Question

For the function $f\left(x\right)=\sqrt{4x^2-4x+2}+|x|$ , which of the following holds good

• A. The least value of the function is $\frac{\sqrt{3}+1}{2\sqrt{2}}$
  
• B. The least possible value of function is $\frac{\sqrt{3}+1}{\sqrt{2}}$
• C. The function takes its least value at $x=\frac{3-\sqrt{3}}{6}$
• D. Function has exactly one point of minima

Answer 1

Answer: (C), (D)

The correct options are
C The function takes its least value at $x=\frac{3-\sqrt{3}}{6}$
D Function has exactly one point of minima

$y=\sqrt{4x^2-4x+2}+|x|$
=$\sqrt{(2x-1{)}^2+1}+|x|$
if $x<0$
$y=\sqrt{(2x-1{)}^2+1}-x$
$y^{\prime }=\frac{(2x-1)\cdot 2}{\sqrt{(2x-1{)}^2}+1}-1\Rightarrow y^{\prime }<0\forall x<0$
if $x>0$
$y=\sqrt{(2x-1{)}^2+1}+x$
$y^{\prime }=\frac{2(2x-1)}{\sqrt{(2x-1{)}^2+1}}+1$
$\Rightarrow y^{\prime }>0\Rightarrow \frac{\sqrt{3}-1}{2\sqrt{3}}<x<\infty$
$y^{\prime }<0\Rightarrow 0<x<\frac{\sqrt{3}-1}{2\sqrt{3}}$


minima at $x=\frac{\sqrt{3}-1}{2\sqrt{3}}$