A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
As. \[x=\frac{6pq}{p+q}\] ...(i) |
On dividing both sides by 3p, |
\[\frac{x}{3p}=\frac{2q}{p+q}\] |
By componendo and dividendo rule, |
\[\frac{x+3p}{x-3q}=\frac{2q+p+q}{2q-p-q}=\frac{p+3q}{q-p}\] ?(ii) |
Again, dividing Eq. (i) by 3g, |
\[\frac{x}{3q}=\frac{2p}{p+q}\] |
By componendo and dividendo, |
\[\frac{x+3q}{x-3q}=\frac{2p+(p+q)}{2p-(p+q)}=\frac{3p+q}{p-q}\] ?(iii) |
Adding Eqs. (ii) and (iii), |
\[\frac{x+3p}{x-3p}+\frac{x+3q}{x-3q}=\frac{p+3q}{q-p}+\frac{3p+q}{p-q}\] |
\[=\frac{p+3q-3p-q}{q-p}=\frac{2(q-p)}{(q-p)}=2\] |
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