Question and Answers Forum
All Questions Topic List
AlgebraQuestion and Answers: Page 341
Question Number 31915 Answers: 1 Comments: 3
$$\boldsymbol{{If}}\::\: \\ $$$$\left({x}^{\mathrm{2}} +{x}+\mathrm{2}\right)^{\mathrm{2}} −\left({a}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{2}\right) \\ $$$$+\:\left({a}−\mathrm{4}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:{has}\:{at}\:{least}\: \\ $$$${one}\:{root}\:,\:{then}\:{find}\:{complete}\:{set}\:{of}\: \\ $$$${values}\:{of}\:{a}. \\ $$
Question Number 31895 Answers: 0 Comments: 3
Question Number 32372 Answers: 1 Comments: 0
Question Number 31868 Answers: 0 Comments: 0
$${Let}\:{a}>{b}>\mathrm{1}\:{be}\:{positive}\:{integers}\:{with}\:{b}\:{odd}. \\ $$$${Let}\:{n}\:{be}\:{a}\:{positive}\:{integer}\:{as}\:{well}.\:{If}\:\:{b}^{{n}} \:{divides} \\ $$$${a}^{{n}} −\mathrm{1},\:{prove}\:{that}\:{a}^{{b}} \:>\:\frac{\mathrm{3}^{{n}} }{{n}}. \\ $$$${Solution}\:{please}.\:{Thanks}\:{in}\:{advance}!! \\ $$
Question Number 31865 Answers: 1 Comments: 6
$${Range}\:{of}\:{function}\:: \\ $$$${f}\left({x}\right)=\:\mathrm{6}^{{x}} +\mathrm{3}^{{x}} +\mathrm{6}^{−{x}} +\mathrm{3}^{−{x}} +\mathrm{2}. \\ $$
Question Number 31864 Answers: 1 Comments: 0
$${Let}\:{S}_{{n}} =\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}{n}} {\sum}}\left(−\mathrm{1}\right)^{\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}} {k}^{\mathrm{2}} . \\ $$$${Then}\:{S}_{{n}} \:{can}\:{take}\:{the}\:{value}\left({s}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{1056} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{1088} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{1120} \\ $$$$\left.\mathrm{4}\right)\:\mathrm{1332}. \\ $$
Question Number 31804 Answers: 1 Comments: 0
$${A}\:{quadratic}\:{equation}\:{p}\left({x}\right)=\mathrm{0}\:{having} \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{2}} \:{unity}\:{is}\:{such}\:{that} \\ $$$${p}\left({x}\right)=\mathrm{0}\:{and}\:{p}\left({p}\left({p}\left({x}\right)\right)\right)=\mathrm{0}\:{have}\:{a}\: \\ $$$${common}\:{root}\:{then}, \\ $$$${prove}\:{that}\::\:\:{p}\left(\mathrm{0}\right)×{p}\left(\mathrm{1}\right)=\mathrm{0}. \\ $$
Question Number 31771 Answers: 1 Comments: 1
$${Consider}\:{a}\:{sequence}\:{in}\:{the}\:{form}\:{of} \\ $$$${groups}\:\left(\mathrm{1}\right),\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{3},\mathrm{3},\mathrm{3}\right),\left(\mathrm{4},\mathrm{4},\mathrm{4},\mathrm{4}\right), \\ $$$$\left(\mathrm{5},\mathrm{5},\mathrm{5},\mathrm{5},\mathrm{5}\right),............ \\ $$$${then}\:{the}\:\mathrm{2000}{th}\:{term}\:{of}\:{the}\:{above}\: \\ $$$${sequence}\:{is}\::\:? \\ $$
Question Number 31763 Answers: 3 Comments: 0
$${please}\:{find}\:{the}\:{integral}\:{solutions}\:\left({x}\:{and}\:{y}\right)\: \\ $$$$\left({xy}−\mathrm{7}\right)^{\mathrm{2}} \:={x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \\ $$
Question Number 31726 Answers: 1 Comments: 3
$${Let}\:{a}_{\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:,\:{a}_{{k}+\mathrm{1}} ={a}_{{k}} ^{\mathrm{2}} +{a}_{{k}} \forall\:{k}\geqslant\:\mathrm{1}. \\ $$$${then}\:{a}_{\mathrm{101}} \:\:{is}\:{greater}\:{than} \\ $$$$\left.{a}\right)\:\mathrm{1}\: \\ $$$$\left.{b}\right)\:\mathrm{2} \\ $$$$\left.{c}\right)\:\mathrm{3} \\ $$$$\left.{d}\right)\:\mathrm{4}\:. \\ $$
