Question

If $g\left(x\right)=k$ has exactly two solutions then k belong to

• A. $(-2,-1)$
• B. $(-1,0)$
• C. $(-3,-2)$
• D. None of these
  

Answer 1

The correct option is D None of these

Let $f\left(x\right)=u\left(x\right)-2v\left(x\right)$
where $u\left(x\right)=min(x^2,\left|x\right|);\left|x\right|\le 1max(x^2,\left|x\right|);\left|x\right|>1$
from the figure
Clearly $u\left(x\right)=x^2$
$v\left(x\right)=min(x^2,\left|x\right|);\left|x\right|>1max(x^2,\left|x\right|);\left|x\right|\le 1$
Clearly, $v\left(x\right)=|x|$
and $g\left(x\right)=minf\left(t\right);-3\le t\le x;-3\le x\le 0maxf\left(t\right);0\le t\le x;0\le x\le 3$
$f\left(x\right)=u\left(x\right)-2v\left(x\right)=x^2-2|x|$
$g\left(x\right)=min\left(f\right(t\left)\right)=t^2-2|t|=t^2+2t,$ if $-3<t<x$
$g\left(x\right)=x^2+2x$ if x < -1, else $g\left(x\right)=-1$ for x>-1
$g\left(x\right)=max\left(f\right(t\left)\right)=t^2-2|t|=t^2-2t,$ if $0<t<x$
$g\left(x\right)=0$ if x<2, else $g\left(x\right)=x^2-2x$ for x>2