A) If x>y |
B) \( \Large If \ x\ge y \) |
C) lfx |
D) If x = y or relation cannot be established |
D) If x = y or relation cannot be established |
\( \Large \frac{4}{\sqrt{x}}+\frac{7}{\sqrt{x}}=\sqrt{x} = \frac{11}{\sqrt{x}} = \sqrt{x}\)
Therefore, x = 11
and \( \Large y^{2} - \frac{ \left(11\right)^{5/2} }{\sqrt{y}} = 0 = y^{2} = \frac{ \left(11\right)^{5/2} }{ \left(y\right)^{1/2} } \)
= \( \Large y^{2} \times y^{1/2} = \left(11\right)^{5/2} \)
= \( \Large \left(y\right)^{5/2} = \left(11\right)^{5/2} \)
Therefore, y = 11
Therefore, x = y