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Monotonically Increasing Functions

Let fx be a...

Question

Let $f (x)$ be an increasing function defined on $(0, \infty)$ . If $f (2 a^{2} + a + 1) > f (3 a^{2} - 4 a + 1)$ , then the possible integral value (s) of $a$ is/are

A

1

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B

5

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C

3

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D

6

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Solution

The correct option is B $1$
$\begin{matrix} f (2 a^{2} + a + 1) > f (3 a^{2} - 4 a + 1) = 2 a^{2} + a + 1 > 3 a^{2} - 4 a + 1 = - a^{2} + 5 a > 0 = a^{2} - 5 a < 0 = a (a - 5) < 0 = a \in (0, 5) a = 1, 3 \end{matrix}$
therefore $a = 1$ or $a = 3$

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