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AlgebraQuestion and Answers: Page 226

Question Number 119939 Answers: 3 Comments: 0

Question Number 119921 Answers: 2 Comments: 0

solving the following system of equations { ((((3x−y)/(x−3y))=x^2 )),((((3y−z)/(y−3z))=y^2 )),((((3z−x)/(z−3x))=z^2 )) :}

$${solving}\:{the}\:{following}\:{system}\:{of}\:{equations} \\ $$$$\begin{cases}{\frac{\mathrm{3}{x}−{y}}{{x}−\mathrm{3}{y}}={x}^{\mathrm{2}} }\\{\frac{\mathrm{3}{y}−{z}}{{y}−\mathrm{3}{z}}={y}^{\mathrm{2}} }\\{\frac{\mathrm{3}{z}−{x}}{{z}−\mathrm{3}{x}}={z}^{\mathrm{2}} }\end{cases} \\ $$

Question Number 119894 Answers: 2 Comments: 0

Question Number 119849 Answers: 1 Comments: 0

Find all pair(x,y) of real numbers that are the solutions to the system { ((x^4 +2x^3 −y=−(1/4)+(√3))),((y^4 +2y^3 −x=−(1/4)−(√3))) :}

$${Find}\:{all}\:{pair}\left({x},{y}\right)\:{of}\:{real}\:{numbers} \\ $$$${that}\:{are}\:{the}\:{solutions}\:{to}\:{the}\:{system} \\ $$$$\begin{cases}{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −{y}=−\frac{\mathrm{1}}{\mathrm{4}}+\sqrt{\mathrm{3}}}\\{{y}^{\mathrm{4}} +\mathrm{2}{y}^{\mathrm{3}} −{x}=−\frac{\mathrm{1}}{\mathrm{4}}−\sqrt{\mathrm{3}}}\end{cases} \\ $$

Question Number 119848 Answers: 1 Comments: 0

Solve in real numbers the equation (x)^(1/(3 )) + ((x−1))^(1/(3 )) + ((x+1))^(1/(3 )) = 0

$${Solve}\:{in}\:{real}\:{numbers}\:{the}\:{equation} \\ $$$$\sqrt[{\mathrm{3}\:}]{{x}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}−\mathrm{1}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}+\mathrm{1}}\:=\:\mathrm{0} \\ $$

Question Number 119832 Answers: 0 Comments: 0

evaluate: I = ∫_0 ^( 1) (((x+1)/x))^(x!) dx

$$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{evaluate}:\:\:\:\:{I}\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{{x}}\right)^{{x}!} {dx} \\ $$$$ \\ $$$$ \\ $$

Question Number 119831 Answers: 0 Comments: 1

evaluate: I = ∫_1 ^( ∞) ((1/x))^x dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{evaluate}:\:\:\:{I}\:\:=\:\int_{\mathrm{1}} ^{\:\infty} \left(\frac{\mathrm{1}}{{x}}\right)^{{x}} {dx} \\ $$$$\: \\ $$$$\: \\ $$

Question Number 119807 Answers: 1 Comments: 0

Let f be a real-valued function defined on the inte- rval [−1, 1]. If the area of the equilateral triangle with (0, 0) and (x, f(x)) as two vertices is (√3)/4, then f(x) is equal to (A) (√(1−x^2 )) (B) (√(1+x^2 )) (C) −(√(1−x^2 )) (D) −(√(1+x^2 ))

$$\mathrm{Let}\:{f}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}-\mathrm{valued}\:\mathrm{function}\:\mathrm{defined}\:\mathrm{on}\:\mathrm{the}\:\mathrm{inte}- \\ $$$$\mathrm{rval}\:\left[−\mathrm{1},\:\mathrm{1}\right].\:\mathrm{If}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{with} \\ $$$$\left(\mathrm{0},\:\mathrm{0}\right)\:\mathrm{and}\:\left(\mathrm{x},\:{f}\left(\mathrm{x}\right)\right)\:\mathrm{as}\:\mathrm{two}\:\mathrm{vertices}\:\mathrm{is}\:\sqrt{\mathrm{3}}/\mathrm{4},\:\mathrm{then}\:{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{A}\right)\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{C}\right)\:−\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:−\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$

