Vincen wrote:If for any number N, f(N) is defined as the least integer that is greater than or equal to N^2, then f(-1.1) =

(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

The OA is E.

Experts, I don't understand how f works. Can you give me some examples and then explain to me how to solve this PS question? Please.

Say we had f(1.3) If f(N) is defined as the least integer that is greater than or or equal to N^2, then f(1.3) would be the least integer greater than or equal to 1.3^2, or 1.69. The least integer greater than or equal to 1.69 is 2.

Or say we had f(.5). The least integer greater than or equal to .5^2, or .25 is 1.

Effectively, we're squaring the value in parentheses and then rounding up to the nearest integer.

So if we have f(-1.1), then (-1.1)^2 will be 1.21. (It's enough to see that this value will be greater than 1 and less than 2.) The least integer greater than or equal to 1.21 is 2. The answer is E

If for any number N, f(N) is defined as the least integer that is greater than or equal to N^2, then f(-1.1) =

(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

The OA is E.

Experts, I don't understand how f works. Can you give me some examples and then explain to me how to solve this PS question? Please.

Hi Vincen,
Let's take a look at your question.

The question states:
f(N) is defined as the least integer that is greater than or equal to N^2.
Therefore,
$$f\left(-1.1\right)\ =\ \left(-1.1\right)^2$$
$$f\left(-1.1\right)\ =\ 1.21$$

The least integer that is greater than or equal to 1.21 is 2.
Therefore, Option E is correct.

Hope it helps.
I am available if you'd like any follow up.