Question

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