Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B), (C), (D)

Since $f$ is defined on $(0, \infty)$
Therefore, $2 a^{2}+a+1>0$
which is true as $D< 0$ also $3 a^{2}-4 a+1>0$
$(3 a-1)(a-1)>0 $
$\Rightarrow a<1 / 3$ or $a >1 \,\,\, ...(i)$
as $f$ is increasing hence
$f\left(2 a^{2}+a+1\right)>f\left(3 a^{2}-4 a+1\right) $
$\Rightarrow 2 a^{2}+a+1>3 a^{2}-4 a+1 $
$\Rightarrow 0>a^{2}-5 a $
$\Rightarrow a(a-5)<0 $
$\Rightarrow (0,5)\,\,\, ...(ii)$
from (i) and (ii), we get
hence, $a \in(0,1 / 3) \cup(1,5)$
Therefore, possible integers are $\{2,3,4\}$.