Question

Solve the equation

4{x} = x + [x], where {} represents smallest integer function and [ ] is greatest integer function.

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Answer 1

Answer:

f(x)=[x]

x= any number

[x]= smallest integer

[x]=x if x= integer

[x]>x if x



= integer

Like we put the value x=1 than function return same value Because 1 is a integer number

f(1)=[1]=1

But when we put x=1.5, then its return 2. Because 2 is grater than 1.5.

here smallest integer means all the greater number (which is entered) we select smallest one.

For example there are hell number of integer which is greater than 15 but among those all integer the smallest integer is 2. So smallest integer of 1.5 is 2.

solution

Answer 2

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\:4 \{ x\} = x + [x]$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: x = [x] + \{ x\}}} - - - (1)$

Using this in given equation, we get

$\rm :\longmapsto\:4\{ x\} = [x] + \{ x\} + [x]$

$\rm :\longmapsto\:4\{ x\} - \{ x\} = 2[x]$

$\rm :\longmapsto\:3\{ x\} = 2[x]$

$\bf\implies \:[x] = \dfrac{3}{2}\{ x\}$

We know,

$\rm :\longmapsto\:0 \leqslant \{ x\} < 1$

$\rm :\longmapsto\:0 \leqslant \dfrac{3}{2} \{ x\} < \dfrac{3}{2}$

$\bf\implies \:0 \leqslant [x] < \dfrac{3}{2}$

$\bf\implies \:0 \leqslant [x] < 1.5$

$\bf\implies \:[x] = 0 \: \: or \: \: [x] = 1$

As,

$\rm :\longmapsto\: \:[x] = \dfrac{3}{2}\{ x\}$

$\bf\implies \:\{ x\} = \dfrac{2}{3} [x]$

So,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf [x] & \bf \{ x\} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 1 & \sf \dfrac{2}{3} \end{array}} \\ \end{gathered}$

Now,

$\rm :\longmapsto\:x = [x] + \{ x\}$

$\bf\implies \:x = 1 + \dfrac{2}{3} \: \: or \: \: x = 0 + 0$

$\bf\implies \:x = \dfrac{5}{3} \: \: or \: \: x = 0$