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Question Number 28211 by ajfour last updated on 22/Jan/18

Commented by math solver last updated on 22/Jan/18

$t h a n k y o u s i r!$
Commented by math solver last updated on 22/Jan/18

$s i r, i t h i n k t h e y a r e r e f e r r i n g$ $m a x . a m p l i t u d e a s m a x . a r g u m e n t .$
Commented by ajfour last updated on 22/Jan/18

$Q . 28189 (s o l u t i o n)$ $t h e b l u e r e g i o n i s A \cap B \cap C .$ $6 p o i n t s w i t h i n t e g r a l c o o r d i n a t e s,$ $i s f i n e b u t p o i n t h a v i n g m a x i m u m$ $a m p l i t u d e (n o t m a g n i t u d e) w o u l d$ $b e . . . ? (y o u c a n s e e f o r y o u r s e l f$ $m a t h s o l v e r) .$
Commented by math solver last updated on 22/Jan/18

$c a n u e l a b o r a t e a l i t t l e h o w y o u$ $d i d d r a w t h e g r a p h s . . . .$
Commented by ajfour last updated on 22/Jan/18

$o n e r e d d o t a t (- 1, 0) i n s t e a d a t$ $(- 3 / 2, 0) (e r r o r)$
Commented by ajfour last updated on 22/Jan/18

${(x + 1)}^{2} + y^{2} ⩽ {(2 + x)}^{2}$ $\Rightarrow y^{2} ⩽ 2 (x + \frac{3}{2})$ $t h e r e g i o n d e p i c t s c o n c a v e s i d e$ $v e r t e x a t (- \frac{3}{2}, 0)$ $f o r x = 0, y = \pm \sqrt{3}$ $f o r x = - 1, y = \pm 1$ $∣ \frac{z - 1}{z + 1} ∣⩾ 1 \Rightarrow o n o r l e f t o f I m a x i s$ $∣ z - 1 ∣⩾ 1 \Rightarrow o u t s i d e o r o n c i r c l e$ $w i t h c e n t r e (1, 0) a n d r a d i u s 1 .$ $h e n c e A \cap B \cap C i s t h e b l u e r e g i o n .$
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