A) rational
B) irrational of the form \[\sqrt{p}\]
C) irrational of the form \[\frac{\sqrt{p}-1}{4},\] where p is an odd integer
D) irrational of the form \[\frac{\sqrt{p}+1}{4},\] where p is an even integer
Correct Answer: A
Solution :
We have, \[\sin \pi ({{x}^{2}}+x)=\sin \pi {{x}^{2}}\] \[\Rightarrow \,\,\pi ({{x}^{2}}+x)=n\pi +{{(-1)}^{n}}\pi {{x}^{2}}\] \[\therefore \] Either \[{{x}^{2}}+x=2m+{{x}^{2}}\Rightarrow x=2m\in I\] or \[{{x}^{2}}+x=k-{{x}^{2}},\] where k is an odd integer \[\Rightarrow \,\,2{{x}^{2}}+x-k=0\Rightarrow x=\frac{-1\pm \sqrt{1+8k}}{4}\] For least positive non-integral solution is \[x=\frac{1}{2},\] when \[k=1\]
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