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AlgebraQuestion and Answers: Page 223

Question Number 118435 Answers: 2 Comments: 0

If 4 (x^9 )^(1/(4 )) −9 (x^9 )^(1/(8 )) + 4 = 0 , then (x^9 )^(1/(4 )) + (x^(−9) )^(1/(4 )) =?

$${If}\:\mathrm{4}\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{9}} }\:−\mathrm{9}\:\sqrt[{\mathrm{8}\:}]{{x}^{\mathrm{9}} }\:+\:\mathrm{4}\:=\:\mathrm{0}\:,\:{then}\: \\ $$$$\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{9}} }\:+\:\sqrt[{\mathrm{4}\:}]{{x}^{−\mathrm{9}} }\:=? \\ $$

Question Number 118391 Answers: 2 Comments: 0

show that if ^ a^2 +b^2 can be divised by 7, a+b can also be divised by 7.

$${show}\:{that}\:{if}\:\:^{} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:{can}\:{be}\:{divised} \\ $$$${by}\:\mathrm{7},\:{a}+{b}\:{can}\:{also}\:{be}\:{divised}\:{by}\:\mathrm{7}. \\ $$

Question Number 118371 Answers: 0 Comments: 1

old problem question 118120 tan tan x =tan 3x −tan 2x let t=tan x (1) tan t =((t^5 +2t^3 +t)/(3t^4 −4t^2 +1)) for t≥0 we get (approximating) t_0 =0 t_1 ≈1.28941477 t_2 ≈4.17629616 t_3 ≈7.49316173 t_4 ≈10.7303610 t_5 ≈13.9285293 ... x=nπ+arctan t let n=0 to stay in the first period 0≤t<+∞ ⇒ 0≤arctan t <(π/2) ⇒ (1) has infinite solutions for 0≤x<(π/2) graphically this is easy to see, plot these: f_1 (t)=tan t f_2 (t)=((t(t^4 +2t^2 +1))/(3t^4 −4t^2 +1))=(t/3)+((2t(5t^2 +1))/(3(3t^4 −4t^2 +1))) ⇒ g(t)=(1/3)t is asymptote of f_1 (t) and obviously tan t =at with a∈R has infinite solutions

$$\mathrm{old}\:\mathrm{problem}\:{question}\:\mathrm{118120} \\ $$$$\mathrm{tan}\:\mathrm{tan}\:{x}\:=\mathrm{tan}\:\mathrm{3}{x}\:−\mathrm{tan}\:\mathrm{2}{x} \\ $$$$\mathrm{let}\:{t}=\mathrm{tan}\:{x} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\mathrm{tan}\:{t}\:=\frac{{t}^{\mathrm{5}} +\mathrm{2}{t}^{\mathrm{3}} +{t}}{\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{for}\:{t}\geqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\left(\mathrm{approximating}\right) \\ $$$${t}_{\mathrm{0}} =\mathrm{0} \\ $$$${t}_{\mathrm{1}} \approx\mathrm{1}.\mathrm{28941477} \\ $$$${t}_{\mathrm{2}} \approx\mathrm{4}.\mathrm{17629616} \\ $$$${t}_{\mathrm{3}} \approx\mathrm{7}.\mathrm{49316173} \\ $$$${t}_{\mathrm{4}} \approx\mathrm{10}.\mathrm{7303610} \\ $$$${t}_{\mathrm{5}} \approx\mathrm{13}.\mathrm{9285293} \\ $$$$... \\ $$$${x}={n}\pi+\mathrm{arctan}\:{t} \\ $$$$\mathrm{let}\:{n}=\mathrm{0}\:\mathrm{to}\:\mathrm{stay}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{period} \\ $$$$\mathrm{0}\leqslant{t}<+\infty\:\Rightarrow\:\mathrm{0}\leqslant\mathrm{arctan}\:{t}\:<\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\mathrm{1}\right)\:\mathrm{has}\:\mathrm{infinite}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{0}\leqslant{x}<\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{graphically}\:\mathrm{this}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see},\:\mathrm{plot}\:\mathrm{these}: \\ $$$${f}_{\mathrm{1}} \left({t}\right)=\mathrm{tan}\:{t} \\ $$$${f}_{\mathrm{2}} \left({t}\right)=\frac{{t}\left({t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}=\frac{{t}}{\mathrm{3}}+\frac{\mathrm{2}{t}\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\Rightarrow\:{g}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{3}}{t}\:\mathrm{is}\:\mathrm{asymptote}\:\mathrm{of}\:{f}_{\mathrm{1}} \left({t}\right) \\ $$$$\mathrm{and}\:\mathrm{obviously}\:\mathrm{tan}\:{t}\:={at}\:\mathrm{with}\:{a}\in\mathbb{R}\:\mathrm{has} \\ $$$$\mathrm{infinite}\:\mathrm{solutions} \\ $$

