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Match the following columns:
Column-I Column-II
a) If a, b, c and d are 4 non-zero real numbers such that ${(d + a - b)}^{2} + {(d + b - c)}^{2} = 0$ and the roots of the equation $a (b - c) x^{2} + b (c - a) x + c (a - b) = 0$ are real and equal, then p) $a + b + c = 0$
b) If the roots of the equation $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0$ are real and equal, then q) a,b,c are in A.P
c) If the equation $a x^{2} + b x + c = 0$ and $x^{3} - 3 x^{2} + 3 x - 1 = 0$ have a common real root, then r) a,b,c are in G.P
d) d) Let a, b, c be positive real numbers such that the expression $b x^{2} + (\sqrt{{(a + c)}^{2} + 4 b^{2}}) x + (a + c)$ is non-negative, $\forall x \in R,$ then s) a,b,c are in H.P

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a – p, b – r, c – q, d – s

a – q,r, b – r, c – p, d – q

a – q, b – s, c – q, d – p

a – q,r, b – r, c – q, d – p

answer is B.

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Detailed Solution

a) $d + a - b = 0 = d + b - c \Rightarrow 2 b = a + c$
$x = 1$ is root and equal roots, POR=1
$\Rightarrow \frac{c (a - b)}{a (b - c)} = 1 \Rightarrow a c - b c = a b - a c$
$\Rightarrow 2 a c = b (a + c) \Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$
a, b, c are in H.P
b) ${(a x - b)}^{2} + {(b x - c)}^{2} = 0 \Rightarrow x = \frac{b}{a} = \frac{c}{b} \Rightarrow b^{2} = a c$
a, b, c are in G.P
c) $x = 1 \Rightarrow a + b + c = 0$
d) $Δ \leq 0 \Rightarrow {(a + c)}^{2} + 4 b^{2} - 4 b (a + c) \leq 0$
$\Rightarrow {(a + c - 2 b)}^{2} \leq 0 \Rightarrow a + c - 2 b = 0 \Rightarrow a + c = 2 b$
a, b, c are in A.P

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