If 1x+1x+a=1m+1m+a has roots equal in magnitude but opposite in sign, then

Given Equation 1x+1x+a=1m+1m+aIt can be observed that one root is 'm'∴ Other root will be '−m'⇒ 1−m+1m+a=1m+1m+a⇒ 1a−m−1a−m=2m⇒ a+m−a+ma2−m2=2m⇒ 2ma2−m2= 2m⇒ m2= a2−m2⇒ 2m2=a2

If 1x+1x+a=1m+1m+a has roots equal in magnitude but opposite in sign, then

Given Equation 1x+1x+a=1m+1m+aIt can be observed that one root is 'm'∴ Other root will be '−m'⇒ 1−m+1m+a=1m+1m+a⇒ 1a−m−1a−m=2m⇒ a+m−a+ma2−m2=2m⇒ 2ma2−m2= 2m⇒ m2= a2−m2⇒ 2m2=a2

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$I f \frac{1}{x} + \frac{1}{x + a} = \frac{1}{m} + \frac{1}{m + a} h a s r o o t s e q u a l i n m a g n i t u d e b u t o p p o s i t e i n s i g n, t h e n$

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$a^{2} = 4 m^{2}$

$a = 2 m$

$a^{2} = 2 m^{2}$

$a^{2} = 4 m$

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n \frac{1}{x} + \frac{1}{x + a} = \frac{1}{m} + \frac{1}{m + a} \\ I t c a n b e o b s e r v e d t h a t o n e r o o t i s' m' \\ ∴ O t h e r r o o t w i l l b e' - m' \\ \Rightarrow \frac{1}{- m} + \frac{1}{m + a} = \frac{1}{m} + \frac{1}{m + a} \\ \Rightarrow \frac{1}{a - m} - \frac{1}{a - m} = \frac{2}{m} \\ \Rightarrow \frac{a + m - a + m}{a^{2} - m^{2}} = \frac{2}{m} \\ \Rightarrow \frac{2 m}{a^{2} - m^{2}} = \frac{2}{m} \\ \Rightarrow m^{2} = a^{2} - m^{2} \\ \Rightarrow 2 m^{2} = a^{2} \end{array}$