Question Number 119800 Answers: 0 Comments: 2

Examples of functions such that f(x+y)=f(x)+f(y) for all x,y∈R

$$\mathrm{Examples}\:\mathrm{of}\:\mathrm{functions}\:\mathrm{such}\:\mathrm{that} \\ $$$${f}\left(\mathrm{x}+\mathrm{y}\right)={f}\left(\mathrm{x}\right)+{f}\left(\mathrm{y}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{x},\mathrm{y}\in\mathbb{R} \\ $$

Question Number 119801 Answers: 0 Comments: 0

If f:R→R is a function such that f(0)=1 and f(x+f(y))= f(x)+y for all x, y∈R, then (A) 1 is a period of f (B) f(n)=1 for all integers n (C) f(n)=n for all integers n (D) f(−1)=0

$$\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{a}\:\mathrm{function}\:\mathrm{such}\:\mathrm{that}\:{f}\left(\mathrm{0}\right)=\mathrm{1}\:\mathrm{and}\:{f}\left(\mathrm{x}+{f}\left(\mathrm{y}\right)\right)= \\ $$$${f}\left(\mathrm{x}\right)+\mathrm{y}\:\mathrm{for}\:\mathrm{all}\:\mathrm{x},\:\mathrm{y}\in\mathbb{R},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f} \\ $$$$\left(\mathrm{B}\right)\:{f}\left({n}\right)=\mathrm{1}\:\mathrm{for}\:\mathrm{all}\:\mathrm{integers}\:{n} \\ $$$$\left(\mathrm{C}\right)\:{f}\left({n}\right)={n}\:\mathrm{for}\:\mathrm{all}\:\mathrm{integers}\:{n} \\ $$$$\left(\mathrm{D}\right)\:{f}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$

Question Number 119797 Answers: 0 Comments: 0

Q1 Let M_2 be the set of square matrices of order 2 over the real number system and R={(A,B)∈M_2 ×M_2 ∣A=P^( T) BP for some non-singular P ∈M} Then R is (A) symmetric (B) transitive (C) reflexive on M_2 (D) not an equivalence relation on M_2 Q2 For any integer n, let I_n be the interval (n, n+1). Define R={(x, y)∈R∣both x, y ∈ I_n for some n∈Z} Then R is (A) reflexive on R (B) symmetric (C) transitive (D) an equivalence relation

Question Number 119795 Answers: 4 Comments: 0

Solve in real numbers the system of equations { (((3x+y)(x+3y)(√(xy)) =14)),(((x+y)(x^2 +14xy+y^2 )= 36)) :}

$${Solve}\:{in}\:{real}\:{numbers}\:{the}\:{system}\:{of} \\ $$$${equations}\:\begin{cases}{\left(\mathrm{3}{x}+{y}\right)\left({x}+\mathrm{3}{y}\right)\sqrt{{xy}}\:=\mathrm{14}}\\{\left({x}+{y}\right)\left({x}^{\mathrm{2}} +\mathrm{14}{xy}+{y}^{\mathrm{2}} \right)=\:\mathrm{36}}\end{cases}\: \\ $$

Question Number 119774 Answers: 1 Comments: 0

Question Number 119757 Answers: 0 Comments: 0

For any integer n, let I_n be the interval (n, n+1). Define R={(x, y)∈R∣both x, y ∈ I_n for some n∈Z} Then R is (A) reflexive on R (B) symmetric (C) transitive (D) an equivalence relation

$$\mathrm{For}\:\mathrm{any}\:\mathrm{integer}\:{n},\:\mathrm{let}\:{I}_{{n}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{interval}\:\left({n},\:{n}+\mathrm{1}\right). \\ $$$$\mathrm{Define} \\ $$$$\:\:\:\:\:\:\:\mathcal{R}=\left\{\left(\mathrm{x},\:\mathrm{y}\right)\in\mathbb{R}\mid\mathrm{both}\:\mathrm{x},\:\mathrm{y}\:\in\:{I}_{{n}} \:\mathrm{for}\:\mathrm{some}\:{n}\in\mathbb{Z}\right\} \\ $$$$\mathrm{Then}\:\mathcal{R}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{reflexive}\:\mathrm{on}\:\mathbb{R} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{symmetric} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{transitive} \\ $$$$\left(\mathrm{D}\right)\:\mathrm{an}\:\mathrm{equivalence}\:\mathrm{relation} \\ $$

