SATHEE: Quantitative Aptitude Ques 1310

Quantitative Aptitude Ques 1310

Question: If $ x+y+z=0, $ then the value of $ \frac{1}{x^{2}+y^{2}-z^{2}}+\frac{1}{x^{2}+z^{2}-y^{2}}+\frac{1}{y^{2}+z^{2}-x^{2}} $ is

Options:

A) $ -2 $

B) $ -\frac{1}{2} $

C) $ 0 $

D) $ \frac{1}{2} $

Show Answer

Answer:

Correct Answer: C

Solution:

Expression is $ \frac{1}{x^{2}+y^{2}-z^{2}} $ $ +\frac{1}{x^{2}+z^{2}-y^{2}}+\frac{1}{y^{2}+z^{2}-x^{2}} $ Given that, $ x+y+z=0 $

$ \therefore $ $ x=-(y+z), $ $ y=-(x+z) $ and $ z=-(x+y) $ Putting the values of x, y and z in the expression $ =\frac{1}{x^{2}+y^{2}-{{{-(x+y)}}^{2}}}+\frac{1}{x^{2}+z^{2}-{{{-(x+z)}}^{2}}} $ $ +\frac{1}{y^{2}+z^{2}-{{{-(y+z)}}^{2}}} $ $ =\frac{1}{x^{2}+y^{2}-x^{2}-y^{2}-2xy}+ $ $ \frac{1}{x^{2}+z^{2}-x^{2}-z^{2}-2xz}+\frac{1}{y^{2}+z^{2}-y^{2}-z^{2}-2yz} $ $ =\frac{-1}{2xy}-\frac{1}{2xz}-\frac{1}{2yz}=-\frac{1}{2}( \frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz} ) $ $ =-\frac{1}{2}( \frac{x+y+z}{xyz} )=-\frac{1}{2}( \frac{0}{xyz} ) $ $ [\because x+y+z=0\text{(given) }] $ $ =0 $

Quantitative Aptitude Ques 1309

Quantitative Aptitude Ques 1311

Quantitative Aptitude Ques 1310

Question: If $ x+y+z=0, $ then the value of $ \frac{1}{x^{2}+y^{2}-z^{2}}+\frac{1}{x^{2}+z^{2}-y^{2}}+\frac{1}{y^{2}+z^{2}-x^{2}} $ is

Options:

Answer:

Solution:

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