Question Number 118346 Answers: 0 Comments: 0

Let 30 furniture sets arrive at two city stations A and B ,15 sets for each station All funiture sets need to be delivered to two furniture stores C and D,and 10 sets must be delivered to store C,and to store D−20.It is known that delivery one furniture set from the station A to the stores C and D costs 1 and 3 monetary units,and from station B−2 and 5 units respectively,.It is necessary to draw out such a transportation plan that the cost of transportation is the lowest

$$\mathrm{Let}\:\mathrm{30}\:\mathrm{furniture}\:\mathrm{sets}\:\mathrm{arrive}\:\mathrm{at}\:\mathrm{two}\:\mathrm{city} \\ $$$$\mathrm{stations}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:,\mathrm{15}\:\mathrm{sets}\:\mathrm{for}\:\mathrm{each}\:\mathrm{station} \\ $$$$\mathrm{All}\:\mathrm{funiture}\:\mathrm{sets}\:\mathrm{need}\:\mathrm{to}\:\mathrm{be}\:\mathrm{delivered} \\ $$$$\mathrm{to}\:\mathrm{two}\:\mathrm{furniture}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D},\mathrm{and}\:\mathrm{10} \\ $$$$\mathrm{sets}\:\mathrm{must}\:\mathrm{be}\:\mathrm{delivered}\:\mathrm{to}\:\mathrm{store}\:\mathrm{C},\mathrm{and} \\ $$$$\mathrm{to}\:\mathrm{store}\:\mathrm{D}−\mathrm{20}.\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{delivery} \\ $$$$\mathrm{one}\:\mathrm{furniture}\:\mathrm{set}\:\mathrm{from}\:\mathrm{the}\:\mathrm{station}\:\mathrm{A}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D}\:\mathrm{costs}\:\mathrm{1}\:\mathrm{and}\:\mathrm{3}\:\mathrm{monetary} \\ $$$$\mathrm{units},\mathrm{and}\:\mathrm{from}\:\mathrm{station}\:\mathrm{B}−\mathrm{2}\:\mathrm{and}\:\mathrm{5}\:\mathrm{units} \\ $$$$\:\mathrm{respectively},.\mathrm{It}\:\mathrm{is}\:\mathrm{necessary}\:\mathrm{to}\:\mathrm{draw} \\ $$$$\mathrm{out}\:\mathrm{such}\:\mathrm{a}\:\mathrm{transportation}\:\mathrm{plan}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{cost}\:\mathrm{of}\:\mathrm{transportation}\:\mathrm{is}\:\mathrm{the}\:\mathrm{lowest} \\ $$

Question Number 118322 Answers: 2 Comments: 2

solve (1/x)+(1/y)+(1/z)=(3/4) with x,y,z∈N

$${solve} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${with}\:{x},{y},{z}\in\mathbb{N} \\ $$

Question Number 118298 Answers: 0 Comments: 1

b, x, y, c are consecutive terms of a G.P. ∴ x = br, y = br^2 , c = br^3 A.M. between b and c is a ∴ ((b+c)/2) = a ⇒ 2a = b+c 2abc = (b+c)bc = b^2 c + bc^2 = b^2 (br^3 ) + b(br^3 )^2 = b^3 r^3 + b^3 r^6 = (br)^3 + (br^2 )^3 = x^3 + y^3 ∴ x^3 + y^3 = 2abc ←

