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

$\because$‘f’ is defined on $(0,\infty )$

$\therefore 2a^2+a+1>0$which is true as $D<0$

Also$3a^2-4a+1>0$

$(3a-1)(a-1)>0$

$a<\frac{1}{3}ora>1$

As ‘f’ is increasing hence

$f(2a^2+a+1)>f(3a^2-4a+1)$

$2a^2+a+1>3a^2-4a+1$

$0>a^2-5a$

$a^2-5a<0$

$0<a<5$

Possible integers are 2, 3, 4.