Match the following for the equation x²+a|x|+1=0 where ‘a’ is parameter

COLUMN-I

COLUMN-II
a) No real roots p) a<-2
b) Two real roots q) $ϕ$
c) Three real roots r) a=-2
d) Four distinct real roots s) $a \geq 0$

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$a \to q, b \to p, c \to r, d \to s$

$a \to s, b \to r, c \to q, d \to p$

$a \to p, b \to r, c \to q, d \to s$

$a \to r, b \to s, c \to q, d \to p$

answer is B.

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Detailed Solution

$a \geq 0 \Rightarrow x^{2} + a | x | + 1 = 0 \begin{array}{l} \Rightarrow 3 terms are positive \forall x \in R \\ \Rightarrow No Real Roots \\ If a=-2 \Rightarrow x^{2} -2|x|+1=0 \\ {(|x|-1)}^{2} = 0 \Rightarrow x = \pm 1 \\ \Rightarrow Two real roots \\ If a<-2 then roots |x|= \frac{-a \pm \sqrt{a^{2} -4}}{2} > 0 \\ \Rightarrow F o u r real roots \end{array} T h e e q u a t i o n h a s n o 3 r e a l r o o t s f o r a n y v a l u e o f a$