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Question Number 37224 Answers: 0 Comments: 0

let p(x)=(1+jx)^n −(1−jx)^n with j=e^(i((2π)/3)) find p at form r(x)e^(iθ(x)) 2) calculate ∫_0 ^1 r(x) e^(iθ(x)) dx .

$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{jx}\right)^{{n}} \:−\left(\mathrm{1}−{jx}\right)^{{n}} \:{with} \\ $$$${j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:\:{find}\:{p}\:{at}\:\:{form}\:{r}\left({x}\right){e}^{{i}\theta\left({x}\right)} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {r}\left({x}\right)\:{e}^{{i}\theta\left({x}\right)} {dx}\:. \\ $$

Question Number 37221 Answers: 0 Comments: 8

Hi! MrW is back! I′m sorry for having been absent for months without giving you a message. It′s alright with me. The last few months I was very very busy with our new house and I had no time for other things. I changed my smartphone and lost my old ID. Sorry and thanks to all those who have cared about me! Sir Rasheed, your post #35656 has moved me especially, thank you! I have viewed some new posts and I′m happy to see that a lot of old friends are still active in the forum. I′ll try to find time to read more new posts.

$${Hi}!\:{MrW}\:{is}\:{back}! \\ $$$${I}'{m}\:{sorry}\:{for}\:{having}\:{been}\:{absent}\:{for} \\ $$$${months}\:{without}\:{giving}\:{you}\:{a}\:{message}. \\ $$$${It}'{s}\:{alright}\:{with}\:{me}.\: \\ $$$${The}\:{last}\:{few}\:{months}\:{I}\:{was}\:{very}\:{very} \\ $$$${busy}\:{with}\:{our}\:{new}\:{house}\:{and}\:{I}\:{had}\:{no} \\ $$$${time}\:{for}\:{other}\:{things}.\:{I}\:{changed}\:{my} \\ $$$${smartphone}\:{and}\:{lost}\:{my}\:{old}\:{ID}. \\ $$$${Sorry}\:{and}\:{thanks}\:{to}\:{all}\:{those}\:{who}\:{have} \\ $$$${cared}\:{about}\:{me}!\: \\ $$$${Sir}\:{Rasheed},\:{your}\:{post}\:#\mathrm{35656}\:{has} \\ $$$${moved}\:{me}\:{especially},\:{thank}\:{you}! \\ $$$${I}\:{have}\:{viewed}\:{some}\:{new}\:{posts}\:{and} \\ $$$${I}'{m}\:{happy}\:{to}\:{see}\:{that}\:{a}\:{lot}\:{of}\:{old}\:{friends} \\ $$$${are}\:{still}\:{active}\:{in}\:{the}\:{forum}. \\ $$$${I}'{ll}\:{try}\:{to}\:{find}\:{time}\:{to}\:{read}\:{more}\:{new} \\ $$$${posts}. \\ $$

Question Number 37220 Answers: 1 Comments: 0

Question Number 37218 Answers: 1 Comments: 0

Proove that a) ((1−cosA + cos B − cos(A+B))/(1+cos A − cosB−cos(A+B)))= tan(A/2).cot (B/2) b) cosα cos(60−α)cos(60+α)= (1/4)cos3α

$$\:{Proove}\:{that} \\ $$$$\left.{a}\right)\:\:\frac{\mathrm{1}−{cosA}\:+\:{cos}\:{B}\:−\:{cos}\left({A}+{B}\right)}{\mathrm{1}+{cos}\:{A}\:−\:{cosB}−{cos}\left({A}+{B}\right)}=\:{tan}\frac{{A}}{\mathrm{2}}.{cot}\:\frac{{B}}{\mathrm{2}} \\ $$$$\left.{b}\right)\:{cos}\alpha\:{cos}\left(\mathrm{60}−\alpha\right){cos}\left(\mathrm{60}+\alpha\right)=\:\frac{\mathrm{1}}{\mathrm{4}}{cos}\mathrm{3}\alpha \\ $$$$ \\ $$

