A) \[x=p+q+r\]
B) \[x=p-q+r\]
C) \[x=\frac{p+q}{q+r}\]
D) \[x=\frac{p}{q}+r\]
Correct Answer: A
Solution :
We have \[\frac{p+q-x}{r}+\frac{q+r-x}{p}+\frac{r+p-x}{q}\]= \[\frac{-4x}{p+q+r}\] \[\frac{p+q+r-x}{r}+\frac{p+q+r-x}{p}+\frac{p+q+r-x}{q}\] =\[4-\frac{4x}{p+q+r}\] \[\Rightarrow \] \[(p+q+r-x)\,\left( \frac{1}{p}+\frac{1}{q}+\frac{1}{r} \right)\,=4\,\left( \frac{p+q+r-x}{p+q+r} \right)\] \[\Rightarrow \] \[(p+q+r-x)\,\left[ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}-\frac{4}{p+q+r} \right]=0\] \[\Rightarrow \]\[x=p+q+r\].
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