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
Verified Correct Answer: C.
Step-by-Step Mathematical Solution
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
Source: Myschool Past Questions Archive (Record #32491)