I. $x^{2} - 13 x + 42 = 0$
II. $4 y^{2} = 121$

if x > y

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if x < y

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if x <= y

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if x = y or relationship between x and y cannot be determined

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Solution

The correct option is A if x > y
$x^{2} - 13 x + 42 = 0 x^{2} - 7 x - 6 x + 42 = 0 x (x - 7) - 6 (x - 7) = 0 (x - 7) (x - 6) = 0 x = 6, 7 4 y^{2} = 121 \Rightarrow 4 y^{2} - 121 = 0 (2 y - 11) (2 y + 11) = 0 y = \frac{- 11}{2}, \frac{11}{2}$
$∴$ x > y

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