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Question Number 16082 by Tinkutara last updated on 17/Jun/17
Solution of equation 2[x] + 4{x} = 3x  + 2 (where {∙} denotes fractional  function and [∙] denotes G.I.F) is  (1) {−2}  (2) {−2, − (3/2)}  (3) φ  (4) R
$$\mathrm{Solution}\:\mathrm{of}\:\mathrm{equation}\:\mathrm{2}\left[{x}\right]\:+\:\mathrm{4}\left\{{x}\right\}\:=\:\mathrm{3}{x} \\ $$$$+\:\mathrm{2}\:\left(\mathrm{where}\:\left\{\centerdot\right\}\:\mathrm{denotes}\:\mathrm{fractional}\right. \\ $$$$\left.\mathrm{function}\:\mathrm{and}\:\left[\centerdot\right]\:\mathrm{denotes}\:\mathrm{G}.\mathrm{I}.\mathrm{F}\right)\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\left\{−\mathrm{2}\right\} \\ $$$$\left(\mathrm{2}\right)\:\left\{−\mathrm{2},\:−\:\frac{\mathrm{3}}{\mathrm{2}}\right\} \\ $$$$\left(\mathrm{3}\right)\:\phi \\ $$$$\left(\mathrm{4}\right)\:{R} \\ $$
Commented by Tinkutara last updated on 17/Jun/17
I got x = −2. I have checked that it  satisfies the equation. But answer  given in my book is φ. How is it  possible?
$$\mathrm{I}\:\mathrm{got}\:{x}\:=\:−\mathrm{2}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{checked}\:\mathrm{that}\:\mathrm{it} \\ $$$$\mathrm{satisfies}\:\mathrm{the}\:\mathrm{equation}.\:\mathrm{But}\:\mathrm{answer} \\ $$$$\mathrm{given}\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}\:\mathrm{is}\:\phi.\:\mathrm{How}\:\mathrm{is}\:\mathrm{it} \\ $$$$\mathrm{possible}? \\ $$
Commented by prakash jain last updated on 17/Jun/17
x=[x]+{x}  2[x]+4{x}=3[x]+3{x}+2  [x]={x}−2  0≤{x}<1  [x] and 2 are integers so {x}=0  [x]=−2  ⇒x=−2  this is the only solutiin.
$${x}=\left[{x}\right]+\left\{{x}\right\} \\ $$$$\mathrm{2}\left[{x}\right]+\mathrm{4}\left\{{x}\right\}=\mathrm{3}\left[{x}\right]+\mathrm{3}\left\{{x}\right\}+\mathrm{2} \\ $$$$\left[{x}\right]=\left\{{x}\right\}−\mathrm{2} \\ $$$$\mathrm{0}\leqslant\left\{{x}\right\}<\mathrm{1} \\ $$$$\left[{x}\right]\:\mathrm{and}\:\mathrm{2}\:\mathrm{are}\:\mathrm{integers}\:\mathrm{so}\:\left\{{x}\right\}=\mathrm{0} \\ $$$$\left[{x}\right]=−\mathrm{2} \\ $$$$\Rightarrow{x}=−\mathrm{2} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solutiin}. \\ $$

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