\( \Large x^{2}-32=112; y-\sqrt{169}=0 \)

A) If x>y

B) \( \Large If x\ge y \)

C) lfx

D) If x = y or relation cannot be established

Correct answer:
D) If x = y or relation cannot be established

Description for Correct answer:

\( \Large x^{2} - 32 = 112 \)

=> \( \Large x^{2} = 112 + 32 => x^{2} = 144 \)

Therefore, x = \( \Large \pm \) 12

and \( \Large y - \sqrt{169} = 0 \)

=>\( \Large y = \sqrt{169} => y = \pm 13 \)

Therefore, Relation cannot be established



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