Question Number 119713 Answers: 1 Comments: 3

If 3x+(1/(2x))=6 find 8x^3 +(1/(27x^3 ))

$$\mathrm{If}\:\mathrm{3x}+\frac{\mathrm{1}}{\mathrm{2x}}=\mathrm{6}\:\mathrm{find}\:\mathrm{8x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{27x}^{\mathrm{3}} } \\ $$

Question Number 119635 Answers: 1 Comments: 0

If a function f:R→R satisfies the relation f(x+1)+f(x−1)=(√3)f(x) for all x∈R then a period of f is (A) 10 (B) 12 (C) 6 (D) 4

$$\mathrm{If}\:\mathrm{a}\:\mathrm{function}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:\:\:\:\:\:{f}\left(\mathrm{x}+\mathrm{1}\right)+{f}\left(\mathrm{x}−\mathrm{1}\right)=\sqrt{\mathrm{3}}{f}\left(\mathrm{x}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{10}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{12}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4} \\ $$

Question Number 119634 Answers: 2 Comments: 0

Q1 If f:R→R is defined by f(x)=[x]+[x+(1/2)]+[x+(2/3)]−3x+5 where [x] is the integral part of x, then a period of f is (A) 1 (B) 2/3 (C) 1/2 (D) 1/3 Q2 Let a<c<b such that c−a=b−c. If f:R→R is a function satisfying the relation f(x+a)+f(x+b)=f(x+c) for all x∈R then a period of f is (A) (b−a) (B) 2(b−a) (C) 3(b−a) (D) 4(b−a)

Question Number 119580 Answers: 3 Comments: 0

Given k ∈ N. 1) justify these relations: 3^(2k) +1≡2[8] and 3^(2k+1) +1≡4[8]. 2) Given (E): 2^n −3^m =1. n and m are unknowed. • Show that if m is even , (E) does not have solution. ■ Deduct from the first question 1) that the couple (2;1) is the only solution of (E).

$$\mathrm{Given}\:\mathrm{k}\:\in\:\mathbb{N}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{justify}\:\mathrm{these}\:\mathrm{relations}: \\ $$$$\mathrm{3}^{\mathrm{2k}} +\mathrm{1}\equiv\mathrm{2}\left[\mathrm{8}\right]\:\mathrm{and}\:\mathrm{3}^{\mathrm{2k}+\mathrm{1}} +\mathrm{1}\equiv\mathrm{4}\left[\mathrm{8}\right]. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Given}\:\left(\mathrm{E}\right):\:\mathrm{2}^{\mathrm{n}} −\mathrm{3}^{\mathrm{m}} =\mathrm{1}.\:\mathrm{n}\:\mathrm{and}\:\mathrm{m}\:\mathrm{are}\:\mathrm{unknowed}. \\ $$$$\bullet\:\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{m}\:\mathrm{is}\:\mathrm{even}\:,\:\left(\mathrm{E}\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\: \\ $$$$\mathrm{solution}. \\ $$$$\left.\blacksquare\:\mathrm{Deduct}\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{question}\:\mathrm{1}\right)\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{couple}\:\left(\mathrm{2};\mathrm{1}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{of}\:\left(\mathrm{E}\right). \\ $$

Question Number 119576 Answers: 2 Comments: 0

Question Number 119540 Answers: 1 Comments: 0

Question Number 119538 Answers: 1 Comments: 1

Question Number 119490 Answers: 0 Comments: 0

Let a<c<b such that c−a=b−c. If f:R→R is a function satisfying the relation f(x+a)+f(x+b)=f(x+c) for all x∈R then a period of f is (A) (b−a) (B) 2(b−a) (C) 3(b−a) (D) 4(b−a)

$$\mathrm{Let}\:{a}<\mathrm{c}<\mathrm{b}\:\mathrm{such}\:\mathrm{that}\:\mathrm{c}−{a}=\mathrm{b}−\mathrm{c}.\:\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:{f}\left(\mathrm{x}+{a}\right)+{f}\left(\mathrm{x}+\mathrm{b}\right)={f}\left(\mathrm{x}+\mathrm{c}\right)\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\left(\mathrm{b}−{a}\right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{3}\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4}\left(\mathrm{b}−{a}\right) \\ $$

