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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}$$,k].The value of k is \begin{array}{1 1}(A)\;1&(B)\;2\\(C)\;3&(D)\;\text{None}\end{array} Can you answer this question? ## 1 Answer 0 votes x+y=4-z x^2+y^2=6-z^2 xy=(x+y)^2-(x^2+y^2) \Rightarrow (4-z)^2-(6-z^2) \Rightarrow 2z^2-8z+10 The quadratic equation whose roots are x & y are t^2-(x+y)t+xy=0 \Rightarrow t^2-(4-z)t+(2z^2-8z+10)=0 Since x and y are real \therefore D \geq 0 \Rightarrow (4-z)^2-4(2z^2-8z+10)\geq 0 \Rightarrow 3z^2-8z+4\leq 0 \Rightarrow (3z-2)(z-2)\leq 0 \Rightarrow z\in [\large\frac{2}{3},$$2]$
Hence k=2
Hence (B) is the correct answer.