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Updated on: 09 Sep 2019, 04:07
Originally posted by freedom128 on 04 Sep 2019, 18:46.
Last edited by freedom128 on 09 Sep 2019, 04:07, edited 12 times in total.
Last edited by freedom128 on 09 Sep 2019, 04:07, edited 12 times in total.
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If a and b are integers and |b| < a, is \(\sqrt{a^2 + b^2} = a + b\) ?
(1) \(b^a<0\)
(2) |ab| is not a square of an integer
posting an efficient solution is much appreciated
(1) \(b^a<0\)
(2) |ab| is not a square of an integer
posting an efficient solution is much appreciated
04 Sep 2019, 20:19
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chondro48
All 6 responses have gone wrong as per timer, so let me try to give a solution.
Always try to simplify the given main statement..
If a and b are integers and |b| < a, MEANS a>0 as |b| will be atleast 0.
is \(\sqrt{a^2 + b^2} = a + b\)
Square both sides.. \(a^2+b^2=(a+b)^2=a^2+b^2+2ab.....2ab=0 or ab=0\)
So we are asked -- Is ab=0?, but a>0, so finally the question becomes is b=0?
Now let us see if any statement can give us whether b is 0 or not?
(1) \(b^a<0\)
This tells us that a is ODD and b is NEGATIVE, so our answer is NO, b is not 0..
Sufficient
(2) |ab| is not a square of an integer
This tells us that \(|ab|\neq{0}\), as 0 is taken as a square since \(0=0^2\)
Again sufficient
D
If a and b are integers and \(|b| < a\), is \(√a^2+b^2 = a+b\) ?
(1) \(b^a<0\)
(2) |ab| is not a square of an integer
\(|b| < a\) implies that \(a > 0\) where either \(b < 0\) or \(0 < b < a\) .
\(√a^2+b^2 = a+b\)
There \(a+b > 0\) where \(b < 0 < a\) or \(b > 0\).
So, we have \(a > 0\) and \(b < a\) (\(b < 0\) or \(b > 0\))
(1) \(b^a<0\)
implies that \(b < 0\) since \(a > 0\) only if 'a' is odd i.e. 3, 5 and so on.
'b' can take values -1, -2 -3, -4 and so on.
Minimum value of \(a + b = 3 - 1 = 2\)
minimum value of \(√a^2+b^2 = √3^2+(-1)^2 = √10\) Hence NO.
Further calculations would reveal that \(√a^2+b^2\) is always bigger than because \(a^2 > a + b\) always.
Hence \(√a^2+b^2 ≠ a + b\) always. Thus, always a NO answer.
SUFFICIENT.
(2) |ab| is not a square of an integer
\(|ab| ≠ integer^2\) since that would make \(a = -b\) which violates \(|b| < a\)
So |ab| = any integer i.e. 1, 2, 3, 5, 6, 7, 8, 10... excepts perfect squares.
\(|ab| ≠ 1\) since that violates \(|b| < a\) [a = 1 and b = -1]
Thus minimum value of \(|ab| = 2\) where a = 2 and b = -1. So \(√a^2+b^2 = √2^2+(-1)^2 = √5\) and \(a + b = 2 - 1 = 1\). Hence NO.
Or \(|ab| = 3\) where a = 3 and b = -1 So \(√a^2+b^2 = √3^2+(-1)^2 = √10\) and \(a + b = 3 - 1 = 2\). Hence NO.
Thus, \(√a^2+b^2 ≠ a+b\) always.
HENCE NO.
SUFFICIENT.
Answer (D).
(1) \(b^a<0\)
(2) |ab| is not a square of an integer
\(|b| < a\) implies that \(a > 0\) where either \(b < 0\) or \(0 < b < a\) .
\(√a^2+b^2 = a+b\)
There \(a+b > 0\) where \(b < 0 < a\) or \(b > 0\).
So, we have \(a > 0\) and \(b < a\) (\(b < 0\) or \(b > 0\))
(1) \(b^a<0\)
implies that \(b < 0\) since \(a > 0\) only if 'a' is odd i.e. 3, 5 and so on.
'b' can take values -1, -2 -3, -4 and so on.
Minimum value of \(a + b = 3 - 1 = 2\)
minimum value of \(√a^2+b^2 = √3^2+(-1)^2 = √10\) Hence NO.
Further calculations would reveal that \(√a^2+b^2\) is always bigger than because \(a^2 > a + b\) always.
Hence \(√a^2+b^2 ≠ a + b\) always. Thus, always a NO answer.
SUFFICIENT.
(2) |ab| is not a square of an integer
\(|ab| ≠ integer^2\) since that would make \(a = -b\) which violates \(|b| < a\)
So |ab| = any integer i.e. 1, 2, 3, 5, 6, 7, 8, 10... excepts perfect squares.
\(|ab| ≠ 1\) since that violates \(|b| < a\) [a = 1 and b = -1]
Thus minimum value of \(|ab| = 2\) where a = 2 and b = -1. So \(√a^2+b^2 = √2^2+(-1)^2 = √5\) and \(a + b = 2 - 1 = 1\). Hence NO.
Or \(|ab| = 3\) where a = 3 and b = -1 So \(√a^2+b^2 = √3^2+(-1)^2 = √10\) and \(a + b = 3 - 1 = 2\). Hence NO.
Thus, \(√a^2+b^2 ≠ a+b\) always.
HENCE NO.
SUFFICIENT.
Answer (D).