Question Number 119483 Answers: 1 Comments: 0

Let a>0 and f:R→R a function satisfying f(x+a)=1+[2−3f(x)+3f(x)^2 −f(x)^3 ]^(1/3) for all x∈R. Then a period of f(x) is ka where k is a positive integer whose value is (A)1 (B)2 (C)3 (D)4

$$\mathrm{Let}\:{a}>\mathrm{0}\:\mathrm{and}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{a}\:\mathrm{function}\:\mathrm{satisfying} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{x}+{a}\right)=\mathrm{1}+\left[\mathrm{2}−\mathrm{3}{f}\left(\mathrm{x}\right)+\mathrm{3}{f}\left(\mathrm{x}\right)^{\mathrm{2}} −{f}\left(\mathrm{x}\right)^{\mathrm{3}} \right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R}.\:\mathrm{Then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\left(\mathrm{x}\right)\:\mathrm{is}\:{ka}\:\mathrm{where}\:{k}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{whose}\:\mathrm{value}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\mathrm{4} \\ $$

Question Number 119396 Answers: 3 Comments: 1

If the roots of the equation 24x^4 −52x^3 +18x^2 +13x−6=0 are α , −α , β and (1/β). Find the value of α and β.

$${If}\:{the}\:{roots}\:{of}\:{the}\:{equation}\: \\ $$$$\mathrm{24}{x}^{\mathrm{4}} −\mathrm{52}{x}^{\mathrm{3}} +\mathrm{18}{x}^{\mathrm{2}} +\mathrm{13}{x}−\mathrm{6}=\mathrm{0}\:{are}\: \\ $$$$\alpha\:,\:−\alpha\:,\:\beta\:{and}\:\frac{\mathrm{1}}{\beta}.\:{Find}\:{the}\:{value}\:{of}\: \\ $$$$\alpha\:{and}\:\beta. \\ $$

Question Number 119303 Answers: 3 Comments: 0

let x,y,z be positive real numbers such that x+y+z=1. Determine the minimum value of (1/x)+(4/y)+(9/z).

$$\:{let}\:{x},{y},{z}\:{be}\:{positive}\:{real}\:{numbers}\: \\ $$$${such}\:{that}\:{x}+{y}+{z}=\mathrm{1}.\:{Determine}\: \\ $$$${the}\:{minimum}\:{value}\:{of}\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{4}}{{y}}+\frac{\mathrm{9}}{{z}}. \\ $$

Question Number 119204 Answers: 0 Comments: 0

40−misolning yechimi: y=(x^3 /3)+2x^2 −5x+7 Kritik nuqtalarini topish uchun: 1. Funksiyadan hosila olamiz 2. Funksiya hosilasini nolga tenglab, tenglamani yechamiz. y′=x^2 +4x−5=0 ⇒ x_1 =1; x_2 =−5 x_1 +x_2 =1−5=−4 Javob: −4

$$\mathrm{40}−\mathrm{misolning}\:\:\:\mathrm{yechimi}: \\ $$$$\mathrm{y}=\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{7} \\ $$$$\mathrm{Kritik}\:\mathrm{nuqtalarini}\:\mathrm{topish}\:\mathrm{uchun}:\: \\ $$$$\mathrm{1}.\:\mathrm{Funksiyadan}\:\mathrm{hosila}\:\mathrm{olamiz} \\ $$$$\mathrm{2}.\:\mathrm{Funksiya}\:\mathrm{hosilasini}\:\mathrm{nolga}\:\mathrm{tenglab},\: \\ $$$$\mathrm{tenglamani}\:\mathrm{yechamiz}. \\ $$$$\mathrm{y}'=\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{5}=\mathrm{0}\:\:\:\Rightarrow\:\:\:\mathrm{x}_{\mathrm{1}} =\mathrm{1};\:\:\mathrm{x}_{\mathrm{2}} =−\mathrm{5} \\ $$$$\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =\mathrm{1}−\mathrm{5}=−\mathrm{4}\:\:\:\:\:\mathrm{Javob}:\:\:−\mathrm{4}\:\:\:\: \\ $$

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