[|2x-3/4|] = 5[/m] , where [x] denotes the greatest integer that is s

Question

= 5

, where   denotes the greatest integer that is smaller than or equal to x. Find the number of integers that satisfy the above equation.

A. 0
B. 1
C. 2
D. 3
E. 4

RamseyGooner · Accepted Answer

I have always had trouble with Modulus and inequalities, here goes my translation -

denotes the greatest integer that is  smaller than or equal to x

= 6,   = 5,  = 5 etc

5 <= |(2x-3)/4| < 6
20 <= |2x-3| < 24

Now taking mod as +,
23 <= 2x < 27; x can be 12 or 13

Now taking mod as -,
17 <= -2x < 21; x can be -9 or -10

So , x can be -9,-10, 12, 13

Hope this helps!

ntngocanh19 · Answer

=>4 < |(2x-3)/4| <=5 => 16 < |2x-3|<= 20 So , x can be -7,-8, 10, 11 => hence choice E Posted from my mobile device

joyousjoyce · Answer

Hello, could I get your help to share why is >4 the lower limit?

Thank you!

ntngocanh19 · Answer

It's because the definition of is the greatest integer that is smaller than or equal to x. Here we have = 5 (A=|(2x-3)/4|) Then if A<=4 <=> =4 => A must be greater than 4 Hope this help!

joyousjoyce · Answer

Thanks for helping to explain in more details. I think I am having challenges understanding the limit... let me try to share my thinking.

is the greatest integer that is smaller than or equal to x
x = 2.5,   = 2

In the question above   = 5, so shouldn't it be that A = 5~5.9?
  = 5
  = 5

For A<=4,   = 4, and   = 4. Hence A has to be at least 5.

Could you share with me the flaw in my thinking? Thank you..

ntngocanh19 · Answer

I think you are right. A I found out that I'm confused between and x. Thank you for pointing out

Sevil92 · Answer

Hello everyone!

Unfortunately, explanations did not help me get the logic here. I just solved the equation in traditional manner (left side with positive x equals to 5; left side with negative x equals to 5) and got   and   respectively, and ended up that x equals 11 and -9 (smaller integers than x itself). Can anyone correct me? I cannot see why we take 4 as a limit.

pankaj92 · Answer

denotes the greatest integer that is smaller than or equal to x = 5, = 5, = 5 etc Hence, we can write terms inside as 5 <= |(2x-3)/4| < 6 Now we solve for both + and - case of modulus: Case 1 : |(2x-3)/4|>0 i.e x>3/2 5 <= (2x-3)/4 < 6 i.e. 11.5<=x<13.5 as x>3/2 so we get 2 integer as solution i.e. 12 and 13 Now, Case 2: |(2x-3)/4|<0 i.e x>3/2 5 <= -(2x-3)/4 < 6 i.e. -10.5<=x<-8.5 as x<3/2 so we get 2 integer as solution i.e. -10 and -9 Hence total 4 integer solutions are possible. Answer is option E

[|2x-3/4|] = 5[/m] , where [x] denotes the greatest integer that is s

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[|2x-3/4|] = 5[/m] , where [x] denotes the greatest integer that is s

Prep Toolkit

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My Rewards