Question Number 118286 Answers: 2 Comments: 0

What condition should be satisfied by the vectors a and b for the following relations to hold true :(a)∣a+b∣=∣a−b∣ ;(b)∣ a+b∣>∣ a−b∣;(c)∣a+ b∣< ∣a−b∣

Question Number 118280 Answers: 0 Comments: 1

(26) OA^(→) = a^→ = (((4.8)),((3.6)) ) , OB^(→) = b^→ = ((( 8)),((15)) ) a^→ . b^→ = (4.8)(8) + (3.6)(15) = 92.4^(→ ) a = (√(4.8^2 +3.6^2 )) = (√(23.04+12.96)) = (√(36)) = 6 b = (√(8^2 +15^2 )) = (√(64+225)) = (√(289)) = 17 a.b = 6 . 17 = 102 cos AOB = ((a^→ .b^→ )/(a.b)) = ((92.4)/(102)) = 0.9059 = cos 25.06° ∴ ∠AOB = 25.06°

$$\left(\mathrm{26}\right)\:\:\:\overset{\rightarrow} {\mathrm{OA}}\:=\:\overset{\rightarrow} {\mathrm{a}}\:=\:\begin{pmatrix}{\mathrm{4}.\mathrm{8}}\\{\mathrm{3}.\mathrm{6}}\end{pmatrix}\:\:,\:\:\overset{\rightarrow} {\mathrm{OB}}\:=\:\overset{\rightarrow} {\mathrm{b}}\:=\:\begin{pmatrix}{\:\:\mathrm{8}}\\{\mathrm{15}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}.\:\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{4}.\mathrm{8}\right)\left(\mathrm{8}\right)\:+\:\left(\mathrm{3}.\mathrm{6}\right)\left(\mathrm{15}\right)\:=\:\mathrm{92}.\overset{\rightarrow\:} {\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\sqrt{\mathrm{4}.\mathrm{8}^{\mathrm{2}} +\mathrm{3}.\mathrm{6}^{\mathrm{2}} }\:=\:\:\sqrt{\mathrm{23}.\mathrm{04}+\mathrm{12}.\mathrm{96}}\:=\:\sqrt{\mathrm{36}}\:=\:\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:=\:\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{64}+\mathrm{225}}\:=\:\sqrt{\mathrm{289}}\:=\:\mathrm{17} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}.\mathrm{b}\:=\:\mathrm{6}\:.\:\mathrm{17}\:=\:\mathrm{102} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\mathrm{AOB}\:=\:\frac{\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}}{\mathrm{a}.\mathrm{b}}\:=\:\frac{\mathrm{92}.\mathrm{4}}{\mathrm{102}}\:=\:\mathrm{0}.\mathrm{9059}\:=\:\mathrm{cos}\:\mathrm{25}.\mathrm{06}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\angle\mathrm{AOB}\:=\:\mathrm{25}.\mathrm{06}° \\ $$

Question Number 118277 Answers: 0 Comments: 0

(23) Let a^→ = ((( 7)),((24)) ) and b^→ = ((( 3)),((−4)) ) a^→ .b^→ = (7)(3) + (24)(−4) = 21−96 = −75 a = (√(7^2 +24^2 )) = (√(49+576)) = (√(625)) = 25 b = (√(3^2 +(−4)^2 )) = (√(9+16)) = (√(25)) = 5 a.b = 25 . 5 = 125 Let θ be the angle between a^→ and b^→ . cosθ = ((a^→ .b^→ )/(a.b)) = ((−75)/(125)) = −0.6 = cos126.87° ∴ θ = 126.87°

