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Question Number 31493 Answers: 0 Comments: 0
$${if}\:\left({xcos}\theta\:+{sint}\right)^{{n}} \:={Q}\left({x}^{\mathrm{2}} +\mathrm{1}\right)\:+{R}\:\:{find}\:{tbe}\:{polynomialR} \\ $$
Question Number 31492 Answers: 0 Comments: 0
$${find}\:{all}\:{polynomial}\:{p}\left({x}\right)\:{wich}\:{verify}\: \\ $$$$\forall{k}\in{Z}\:\:\:\int_{{k}} ^{{k}+\mathrm{1}} {p}\left({x}\right){dx}={k}+\mathrm{1}. \\ $$
Question Number 31491 Answers: 0 Comments: 0
$${let}\:{p}\left({x}\right)=\:{x}^{{n}} \:+{a}_{{n}−\mathrm{1}} {x}^{{n}−\mathrm{1}} \:+....\:{a}_{\mathrm{1}} {x}\:+{a}_{{o}} \\ $$$${if}\:\:\xi\:\:{is}\:{roots}\:{of}\:{p}\left({x}\right)\:{prove}\:{that}\:\mid\xi\mid\:\leqslant\:\mathrm{1}+{max}_{\mathrm{0}\leqslant{i}\leqslant{n}−\mathrm{1}} \:\mid{a}_{{i}} \mid \\ $$
Question Number 31490 Answers: 0 Comments: 0
$${simplify}\:{p}\left({x}\right)=\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)....\left(\mathrm{1}+{x}^{\mathrm{2}{n}} \right)\:{with}\:{n}\:{fromN} \\ $$$${then}\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right). \\ $$
Question Number 31485 Answers: 2 Comments: 0
Question Number 31456 Answers: 0 Comments: 2
$$\mathbb{F}{ind}\:{sum}\:{of} \\ $$$${S}=\:\frac{\mathrm{2}}{\mathrm{3}}\:+\:\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{6}}{\mathrm{3}^{\mathrm{3}} }\:+\:\frac{\mathrm{8}}{\mathrm{3}^{\mathrm{4}} }\:+......+\infty\:? \\ $$
Question Number 31409 Answers: 1 Comments: 0
$${The}\:{maximum}\:{area}\:{of}\:{the}\:{triangle} \\ $$$${whose}\:{sides}\:{a},{b}\:{and}\:{c}\:{satisfy}\: \\ $$$$\mathrm{0}\leqslant{a}\leqslant\mathrm{1}\:,\:\mathrm{1}\leqslant{b}\leqslant\mathrm{2}\:,\:\mathrm{2}\leqslant{c}\leqslant\mathrm{3}\:{is}\:: \\ $$$$\left.{A}\right)\:\mathrm{1} \\ $$$$\left.{B}\right)\:\mathrm{2} \\ $$$$\left.{C}\right)\:\mathrm{1}.\mathrm{5} \\ $$$$\left.{D}\right)\:\mathrm{0}.\mathrm{5}\:\:\:\:\:\:\:? \\ $$
Question Number 31336 Answers: 0 Comments: 1
$${Find}\:{the}\:{principal}\:{value}\:{of} \\ $$$${z}=\left(\mathrm{1}−{i}\right)^{\mathrm{1}+{i}} .{Hence}\:{find}\:{the} \\ $$$${modulus}\:{of}\:{the}\:{result}. \\ $$
Question Number 31335 Answers: 0 Comments: 3
$${Find}\:{the}\:{pricipal}\:{value}\:{of}\: \\ $$$${z}=\left(\mathrm{1}−{i}\right)^{{i}} \\ $$
Question Number 31324 Answers: 1 Comments: 0
Question Number 31323 Answers: 1 Comments: 0
Question Number 31320 Answers: 1 Comments: 0
$$\mathrm{Let}\:{p}\:\mathrm{and}\:{q}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{2}{mx}\:−\:\mathrm{5}{n}\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:{m}\:\mathrm{and}\:{n}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{2}{px}\:−\:\mathrm{5}{q}\:=\:\mathrm{0} \\ $$$$\mathrm{If}\:{p}\:\neq\:{q}\:\neq\:{m}\:\neq\:{n},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${p}\:+\:{q}\:+\:{m}\:+\:{n}\:\mathrm{is}\:... \\ $$
Question Number 31317 Answers: 0 Comments: 1
Question Number 31286 Answers: 1 Comments: 1
