A) \[\sqrt{5}-1\]
B) \[\sqrt{3}\]
C) 2
D) none of these
Correct Answer: C
Solution :
| \[\,x+y=4-z,{{x}^{2}}+{{y}^{2}}=6-{{z}^{2}},\] |
| \[\therefore \]\[2xy={{(x+y)}^{2}}-({{x}^{2}}+{{y}^{2}})\] |
| \[={{\left( 4-z \right)}^{2}}-\left( 6-{{z}^{2}} \right)=2{{z}^{2}}-8z+10\] |
| The quadratic equation whose roots are x and y is \[{{\operatorname{t}}^{2}}-\left( x+y \right)t+xy=0\] or \[{{\operatorname{t}}^{2}}-\left( 4-z \right)t+{{z}^{2}}-4z+5=0\] |
| Since x and y are real \[{{\left( 4-z \right)}^{2}}-4\left( {{z}^{2}}-4z+5 \right)\ge 0\]\[\Leftrightarrow 3{{z}^{2}}-8z+4\le 0\Leftrightarrow \left( 3z-2 \right)\left( z-2 \right)\le 0\]\[\Leftrightarrow 2/3\le z\le 2.\] |
| That z can take value 2 where\[x=y=1\]. |
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