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If the equation `(1)/(x) + (1)/(x+a)=(1)/(lambda)+(1)/(lambda+a)` has real roots that are equal in magnitude and opposite in sign, thenA. `lambda^(2) = 3a^(2)`B. `lambda^(2) = 2a^(2)`C. `lambda^(2) = a^(2)`D. `a^(2) = 2 lambda^(2)` |
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Answer» Correct Answer - D We have, `(1)/(x)+(1)/(x+a)=(1)/(lambda)+(1)/(lambda+a)" "...(i)` Clearly, `x = lambda` is a root of this equation. It is given that equation (i) has real roots that are equal in magnitude and opposite in sign. Therefore, `x = - lambda` is a root of equation (i). `therefore" "-(1)/(lambda)+(1)/(a-lambda)=(1)/(lambda)+(1)/(lambda + a)` `rArr" "(2)/(lambda)=(1)/(a-lambda)-(1)/(a+lambda)` `rArr" "(2)/(lambda)=(2 lambda)/(a^(2)-lambda^(2)) rArr a^(2) - lambda^(2) = lambda^(2) rArr a^(2) = 2 lambda^(2)` |
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