$${Find}\:{all}\:{set}\:{of}\:{ordered}\:{triple}/{s}\:\left({x},{y},{z}\right),\:\:{x},{y},{z}\in\Re,\:{such}\:{that} \\ $$$${x}−{y}=\mathrm{1}−{z} \\ $$$$\mathrm{3}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{5}\left(\mathrm{1}−{z}^{\mathrm{2}} \right) \\ $$$$\mathrm{7}\left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right)=\mathrm{19}\left(\mathrm{1}−{z}^{\mathrm{3}} \right). \\ $$$${Please}\:{show}\:{your}\:{solution}. \\ $$
Question Number 31290 Answers: 1 Comments: 1
Question Number 31259 Answers: 0 Comments: 4
Question Number 31214 Answers: 2 Comments: 2
Question Number 31194 Answers: 2 Comments: 1
$${Find}\:{the}\:{remainder}\:{when}\:{x}^{\mathrm{203}} −\mathrm{1} \\ $$$${is}\:{divided}\:{by}\:{x}^{\mathrm{4}} −\mathrm{1}. \\ $$
Question Number 31047 Answers: 0 Comments: 0
$${let}\:\Delta=\left\{\left({x},{y}\right)\in{N}^{\mathrm{2}} \:/{x}+{y}={n}\:,\:{n}\in{N}\right\}\:{find}\:{card}\Delta \\ $$$$\left.\mathrm{2}\right)\:{let}\:{A}=\:\left\{\left({x},{y}\right)\in{N}^{\mathrm{2}} /\:{x}+\mathrm{2}{y}={n}\right\}\:{find}\:{card}\:{A}. \\ $$
Question Number 31046 Answers: 0 Comments: 0
$${prove}\:{that}\:{C}_{{n}} ^{{o}} \:{C}_{{n}} ^{{p}} \:+{C}_{{n}} ^{\mathrm{1}} \:{C}_{{n}−\mathrm{1}} ^{{p}−\mathrm{1}} \:+...{C}_{{n}} ^{{p}} \:{C}_{{n}−{p}} ^{\mathrm{0}} =\mathrm{2}^{{p}} \:{C}_{{n}} ^{{p}} \:\:\:. \\ $$
Question Number 31042 Answers: 0 Comments: 0
$${solve}\:{in}\:{N}^{\mathrm{2}} \:\:\mathrm{9}{y}^{\mathrm{2}} \:−\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{32}\:. \\ $$
Question Number 31041 Answers: 0 Comments: 0
$${solve}\:{in}\:{Z}^{\mathrm{2}} \:\mathrm{11}{x}\:−\mathrm{5}{y}\:=\mathrm{14} \\ $$
Question Number 31040 Answers: 0 Comments: 0
$${solve}\:{in}\:{Z}^{\mathrm{2}} \:{x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} =\mathrm{1969} \\ $$
Question Number 31039 Answers: 0 Comments: 0
$${find}\:{the}\:{sum} \\ $$$${s}_{\mathrm{0}} =\:{C}_{{n}} ^{{o}} \:+{C}_{{n}} ^{\mathrm{4}} \:+{C}_{{n}} ^{\mathrm{8}} \:+... \\ $$$${s}_{\mathrm{1}} =\:{C}_{{n}} ^{\mathrm{1}} \:+{C}_{{n}} ^{\mathrm{5}} \:+{C}_{{n}} ^{\mathrm{9}} \:+.... \\ $$$${s}_{\mathrm{3}} =\:{C}_{{n}} ^{\mathrm{2}} \:\:+{C}_{{n}} ^{\mathrm{6}} \:+\:{C}_{{n}} ^{\mathrm{10}} \:+.... \\ $$
Question Number 31038 Answers: 0 Comments: 0
$$\left.{let}\:{give}\:\alpha\:\in\right]−\pi\:,\pi\left[\right. \\ $$$$\left.\mathrm{1}\right){prove}\:{that}\:\:{sin}^{\mathrm{2}} \alpha\:−\mathrm{2}\left(\mathrm{1}+{cos}\alpha\right)\:=−\mathrm{4}{cos}^{\mathrm{4}} \left(\frac{\alpha}{\mathrm{2}}\right) \\ $$$$\left.\mathrm{2}\right){solve}\:{inside}\:{C}\:\:{z}^{\mathrm{2}} \:−\mathrm{2}{z}\:{sin}\alpha\:+\mathrm{2}\left(\mathrm{1}+{cos}\alpha\right)=\mathrm{0}\:{find} \\ $$$${the}\:{module}\:{and}\:{arg}\:{of}\:{the}\:{roots}. \\ $$
Question Number 31037 Answers: 0 Comments: 0
$${let}\:\:\theta\in\left[\mathrm{0},\pi\right]\:\:{and}\:{Z}=\mathrm{1}+{cos}\theta\:+{isin}\theta \\ $$$${Z}^{'} ={cos}\theta\:+\left(\mathrm{1}+{sin}\theta\right){i}\:\:{find}\:\mid{ZZ}^{'} \mid\:\:{arg}\left({ZZ}^{'} \right) \\ $$$$\mid\frac{{Z}}{{Z}^{'} }\mid\:,\:{arg}\left(\frac{{Z}}{{Z}^{,} }\right)\:\:,\:\:\mid{Z}−{Z}^{,} \mid\:{and}\:{arg}\left({Z}−{Z}^{,} \right). \\ $$
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