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Question Number 96712 by john santu last updated on 04/Jun/20

	Answered by bobhans last updated on 04/Jun/20
	$since f(x) is defined on (0,+∞), from { ((2a^2 +a+1=2(a+(1/4))^2 +(7/8)>0)),((3a^2 −4a+1=(3a−1)(a−1)>0)) :} we get a<(1/3) ∪ a>1...(i) since f(x) is a decreasing function on (0, +∞) so 2a^2 +a+1 >3a^2 −4a+1 ⇒a^2 −5a < 0 , thus 0<a<5...(ii) combining this with (i) we have 0<a<(1/3) ∪ 1<a<5$
	$since f (x) is defined on (0, + \infty), from$ ${\begin{cases} 2 a^{2} + a + 1 = 2 {(a + \frac{1}{4})}^{2} + \frac{7}{8} > 0 \\ 3 a^{2} - 4 a + 1 = (3 a - 1) (a - 1) > 0 \end{cases}$ $we get a < \frac{1}{3} \cup a > 1 . . . (i)$ $s i n c e f (x) i s a d e c r e a s i n g f u n c t i o n$ $o n (0, + \infty) so 2 a^{2} + a + 1 > 3 a^{2} - 4 a + 1$ $\Rightarrow a^{2} - 5 a < 0, thus 0 < a < 5 . . . (i i)$ $c o m b i n i n g t h i s w i t h (i) we have$ $0 < a < \frac{1}{3} \cup 1 < a < 5$
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