If $a, b, c \in R$ are distinct, then the condition(s) on a, b, c for which the equation $\frac{1}{x - a} + \frac{1}{x - b} + \frac{1}{x - c} + \frac{1}{x - b - c + a} = 0$ has real roots is

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none of these

$a + b + c \neq 0, a \neq 0$

$2 a = b + c$

$a - b - c \neq 0, a \neq 0$

answer is D.

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Detailed Solution

$\frac{1}{x - a} + \frac{1}{x - b} + \frac{1}{x - c} + \frac{1}{x - b - c + a} = 0 \frac{1}{x - b} + \frac{1}{x - c} + \frac{1}{x - a} + \frac{1}{x - b - c + a} = 0 \frac{2 x - (b + c)}{(x - b) (x - c)} + \frac{2 x - (b + c)}{(x - a) (x - b - c + a)} = 0 2 x - (b + c) = 0 \frac{x^{2} - a^{2} - x b - x c + a b + a c + x^{2} - x b - x c + b c}{(x - a) (x - b) (x - c) (x - b - c + a)} = 0 2 x^{2} - 2 x (b + c) + a b + b c + a c - a^{2} = 0$

$∆ = {(b + c)}^{2} - 2 (a b + b c + c a - a^{2}) = a^{2} + c^{2} - 2 a c + a^{2} + b^{2} - 2 a b = {(a - c)}^{2} + {(a - b)}^{2} \geq 0$
Clearly $Δ \geq 0$ . It has always real roots, $\forall a, b, c$ .