If \(3x - \frac{1}{4})^{\frac{1}{2}} > \frac{1}{4} - x \), then the interval of values of x is

A.

x > \(\frac{1}{3}\)

B.

x < \(\frac{1}{3}\)

C.

x < \(\frac{1}{4}\)

D.

x < \(\frac{9}{16}\)

E.

x > \(\frac{9}{16}\)

Correct answer is E

\(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x\)

= \(3x - 4^{\frac{1}{2}} > \frac{1}{4} - x\)

= \(3x - 2 > \frac{1}{4} - x\)

= \(3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}\)

\(x > \frac{9}{16}\)