A) \[\underline{+}\,\frac{1}{3}\]
B) \[\underline{+}\,\frac{1}{4}\]
C) \[\underline{+}\,\frac{1}{8}\]
D) \[\underline{+}\,\frac{1}{2}\]
Correct Answer: C
Solution :
\[p+\frac{1}{4}\sqrt{p}+{{k}^{2}}\] \[={{(\sqrt{p})}^{2}}+2\sqrt{p.}\frac{1}{8}+{{\left( \frac{1}{8} \right)}^{2}}-{{\left( \frac{1}{8} \right)}^{2}}+{{k}^{2}}\] \[\Rightarrow \] \[{{k}^{2}}={{\left( \frac{1}{8} \right)}^{2}}\Rightarrow k=\underline{+}\frac{1}{8}\]
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