Question

The value of $k$ for which the equation $|x-2|+|x-6|-|x+1|=k$ has atleast one solution

• A. $(-\infty ,-1)$
• B. $[-3,\infty )$
• C. $(-\infty ,3)$
• D. $(1,\infty )$

Answer 1

The correct option is B $[-3,\infty )$

$|x-2|+|x-6|-|x+1|=k$
Here, the turning points are $-1,2,6$

Let $f\left(x\right)=|x-2|+|x-6|-|x+1|$
$f\left(x\right)=\{\begin{array}{cc}-(x-2)-(x-6)+(x+1),& x<-1\\ -(x-2)-(x-6)-(x+1),& -1\le x<2\\ (x-2)-(x-6)-(x+1),& 2\le x<6\\ (x-2)+(x-6)-(x+1),& x\ge 6\end{array}$

$\Rightarrow f\left(x\right)=\{\begin{array}{cc}9-x,& x<-1\\ 7-3x,& -1\le x<2\\ 3-x,& 2\le x<6\\ x-9,& x\ge 6\end{array}$

Graph will be

For atleast one solution $k\in [-3,\infty )$