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AllQuestion and Answers: Page 66

Question Number 215032 Answers: 1 Comments: 1

Question Number 215020 Answers: 1 Comments: 2

f: [0 , 1] →R is given. f ′′ is continuous . by the way f(0)=f(1). prove that : determinant (((∫_0 ^( 1) ( f ′′ (x))^( 2) dx ≥ 3(f ′(1))^2 )))

$$ \\ $$$$\:\:\:\:{f}:\:\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:\rightarrow\mathbb{R}\:{is}\:{given}. \\ $$$$\:\:\:\:{f}\:''\:\:\:\:{is}\:{continuous}\:. \\ $$$$\:\:\:\:{by}\:{the}\:{way}\:\:{f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right). \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$\:\:\:\: \\ $$$$\begin{array}{|c|}{\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\:{f}\:''\:\left({x}\right)\right)^{\:\mathrm{2}} {dx}\:\geqslant\:\mathrm{3}\left({f}\:'\left(\mathrm{1}\right)\right)^{\mathrm{2}} }\\\hline\end{array} \\ $$$$ \\ $$

Question Number 215017 Answers: 2 Comments: 0

Re^ soudre dans C l′e^ quation : sin(z) = 2.

$${R}\acute {{e}soudre}\:{dans}\:\mathbb{C}\:{l}'\acute {{e}quation}\:: \\ $$$${sin}\left({z}\right)\:=\:\mathrm{2}. \\ $$

Question Number 215011 Answers: 0 Comments: 0

Question Number 215005 Answers: 1 Comments: 0

a,b,c,d∈R such that, (a+b)(c+d)=2 (a+c)(b+d)=3 (a+d)(b+c)=4 find: (a^2 +b^2 +c^2 +d^2 )_(minimum.)

$$\:{a},{b},{c},{d}\in{R}\:{such}\:{that}, \\ $$$$\:\left({a}+{b}\right)\left({c}+{d}\right)=\mathrm{2} \\ $$$$\:\left({a}+{c}\right)\left({b}+{d}\right)=\mathrm{3} \\ $$$$\:\left({a}+{d}\right)\left({b}+{c}\right)=\mathrm{4}\: \\ $$$$\:{find}:\:\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)_{{minimum}.} \\ $$

Question Number 215004 Answers: 0 Comments: 3

Question Number 214997 Answers: 1 Comments: 0

Σ_(k=−∞) ^(+∞) ((k^2 +k+1)/(k^4 +1))=?

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{k}=−\infty} {\overset{+\infty} {\sum}}\frac{{k}^{\mathrm{2}} +{k}+\mathrm{1}}{{k}^{\mathrm{4}} +\mathrm{1}}=? \\ $$$$ \\ $$

Question Number 214977 Answers: 2 Comments: 0

Solve: (((3x−5)^5 −(2x−3)^5 −(x−2)^5 )/((3x−5)^3 −(2x−3)^3 −(x−2)^3 ))=65

$$\:{Solve}: \\ $$$$\:\frac{\left(\mathrm{3}{x}−\mathrm{5}\right)^{\mathrm{5}} −\left(\mathrm{2}{x}−\mathrm{3}\right)^{\mathrm{5}} −\left({x}−\mathrm{2}\right)^{\mathrm{5}} }{\left(\mathrm{3}{x}−\mathrm{5}\right)^{\mathrm{3}} −\left(\mathrm{2}{x}−\mathrm{3}\right)^{\mathrm{3}} −\left({x}−\mathrm{2}\right)^{\mathrm{3}} }=\mathrm{65} \\ $$

Question Number 214973 Answers: 0 Comments: 1

Question Number 214964 Answers: 2 Comments: 0

Question Number 214961 Answers: 1 Comments: 2

Hi guys I am new

$$\mathrm{Hi}\:\mathrm{guys}\:\mathrm{I}\:\mathrm{am}\:\mathrm{new} \\ $$

Question Number 214941 Answers: 2 Comments: 1

Cannot we find m? Given (m^2 −1)^2 =2km(m−(1/k)) ∀ k∈R

$${Cannot}\:{we}\:{find}\:{m}?\:{Given} \\ $$$$\:\:\left({m}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2}{km}\left({m}−\frac{\mathrm{1}}{{k}}\right)\:\:\:\:\forall\:{k}\in\mathbb{R} \\ $$

Question Number 214936 Answers: 0 Comments: 17

Can we expand the Descartes′ Theorem to more than 4 circles?

$${Can}\:{we}\:{expand}\:{the}\:{Descartes}'\:{Theorem}\:{to}\:{more}\:{than}\:\mathrm{4}\:{circles}? \\ $$

Question Number 214928 Answers: 1 Comments: 2

Question Number 214925 Answers: 2 Comments: 1

Question Number 214923 Answers: 0 Comments: 0

Question Number 214917 Answers: 2 Comments: 0

Question Number 214916 Answers: 2 Comments: 0

Question Number 214915 Answers: 2 Comments: 0

Question Number 214888 Answers: 2 Comments: 2

Question Number 214905 Answers: 4 Comments: 2

Question Number 214900 Answers: 2 Comments: 0

determinant ()

Question Number 214898 Answers: 0 Comments: 0

Question Number 214876 Answers: 2 Comments: 3

Question Number 214860 Answers: 3 Comments: 1

Find: 1+ (2^3 /(3!)) + (3^3 /(3!)) + ... + (n^3 /(n!)) = ?

$$\mathrm{Find}: \\ $$$$\mathrm{1}+\:\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}!}\:\:+\:\:\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!}\:\:+\:\:...\:\:+\:\:\frac{\mathrm{n}^{\mathrm{3}} }{\mathrm{n}!}\:\:=\:\:? \\ $$

Question Number 214859 Answers: 2 Comments: 0

ABC is a triangle such that AB = AC. D is a point on side AC such that BC^2 = AC.CD. Prove that BD = BC.

$$\mathrm{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{such}\:\mathrm{that}\:\mathrm{AB}\:=\:\mathrm{AC}.\: \\ $$$$\mathrm{D}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{side}\:\mathrm{AC}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{BC}^{\mathrm{2}} \:=\:\mathrm{AC}.\mathrm{CD}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{BD}\:=\:\mathrm{BC}. \\ $$

Question Number 214857 Answers: 1 Comments: 0

∫_0 ^1 ln(x)ln(1+x)ln(1−x)dx=?

$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left({x}\right)\mathrm{ln}\left(\mathrm{1}+{x}\right)\mathrm{ln}\left(\mathrm{1}−{x}\right){dx}=? \\ $$

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