$$\left(\mathrm{23}\right)\:\mathrm{Let}\:\overset{\rightarrow} {\mathrm{a}}\:=\:\begin{pmatrix}{\:\:\mathrm{7}}\\{\mathrm{24}}\end{pmatrix}\:\:\:\mathrm{and}\:\:\overset{\rightarrow} {\mathrm{b}}\:=\:\begin{pmatrix}{\:\:\:\mathrm{3}}\\{−\mathrm{4}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{7}\right)\left(\mathrm{3}\right)\:+\:\left(\mathrm{24}\right)\left(−\mathrm{4}\right)\:=\:\mathrm{21}−\mathrm{96}\:=\:−\mathrm{75} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\:\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{49}+\mathrm{576}}\:=\:\sqrt{\mathrm{625}}\:=\:\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:=\:\:\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{4}\right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{9}+\mathrm{16}}\:=\:\sqrt{\mathrm{25}}\:=\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}.\mathrm{b}\:=\:\mathrm{25}\:.\:\mathrm{5}\:=\:\:\mathrm{125}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Let}\:\theta\:\mathrm{be}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\overset{\rightarrow} {\mathrm{a}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{b}}. \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{cos}\theta\:=\:\frac{\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}}{\mathrm{a}.\mathrm{b}}\:=\:\frac{−\mathrm{75}}{\mathrm{125}}\:=\:−\mathrm{0}.\mathrm{6}\:=\:\mathrm{cos126}.\mathrm{87}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\theta\:=\:\mathrm{126}.\mathrm{87}°\:\:\:\: \\ $$

Question Number 118275 Answers: 1 Comments: 1

Question Number 118263 Answers: 0 Comments: 0

Question Number 118227 Answers: 1 Comments: 0

If x = (√(42−(√(42−(√(42−...)))))) y = (√(x+(√(x+(√(x+...)))))) z=(√(y.(√(y.(√(y.(√(y...)))))))) . Find x+y+z .

$${If}\:{x}\:=\:\sqrt{\mathrm{42}−\sqrt{\mathrm{42}−\sqrt{\mathrm{42}−...}}} \\ $$$${y}\:=\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...}}} \\ $$$${z}=\sqrt{{y}.\sqrt{{y}.\sqrt{{y}.\sqrt{{y}...}}}}\:.\:{Find}\:{x}+{y}+{z}\:. \\ $$

Question Number 118196 Answers: 1 Comments: 0

solve in N: b^3 (2b^2 +2b+1)=18360

$${solve}\:{in}\:\mathbb{N}: \\ $$$${b}^{\mathrm{3}} \left(\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{b}+\mathrm{1}\right)=\mathrm{18360} \\ $$

Question Number 118193 Answers: 1 Comments: 0

Given A=n^2 −2n+2 , B=n^2 +2n+2 n ∈ N^∗ −{1}. Show that ∀ divisor of A which divise n can also divise 2. Show that all common divisor of A and B can divise 4n.

$${Given}\:{A}={n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{2}\:,\:{B}={n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{2} \\ $$$${n}\:\in\:\mathbb{N}^{\ast} −\left\{\mathrm{1}\right\}. \\ $$$${Show}\:{that}\:\forall\:{divisor}\:{of}\:{A}\:{which}\:{divise} \\ $$$${n}\:{can}\:{also}\:{divise}\:\mathrm{2}. \\ $$$${Show}\:{that}\:{all}\:{common}\:{divisor}\:{of}\: \\ $$$${A}\:{and}\:{B}\:{can}\:{divise}\:\mathrm{4}{n}. \\ $$

Question Number 118184 Answers: 2 Comments: 1

factorise x^4 +4

$${factorise}\:{x}^{\mathrm{4}} +\mathrm{4} \\ $$

Question Number 118181 Answers: 1 Comments: 1

find all numbers >1 from N which their cube are <18360

$${find}\:{all}\:{numbers}\:>\mathrm{1}\:{from}\:\mathbb{N}\:{which} \\ $$$${their}\:{cube}\:{are}\:<\mathrm{18360} \\ $$

Question Number 118180 Answers: 1 Comments: 0

show that if n is odd , n(n^2 +3) is even.

$${show}\:{that}\:{if}\:{n}\:{is}\:{odd}\:,\:{n}\left({n}^{\mathrm{2}} +\mathrm{3}\right)\:{is}\:{even}. \\ $$

Question Number 118172 Answers: 1 Comments: 0

Find the area of a rhombus with side 8 cm

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rhombus}\:\mathrm{with}\:\mathrm{side}\:\:\mathrm{8}\:\mathrm{cm} \\ $$

