Question

Consider the equation $x^2+a|x|+1=0$, where $a$ is a parameter.

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{No real roots}& \text{(P)}& a<-2\\ \text{(B)}& \text{Two real roots}& \text{(Q)}& \varphi \\ \text{(C)}& \text{Three real roots}& \text{(R)}& a=-2\\ \text{(D)}& \text{Four distinct real roots}& \text{(S)}& a\ge 0\\ & & \text{(T)}& a=2\end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(A)}\to \text{(P)}$
• B. $\text{(A)}\to \text{(Q)}$
• C. $\text{(B)}\to \text{(T)}$
• D. $\text{(B)}\to \text{(R)}$

Answer 1

The correct option is D $\text{(B)}\to \text{(R)}$

Since $|x{|}^2=x^2,\forall x\in R$
$x^2+a|x|+1=0=|x{|}^2+a|x|+1\cdots \left(1\right)$
Let $|x|=t$
Then, equation $\left(1\right)$ becomes
$t^2+at+1=0\cdots \left(2\right)$

$\left(A\right)$
$|x{|}^2+a|x|+1=0$
Clearly, if $a\ge 0$, then there are no real roots as all the terms in L.H.S. will be positive.

$\left(B\right)$
$t=0$ is not a root of equation $\left(2\right)$.
So, for two real roots of equation $\left(1\right)$,
equation $\left(2\right)$ should have equal positive roots.
$\Rightarrow D=a^2-4=0$ and $\frac{-a}{2}>0$
$\Rightarrow a=\pm 2$ and $a<0$
$\Rightarrow a=-2$

$\left(C\right)$
For three real roots of equation $\left(1\right)$, one root of equation $\left(2\right)$ should be $0$ and other should be positive.
But $t=0$ is not a root of equation $\left(2\right)$.
$\therefore$ equation $\left(1\right)$ cannot have three real roots for any value of $a$.

$\left(D\right)$
For four distinct real roots of equation $\left(1\right)$, equation $\left(2\right)$ should have two positive real solutions in $t$.
So, $\frac{-a}{2}>0,$ ($x$-coordinate of vertex)
and $a^2-4>0$ $(D>0)$
$\Rightarrow a<0,$ and $a\in (-\infty ,-2)\cup (2,\infty )$
$\Rightarrow a<-2$

$\left(A\right)\to \left(S\right),\left(B\right)\to \left(R\right),\left(C\right)\to \left(Q\right),\left(D\right)\to \left(P\right)$