Question

The number of real x satisfying the equation
(x-2)[x]={x}-1

Answer 1

I m getting only 1 value of x which is 1

Answer 2

@priyar @hartnn @ganeshie8 @wio
@michele_laino

Answer 3

Here ,{x} stands for fractional part function
And [x] is greatest integer function

Answer 4

\(x=1\) is obviously a solution. Other than that, we know that the right hand side is negative and lies in \([-1,0)\). So \(x-2\) and \([x]\) have opposite signs. That restricts our domain to \([0,2]\).
\([x]\) is obviously not zero.
\([x]\) can be one. We've already found one such solution. Let's try to find others. \(x-2 = \{x \} -1 \Rightarrow \lfloor x\rfloor =1\). This already tells us that there are infinite solutions.

Answer 5

All reals in the interval \([1, 2)\) satisfy that, so the answer is infinity.

Answer 6

Okk
Shall i tell you my way now..which strictly bans over [x] to be equal to 1

Answer 7

We can write x as [x]+{x}
Now lhs would become [x]^2+[x]{x}-2[x]
Equating , we will be left with
(1-[x])^2={x}(1-[x])
Now , {x}= (1-[x])^2/1-[x]

Answer 8

Now this clearly tells that [x] can never be equal to 1

Answer 9

Well, i am getting x belongs to [1,infinity)..is it correct?

Answer 10

It's not.

Answer 11

Yep !!..he's correct...

Answer 12

hi

Answer 13

do u need help

Answer 14

Yep ! Will you help?

Answer 15

sure

Answer 16

Do so then!

Answer 17

(x - 2)[x] = {x} - 1

x = 10
8*10 = 0 - 1?

The answer is not 1 to infinity...

Answer 18

@parthkohli what is the fault in method?

Answer 19

My method*

Answer 20

The fault is that you get the wrong solution. =P

Answer 21

Like seriously!!
What is that thing in my method which is resulting in that fault?

Answer 22

In the second-last line, you wrote this:\[(1-|x|)^2 = \{ x \}(1 - |x|)\]This equation says that whenever \(|x| = 1\), this equation will be satisfied. This itself tells us that the equation will have infinite solutions.

Answer 23

Bt if u solve this to take the value for {x}
U get {x}=(1-[x])^2/1-[x]

Answer 24

and for it to hold true we must say that [x] is not equal to 1

Answer 25

That's very illegal in maths. You just changed the equation to something that it's not.\[(1-x)^2 = k(1-x)\]Are you saying that \(x = 1\) does not satisfy the above equation just because if we solve for \(k\)\[k = \frac{(1-x)^2}{1-x}\]it does not allow \(x = 1\)? But \(x = 1\) does satisfy the original equation and the original equation is all we care about.

Answer 26

Okay!!

Answer 27

Ty @parthkohli

Answer 28

<3

Answer 29

if I add \([x]\) to both side, I get:
\[\Large \begin{gathered}
\left( {x - 2} \right)\left[ x \right] + \left[ x \right] = \left\{ x \right\} - 1 + \left[ x \right] \hfill \\
\hfill \\
\left( {x - 1} \right)\left[ x \right] = x - 1 \hfill \\
\end{gathered} \]
being:
\[\Large \left\{ x \right\} + \left[ x \right] = x\]