Question Number 118023 Answers: 1 Comments: 0

..calculus.. x,y,z ∈R^+ and x^2 +y^2 +z^2 =1 find min_(x,y,z∈R^(+ ) ) ((((yz)/x)+((xz)/y)+((xy)/z)) )=? m.n.1970..

$$ \\ $$$$\:\:\:\:\:\:\:..{calculus}.. \\ $$$$\:\:{x},{y},{z}\:\in\mathbb{R}^{+} \:\:{and}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:=\mathrm{1} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{find}\:\:\:\:\:\: \\ $$$$\:\:\:\:{min}_{{x},{y},{z}\in\mathbb{R}^{+\:\:\:\:} } \left(\left(\frac{{yz}}{{x}}+\frac{{xz}}{{y}}+\frac{{xy}}{{z}}\right)\:\right)=? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970}.. \\ $$

Question Number 118011 Answers: 4 Comments: 0

Question Number 118002 Answers: 1 Comments: 0

Question Number 117984 Answers: 1 Comments: 0

If f(x) is a polynomial function satisfying the relation f(x)+f((1/x))=f(x)f((1/x)) for all 0≠x∈R and if f(2)=9, then f(6) is (A) 216 (B) 217 (C) 126 (D) 127

$$\mathrm{If}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)={f}\left({x}\right){f}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{0}\neq{x}\in\mathbb{R}\:\mathrm{and}\:\mathrm{if}\:{f}\left(\mathrm{2}\right)=\mathrm{9},\:\mathrm{then}\:\mathrm{f}\left(\mathrm{6}\right)\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{216}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{217}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{126}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{127} \\ $$

Question Number 117972 Answers: 2 Comments: 0

The number of surjections of {1,2,3,4} onto {x,y} is (A) 16 (B) 8 (C) 14 (D) 6

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{surjections}\:\mathrm{of}\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\right\}\:\mathrm{onto}\:\left\{\mathrm{x},\mathrm{y}\right\}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{16}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{14}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{6} \\ $$

Question Number 117934 Answers: 1 Comments: 0

Let f : [1,∞)→[2,∞) be the function defined by f(x)=x+(1/x) If g : [2,∞)→[1,∞), is a function such that (g○f)(x)=x for all x≥1. Show that g(t)=((t+(√(t^2 −4)))/2)

$$\mathrm{Let}\:{f}\::\:\left[\mathrm{1},\infty\right)\rightarrow\left[\mathrm{2},\infty\right)\:\mathrm{be}\:\mathrm{the}\:\mathrm{function}\:\mathrm{defined}\:\mathrm{by} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)={x}+\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{If}\:\mathrm{g}\::\:\left[\mathrm{2},\infty\right)\rightarrow\left[\mathrm{1},\infty\right),\:\mathrm{is}\:\mathrm{a}\:\mathrm{function}\:\mathrm{such}\:\mathrm{that}\:\left(\mathrm{g}\circ{f}\right)\left({x}\right)={x} \\ $$$$\mathrm{for}\:\mathrm{all}\:{x}\geqslant\mathrm{1}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{g}\left({t}\right)=\frac{{t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$

Question Number 117828 Answers: 1 Comments: 0

1)((√3)−1)((√3)+1)=(√3)×(√3)−(√3)−1 =3−(√3)−1 =2−(√3) 2)(2x+(√3))(2x−(√3))=(2x)^2 −2x(√3)+2x(√3)−3 =4x^2 −3

$$\left.\mathrm{1}\right)\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)=\sqrt{\mathrm{3}}×\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$$=\mathrm{3}−\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$$=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right)\left(\mathrm{2}{x}+\sqrt{\mathrm{3}}\right)\left(\mathrm{2}{x}−\sqrt{\mathrm{3}}\right)=\left(\mathrm{2}{x}\right)^{\mathrm{2}} −\mathrm{2}{x}\sqrt{\mathrm{3}}+\mathrm{2}{x}\sqrt{\mathrm{3}}−\mathrm{3} \\ $$$$=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{3} \\ $$

Question Number 117827 Answers: 0 Comments: 5

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