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Question Number 112356 Answers: 3 Comments: 1

Question Number 112338 Answers: 0 Comments: 0

Question Number 112334 Answers: 0 Comments: 0

Question Number 112331 Answers: 0 Comments: 2

Question Number 112328 Answers: 1 Comments: 0

find the all root (m^3 −6m+9)=0

$${find}\:{the}\:{all}\:{root}\:\left({m}^{\mathrm{3}} −\mathrm{6}{m}+\mathrm{9}\right)=\mathrm{0} \\ $$

Question Number 112313 Answers: 5 Comments: 2

solve ∫_0 ^1 ((lnx)/(√(1+x^2 )))dx

$${solve} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$

Question Number 112323 Answers: 0 Comments: 1

_≤^≥ = SOLVE the EQUATION_ ^ _( •) n−⌊(√n)⌋−⌊(n)^(1/3) ⌋+⌊(n)^(1/6) ⌋=2016

$$\:_{\leqslant} ^{\geqslant} =\:\:\mathbb{SOLVE}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathcal{EQUATION}}_{} ^{\:_{\:\:\bullet} } \\ $$$$\:\:\:\boldsymbol{\mathrm{n}}−\lfloor\sqrt{\boldsymbol{\mathrm{n}}}\rfloor−\lfloor\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{n}}}\rfloor+\lfloor\sqrt[{\mathrm{6}}]{\boldsymbol{\mathrm{n}}}\rfloor=\mathrm{2016} \\ $$$$ \\ $$

Question Number 112320 Answers: 2 Comments: 1

Question Number 112311 Answers: 1 Comments: 2

lim_(x→0) (√((3sin^2 x+cos 2x−1)/(x.tan 2x))) ?

$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt{\frac{\mathrm{3sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{2x}−\mathrm{1}}{\mathrm{x}.\mathrm{tan}\:\mathrm{2x}}}\:? \\ $$

Question Number 112310 Answers: 4 Comments: 0

Question Number 112308 Answers: 0 Comments: 0

Question Number 112301 Answers: 0 Comments: 1

Question Number 112280 Answers: 2 Comments: 0

If 15 ≤ ∣10−(1/3)a∣ < 20 find a ∈ R

$$\:\:\mathrm{If}\:\mathrm{15}\:\leqslant\:\mid\mathrm{10}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{a}\mid\:<\:\mathrm{20}\: \\ $$$$\mathrm{find}\:\mathrm{a}\:\in\:\mathbb{R} \\ $$

Question Number 112271 Answers: 2 Comments: 0

(√(bemath)) ∫ sin x (√(1−sin x)) dx ?

$$\:\:\:\:\sqrt{\mathrm{bemath}} \\ $$$$\:\:\:\int\:\mathrm{sin}\:\mathrm{x}\:\sqrt{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}\:\mathrm{dx}\:? \\ $$

Question Number 112266 Answers: 1 Comments: 0

Let Ω denote the circumcircle of ABC. The tangent to Ω at A meets BC at X. Let the angle bisectors of ∠AXB meet AC and AB at E and F respectively. D is the foot of the angle bisector from ∠BAC on BC. Let AD intersect EF at K and Ω again at L(other than A). Prove that AEDF is a rhombus and further prove that the circle defined by triangle KLX passes through the midpoint of line segment BC.

