We must have $2a^2+a+1 > 3a^2-4a+1$ $\implies 0>a^2-5a$ $\implies 0>a(a-5)$ $\implies a$ and $a-5$ have opposite signs. Thus $a \in (0,5)$

adityaguharoy wrote: We must have $2a^2+a+1 > 3a^2-4a+1$ $\implies 0>a^2-5a$ $\implies 0>a(a-5)$ $\implies a$ and $a-5$ have opposite signs. Thus $a \in (0,5)$ This is the first interval. 2)For domain we must have $2a^2+a+1>0$ which is true because $\Delta <0$ $\implies$ $a\in (-\infty ,+\infty )$ 3)For domain we must have $3a^2-4a+1>0 \iff 3(a-1)(a-\frac 13)>0 \iff $ $a\in (-\infty ,\frac 13)\cup (1,+\infty )$ So the answer is$a\in (0,5)\cap (-\infty ,+\infty )\cap ((-\infty ,\frac 13)\cup (1,+\infty ))$ so $\boxed{a\in (0,\frac 13)\cup (1,5)}$

nicky-glass wrote: adityaguharoy wrote: We must have $2a^2+a+1 > 3a^2-4a+1$ $\implies 0>a^2-5a$ $\implies 0>a(a-5)$ $\implies a$ and $a-5$ have opposite signs. Thus $a \in (0,5)$ This is the first interval. 2)For domain we must have $2a^2+a+1>0$ which is true because $\Delta <0$ $\implies$ $a\in (-\infty ,+\infty )$ 3)For domain we must have $3a^2-4a+1>0 \iff 3(a-1)(a-\frac 13)>0 \iff $ $a\in (-\infty ,\frac 13)\cup (1,+\infty )$ So the answer is$a\in (0,5)\cap (-\infty ,+\infty )\cap ((-\infty ,\frac 13)\cup (1,+\infty ))$ so $\boxed{a\in (0,\frac 13)\cup (1,5)}$ Yes thank you.