| Directions: In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer. [IBPS (RRB) Grade A 2012] |
| I. \[(6{{x}^{2}}+17)-(3{{x}^{2}}+20)=0\] |
| II. \[(5{{y}^{2}}-12)-(9{{y}^{2}}-16)=0\] |
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. \[(6{{x}^{2}}+17)-(3{{x}^{2}}+20)=0\] |
| \[\Rightarrow \]\[6{{x}^{2}}-3{{x}^{2}}+17-20=0\] |
| \[\Rightarrow \]\[3{{x}^{2}}-3=0\] |
| \[\Rightarrow \]\[{{x}^{2}}=1\]\[\Rightarrow \]\[x=\pm \,\,1\] |
| II. \[(5{{y}^{2}}-12)-(9{{y}^{2}}-16)=0\] |
| \[\Rightarrow \]\[5{{y}^{2}}-9{{y}^{2}}-12+16=0\] |
| \[\Rightarrow \]\[-\,\,4{{y}^{2}}+4=0\] |
| \[\Rightarrow \]\[{{y}^{2}}=1\]\[\Rightarrow \]\[y=\pm \,\,1\] |
| Hence, \[x=y\] |
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