TS ICET 24 May 2018 Shift 1 Paper

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Question : 76 of 200

Marks: +1, -0

If

\x/\a=\y/\b=\z/\c

and

√{a x}+√{b y}+√{c z}=K √{(a+b+c)(x+y+x)}

then k =

1
0
-1
$√{abc$

Solution:

Let

\;{x}/{a}=\;{y}/{h}=\;{z}/{c}=P

⇒ x=a P ; y=b P ; z=c P

⇒ x+y+z=(a+b+c) P

K=\;{√{a x}+√{b y}+√{c z}}/{√{(a+b+c)(x+y+z)}}=\;{√{a^{2} P}+√{b^{2} P}+√{c^{2} P}}/{√{(a+b+c)(a P+b P+c P)}}

=\;{(a+b+c) √{P}}/{(a+b+c) √{P}}=1

Note: Correction in the question

√{a x}+√{b y}+√{c z}=k √{(a+b+c)(x+y+z)}

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