B } \leq \sin \mathrm { A } \sin \mathrm { B } + \cos \mathrm { A } \cos \mathrm { B } = \cos ( \mathrm { A } - \mathrm { B } )$
$$\Rightarrow \quad \cos ( \mathrm { A } - \mathrm { B } ) \geq 1 \Rightarrow \cos ( \mathrm {~A} - \mathrm { B } ) = 1 \Rightarrow \sin \mathrm { C } = 1$$

1. Match the following
   (i) $\sum _ { i = 1 } ^ { \infty } \tan ^ { - 1 } \left( \frac { 1 } { 2 i ^ { 2 } } \right) = t$, then $\tan t =$
   (A) 0
   (ii) Sides $\mathrm { a } , \mathrm { b } , \mathrm { c }$ of a triangle ABC are in AP and

$$\cos \theta _ { 1 } = \frac { \mathrm { a } } { \mathrm {~b} + \mathrm { c } } , \cos \theta _ { 2 } = \frac { \mathrm { b } } { \mathrm { a } + \mathrm { c } } , \cos \theta _ { 3 } = \frac { \mathrm { c } } { \mathrm { a } + \mathrm { b } } , \text { then } \tan ^ { 2 } \left( \frac { \theta _ { 1 } } { 2 } \right) + \tan ^ { 2 } \left( \frac { \theta _ { 3 } } { 2 } \right) = \quad \text { (B) } 1$$
(iii) A line is perpendicular to $x + 2 y + 2 z = 0$ and passes through $( 0,1,0 )$.
(C) $\frac { \sqrt { 5 } } { 3 }$
The perpendicular distance of this line from the origin is
(D) $2 / 3$
(iv) Data could not be retrieved.
Sol. (i) $\quad \sum _ { \mathrm { i } = 1 } ^ { \infty } \tan ^ { - 1 } \left[ \frac { 1 } { 2 \mathrm { i } ^ { 2 } } \right] = \mathrm { t }$
$$\begin{aligned} \text { Now } & ; \sum _ { \mathrm { i } = 1 } ^ { \infty } \tan ^ { - 1 } \left[ \frac { 2 } { 4 \mathrm { i } ^ { 2 } - 1 + 1 } \right] \\ = & \sum _ { \mathrm { i } = 1 } ^ { \infty } \left[ \tan ^ { - 1 } ( 2 \mathrm { i } + 1 ) - \tan ^ { - 1 } ( 2 \mathrm { i } - 1 ) \right] \\ = & { \left[ \left( \tan ^ { - 1 } 3 - \tan ^ { - 1 } 1 \right) + \left( \tan ^ { - 1 } 5 - \tan ^ { - 1 } 3 \right) + \cdots + \tan ^ { - 1 } ( 2 \mathrm { n } + 1 ) - \tan ^ { - 1 } ( 2 \mathrm { n } - 1 ) \ldots . . \infty \right] } \\ & \mathrm { t } = \tan ^ { - 1 } ( 2 \mathrm { n } + 1 ) - \tan ^ { - 1 } 1 = \lim _ { \mathrm { n } \rightarrow \infty } \tan ^ { - 1 } \frac { 2 \mathrm { n } } { 1 + ( 2 \mathrm { n } + 1 ) } \\ \Rightarrow \quad & \tan \mathrm { t } = \lim _ { \mathrm { n } \rightarrow \infty } \frac { \mathrm { n } } { \mathrm { n } + 1 } \Rightarrow \mathrm { t } = \frac { \pi } { 4 } \end{aligned}$$
(ii) We have $\cos \theta _ { 1 } = \frac { 1 - \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } } { 1 + \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } } = \frac { \mathrm { a } } { \mathrm { b } + \mathrm { c } } \Rightarrow \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } = \frac { \mathrm { b } + \mathrm { c } - \mathrm { a } } { \mathrm { b } + \mathrm { c } + \mathrm { a } }$
Also, $\cos \theta _ { 3 } = \frac { 1 - \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } } { 1 + \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } } = \frac { \mathrm { c } } { \mathrm { a } + \mathrm { b } } \Rightarrow \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } = \frac { \mathrm { a } + \mathrm { b } - \mathrm { c } } { \mathrm { a } + \mathrm { b } + \mathrm { c } }$ $\therefore \quad \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } + \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } = \frac { 2 \mathrm {~b} } { 3 \mathrm {~b} } = \frac { 2 } { 3 }$
(iii) Line through $( 0,1,0 )$ and perpendicular to plane $x + 2 y + 2 z = 0$ is given by $\frac { x - 0 } { 1 } = \frac { y - 1 } { 2 } = \frac { z - 1 } { 2 } = r$.
Let $\mathrm { P } ( \mathrm { r } , 2 \mathrm { r } + 1,2 \mathrm { r } )$ be the foot of perpendicular on the straight line then
$$r \times 1 + ( 2 r + 1 ) 2 + 2 \times 2 r = 0 \Rightarrow r = - \frac { 2 } { 9 }$$
$\therefore \quad$ Point is given by $\left( - \frac { 2 } { 9 } , \frac { 5 } { 9 } , - \frac { 4 } { 9 } \right)$ $\therefore \quad$ Required perpendicular distance $= \sqrt { \frac { 4 + 25 + 16 } { 81 } } = \frac { \sqrt { 5 } } { 3 }$ units.
(iv) Data could not be retrieved.

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