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# The number of solutions of ∣[x]−2x∣=4, where [x] is the greatest integer ≤x, is:A. 2B. 4C. 1D. Infinite  Hint: In order to solve this problem one should know that the greatest integer function is a piecewise defined function. If the number is an integer, use that integer. If the number is not an integer, use the next smaller integer. And by knowing if we are opening a modulus function then the terms on the other side will be multiplied by $\pm$. By using these properties you will surely get the right answer.

The given equation is ∣[x]−2x∣=4 where [x] is the greatest integer $\leqslant$x.
∣[x]−2x∣=4
We will open modulus from the LHS of the equation then there will be $\pm 4$ in RHS of the equation.

⇒[x] − 2x = ±4 ……(1)
We know that if a number ‘n’ is in greatest integer function [] such that: [n]
Then n = n + {n} where n is an integer part of ‘n’ and {n} is the part whose value belongs to [0,1). So {n} can be 0 but less than 1 and cannot be less than 0.

So, we can say x = x +{x} and [x] = x ({x}=0 when x is an integer)

On putting the value of x and [x] in the equation (1) we get the new equation as,
$\Rightarrow$x -2 {x} -2x = ±4
$\Rightarrow$x-2{x}-2x= ±4
$\Rightarrow$2{x}-x = ±4 (2)
So we will now consider two cases:
Case 1: When x is an integer.
Then {x} = 0
So, x = $\mp 4$
Then the value of x will be -4, +4

Case 2: When x is not an integer.
Then {x} $\in (0,1)$.
So, the equation (2) becomes
x = ±4 - 2{x}
In this case x will be an integer only when x is $\dfrac{1}{2}$
Then the equation becomes,
x = $\pm$4 - 2$\left( {\dfrac{1}{2}} \right)$
x = $\pm$4 – 1
x = 3, -5
The possible values of x are 3, -5, -4, 4.

Note: Whenever you face such types of problems you have to use the concepts of greatest integer function and modulus function. The greatest integer function is a piecewise defined function. If the number is an integer, use that integer. If the number is not an integer, use the next smaller integer. And by knowing if we are opening a modulus function then the terms on the other side will be multiplied by $\pm$. By using these concepts you will get the correct solution.
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