| Directions: In each question below one or more equation (s) is / are provided. On the basis of these you have to find out relation between p, q and give answer. [SBI (PO) 2000] |
| I. \[6{{q}^{2}}+\frac{1}{2}=\frac{7}{2}q\] |
| II. \[12{{p}^{2}}+2=10p\] |
A) If \[p=q\]
B) If \[p>q\]
C) If \[p<q\]
D) If \[p\ge q\]
Correct Answer: D
Solution :
| I. \[6{{q}^{2}}+\frac{1}{2}=\frac{7}{2}q\]\[\Rightarrow \]\[\frac{12{{q}^{2}}+1}{2}=\frac{7}{2}q\] |
| \[\Rightarrow \] \[12{{q}^{2}}+1=7q\] |
| \[\Rightarrow \] \[12{{q}^{2}}-7q+1=0\] |
| \[\Rightarrow \]\[12{{q}^{2}}-3q-4q+1=0\] |
| \[\Rightarrow \]\[3q\,\,(4q-1)-(4q-1)=0\] |
| \[\Rightarrow \] \[(4q-1)(3q-1)=0\]\[\Rightarrow \]\[q=\frac{1}{3},\]\[\frac{1}{4}\] |
| II. \[12{{p}^{2}}+2-10p=0\] |
| \[\Rightarrow \]\[12{{p}^{2}}-10p+2=0\] |
| \[\Rightarrow \]\[12{{p}^{2}}-6p-4p+2=0\] |
| \[\Rightarrow \]\[6p\,\,(2p-1)(2p-1)=0\] |
| \[\Rightarrow \]\[(6p-2)(2p-1)=0\] |
| \[\Rightarrow \] \[p=\frac{1}{2},\]\[\frac{1}{3}\] |
| \[\therefore \]\[p\ge q\] |
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