Question

The number of solution of $|\left[x\right]-2x|=4$, where $\left[x\right]$ denotes the greatest integer less than $x$ is

• A. infinite
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (B)

4

$\mid \left[x\right]-2x\mid =4$

$Case1:Ifx\in Z$

$Then,\mid \left[x\right]-2x\mid =4$

$⟹\mid x-2x\mid =4$

$⟹\mid -x\mid =4$

$⟹\mid x\mid =4$

$⟹x=\pm 4$

$Case2:Ifx\notin Z$

$Let,x=I+f$

$where,IisInteger$

$and,fisfractions.t.f\in [0,1)$

$Then,\mid \left[x\right]-2x\mid =4$

$⟹\mid [I+f]-2(I+f)\mid =4$

$⟹\mid I-2(I+f)\mid =4$

$⟹\mid -(I+2f)\mid =4$

$⟹\mid (I+2f)\mid =4$

In the above equation, R.H.S. is Integer. So, to make L.H.S. Integer $2f$ must be Integer.

$⟹f=\frac{1}{2}$

So, $\mid (I+2f)\mid =4$

$⟹\mid (I+2\times \frac{1}{2})\mid =4$

$⟹\mid (I+1)\mid =4$

$⟹I+1=\pm 4$

$⟹I=3or-5$

$So,x=I+f$

$⟹x=3+\frac{1}{2}orx=-5+\frac{1}{2}$

$⟹x=\frac{7}{2},\frac{-9}{2}$

Hence, $x=\pm 4,\frac{7}{2},\frac{-9}{2}$

Hence, $x$ has 4 solutions.

$\therefore$ option B is correct.