| Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer. |
| I. \[\frac{3}{\sqrt{x}}+\frac{4}{\sqrt{x}}=\sqrt{x}\] |
| II. \[{{y}^{3}}-\frac{{{(7)}^{7/2}}}{\sqrt{y}}=0\] |
A) If \[x>y\]
B) If \[x\ge y\]
C) If \[x<y\]
D) If \[x\le y\]
E) If x = y or the relationship cannot be established
Correct Answer: E
Solution :
| I. \[\frac{3}{\sqrt{x}}+\frac{4}{\sqrt{x}}=\sqrt{x}\] |
| \[\Rightarrow \] \[\frac{3+4}{\sqrt{x}}=\sqrt{x}\]\[\Rightarrow \]\[x=7\] |
| II. \[{{y}^{3}}-\frac{{{(7)}^{7/2}}}{\sqrt{y}}=0\] |
| \[\Rightarrow \] \[\frac{{{(y)}^{3}}\,\,{{(y)}^{1/2}}-{{(7)}^{7/2}}}{\sqrt{y}}=0\] \[\Rightarrow \] \[{{(y)}^{7/2}}-{{(7)}^{7/2}}=0\] |
| \[\Rightarrow \] \[{{(y)}^{7/2}}={{(7)}^{7/2}}\] |
| \[\Rightarrow \] \[y=7\] |
| Hence, \[x=y\] |
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