Question

For the equation $|x-1|+|x-3|-2|x-2|=\lambda x$, which of the following is/are correct

• A. For $\lambda =1$, then equation has $2$ distinct real solutions.
• B. For $\lambda \in (0,1)$, the equation has $3$ distinct real roots
• C. For $\lambda \in (1,2)$, the equation has $3$ distinct real roots
• D. For $\lambda =0$, the equation has infinite solutions.

Answer 1

The correct option is D For $\lambda =0$, the equation has infinite solutions.

$\text{Solve: Given,}$

$|x-1|+|x-3|-2|x-2|=\lambda x$

For $x⩾3$

$\therefore \lambda x=x-1+x-3-2(x-2)=2x-4-2x+4$

$\{\begin{array}{cc}|x|& =xx⩾0\\ |x|& =-x-x<0\end{array}$

$\Rightarrow \lambda x=0$

for $2⩽x<3$

$\Rightarrow \lambda x=x-1+3-x-2x+4$

$\Rightarrow \lambda x=-2x+6$

for $1⩽x<2$

$\Rightarrow \lambda x=x-1+3-x+2(x-2)$

$=>\lambda x=2x-4+2=2x-2$

and for $x<1$

$\lambda x=1-x+3-x+2(x-2)$

$\Rightarrow \lambda x=4-2x+2x-4=0$

$\lambda x=\{\begin{array}{cc}0& x<1\\ 2x-2& 1⩽x<2\\ -2x+6& 2⩽x<3\\ 0& x⩾3\end{array}$

So, for $\lambda =0$ we have

$\begin{array}{cc}x<1& \text{and}x⩾3\\ x=1& \text{and}x=3\end{array}\}$, It has Infinite

solution.