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AlgebraQuestion and Answers: Page 228

Question Number 131084 Answers: 2 Comments: 0

let solve this in R : x^4 −x^3 −4x^2 +x+1=0.

$$\mathrm{let}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{in}\:\mathbb{R}\::\: \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}. \\ $$

Question Number 131075 Answers: 1 Comments: 0

If a−3=−b−4=−c−5=d+6=e+7= a−b−c+d+e+8 then a−b−c+d+e =?

$$\mathrm{If}\:\mathrm{a}−\mathrm{3}=−\mathrm{b}−\mathrm{4}=−\mathrm{c}−\mathrm{5}=\mathrm{d}+\mathrm{6}=\mathrm{e}+\mathrm{7}= \\ $$$$\mathrm{a}−\mathrm{b}−\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{8}\:\mathrm{then}\:\mathrm{a}−\mathrm{b}−\mathrm{c}+\mathrm{d}+\mathrm{e}\:=? \\ $$

Question Number 131064 Answers: 3 Comments: 0

Question Number 131016 Answers: 0 Comments: 0

Question Number 130853 Answers: 1 Comments: 0

Question Number 130846 Answers: 1 Comments: 1

prove that 3sec^(−1) ((√2))−4csc^(−1) ((√2))+5cot^(−1) (2)=1.533

$${prove}\:{that} \\ $$$$\mathrm{3}{sec}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)−\mathrm{4}{csc}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)+\mathrm{5}{cot}^{−\mathrm{1}} \left(\mathrm{2}\right)=\mathrm{1}.\mathrm{533} \\ $$

Question Number 130806 Answers: 1 Comments: 0

Question Number 130802 Answers: 2 Comments: 0

Question Number 130800 Answers: 0 Comments: 1

Question Number 130733 Answers: 1 Comments: 1

Question Number 130719 Answers: 0 Comments: 2

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Question Number 130673 Answers: 0 Comments: 0

Question Number 130605 Answers: 2 Comments: 0

0^0 =??

$$\mathrm{0}^{\mathrm{0}} =?? \\ $$

Question Number 130593 Answers: 1 Comments: 3

if a_0 =1, a_1 =2 and a_(n+1) =(√(a_n a_(n−1) )) find a_n in terms of n.

$${if}\:{a}_{\mathrm{0}} =\mathrm{1},\:{a}_{\mathrm{1}} =\mathrm{2}\:{and}\:{a}_{{n}+\mathrm{1}} =\sqrt{{a}_{{n}} {a}_{{n}−\mathrm{1}} } \\ $$$${find}\:{a}_{{n}} \:{in}\:{terms}\:{of}\:{n}. \\ $$

Question Number 130496 Answers: 3 Comments: 0

(d/dx)(x!)=?

$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}!\right)=? \\ $$

Question Number 130476 Answers: 1 Comments: 0

Question Number 130472 Answers: 2 Comments: 0

Question Number 130455 Answers: 0 Comments: 0

f(x)=x+ln∣e^x −2∣ { ((f(x)=x+ln(e^x −2) x∈]ln2;+∞[)),((f(x)=x+ln(−e^x +2) x∈]−∞;ln2[)) :} lim_(x→+∞) f(x)=lim_(x→+∞) (x+ln(e^x −2))=+∞ lim_(x→−∞) f(x)=lim_(x→−∞) (x+ln(−e^x +2))=−∞ lim_(x→−∞) e^x =0 lim_(x→−∞) ln(−e^x +2)=ln2 lim_(x→−∞) (x+ln(−e^x +2)=−∞ lim_(x→^> ln2) f(x)=lim_(x→^> ln2) (x+ln(e^x −2))=−∞ lim_(x→^< ln2) f(x)=lim_(x→^< ln2) (x+ln(−e^x +2))=−∞

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{ln}\mid\mathrm{e}^{\mathrm{x}} −\mathrm{2}\mid \\ $$$$\begin{cases}{\left.\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{2}\right)\:\:\:\:\:\:\mathrm{x}\in\right]\mathrm{ln2};+\infty\left[\right.}\\{\left.\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)\:\:\:\mathrm{x}\in\right]−\infty;\mathrm{ln2}\left[\right.}\end{cases} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}f}\left(\mathrm{x}\right)=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{2}\right)\right)=+\infty \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}f}\left(\mathrm{x}\right)=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)\right)=−\infty \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}e}^{\mathrm{x}} =\mathrm{0} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)=\mathrm{ln2} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)=−\infty\right. \\ $$$$\underset{{x}\overset{>} {\rightarrow}\mathrm{ln2}} {\mathrm{lim}f}\left(\mathrm{x}\right)=\underset{{x}\overset{>} {\rightarrow}\mathrm{ln2}} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{2}\right)\right)=−\infty \\ $$$$\underset{{x}\overset{<} {\rightarrow}\mathrm{ln2}} {\mathrm{lim}f}\left(\mathrm{x}\right)=\underset{{x}\overset{<} {\rightarrow}\mathrm{ln2}} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)\right)=−\infty \\ $$

Question Number 130442 Answers: 2 Comments: 0

g(x)=3x−2. determine a et b such that a ≤ g(x) ≤ b.

$${g}\left({x}\right)=\mathrm{3}{x}−\mathrm{2}. \\ $$$$\mathrm{determine}\:{a}\:\mathrm{et}\:{b}\:\mathrm{such}\:\mathrm{that}\:{a}\:\leqslant\:{g}\left({x}\right)\:\leqslant\:{b}. \\ $$

