Question

If the equation $|x+1|+|x-3|=k$ has exactly two solution then $k$ lies in

• A. $(4,\infty )$
• B. $[4,\infty )$
• C. $R$
• D. $(-\infty ,4)$

Answer 1

The correct option is A $(4,\infty )$

$y=|x+1|+|x-3|$
Here the turning points are $-1,3$
$y=\{\begin{array}{cc}-(x+1)-(x-3)& x<-1\\ (x+1)-(x-3)& -1\le x<3\\ (x+1)+(x-3)& x\ge 3\end{array}\Rightarrow y=\{\begin{array}{cc}2-2x& x<-1\\ 4& -1\le x<3\\ 2x-2& x\ge 3\end{array}$
Graph will be


So when $k>4$, then the given expression has exactly two solutions.
Hence, $k\in (4,\infty )$