Question

Consider the equation $x^2-4|x|+3-|k-1|=0$, where $k$ is a real parameter.
$$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(A)}& \text{No real roots}& \text{(P)}& k\in (-2,4)\\ \text{(B)}& \text{Two distinct real roots}& \text{(Q)}& k\in \{-2,4\}\\ \text{(C)}& \text{Three distinct real roots}& \text{(R)}& k\in (-\infty ,-2)\cup (4,\infty )\\ \text{(D)}& \text{Four distinct real roots}& \text{(S)}& k\in \varphi \\ & & \text{(T)}& k\in [-2,4]\\ & & \text{(U)}& k\in (-\infty ,-2]\cup [4,\infty )\end{array}$$
Which of the following is the only CORRECT combination?

• A. $\left(C\right)\to \left(Q\right)$
• B. $\left(D\right)\to \left(S\right)$
• C. $\left(B\right)\to \left(T\right)$
• D. $\left(A\right)\to \left(U\right)$

Answer 1

The correct option is A $\left(C\right)\to \left(Q\right)$

$x^2-4|x|+3-|k-1|=0\cdots \left(1\right)$
Let $t=|x|\Rightarrow t^2=|x{|}^2=x^2$
So, equation ($1$) reduces to$t^2-4t+3-|k-1|=0\cdots \left(2\right)$
Let the roots of the equation $\left(2\right)$ be $t_1,t_2$
Now, as $t=|x|$
(i) For every $t>0$, there exist $2$ possible values of $x.$
(ii) For every $t<0$, there exists no value of $x.$

$\text{(A)}$ For equation $\left(1\right)$ to have no real roots, both the roots of equation $\left(2\right)$ should be $<0$ or non-real (i.e., $D<0$)
Both roots $t_1,t_2<0$
Required conditions are
$\left(i\right)D\ge 0\Rightarrow 16-4(3-|k-1|)\ge 0\Rightarrow 4+|k-1|\ge 0\Rightarrow k\in R\left(ii\right)-\frac{b}{2a}<0\Rightarrow \frac{4}{2}<0$
which is not possible.
So, $k\in \varphi$
$\left(A\right)\to \left(S\right)$

$\text{(B)}$ For two distinct real roots,
Case $1:$ One root of equation $\left(2\right)$ is positive and other root is negative
So, $0$ lies in between the roots.
$f\left(0\right)<0\Rightarrow 3-|k-1|<0\Rightarrow |k-1|>3\Rightarrow k-1>3,k-1<-3\Rightarrow k>4,k<-2\Rightarrow k\in (-\infty ,-2)\cup (4,\infty )$
Case $2:$ Equation $2$ have a positive repeated root
So, $D=0,-\frac{b}{2a}>0$
$D=0\Rightarrow 16-4(3-|k-1|)=0\Rightarrow 4+|k-1|=0$
Not possible.
Therefore, from both the cases,
$k\in (-\infty ,-2)\cup (4,\infty )$
$\left(B\right)\to \left(R\right)$

$\text{(C)}$ For three distinct real roots,
one root of equation $\left(2\right)$ should be $0$ and another root should be positive.
$\left(i\right)f\left(0\right)=0\Rightarrow 3-|k-1|=0\Rightarrow |k-1|=3\Rightarrow k=-2,4\left(ii\right)-\frac{b}{2a}>0\Rightarrow \frac{4}{2}>0\Rightarrow k\in R\therefore k\in \{-2,4\}$
$\left(C\right)\to \left(Q\right)$

$\text{(D)}$ For four distinct real roots,
both roots of the equation $\left(2\right)$ should be positive.
$\left(i\right)D>0\Rightarrow 16-4(3-|k-1|)>0\Rightarrow 4+|k-1|>0\Rightarrow k\in R\left(ii\right)-\frac{b}{2a}>0\Rightarrow k\in R\left(iii\right)f\left(0\right)>0\Rightarrow 3-|k-1|>0\Rightarrow |k-1|<3\Rightarrow k\in (-2,4)$
$\left(D\right)\to \left(P\right)$

Hence, the correct pairs are,
$\left(A\right)\to \left(S\right)$
$\left(B\right)\to \left(R\right)$
$\left(C\right)\to \left(Q\right)$
$\left(D\right)\to \left(P\right)$