Question Number 37217 Answers: 0 Comments: 0

Question Number 37204 Answers: 0 Comments: 0

Question Number 37189 Answers: 1 Comments: 0

If ∫ ((cos 4x+1)/(cot x−tan x)) dx=A cos 4x+B, then find A,B?

$$\mathrm{If}\:\int\:\:\frac{\mathrm{cos}\:\mathrm{4}{x}+\mathrm{1}}{\mathrm{cot}\:{x}−\mathrm{tan}\:{x}}\:{dx}={A}\:\mathrm{cos}\:\mathrm{4}{x}+{B},\:\mathrm{then}\: \\ $$$$\mathrm{find}\:\mathrm{A},\mathrm{B}? \\ $$

Question Number 37188 Answers: 2 Comments: 0

∫ (√((cos x−cos^3 x)/(1−cos^3 x))) dx =

$$\int\:\sqrt{\frac{\mathrm{cos}\:{x}−\mathrm{cos}^{\mathrm{3}} {x}}{\mathrm{1}−\mathrm{cos}^{\mathrm{3}} {x}}}\:{dx}\:= \\ $$

Question Number 37209 Answers: 1 Comments: 3

Question Number 37179 Answers: 0 Comments: 4

∫ ((3x^2 +1)/((x^2 −1)^3 )) dx = ?

$$\int\:\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }\:{dx}\:=\:? \\ $$

Question Number 37178 Answers: 0 Comments: 0

suppose T is a matrix of enlargement by factor k find the matrix S such that ST[(x,y)]=(x,y).

$${suppose}\:{T}\:{is}\:{a}\:{matrix}\:{of}\:{enlargement} \\ $$$$ \\ $$$${by}\:{factor}\:{k}\:{find}\:{the}\:{matrix}\:{S}\:{such}\:{that} \\ $$$${ST}\left[\left({x},{y}\right)\right]=\left({x},{y}\right). \\ $$

Question Number 37171 Answers: 0 Comments: 0

Question Number 37168 Answers: 0 Comments: 0

A stone is thrown into a circular pond of radius 1m.Suppose the stone falls uniformly at random on the area of the pond.What will be the expected distance od the stone from the centre of the pond. a)1/3 b)1/2 c)2/3 d)1/(√2)

$${A}\:{stone}\:{is}\:{thrown}\:{into}\:{a}\:{circular} \\ $$$${pond}\:{of}\:{radius}\:\mathrm{1}{m}.{Suppose}\:{the} \\ $$$${stone}\:{falls}\:{uniformly}\:{at}\:{random} \\ $$$${on}\:{the}\:{area}\:{of}\:{the}\:{pond}.{What} \\ $$$${will}\:{be}\:{the}\:{expected}\:{distance}\:{od} \\ $$$${the}\:{stone}\:{from}\:{the}\:{centre}\:{of}\:{the} \\ $$$${pond}. \\ $$$$ \\ $$$$\left.{a}\left.\right)\left.\mathrm{1}\left./\mathrm{3}\:{b}\right)\mathrm{1}/\mathrm{2}\:{c}\right)\mathrm{2}/\mathrm{3}\:{d}\right)\mathrm{1}/\sqrt{\mathrm{2}} \\ $$

Question Number 37166 Answers: 0 Comments: 2

if 3x^2 +2αxy+2y^2 +2ax−4y+1 can be resolved into two linear factors, prove that ′α′ is a root of the equation x^2 +4ax+2a^2 +6=0

$${if}\:\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\alpha{xy}+\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{4}{y}+\mathrm{1} \\ $$$${can}\:{be}\:{resolved}\:\:{into}\:\:{two}\:\:{linear} \\ $$$${factors},\:\:{prove}\:\:{that}\:\:'\alpha'\:\:{is}\:{a}\:{root}\: \\ $$$${of}\:{the}\:{equation}\:{x}^{\mathrm{2}} +\mathrm{4}{ax}+\mathrm{2}{a}^{\mathrm{2}} +\mathrm{6}=\mathrm{0} \\ $$

Question Number 37155 Answers: 0 Comments: 4

3cos x−4sin x=tan^2 x locate x.

