Question

1. an incorrect definition of absolute value is

a. $|x|=\

$$\begin{cases}x\\text{ for }x>0\\\\0\\text{ for }x = 0\\\\-x\\text{ for }x<0\\end{cases}$$

$
b. $|x|=\

$$\begin{cases}x\\text{ for }x\\geq0\\\\-x\\text{ for }x<0\\end{cases}$$

$
c. $|x|=\

$$\begin{cases}x\\text{ for }x>0\\\\-x\\text{ for }x\\leq0\\end{cases}$$

$
d. $|x|=\

$$\begin{cases}x\\text{ for }x>0\\\\-x\\text{ for }x<0\\end{cases}$$

$

1. if the graph of $y = f(x),x\in\mathrm{r}$ is the same as $y = |f(x)|,x\in\mathrm{r}$, then

a. $f(x)$ may be of the form $ax + b,a\
eq0$ or of the form $ax^2 + bx + c,a\
eq0$
b. $f(x)$ may be of the form $ax + b,a\
eq0$ but cannot be of the form $ax^2 + bx + c,a\
eq0$
c. $f(x)$ may be of the form $ax^2 + bx + c,a\
eq0$, but cannot be of the form $ax + b,a\
eq0$
d. $f(x)$ cannot be of the form $ax + b,a\
eq0$ or $ax^2 + bx + c,a\
eq0$

Explanation:

For Question 4:

Step1: Recall absolute value definition

The absolute value $|x|$ is the non-negative value of $x$: $|0|=0$, $|x|=x$ for $x>0$, $|x|=-x$ for $x<0$.

Step2: Analyze each option

• Option A: Covers all cases ($x>0, x=0, x<0$) correctly.
• Option B: Combines $x=0$ with $x>0$ (since $|0|=0=0$), correct.
• Option C: Combines $x=0$ with $x<0$ (since $|0|=0=-0$), correct.
• Option D: Does not define $|x|$ for $x=0$, so it is incorrect.

Step1: Interpret the condition

If $y=f(x)$ and $y=|f(x)|$ have identical graphs, then $f(x) \geq 0$ for all $x \in \mathbb{R}$.

Step2: Analyze linear functions

A linear function $f(x)=ax+b, a
eq0$ is a line with slope $a
eq0$, so it will take negative values for some $x \in \mathbb{R}$. Thus it cannot satisfy the condition.

Step3: Analyze quadratic functions

A quadratic function $f(x)=ax^2+bx+c, a
eq0$ can be non-negative for all $x \in \mathbb{R}$ if it opens upward ($a>0$) and has no real roots (or a repeated root, i.e., discriminant $\Delta=b^2-4ac \leq0$). So it can satisfy the condition.

Answer:

D. $|x|=

$$\begin{cases} x\text{ for }x > 0 \\ -x\text{ for }x < 0 \end{cases}$$

$

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For Question 5: