What is the least possible value of \(\frac{9}{1 + 2x^2}\) if 0 \(\geq\) x \(\geq\) 2?

A.

9

B.

5

C.

1

D.

2

Correct answer is C

0 \(\geq\) x \(\geq\) 2 \(\to\) 0, 1, 2

If x = 0, \(\frac{9}{1 + 2x^2}\)

\(\frac{9}{1 + 2(0)^2}\) = \(\frac{9}{1}\)

= 3

If x = 2, \(\frac{9}{1 + 2(1)^2}\)

= \(\frac{9}{3}\)

= 3

If x = 2, \(\frac{9}{1 + 2(2)^2}\)

= \(\frac{9}{9}\)

= 1

The least value of \(\frac{9}{1 + 2x^2}\) is 1 when x = 2