$$\mathrm{3cos}\:{x}−\mathrm{4sin}\:{x}=\mathrm{tan}\:^{\mathrm{2}} {x} \\ $$$${locate}\:{x}. \\ $$

Question Number 37146 Answers: 0 Comments: 0

Question Number 37143 Answers: 0 Comments: 0

Why are following statements wrong? a) There exists a function with domain R satisfying f(x)<0 ∀x , f′(x)>0∀x and f′′(x)>0∀x. b) If f′′(c)=0 then (c,f(c)) is an inflection point.

$$\mathrm{Why}\:\mathrm{are}\:\mathrm{following}\:\mathrm{statements}\:\mathrm{wrong}? \\ $$$$\left.\mathrm{a}\right)\:\mathrm{There}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{function}\:\mathrm{with}\:\mathrm{domain}\: \\ $$$$\mathrm{R}\:\mathrm{satisfying}\:\mathrm{f}\left(\mathrm{x}\right)<\mathrm{0}\:\forall\mathrm{x}\:,\:\mathrm{f}'\left(\mathrm{x}\right)>\mathrm{0}\forall\mathrm{x}\:\mathrm{and} \\ $$$$\mathrm{f}''\left(\mathrm{x}\right)>\mathrm{0}\forall\mathrm{x}. \\ $$$$ \\ $$$$\left.\mathrm{b}\right)\:\mathrm{If}\:\mathrm{f}''\left(\mathrm{c}\right)=\mathrm{0}\:\mathrm{then}\:\left(\mathrm{c},\mathrm{f}\left(\mathrm{c}\right)\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{inflection} \\ $$$$\mathrm{point}. \\ $$

Question Number 37133 Answers: 0 Comments: 0

Question Number 37132 Answers: 0 Comments: 0

Question Number 37131 Answers: 0 Comments: 2

Question Number 37136 Answers: 0 Comments: 1

Question Number 37139 Answers: 2 Comments: 0

if α , β are the roots of the quadratic equation ax^2 +bx+c =0 then find the quadratic equation whose roots are α^(2 ) , β^2

$${if}\:\alpha\:,\:\beta\:\:{are}\:{the}\:{roots}\:{of}\:{the}\:{quadratic} \\ $$$${equation}\:{ax}^{\mathrm{2}} +{bx}+{c}\:=\mathrm{0}\:{then}\:\:{find} \\ $$$${the}\:{quadratic}\:{equation}\:{whose}\:{roots} \\ $$$${are}\:\:\alpha^{\mathrm{2}\:\:\:} ,\:\beta^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

Question Number 37137 Answers: 2 Comments: 4

Find minimum distance between y^2 =8x and x^2 +(y+6)^2 =1.

$$\mathrm{Find}\:\mathrm{minimum}\:\mathrm{distance}\:\mathrm{between} \\ $$$$\mathrm{y}^{\mathrm{2}} =\mathrm{8}{x}\:{and}\:{x}^{\mathrm{2}} +\left({y}+\mathrm{6}\right)^{\mathrm{2}} =\mathrm{1}. \\ $$

Question Number 37129 Answers: 0 Comments: 0

Question Number 37128 Answers: 0 Comments: 0

Question Number 37511 Answers: 0 Comments: 3

To the developer of Tinku Tara: dear sir: for some unknown reasons I don′t get any notification from the app when a post, in which I am involved, has been updated. Where is the problem and how can I solve it? Thank you!

$${To}\:{the}\:{developer}\:{of}\:{Tinku}\:{Tara}: \\ $$$${dear}\:{sir}:\:{for}\:{some}\:{unknown}\:{reasons} \\ $$$${I}\:{don}'{t}\:{get}\:{any}\:{notification}\:{from}\:{the} \\ $$$${app}\:{when}\:{a}\:{post},\:{in}\:{which}\:{I}\:{am}\:{involved}, \\ $$$${has}\:{been}\:{updated}.\:{Where}\:{is}\:{the} \\ $$$${problem}\:{and}\:{how}\:{can}\:{I}\:{solve}\:{it}? \\ $$$${Thank}\:{you}! \\ $$

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