Multiple choice

Statement-I : If $a + b + c > 0$ and $a < 0 < b < c$, then the roots of the equation $a(x - b) (x- c) + b(x- c)(x- a) + c(x- a) (x- b) = 0$ are of both negative. Statement-II : If both roots are negative, then sum of roots $< 0$ and product of roots $> 0$.

  1. Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I

  2. Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I

  3. Statement-I is true, Statement-II is false

  4. Statement-I is false, Statement-II is true
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D Correct answer
Explanation

The equation is a cyclic expression. If x=a, x=b, or x=c, the expression simplifies to a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) = 0. This is a known identity for quadratic equations where the roots are a, b, and c, but since it is quadratic, the roots are actually a, b, and c. However, the statement claims roots are negative; given a < 0 < b < c, the roots are a (negative), b (positive), and c (positive). Thus, Statement-I is false. Statement-II is a standard property of quadratic equations.

AI explanation

Using Vieta's formulas, expanding the given equation gives a sum of roots equal to (ab + bc + ca) / (a + b + c) and a product of roots equal to (3abc) / (a + b + c). With the conditions a < 0 < b < c and a + b + c > 0, the sign of the product depends on the specific values chosen; for instance, if a = -1, b = 2, and c = 3, the product of roots is -18, meaning the roots cannot both be negative. Therefore, Statement-I is false. Statement-II provides the correct algebraic conditions for two negative roots (sum < 0 and product > 0). The result is that only Statement-II is true.

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