A) \[\pm \,2\]
B) 1, 2
C) 2, 3
D) \[\pm \,1\]
Correct Answer: D
Solution :
For \[{{n}^{r}}-tn+\frac{1}{4}\]to be a perfect square,\[r=2\]and \[t=\pm \,1\] Look: \[{{n}^{2}}-n+\frac{1}{4}={{n}^{2}}-2.n.\frac{1}{2}+\frac{1}{4}={{\left( n-\frac{1}{2} \right)}^{2}}\] \[{{n}^{2}}+n+\frac{1}{4}={{n}^{2}}+2.n.\frac{1}{2}+\frac{1}{4}={{\left( n+\frac{1}{2} \right)}^{2}}\]You need to login to perform this action.
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