A) rational
B) irrational of the from \[\sqrt{P}\]
C) irrational of the from \[\frac{\sqrt{P}-1}{4}\]when p is an odd integer
D) irrational of the from \[\frac{\sqrt{P}+1}{4}\] where p is an even integer
Correct Answer: C
Solution :
Given \[\sin \,\pi \,({{x}^{2}}+x)-\sin \,\pi {{x}^{2}}=0\] \[\Rightarrow \] \[2\,\cos \,\pi \left( \frac{2{{x}^{2}}+x}{2} \right)\,\sin \frac{\pi x}{2}=0\] \[\Rightarrow \] \[n\left( \frac{2{{x}^{2}}+x}{2} \right)=2n\pi \pm \frac{\pi }{2}\] \[\Rightarrow \] \[2{{x}^{2}}+x=4n\pm 1\] \[\Rightarrow \] \[2{{x}^{2}}+x-p=0,\] where \[p=4n\pm 1\] is an odd integer \[\Rightarrow \] \[2{{x}^{2}}+x-p=0\] \[\Rightarrow \] \[x=\frac{-1\pm \sqrt{p}}{4}\Rightarrow x=\frac{\sqrt{p}-1}{4}\] \[\left( neglect\,\,x=\frac{-1-\sqrt{p}}{4} \right)\]
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