If [z] denotes the least integer greater than or equal to z and

Official Solution

Given:

• The function [z]
• \([z^2] = 2\)

To find: Can [z] be {2, 1, -2}?

Approach:

• To find which values of [z] are possible and which are not, we need to first know the range of possible values of z. Once we know what z can be, we’ll be able to find what [z] can be.

We’ll get an idea of the possible values of z from the fact that \([z^2] = 2\)

Working Out:

• \([z^2] = 2\)

• This means the least integer that is greater than or equal to \(z^2\) is 2.


• We can write: \(1 < z^2 ≤ 2\)

• The above inequality contains 2 inequalities: \(z^2 > 1\) AND \(z^2 ≤ 2\)

So now, we’ll solve these inequalities one by one, and then find the values of z that satisfy both these inequalities

• Solving \(z^2 > 1\)

• \(z^2 – 1 > 0\)

• \((z+1)(z-1) > 0\)

This means, \(z < - 1\) or \(z > 1\) . . . (1)

• Solving \(z^2 ≤ 2\)

• \(z^2 – 2 ≤ 0\)

• \((z + √2)(z-√2) ≤ 0\)

That is, \(-√2 ≤ z ≤ √2\) . . . (2)

Combining (1) and (2):

Combining them to find their overlap zones and finding those values of z that satisfy both the inequalities

• So, either \(-√2 ≤ z < -1\) or \(1 < z ≤ √2\)


• Either \(-√2 ≤ z < -1\)

• \(-1.4  ≤ z < -1\)


• In this case, \([z] = -1\)


• Or \(1 < z ≤ √2\)

• \(1 < z ≤ 1.4 \)

• \([z] = 2\)

Thus, we see that [z] is either equal to -1 or equal to 2. Out of the 3 given values, only value I (which is 2) is therefore possible.

Looking at the answer choices, we see that the correct answer is Option A


Thanks,
Saquib
Quant Expert
e-GMAT

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts


_________________

\([z^2] = 2\) means that

\(1<z^2\leq{2}\)

Now, Taking Square Root of both the sides.

If Z is positive, then

\(1<z\leq{\sqrt{2}}\)

then [z] = 2

If Z is negative, then

\(-1>z\geq{-\sqrt{2}}\)

then [z] = -1

Answer is A. I only

Given function rounds up a number to the nearest integer. For example, [1.3] = 2, because 2 is the least integer greater than or equal to 1.3.

[\(z^2\)] = 2;

\(1 < z^2 \leq 2\);

\(1 <|z| \leq \sqrt{2}\) (\(\sqrt{2} \approx 1.4\)).

\(1 < z \leq \sqrt{2}\) or \(-1 > z \geq \sqrt{2}\)

[z] can be 2 (if \(1 < z \leq \sqrt{2}\)) or -1 (if \(-1 > z \geq \sqrt{2}\)).

Answer: A.
_________________

Good question...
if [z^2] =2, \(z^2\) lies between 2 and 3
therefore z lies between \(\sqrt{2}=1.4\) and \(\sqrt{3}=1.7\) or between \(-\sqrt{2}=-1.4\) and \(-\sqrt{3}=-1.7\) if z is negative..
so when z is positive it lies between 1.4 and 1.7 and the NEXT greater integer is 2 so [z]=2
but when z is negative it lies between -1.4 and -1.7 and the NEXT greater integer is -1 so [z]=-1.. Be careful here and do not take -2

Only 2 is given
ans A
_________________

I don't understand, where does the z > 1 and z < -1 come from?

Is my reasoning correct:

We know that [z] will take any number, (ex 1.4), and will round it up to the next highest integer, (2 in the case of 1.4) unless Z is already an integer.

Therefore, as others mentioned,

1<z^2 ≤ 2

Which means that √(Z^2), or Z is either larger than √1, which si 1 or -1, or smaller than √2 or -√2 (1.4, or -1.4)

-1 (or 1) <Z ≤ 1.4 ( or -1.4).

The different possibilities for [Z] are thus

1) 0, because if Z> -1 , the closest higher integer is 0,
2) 2, because if Z>1, Z < 1.4 , the closest integers are 2,
3) -1, because if Z> -1.4, the closest highest integer is -1.

The only option that appears in the answer choices is 2, thus the answer is A.

I: If z = 1,2 --> [z] = 2 --> [z^2] = 2

II: If z = 0,5 --> [z] = 1 --> [z^2] = 1

III: If z = -2,2 --> [z] = -2 --> [z^2] = 4

Only I works.

Nice question. Might be good to not follow a heavy paper/pad approach with this one.
_________________

let us assume z^2 = 1.96 then z= 1.4 and greatest integer function will lead to z= 2

now alternatively in the neagtive axis z^2 =1.96 and z=-1.4 then the greatest integer will be -1 ...........a

and we can safely eleminate the other 2 option with the reasoning at a

therefore IMO a