The value of m for which the equation frac{a} | vedclass

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(B) Given equation: $\frac{a}{x + a + m} + \frac{b}{x + b + m} = 1$.Multiplying both sides by $(x + a + m)(x + b + m)$,we get:$a(x + b + m) + b(x + a

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$0$

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(B) Given equation: $\frac{a}{x + a + m} + \frac{b}{x + b + m} = 1$.Multiplying both sides by $(x + a + m)(x + b + m)$,we get:$a(x + b + m) + b(x + a + m) = (x + a + m)(x + b + m)$.$ax + ab + am + bx + ab + bm = x^2 + bx + mx + ax + ab + am + mx + bm + m^2$.Simplifying both sides:$ax + bx + 2ab + am + bm = x^2 + ax + bx + 2mx + ab + am + bm + m^2$.Canceling common terms $(ax + bx + am + bm)$ from both sides:$2ab = x^2 + 2mx + ab + m^2$.Rearranging into standard quadratic form $x^2 + 2mx + m^2 - ab = 0$.For roots to be equal in magnitude but opposite in sign,the sum of the roots must be $0$.Since the sum of roots is $-\frac{	ext{coefficient of } x}{	ext{coefficient of } x^2} = -\frac{2m}{1} = -2m$.Setting $-2m = 0$,we get $m = 0$.

The value of $m$ for which the equation $\frac{a}{x + a + m} + \frac{b}{x + b + m} = 1$ has roots equal in magnitude but opposite in sign is

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