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SSC Sample Paper SSC (Group-C) Sample Test Paper-2

  • question_answer
    If \[\left( {{n}^{r}}-t\,n+\frac{1}{4} \right)\] be a perfect square, then the values of t are:

    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: \[{{n}^{2}}-n+\frac{1}{4}={{n}^{2}}-2.n.\frac{1}{2}+\frac{1}{4}={{\left( n-\frac{1}{2} \right)}^{2}}\] \[{{n}^{2}}+n+\frac{1}{4}={{n}^{2}}+2.n.\frac{1}{2}+\frac{1}{4}={{\left( n+\frac{1}{2} \right)}^{2}}\]


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