If x+y+z=4 and x²+y²+z²=6, then exhaustive set of values of x is
- A. [2/3, 2]
- B. [0, 2/3]
- C. [0, 2]
- D. [-1/3, 2/3]
✅ Correct Answer: [2/3, 2]
Explanation
Using $(x+y+z)^2 = x²+y²+z² + 2(xy+yz+zx)$, get $xy+yz+zx=5$. Solving gives x in [2/3,2].
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