Question Number 130426 Answers: 2 Comments: 0

Question Number 130383 Answers: 1 Comments: 0

If (a−2)+3i=5−bi then a+b=

$${If}\:\left({a}−\mathrm{2}\right)+\mathrm{3}{i}=\mathrm{5}−{bi}\:{then}\:{a}+{b}= \\ $$

Question Number 130362 Answers: 3 Comments: 0

f(x)=((2x+1)/( (√(x^2 −∣2x−3∣)))) Domain D_f = ?

$$\mathrm{f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mid\mathrm{2}{x}−\mathrm{3}\mid}} \\ $$$$\mathrm{Domain}\:\mathrm{D}_{\mathrm{f}} \:=\:? \\ $$

Question Number 130354 Answers: 5 Comments: 0

f(x)=((4x^2 +1)/(2x^2 +1)) prove that 1 ≤ f(x) ≤ 2

$$\mathrm{f}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{1}\:\leqslant\:\mathrm{f}\left({x}\right)\:\leqslant\:\mathrm{2} \\ $$

Question Number 130290 Answers: 1 Comments: 0

Σ_(i=0) ^∞ Σ_(j=0) ^∞ Σ_(k=0) ^∞ (1/3^(i+j+k) ) ? where i≠j≠k

$$\:\underset{\mathrm{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{\mathrm{j}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{i}+\mathrm{j}+\mathrm{k}} }\:?\:\mathrm{where}\:\mathrm{i}\neq\mathrm{j}\neq\mathrm{k} \\ $$

Question Number 130334 Answers: 2 Comments: 0

Complete the square in the expression y^2 +8y+9k and hence find the value of k that makes it a perfect square.

$$\mathrm{Complete}\:\mathrm{the}\:\mathrm{square}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mathrm{y}^{\mathrm{2}} \:+\mathrm{8y}+\mathrm{9k}\:\mathrm{and}\:\mathrm{hence}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{k}\:\mathrm{that}\:\mathrm{makes}\:\mathrm{it}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$

Question Number 130331 Answers: 0 Comments: 0

f(x)=ax^3 +bx^2 +c f ′(x)=3ax^2 +2bx f(0)=−2 ⇒a(0)^3 +b(0)^2 +c=−2 ⇒c=−2 f(−2)=2 f ′(−2)=0 ⇒ { ((a(−2)^3 +b(−2)^2 −2=2)),((3a(−2)^2 +2b(−2)=0)) :} ⇒ { ((−8a+4b=4)),((12a−4b=0)) :} −8a+12a+4b−4b=4 ⇒4a=4 ⇒a=1 −8a+4b=4 ⇒b=1+2a ⇒b=1+2 ⇒b=3 f(x)=x^3 +3x^2 −2 f ′(x)=3x^2 +6x f(0)=(0)^3 +3(0)^2 −2=−2 vrai f(−2)=(−2)^3 +3(−2)^2 −2=−8+12−2=2 vrai f ′(−2)=3(−2)^2 +6(−2)=12−12=0 vrai

$$\mathrm{f}\left(\mathrm{x}\right)={a}\mathrm{x}^{\mathrm{3}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{c} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{3}{a}\mathrm{x}^{\mathrm{2}} +\mathrm{2bx} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=−\mathrm{2} \\ $$$$\Rightarrow{a}\left(\mathrm{0}\right)^{\mathrm{3}} +\mathrm{b}\left(\mathrm{0}\right)^{\mathrm{2}} +\mathrm{c}=−\mathrm{2} \\ $$$$\Rightarrow\mathrm{c}=−\mathrm{2} \\ $$$$\mathrm{f}\left(−\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{f}\:'\left(−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{a}\left(−\mathrm{2}\right)^{\mathrm{3}} +{b}\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{2}}\\{\mathrm{3}{a}\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}{b}\left(−\mathrm{2}\right)=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{−\mathrm{8}{a}+\mathrm{4}{b}=\mathrm{4}}\\{\mathrm{12}{a}−\mathrm{4}{b}=\mathrm{0}}\end{cases} \\ $$$$−\mathrm{8}{a}+\mathrm{12}{a}+\mathrm{4}{b}−\mathrm{4}{b}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{4}{a}=\mathrm{4} \\ $$$$\Rightarrow{a}=\mathrm{1} \\ $$$$−\mathrm{8}{a}+\mathrm{4}{b}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{b}=\mathrm{1}+\mathrm{2}{a} \\ $$$$\Rightarrow\mathrm{b}=\mathrm{1}+\mathrm{2} \\ $$$$\Rightarrow\mathrm{b}=\mathrm{3} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} −\mathrm{2} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{3x}^{\mathrm{2}} +\mathrm{6x} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\left(\mathrm{0}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{0}\right)^{\mathrm{2}} −\mathrm{2}=−\mathrm{2}\:\:\:\:\:\:\mathrm{vrai} \\ $$$$\mathrm{f}\left(−\mathrm{2}\right)=\left(−\mathrm{2}\right)^{\mathrm{3}} +\mathrm{3}\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}=−\mathrm{8}+\mathrm{12}−\mathrm{2}=\mathrm{2}\:\:\:\:\mathrm{vrai} \\ $$$$\mathrm{f}\:'\left(−\mathrm{2}\right)=\mathrm{3}\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{6}\left(−\mathrm{2}\right)=\mathrm{12}−\mathrm{12}=\mathrm{0}\:\:\:\:\:\mathrm{vrai} \\ $$

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