A) 4
B) 2
C) 1
D) \[a+b\]
Correct Answer: B
Solution :
Putting \[H=\frac{2ab}{a+b}\] \[\frac{H+a}{H-a}+\frac{H+b}{H-b}=\frac{2({{H}^{2}}-ab)}{(H-a)(H-b)}=\frac{2\left[ \frac{4ab}{{{(a+b)}^{2}}}-ab \right]}{\left[ \frac{4ab}{{{(a+b)}^{2}}}-ab \right]}=2\]. Trick: Let \[a=1,\ H=\frac{1}{2}\] and \[b=\frac{1}{3}\], then \[\frac{H+a}{H-a}+\frac{H+b}{H-b}=\frac{3/2}{-1/2}+\frac{5/6}{1/6}=2\].
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