Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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.

$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \:\leqslant\:\left(\mathrm{2}+{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{y}^{\mathrm{2}} \:\leqslant\:\mathrm{2}\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${the}\:{region}\:{depicts}\:{concave}\:{side} \\ $$$$\:\:{vertex}\:{at}\:\:\left(−\frac{\mathrm{3}}{\mathrm{2}},\mathrm{0}\right) \\ $$$${for}\:\:{x}=\mathrm{0}\:\:,\:{y}=\pm\sqrt{\mathrm{3}} \\ $$$${for}\:\:{x}=−\mathrm{1},\:\:{y}=\pm\mathrm{1} \\ $$$$\mid\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\mid\geqslant\mathrm{1}\:\:\Rightarrow\:\:{on}\:{or}\:{left}\:{of}\:{Im}\:{axis} \\ $$$$\mid{z}−\mathrm{1}\mid\geqslant\mathrm{1}\:\:\Rightarrow\:{outside}\:{or}\:{on}\:{circle} \\ $$$${with}\:{centre}\:\:\left(\mathrm{1},\mathrm{0}\right)\:{and}\:{radius}\:\mathrm{1}\:. \\ $$$${hence}\:\:{A}\cap{B}\cap{C}\:\:{is}\:{the}\:{blue}\:{region}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com