| Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer. |
| I. \[\frac{18}{{{x}^{2}}}+\frac{6}{x}-\frac{12}{{{x}^{2}}}=\frac{8}{{{x}^{2}}}\] |
| II. \[{{y}^{3}}+9.68+5.64=16.95\] |
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: C
Solution :
| I. \[\frac{18}{{{x}^{2}}}+\frac{6}{x}-\frac{12}{{{x}^{2}}}=\frac{8}{{{x}^{2}}}\]\[\Rightarrow \]\[\frac{6}{{{x}^{2}}}+\frac{6}{x}=\frac{8}{{{x}^{2}}}\] |
| \[\Rightarrow \]\[\frac{6}{x}=\frac{2}{{{x}^{2}}}\]\[\Rightarrow \]\[6{{x}^{2}}=2x\] |
| \[\Rightarrow \]\[2x\,\,(1-3x)=0\]\[\Rightarrow \]\[x=0,\]\[\frac{1}{3}\] |
| II. \[{{y}^{3}}=16.95-(9.68+5.64=16.95-15.32\] |
| \[\Rightarrow \]\[{{y}^{3}}=1.63\]\[\Rightarrow \]\[y=1.176\] |
| \[\therefore \]\[y>x\] |
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