Collection of Algebra and Function Questions

MathematicsFunctions and AlgebraMediumSTEM

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II) choose the best answer

6. Which one of the following is power function? A. $f(x) = 3^x$ B. $f(x) = \log_3 x$ C. $h(x) = 5x^{-2}$ D. $k(x) = \ln x$

7. What is the value of $x$ in floor function $f(x) = \lfloor 2x + 1 \rfloor = 12$.

A. $\frac{11}{2} \le x < 6$ B. $\frac{11}{2} < x < 6$ C. $\frac{11}{2} < x \le 6$ D. $\frac{11}{2} \le x \le 6$

8. Which one of the following is the domain of $f(x) = 5x^4$?

A. $\mathbb{R}$ B. $(-\infty, 0)$ C. $[0, \infty)$ D. $(0, \infty)$

9. Which one of the following is not a function?

A. $R = \{(x, y) : x, y \in \mathbb{R} \text{ and } x^2 - y^2 = 0\}$ C. $R = \{(x, y) : x, y \in \mathbb{R} \text{ and } x^2 + y^2 = 1\}$

B. $R = \{(3, 0), (1, 3), (4, 6), (3, 6)\}$ D. All

10. What is the range of $3sgn(x)$? A. $\{3, 0, -3\}$ B. $\{-1, 0, 1\}$ C. $\mathbb{R}$ D. $[0, \infty)$

11. Which one of the following is the greatest integer of $-\pi$ or $\lfloor -\pi \rfloor$.

A. 4 B. 3 C. -4 D. -5

12. Let $f(x) = 2x^2 + 1$ and $g(x) = \sqrt{x + 1}$, then what is $f \circ g(x)$?

A. $2x + 3$ B. $\sqrt{2x^2 + 2}$ C. $2x - 3$ D. $\sqrt{2x^2 - 2}$

13. Which one of the following are not inverse of each other?

A. $f(x) = \frac{x+2}{x-1}$ and $g(x) = \frac{2x+2}{x-1}$ C. $f(x) = x - 1$ and $g(x) = x + 1$

B. $f(x) = 2x - 3$ and $g(x) = 3 - 2x$ D. $f(x) = 8x$ and $g(x) = \frac{x}{8}$

14. Which one of the following is a polynomial expression?

A. $x^2 + 2x + 1$ B. $2\sqrt{x} + 5x + 1$ C. $3^{-1} + x - 9$ D. All

15. What is the domain of $\frac{2x^2 + 5x + 1}{x^2 + 1}$? A. $\mathbb{R}$ B. $\mathbb{R}^+$ C. $\mathbb{R} \setminus \{-1\}$ D. $\mathbb{Z}$

16. Which one of the following conditions are satisfied for rational inequality

A. $\frac{p(x)}{q(x)} \ge 0$ B. $\frac{p(x)}{q(x)} \le 0$ C. $\frac{p(x)}{q(x)} > 0$ D. All

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Step by Step Written Solution

1
Step 1

Let's find the values of x that satisfy this floor function equation. We are given the function f of x equals the floor of two x plus one, and we are told that this equals twelve.

Solving a Floor Function Equation

$$[2x + 1] = 12$$
2
Step 2

Recall the definition of the floor function. For any integer n, the floor of some expression u equals n if and only if u is greater than or equal to n, but strictly less than n plus one.

3
Step 3

In our problem, the integer n is twelve, and our expression u is two x plus one. So we can rewrite our equation as a double inequality.

$$12 \leq 2x + 1 < 12 + 1$$
4
Step 4

Simplifying the right side, we have twelve is less than or equal to two x plus one, which is less than thirteen.

5
Step 5

Now, we need to isolate x. We'll start by subtracting one from all parts of the inequality.

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About This Question

Subject
Mathematics
Topic
Functions and Algebra
Difficulty
Medium
Exam
STEM
Question Type
Multiple Choice

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