×

# the number of solutions of [2x]-3{2x}=1

the number of solutions of [2x]-3{2x}=1? How to solve such problems

Note by Priyankar Kumar
4 years, 5 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

## Comments

Sort by:

Top Newest

If $$\lfloor 2x \rfloor - 3\{2x\} = 1$$ then $$3\{2x\} = \lfloor 2x\rfloor - 1$$ is an integer, so $$\{2x\} = 0,1/3,2/3$$.

1. If $$\{2x\} = 0$$ then $$\lfloor 2x\rfloor - 1 = 0$$, so $$\lfloor 2x\rfloor = 1$$, and hence $$2x=1$$, so $$x=1/2$$.

2. If $$\{2x\} = 1/3$$ then $$\lfloor 2x\rfloor - 1 = 1$$, so $$\lfloor 2x \rfloor = 2$$, and hence $$2x=7/3$$, so $$x=7/6$$.

3. If $$\{2x\} = 2/3$$ then $$\lfloor 2x\rfloor - 1 = 2$$, so $$\lfloor 2x \rfloor = 3$$, and hence $$2x=11/3$$, so $$x=11/6$$.

- 4 years, 5 months ago

Log in to reply

Thanks

- 4 years, 5 months ago

Log in to reply

(I assume that the square and curly brackets represent floor and fractional part function respectively)

I am not the right person to help out in such problems but the following is worth a try. Add and subtract {2x} in the LHS. The equation now becomes 2x-4{2x}=1 or 2x=1+4{2x}. You can now plot a graph to find the number of solutions.

- 4 years, 5 months ago

Log in to reply

Thanks.

- 4 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...