JAMB - Mathematics (1982 - No. 11)
What is the least possible value of \(\frac{9}{1 + 2x^2}\) if 0 \(\geq\) x \(\geq\) 2?
9
5
1
2
Explanation
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
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
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