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SSC CGL Tier II 25 Oct 2015 Paper 1

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Question : 82 of 100

Marks: +1, -0

If

x^{2}+y^{2}+z^{2}=xy+yz+zx

then the value of

\frac{3x^{4}+7y^{4}+5z^{4}}{5x^{2}y^{2}+7y^{2}z^{2}+3z^{2}x^{2}}

is

2
1
0
–1

Solution:

(2)

x^{2}+y^{2}+z^{2}=xy+yz+zx

\Rightarrow x^{2}+y^{2}+z^{2}-xy-yz-zx=0

\Rightarrow 2x^{2}+2y^{2}+2z^{2}-2xy-2yz-2zx=0

\Rightarrow x^{2}+y^{2}-2xy+y^{2}+z^{2}-2yz+x^{2}+z^{2}-2zx=0

\Rightarrow (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0

\therefore x-y=0\Rightarrow x=y

y-z=0 \Rightarrow y=z

z-x=0 \Rightarrow z=x

\therefore x=y=z

[

If

a^{2}+b^{2}+c^{2}=0,

then

a=0, b=0, c=0 ]

∴ Expression

=\frac{3x^{4}+7y^{4}+5z^{4}}{5x^{2}y^{2}+7y^{2}z^{2}+3z^{2}x^{2}}

=\frac{3x^{4}+7x^{4}+5x^{4}}{5x^{4}+7x^{4}+3x^{4}}

=\frac{15x^{4}}{15x^{4}}=1

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