Determine the value of m for which $$4x+\frac{\sqrt{x}}{6}+\frac{m^2}{4}$$ is a perfect square.
Expression : $$4x+\frac{\sqrt{x}}{6}+\frac{m^2}{4}$$
Let $$\sqrt{x}=y$$
= $$4y^2+\frac{y}{6}+\frac{m^2}{4}$$
= $$(2y)^2+2(2y)(\frac{1}{24})+(\frac{m}{2})^2$$
Using, $$a^2+2ab+b^2=(a+b)^2$$
=> $$\frac{m}{2}=\frac{1}{24}$$
=> $$m=\frac{2}{24}=\frac{1}{12}$$
=> Ans - (B)
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