If $x^{2} + y^{2} + z^{2} = 1$ , then $y z + z x + x y$ lies in the interval

[\frac{1}{2}, 2]

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[- 1, 2]

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[- \frac{1}{2}, 1]

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[- 1, \frac{1}{2}]

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Solution

The correct option is C $[- \frac{1}{2}, 1]$
$x^{2} + y^{2} + z^{2} = 1 \dots [1]$
$(x - y)^{2} + (y - z)^{2} + (z - x)^{2} = 2 (x^{2} + y^{2} + z^{2} - x y - y z - z x)$
$(x - y)^{2} + (y - z)^{2} + (z - x)^{2} = 2 [1 - (x y + y z + z x)]$
$x y + y z + z x = 1 - \frac{1}{2} [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]$
$x y + y z + z x \leq 1 \dots [2]$
As $[(x - y)^{2} + (y - z)^{2} + (z - x)^{2} \geq 0]$

Again,
$(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x)$
$\Rightarrow (x + y + z)^{2} = 1 + 2 (x y + y z + z x)$
$\Rightarrow x y + y z + z x = \frac{1}{2} [(x + y + z)^{2} - 1]$
$\Rightarrow x y + y z + z x \geq - \frac{1}{2} \dots [3]$
$[∵ (x + y + z)^{2} \geq 0]$

From $[2]$ and $[3]$ ,
$x y + y z + z x \in [- \frac{1}{2}, 1]$

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