lahca007 wrote:
Zarrolou wrote:
9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?
(1) ab = 2
(2) 0 < a < b < 2
The question can be seen as (given statement 2):
\([a] + [b] = 1\)
case 1:\(0<(=)a<1\) => \([a]=0\) and \(1(=)<b<2\) => \([b]=1\) so \([a] + [b] = 1\)
or the opposite
case 2: \(0<(=)b<1\) => \([b]=0\) and \(1(=)<a<2\) => \([a]=1\) so \([a] + [b] = 1\)
But as ab=2 we know that one term is \(\frac{1}{2}\) and the other is \(2\)
So we are in one of the two senarios above, IMO C
Though I too think that the correct answer is "C", but the explanation is a little confusing..
as the stmt 1 says that the product of ab=2, so its not necessary that one will be 2 and the other be 1/2. It could be either(-1*-2) or (1/2*4). Hope you got my point here.
From F.S 1 , we know that a,b both are not equal to zero. Also, it is being asked whether, [a]+[b] =1. b = 2/a.
Thus, the question is asking whether, [a] + [2/a] = 1? --> 1-[a] = [2/a]
Notice, for 0<a<1, we have [a] = 0, and [2/a] is atleast 2. Thus, 1-[a] is not equal to [2/a].
Similarly, for 1<a<2, we have [a] = 1, and [2/a] = 1. Thus, 1-[a] is not equal to [2/a].
For a>2, we have 1-[a] as negative and [2/a] as 0. Hence, not equal again.
Similarly, for a<0, we would always have 1-[a] as positive but [2/a] will always be negative and hence not equal.
One can check for a=1,2,-1,-2 and would still find they are not equal. Sufficient.
From F.S 2, we have 0<a<b<2. For a=0.1 and b=1, we have [a]+[b] =1, so a YES. However, for a=1 and b = 1.5, we have [a]+[b] = 2, hence a NO. Insufficient.
A.
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