A) \[{{p}^{2}}+{{q}^{2}}+{{r}^{2}}\]
B) \[{{p}^{2}}+{{q}^{2}}\]
C) \[2({{p}^{2}}+{{q}^{2}})\]
D) \[\frac{{{p}^{2}}+{{q}^{2}}}{2}\]
Correct Answer: B
Solution :
\[\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}\] \[\frac{x+p+x+q}{(x+p)(x+q)}=\frac{1}{r}\] \[(2x+p+q)r={{x}^{2}}+px+qx+pq\] \[{{x}^{2}}+(p+q-2r)x+pq-pr-qr=0\] Let and be the roots. \[\Rightarrow \]\[\alpha +\beta =-(p+q-2r)\] ?[1] \[\Rightarrow \]\[\alpha \beta =pq-pr-qr\] ?[2] Roots are equal in magnitude and opposite in sign \[\Rightarrow \]\[\alpha +\beta =0.\] \[\Rightarrow \]\[-(p+q-2r)=0\] ?[3] \[{{\alpha }^{2}}{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[={{(-(p+q-2r))}^{2}}-2(pq-pr-qr)\] ?(from [1] and [2]) \[={{p}^{2}}+{{q}^{2}}+4{{r}^{2}}+2pq-4pr-4qr-2pq+2pr+2qr\]\[={{p}^{2}}+{{q}^{2}}+2r(2r-p-q)\] ?(from[3]) \[={{p}^{2}}+{{q}^{2}}+0\] \[={{p}^{2}}+{{q}^{2}}\] Hence, answer is option B
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