Question

If $x,y,z$ are real numbers such that $x+y+z=4$ and $x^2+y^2+z^2=6,$ then the range of $x$ is

• A. $[\frac{2}{3},2]$
• B. $(-1,1)$
• C. $[0,2]$
• D. $[2,3]$

Answer 1

The correct option is A $[\frac{2}{3},2]$

Given, $x^2+y^2+z^2=6$ and $x+y+z=4$
Squaring both the sides,
$(x+y+z{)}^2=(4{)}^2$
$\Rightarrow x^2+y^2+z^2+2(xy+yz+zx)=16$
$\Rightarrow 6+2(xy+yz+zx)=16$
$\Rightarrow xy+yz+zx=5$

$y+z=4-x\cdots \left(1\right)$
Using $\left(1\right),$ we get
$y(4-x-y)=5-x(4-x)$
$\Rightarrow 4y-xy-y^2=5-x(4-x)$
$\Rightarrow y^2-(4-x)y+5-x(4-x)=0$

If $y$ is real, then $D\ge 0$
$\Rightarrow (4-x{)}^2-4(5-x(4-x))\ge 0$
$\Rightarrow 16+x^2-8x-20+4x(4-x)\ge 0$
$\Rightarrow 16+x^2-8x-20+16x-4x^2\ge 0$ $\Rightarrow -3x^2+8x-4\ge 0$
$\Rightarrow 3x^2-8x+4\le 0$
$\Rightarrow (3x-2)(x-2)\le 0$
$\Rightarrow x\in [\frac{2}{3},2]$