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AlgebraQuestion and Answers: Page 54

Question Number 196582    Answers: 1   Comments: 0

If x = log_a bc, y = log_b ca and z = log_c ab then prove that x + y + z = xyz − 2.

Ifx=logabc,y=logbcaandz=logcabthenprovethatx+y+z=xyz2.

Question Number 196555    Answers: 0   Comments: 0

In △ABC show that Σ ((1 + cos ∙ (A − B) ∙ cos C)/(h_C ∙ sec C)) = (3/(2 R))

InABCshowthatΣ1+cos(AB)cosChCsecC=32R

Question Number 196614    Answers: 2   Comments: 0

if S_n =(1/(1+5n))+(1/(2+5n))+(1/(3+5n))+...+(1/(6n)), find lim_(n→∞) S_n =?

ifSn=11+5n+12+5n+13+5n+...+16n,findlimnSn=?

Question Number 196612    Answers: 0   Comments: 1

Question Number 196544    Answers: 1   Comments: 0

Question Number 196519    Answers: 1   Comments: 0

if xyz=1, prove ((x/(x−1)))^2 +((y/(y−1)))^2 +((z/(z−1)))^2 ≥1.

ifxyz=1,prove(xx1)2+(yy1)2+(zz1)21.

Question Number 196515    Answers: 1   Comments: 0

Find: ∫_0 ^( +∞) x^𝛑 e^(−x) dx = ?

Find:0+xπexdx=?

Question Number 196483    Answers: 0   Comments: 0

Question Number 196450    Answers: 3   Comments: 0

one solution of the equation (x−a)(x−b)(x−c)(x−d) = 9 is x=2. If a,b,c,d are different integers then a+b+c+d =?

onesolutionoftheequation(xa)(xb)(xc)(xd)=9isx=2.Ifa,b,c,daredifferentintegersthena+b+c+d=?

Question Number 196364    Answers: 1   Comments: 0

xp(x)=x^3 −2x^2 +x−a p(−1)=?

xp(x)=x32x2+xap(1)=?

Question Number 196360    Answers: 2   Comments: 1

Question Number 196355    Answers: 2   Comments: 1

log_5 (√(5(√(5(√(5(√(5.....)))))) ))= ?

log55555.....=?

Question Number 196358    Answers: 0   Comments: 2

Question Number 196343    Answers: 2   Comments: 0

find the minimum value of f(x) f(x) = (√(x^2 −2x +5)) + (√(4x^2 −4x +10 ))

findtheminimumvalueoff(x)f(x)=x22x+5+4x24x+10

Question Number 196325    Answers: 1   Comments: 0

Σ_(n=0) ^∞ arg(n^2 +n+1+i)= π/2

n=0arg(n2+n+1+i)=π/2

Question Number 196300    Answers: 2   Comments: 0

If → n ∈ N and n ≥ 2 Then → tan ((1/(n − 1)) Σ_(k=2) ^n arctan (1/k)) < (2/5) + (𝛄/(n − 1))

IfnNandn2Thentan(1n1nk=2arctan1k)<25+γn1

Question Number 196299    Answers: 1   Comments: 0

If → y = x ! find → (dy/dx)

Ify=x!finddydx

Question Number 196288    Answers: 3   Comments: 1

4^x =(√5^y )=400 ((xy)/(2x+y))=?

4x=5y=400xy2x+y=?

Question Number 196285    Answers: 0   Comments: 1

problem 196258 (please)

problem196258(please)

Question Number 196185    Answers: 1   Comments: 0

Question Number 196183    Answers: 2   Comments: 0

Question Number 196165    Answers: 3   Comments: 0

{: ((x^(x+y) =y^(24) )),((y^(x+y) =x^6 )) }⇒(x,y)=(?,?)

xx+y=y24yx+y=x6}(x,y)=(?,?)

Question Number 196121    Answers: 0   Comments: 0

In forest , a hunter obstains that Every morning a snake eats a mouse Every afternoon a scorpion kills a snake Every night a mouse corrodes a scorpion The 8^(th) day morning , there remains Only one of them , a mouse How many were they, in each species?

Inforest,ahunterobstainsthatEverymorningasnakeeatsamouseEveryafternoonascorpionkillsasnakeEverynightamousecorrodesascorpionThe8thdaymorning,thereremainsOnlyoneofthem,amouseHowmanywerethey,ineachspecies?

Question Number 196119    Answers: 5   Comments: 0

solve (√(100−x^2 ))+(√(64−x^2 ))=12

solve100x2+64x2=12

Question Number 196102    Answers: 0   Comments: 0

Question Number 196086    Answers: 0   Comments: 0

Answer to the question “196008” Σ_(k=1) ^n tan^2 (((kπ)/(2n+1)))=n(2n+1) Ans) according “de moivre” sin(2n+1)α= (((2n+1)),(( 1)) )(cosα)^(2n) sinα− (((2n+1)),(( 3)) )(cosα)^(2n−2) (sinα)^3 +.... =(cosα)^(2n) (sinα)[ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) ) tan^2 α+...] for “α_k =((kπ)/(2n+1)) ; 1≤k≤n ⇒sin(2n+1)α_k =0 ⇒∀ 1≤k≤n→ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) ) tan^2 α_k +...=0 therefore “ x_k =tan^2 α_k ” thr roots of the equation are blowe x^n − (((2n+1)),((2n−1)) ) x^n + (((2n+1)),((2n−3)) )x^(n−1) −...=0 the sume of the roots of the equation is “ s= (((2n+1)),((2n−1)) ) ” ⇒s=Σ_(k=1) ^n tan^2 (((kπ)/(2n+1)))=n(2n+1)✓ the proof of the seconf part is done similarly

Answertothequestion196008nk=1tan2(kπ2n+1)=n(2n+1)Ans)accordingdemoivresin(2n+1)α=(2n+11)(cosα)2nsinα(2n+13)(cosα)2n2(sinα)3+....=(cosα)2n(sinα)[(2n+11)(2n+13)tan2α+...]forαk=kπ2n+1;1knsin(2n+1)αk=01kn(2n+11)(2n+13)tan2αk+...=0thereforexk=tan2αkthrrootsoftheequationareblowexn(2n+12n1)xn+(2n+12n3)xn1...=0thesumeoftherootsoftheequationiss=(2n+12n1)s=nk=1tan2(kπ2n+1)=n(2n+1)theproofoftheseconfpartisdonesimilarly

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