Question Number 31677 Answers: 1 Comments: 0
$$ \\ $$$$\mathrm{24}{x}^{\mathrm{3}} −\mathrm{26}{x}^{\mathrm{2}} +\mathrm{9}{x}−\mathrm{1}=\mathrm{0}\left({solve}\right) \\ $$
Question Number 31670 Answers: 1 Comments: 0
$$\mathrm{how}\:\mathrm{many}\:\mathrm{roots}\:\mathrm{from}\:\mathrm{equation} \\ $$$${ae}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${from}\:{a}>\mathrm{0}\:? \\ $$
Question Number 31642 Answers: 0 Comments: 1
$$\sim\:\mathrm{Equivalence}\:\mathrm{relation} \\ $$$$ \\ $$$$\left(\mathrm{a}\sim\mathrm{b}\:\&\:\mathrm{c}\nsim\mathrm{b}\right)\Rightarrow\mathrm{c}\nsim\mathrm{a} \\ $$$$\mathrm{True}\:\mathrm{or}\:\mathrm{false}\:\mathrm{and}\:\mathrm{why} \\ $$
Question Number 31596 Answers: 2 Comments: 2
Question Number 31595 Answers: 2 Comments: 2
Question Number 31594 Answers: 1 Comments: 0
Question Number 31591 Answers: 1 Comments: 4
Question Number 31542 Answers: 0 Comments: 0
$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)^{{n}} \:−\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)^{{n}} \:\in{C}\left[{x}\right] \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{p}\left({x}\right)\:{inside}\:{C}\left[{x}\right]. \\ $$
Question Number 31499 Answers: 0 Comments: 1
$${find}\:{the}\:{polynial}\:{p}\:{wich}\:{verify}\:{p}\left({x}\right)−{p}^{'} \left({x}\right)={x}^{{n}} \:{then} \\ $$$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {p}\left({x}\right){dx}. \\ $$
Question Number 31498 Answers: 0 Comments: 0
$${find}\:{tbe}\:{value}\:{of}\:\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{k}\pi}{{n}+\mathrm{1}}\right). \\ $$
Question Number 31496 Answers: 0 Comments: 1
$$\left.{let}\:{a}\in\right]\mathrm{0},\pi\left[\:\:\:{and}\:{A}\left({x}\right)=\:{x}^{\mathrm{2}{n}} \:−\mathrm{2}{cos}\left({na}\right){x}^{{n}} \:+\mathrm{1}\right. \\ $$$$\left.\mathrm{1}\right){factorize}\:{inside}\:{C}\left[{x}\right]\:{A}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{inside}\:{R}\left[{x}\right]\:{A}\left({x}\right). \\ $$
Question Number 31495 Answers: 0 Comments: 0
$${let}\:{give}\:{p}\left({x}\right)=\left({x}+{j}\right)^{{n}} \:−\left({x}−{j}\right)^{{n}} \:{with}\:{j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{inside}\:{C}\left[\:{x}\right]\:{p}\left({x}\right) \\ $$$$\left.\mathrm{3}\right){factorize}\:{inside}\:{R}\left[{x}\right]\:{p}\left({x}\right). \\ $$
Question Number 31494 Answers: 0 Comments: 3
$${prove}\:{that}\:{x}^{\mathrm{2}} \:{divide}\:\left({x}+\mathrm{1}\right)^{\underset{} {{n}}} \:−{nx}−\mathrm{1}\:.{nintegr}. \\ $$
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
$${let}\:{p}\left({x}\right)=\:{x}^{{n}} \:+{a}_{{n}−\mathrm{1}} {x}^{{n}−\mathrm{1}} \:+....\:{a}_{\mathrm{1}} {x}\:+{a}_{{o}} \\ $$$${if}\:\:\xi\:\:{is}\:{roots}\:{of}\:{p}\left({x}\right)\:{prove}\:{that}\:\mid\xi\mid\:\leqslant\:\mathrm{1}+{max}_{\mathrm{0}\leqslant{i}\leqslant{n}−\mathrm{1}} \:\mid{a}_{{i}} \mid \\ $$
Pg 336 Pg 337 Pg 338 Pg 339 Pg 340 Pg 341 Pg 342 Pg 343 Pg 344 Pg 345
Terms of Service
Privacy Policy
Contact: info@tinkutara.com