Statement-1 : If $a, b, c > 0$ and the expression
$(a + c) x^{2} + (\sqrt{(a + c)^{2} + 4 b^{2}}) x + b > 0$
$\forall x \in R$ and then $a, b, c$ are in G.P.
Statement-2 : $: a x^{2} + b x + c > 0 \forall x \in R$ R if and only
If $a > 0$ and $b^{2} - 4 a c < 0$

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STATEMENT-1 is True, STATEMENT-2 is True; 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

STATEMENT-1 is True, STATEMENT-2 is False

answer is D.

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Detailed Solution

As $a + c > 0$ we must have

$\begin{array}{l} (a + c)^{2} + 4 b^{2} - 4 (a + c) b < 0 \\ \Rightarrow (2 b - a - c)^{2} < 0 \end{array}$

This is not possible as

$2 b - a - c \in R$

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