Question

If \(\left(3x - \frac{1}{4}\right)^{\frac{1}{2}} > \frac{1}{4} - x \), 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 > \(\frac{9}{16}\)

Answer 1

The correct answer is E. x > \(\frac{9}{16}\)

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

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

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

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

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