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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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