AP ICET 2015 Paper

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Question : 71 of 200

Marks: +1, -0

Given that

x*y=\;\;{x^{2}+4y^{2}}/{xy}

and

x∆y=\;\;{x^{2}}/{y}

for any real numbers x and y. If

a*b=4∆4

, then b =

2a
$a/2$
a
$a/3$

Solution:

Given

\;\;x*y={x^{2}+4y^{2}}/{xy},x∆y={x^{2}}/{y},a*b=4∆4

⇒4∆4={4^{2}}/{4}=4

[Using

x∆y={x^{2}}/{y}]

a*b=4

. . .. . ..(1)

[Q\;a*b=4∆4]

Also

a*b={a^{2}+4b^{2}}/{ab}

.....ii
[Using

x*y={x^{2}+4y^{2}}/{xy}

]

⇒{a^{2}+4b^{2}}/{ab}=4

[From (i) and (ii)]

⇒a^2+4b^2-4ab=0

⇒(a-2b)2=0

⇒a=2b

∴b=a/2

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