Menu Close

Question-28211




Question Number 28211 by ajfour last updated on 22/Jan/18
Commented by math solver last updated on 22/Jan/18
thank you sir!
$${thank}\:{you}\:{sir}! \\ $$
Commented by math solver last updated on 22/Jan/18
sir, i think they are referring   max. amplitude as max. argument.
$${sir},\:{i}\:{think}\:{they}\:{are}\:{referring}\: \\ $$$${max}.\:{amplitude}\:{as}\:{max}.\:{argument}. \\ $$
Commented by ajfour last updated on 22/Jan/18
Q.28189  (solution)  the blue region is A∩B∩C.  6 points with integral coordinates,  is fine but point having maximum  amplitude (not magnitude) would  be ...?  (you can see for yourself  math solver ).
$${Q}.\mathrm{28189}\:\:\left({solution}\right) \\ $$$${the}\:{blue}\:{region}\:{is}\:{A}\cap{B}\cap{C}. \\ $$$$\mathrm{6}\:{points}\:{with}\:{integral}\:{coordinates}, \\ $$$${is}\:{fine}\:{but}\:{point}\:{having}\:{maximum} \\ $$$${amplitude}\:\left({not}\:{magnitude}\right)\:{would} \\ $$$${be}\:…?\:\:\left({you}\:{can}\:{see}\:{for}\:{yourself}\right. \\ $$$$\left.{math}\:{solver}\:\right). \\ $$
Commented by math solver last updated on 22/Jan/18
can u elaborate a little how you   did draw the graphs ....
$${can}\:{u}\:{elaborate}\:{a}\:{little}\:{how}\:{you}\: \\ $$$${did}\:{draw}\:{the}\:{graphs}\:…. \\ $$
Commented by ajfour last updated on 22/Jan/18
one  red dot at (−1,0) instead at  (−3/2,0)   (error)
$${one}\:\:{red}\:{dot}\:{at}\:\left(−\mathrm{1},\mathrm{0}\right)\:{instead}\:{at} \\ $$$$\left(−\mathrm{3}/\mathrm{2},\mathrm{0}\right)\:\:\:\left({error}\right) \\ $$
Commented by ajfour last updated on 22/Jan/18
(x+1)^2 +y^2  ≤ (2+x)^2   ⇒  y^2  ≤ 2(x+(3/2))  the region depicts concave side    vertex at  (−(3/2),0)  for  x=0  , y=±(√3)  for  x=−1,  y=±1  ∣((z−1)/(z+1))∣≥1  ⇒  on or left of Im axis  ∣z−1∣≥1  ⇒ outside or on circle  with centre  (1,0) and radius 1 .  hence  A∩B∩C  is the blue region.
$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \:\leqslant\:\left(\mathrm{2}+{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{y}^{\mathrm{2}} \:\leqslant\:\mathrm{2}\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${the}\:{region}\:{depicts}\:{concave}\:{side} \\ $$$$\:\:{vertex}\:{at}\:\:\left(−\frac{\mathrm{3}}{\mathrm{2}},\mathrm{0}\right) \\ $$$${for}\:\:{x}=\mathrm{0}\:\:,\:{y}=\pm\sqrt{\mathrm{3}} \\ $$$${for}\:\:{x}=−\mathrm{1},\:\:{y}=\pm\mathrm{1} \\ $$$$\mid\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\mid\geqslant\mathrm{1}\:\:\Rightarrow\:\:{on}\:{or}\:{left}\:{of}\:{Im}\:{axis} \\ $$$$\mid{z}−\mathrm{1}\mid\geqslant\mathrm{1}\:\:\Rightarrow\:{outside}\:{or}\:{on}\:{circle} \\ $$$${with}\:{centre}\:\:\left(\mathrm{1},\mathrm{0}\right)\:{and}\:{radius}\:\mathrm{1}\:. \\ $$$${hence}\:\:{A}\cap{B}\cap{C}\:\:{is}\:{the}\:{blue}\:{region}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *