Question

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 integral value(s) of $a$ is/are

• A. $1$
• B. $5$
• C. $3$
• D. $6$

Answer 1

The correct option is C $3$

Increasing function $⟹2a^2+a+1>3a^2-4a+1⟹a^2-5a<0⟹a\in (0,5)⟹a=1,2,3,4$ are the possible integral values.

From domain conditions $2a^2+a+1>0;3a^2-4a+1>0$ we can say that first satisfies but the second one makes it $a\in (0,\frac{1}{3})\bigcup (1,\infty )$.

So possible integral values are $a=2,3,4$