Statement-1: If roots of the equation $x^{2} - b x + c = 0$ are two consecutive integers then $b^{2} - 4 c = 1$ .
Statement-2: If a,b,c are odd integer, then the roots of the equation $4 a b c x^{2} + (b^{2} - 4 a c) x - b = 0$ are real and distinct.

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Statement-1 is True, Statement-2 is False

Statement-1 is True, Statement-2 is Ture; Statement-2 is Not a correct explanation for Statement-1

Statement-1 is False, Statement-2 is True

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

answer is B.

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Detailed Solution

Given that the difference between roots is 1 $\Rightarrow$ $b^{2} - 4 c = 1$

The given equation is $4 a b c x^{2} + (b^{2} - 4 a c) x - b = 0$

$\begin{array}{rcl} {(b^{2} - 4 a c)}^{2} + 16 a b^{2} c & = & b^{4} + 16 a^{2} c^{2} - 8 a b^{2} c + 16 a b^{2} c \\ = & b^{4} + 16 a^{2} c^{2} + 8 a b^{2} c \\ = & {(b^{2} + 4 a c)}^{2} \end{array}$

This is a perfect square, hence the roots are real and distinct.

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