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Question Number 195535 by arkanshh last updated on 04/Aug/23

	Answered by deleteduser1 last updated on 04/Aug/23

	$\frac{N = A \times B}{D} = \frac{(3^{x^{2}} - 3^{x + 6}) (5^{x} - 25)}{4^{x^{2} (3 - x)} - 4^{3 - x}} ⩾ 0 (E q u a l i t y a t x = 2, - 2)$ $W e u s e t h e f a c t t h a t f o r i n e q u a l i t y t o b e t r u e,$ $t h e n N a n d D m u s t h a v e t h e s a m e s i g n (+ v e o r - v e)$ $N o t i c e t h a t x^{2} (3 - x) < 3 - x w h e n x > 3 (D < 0, N > 0)$ $\Rightarrow x > 3 d o e s n o t s a t i s f y t h e i n e q u a l i t y X$ $x = 3 a l s o d o e s n o t s a t i s f y X$ $W h e n 1 < x < 3, D > 0$ $2 < x < 3 \Rightarrow N < 0 \Rightarrow 2 < x < 3 d o e s n o t s a t i s f y X$ $x = 2 s a t i s f i e s ✓$ $1 < x < 2 \Rightarrow N > 0 \Rightarrow 1 < x ⩽ 2 s a t i s f i e s ✓$ $x = 1 d o e s n o t s a t i s f y X$ $W h e n - 1 < x < 1, D < 0 \land N < 0$ $\Rightarrow - 1 < x < 1 s a t i s f y ✓; x = - 1 d o e s n o t s a t i s f y X$ $W h e n - 2 < x < - 1; D > 0, B < 0, A < 0 \Rightarrow N > 0 ✓$ $W h e n x < - 2, D > 0, N < 0 X$ $x \in [- 2, - 1) \cup (- 1, 1) \cup (1, 2]$
	Commented by arkanshh last updated on 04/Aug/23

	$w o o w!$ $w o n d e r f u l l s o l . t h n x f o r y o u r t i m e .$
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