Question

Consider the equation $x^2+a|x|+1=0$, where $a$ is a parameter.
$$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{a. No real roots}& \text{p.}a<-2\\ \text{b. Two real roots}& \text{q.}\varphi \\ \text{c. Three real roots}& \text{r.}a=-2\\ \text{a. Four distinct real roots}& \text{s.}a\ge 0\end{array}$$
Which of the following is the CORRECT combination ?

• A. $a\to s,b\to r,c\to q,d\to p$
• B. $a\to p,b\to r,c\to q,d\to s$
• C. $a\to r,b\to s,c\to p,d\to q$
• D. $a\to s,b\to p,c\to q,d\to r$

Answer 1

The correct option is A $a\to s,b\to r,c\to q,d\to p$

$a.$ $|x{|}^2+a|x|+1=0$
Clearly, if $a\ge 0$, then there are no real roots as all the terms in LHS are positive.

$b.$ For $a=-2$,
$|x{|}^2-2|x|+1=0$
$\Rightarrow (|x|-1{)}^2=0$
$\Rightarrow |x|=1$
$\Rightarrow x=\pm 1$
Therefore, for $a=-2$, the equation has two real roots.

$c.$ $|x|=\frac{-a\pm \sqrt{a^2-4}}{2}$
Obviously, the equation does not have three real roots for any value of $a$

$d.$ If $a<-2$,
then $a^2-4>0$ and $|-a|>\sqrt{a^2-4}$
$\Rightarrow |x|=\frac{-a\pm \sqrt{a^2-4}}{2}>0$
Hence, the equation has four real roots if $a<-2$