Q81. The number of real solutions of the equation $x | x + 5 | + 2 | x + 7 | - 2 = 0$ is $\_\_\_\_$

Q81. The number of distinct real roots of the equation $| x + 1 | | x + 3 | - 4 | x + 2 | + 5 = 0$, is

Q87. The number of distinct real roots of the equation $| x | | x + 2 | - 5 | x + 1 | - 1 = 0$ is $\_\_\_\_$

The number of solution for $\mathrm { x } \in \mathrm { R } , \mathrm { x } | \mathrm { x } - 4 | + | \mathrm { x } - 1 | - 2 = 0$
(B) 2
(C) 3
(D) 4

Q1 Consider the equation
$$(x-1)^2 = |3x-5|.$$
(1) Among all solutions of equation (1), the solutions satisfying $x \geqq \frac{5}{3}$ are $x = \mathbf{A}$ and $x = \mathbf{B}$, where $\mathbf{A} < \mathbf{B}$.
(2) Equation (1) has a total of $\mathbf{C}$ solutions. When the minimum one is denoted by $\alpha$, the integer $m$ satisfying $m-1 < \alpha \leqq m$ is $\mathbf{DE}$.

Consider the expression in $x$
$$P = | x - 1 | + | x - 2 | + | x - a | .$$
We are to find the range of real numbers $a$ such that the value of $P$ is minimized at $x = a$.
First, let us note that the inequality
$$| x - 1 | + | x - 2 | + | x - a | \geqq | x - 1 | + | x - 2 |$$
always holds, and is an equality in the case $x = a$.
When we set
$$y = | x - 1 | + | x - 2 | ,$$
we have
$$y = \begin{cases} - \mathbf { A } & x + \mathbf { B } \\ \mathbf { D } & ( x < \mathbf { C } ) \\ \mathbf { F } & ( \mathbf { C } \leqq x \leqq \mathbf { E } ) \\ \mathbf { G } & ( \mathbf { E } < x ) . \end{cases}$$
When we consider the graph of (1), we see that the minimum value of $y$ is $\mathbf { H }$ and $y$ takes the value $\mathbf{H}$ at every $x$ satisfying $\mathbf{I} \leqq x \leqq \mathbf{J}$.
Thus, for every $a$ satisfying $\mathbf{K} \leqq a \leqq \mathbf{L}$, the value of $P$ is minimized at $x = a$ and its value there is $\mathbf{M}$.

$$f(x) = \sqrt{2-|x+3|}$$
Which of the following is the domain interval of the function?
A) $3 \leq x \leq 5$
B) $-1 \leq x \leq 5$
C) $-3 \leq x \leq 4$
D) $-3 \leq x \leq 0$
E) $-5 \leq x \leq -1$

Let a be a real number. The distance of a from 1 on the number line is $a + 4$ units.
Accordingly, what is $|a|$?
A) $\frac { 3 } { 2 }$
B) $\frac { 5 } { 2 }$
C) $\frac { 7 } { 2 }$
D) $\frac { 7 } { 3 }$
E) $\frac { 8 } { 3 }$

Let $a , b , c$ be real numbers and $a \cdot b \cdot c > 0$ such that
$$\begin{aligned} & a \cdot b = - 2 | a | \\ & \frac { b } { c } = 3 | b | \end{aligned}$$
Given that $\mathbf { a } + \mathbf { b } + \mathbf { c } = \mathbf { 0 }$, what is a?
A) $\frac { 3 } { 2 }$
B) $\frac { 5 } { 2 }$
C) $\frac { 9 } { 2 }$
D) $\frac { 7 } { 3 }$
E) $\frac { 8 } { 3 }$

For real numbers x and y
$$\begin{aligned} & 2 x = 7 - | y | \\ & y = \frac { | x | } { 3 } \end{aligned}$$
Given that, what is the sum $\mathbf { x } + \mathbf { y }$?
A) 12 B) 10 C) 8 D) 6 E) 4

For non-zero real numbers $x$ and $y$
$$\begin{aligned} & | x \cdot y | = - 2 x \\ & \left| \frac { y } { x } \right| = 3 y \end{aligned}$$
the following equalities are given.
Accordingly, what is the sum $x + y$?
A) $\frac { 3 } { 2 }$ B) $\frac { 5 } { 2 }$ C) $\frac { 5 } { 3 }$ D) $\frac { 7 } { 3 }$ E) $\frac { 5 } { 6 }$

For integers $x$ and $y$,
$$| x - 3 | + | 2 x + y | + | 2 x + y - 1 | = 1$$
the equality is satisfied. Accordingly, what is the sum of the values that $y$ can take?
A) $- 12$
B) $- 11$
C) $- 10$
D) $- 9$
E) $- 8$