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Given $x+y+z=4$ and $x^2+y^2+z^2=6$.The range of z is $[\large\frac{2}{3}$$,k]$.The value of k is

$\begin{array}{1 1}(A)\;1&(B)\;2\\(C)\;3&(D)\;\text{None}\end{array} $

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1 Answer

$x+y=4-z$
$x^2+y^2=6-z^2$
$xy=(x+y)^2-(x^2+y^2)$
$\Rightarrow (4-z)^2-(6-z^2)$
$\Rightarrow 2z^2-8z+10$
The quadratic equation whose roots are x & y are
$t^2-(x+y)t+xy=0$
$\Rightarrow t^2-(4-z)t+(2z^2-8z+10)=0$
Since x and y are real
$\therefore D \geq 0$
$\Rightarrow (4-z)^2-4(2z^2-8z+10)\geq 0$
$\Rightarrow 3z^2-8z+4\leq 0$
$\Rightarrow (3z-2)(z-2)\leq 0$
$\Rightarrow z\in [\large\frac{2}{3},$$2]$
Hence k=2
Hence (B) is the correct answer.
answered Apr 16, 2014 by sreemathi.v
 

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