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Updated on: 02 May 2016, 15:48
Originally posted by jjsverbal on 02 May 2016, 15:38.
Last edited by ENGRTOMBA2018 on 02 May 2016, 15:48, edited 2 times in total.
Last edited by ENGRTOMBA2018 on 02 May 2016, 15:48, edited 2 times in total.
Reformatted the question, updated the title and added the OA.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
70% (02:13) correct
30%
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If \(t = a^2 - b^2\) , \(u = a^2 + b^2\) , \(v=2ab\) , what is the value of t, in terms of u and v?
a) \(t = \sqrt { u^2 - v^2}\)
b) \(t = \sqrt {u^2 + v^2}\)
c) \(t = \sqrt {u^2 - v}\)
d) \(t = \sqrt {u + v}\)
e) \(t = \sqrt {u^3 - v^3}\)
a) \(t = \sqrt { u^2 - v^2}\)
b) \(t = \sqrt {u^2 + v^2}\)
c) \(t = \sqrt {u^2 - v}\)
d) \(t = \sqrt {u + v}\)
e) \(t = \sqrt {u^3 - v^3}\)
02 May 2016, 22:15
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mjhoon1004
since we have to get t in terms of u and v..
lets work on u and v..
we can easily see u and v can be converted in form of\((a-b)^2 or (a+b)^2\)..
\(u-v = a^2+b^2-2ab = (a-b)^2\)..
and \(u+v = a^2+b^2+2ab = (a+b)^2\)..
so \((u-v)(u+v) = {(a-b)(a+b)}^2 = (a^2-b^2)^2 = t^2\)..
or \(t = \sqrt { u^2 - v^2}\)..
ans A
another simpler method would be substitute something simple for a and b..
let a = 3 and b = 1..
so \(t = a^2-b^2 = 3^2-1^2 = 8........\\
u= a^2+b^2 =3^2+1^2 = 10............\\
v = 2ab = 2*3*1 = 6..\)
lets substitute value of u and v in all choices and see where we get t as 3..
a) \(t = \sqrt { u^2 - v^2}\)
\(t = \sqrt { 10^2 - 6^2} = 8\).. YES
b) \(t = \sqrt {u^2 + v^2}\)
\(t = \sqrt {10^2 + 6^2} = \sqrt{136}\).. NO
c) \(t = \sqrt {u^2 - v}\)
\(t = \sqrt {10^2 - 4} = \sqrt{96}\)
d) \(t = \sqrt {u + v}\)
\(t = \sqrt {10 + 6} = 4\)..YES
e) \(t = \sqrt {u^3 - v^3}\)
\(t = \sqrt {10^3 - 6^3} = \sqrt{784}\).. NO
ans A
07 Jan 2019, 21:53
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jjsverbal
\((a^2 - b^2)^2\) = \(a^4\) + \(b^4\) - 2\(a^2\)\(b^2\) = \(t^2\)
\((a^2+ b^2)^2\) =\(a^4+b^4+2a^2b^2\) = \(u^2\)
\((2ab)^2 = 4a^2b^2 = v^2\)
It can be seen that \(t^2 = u^2 - v^2\)
So, t = \(\sqrt{u^2-v^2}\)
Choice A