$$\mathrm{Let}\:\Omega\:\mathrm{denote}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of}\:\mathrm{ABC}. \\ $$$$\mathrm{The}\:\mathrm{tangent}\:\mathrm{to}\:\Omega\:\mathrm{at}\:\mathrm{A}\:\mathrm{meets}\:\mathrm{BC}\:\mathrm{at}\:\mathrm{X}. \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{bisectors}\:\mathrm{of}\:\angle\mathrm{AXB}\:\mathrm{meet} \\ $$$$\mathrm{AC}\:\mathrm{and}\:\mathrm{AB}\:\mathrm{at}\:\mathrm{E}\:\mathrm{and}\:\mathrm{F} \\ $$$$\mathrm{respectively}.\:\mathrm{D}\:\mathrm{is}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle} \\ $$$$\mathrm{bisector}\:\mathrm{from}\:\angle\mathrm{BAC}\:\mathrm{on}\:\mathrm{BC}.\:\mathrm{Let}\:\mathrm{AD} \\ $$$$\mathrm{intersect}\:\mathrm{EF}\:\mathrm{at}\:\mathrm{K}\:\mathrm{and}\:\Omega\:\mathrm{again}\:\mathrm{at} \\ $$$$\mathrm{L}\left(\mathrm{other}\:\mathrm{than}\:\mathrm{A}\right).\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{AEDF}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{rhombus}\:\mathrm{and}\:\mathrm{further}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{circle}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{triangle}\:\mathrm{KLX}\:\mathrm{passes} \\ $$$$\mathrm{through}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segment} \\ $$$$\mathrm{BC}. \\ $$

Question Number 112287 Answers: 3 Comments: 0

{ ((x^2 + 3xy + y^2 = −1)),((x^3 + y^3 = 7)) :}

$$\begin{cases}{{x}^{\mathrm{2}} \:+\:\mathrm{3}{xy}\:+\:{y}^{\mathrm{2}} \:=\:−\mathrm{1}}\\{{x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \:=\:\mathrm{7}}\end{cases} \\ $$

Question Number 112254 Answers: 1 Comments: 0

sin ((π/( 7))).sin (((2π)/7)).sin (((3π)/7)) =?

$$\:\:\:\mathrm{sin}\:\left(\frac{\pi}{\:\mathrm{7}}\right).\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right).\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\:=? \\ $$

Question Number 112251 Answers: 1 Comments: 0

∫ (dx/((x^4 −1)(√(x^2 +1))))

$$\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{4}} −\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$

Question Number 112249 Answers: 2 Comments: 1

∫(dx/((αx^2 +px+β)(√(αx^2 +qx+β))))=?

$$\int\frac{{dx}}{\left(\alpha{x}^{\mathrm{2}} +{px}+\beta\right)\sqrt{\alpha{x}^{\mathrm{2}} +{qx}+\beta}}=? \\ $$

Question Number 112248 Answers: 0 Comments: 0

∫_(−((π )/2)) ^(π/2) cos^(2019) (x)cos(2020x)dx pls help

$$\int_{−\frac{\pi\:\:\:}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2019}} \left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{2020x}\right)\mathrm{dx} \\ $$$$\mathrm{pls}\:\mathrm{help} \\ $$

Question Number 112489 Answers: 0 Comments: 0

solve ∫_0 ^(π/2) ln^2 (sinx)dx

$${solve} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{sin}{x}\right){dx} \\ $$

Question Number 112488 Answers: 0 Comments: 0

solve ∫_0 ^∞ ((ln(1+x^2 ))/(1+x^4 ))dx

$${solve} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$

Question Number 112487 Answers: 0 Comments: 0

solve ∫_0 ^∞ ((ln(1+x^3 ))/(1+x^2 ))dx

$${solve} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 112245 Answers: 0 Comments: 0

(1/(1+x^2 ))+(1/(1+x^4 ))+(1/(1+x^8 ))+..... ( x>1)

$$\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{8}} }+.....\:\:\:\:\:\:\:\:\:\:\left(\:{x}>\mathrm{1}\right) \\ $$

Question Number 112231 Answers: 0 Comments: 0

∫z^3 e^(−(4/3)z^3 ) dz

$$\int{z}^{\mathrm{3}} {e}^{−\frac{\mathrm{4}}{\mathrm{3}}{z}^{\mathrm{3}} } {dz} \\ $$

Question Number 112217 Answers: 0 Comments: 2

proporsed by m.n july 1970 evaluate Σ_(n=1) ^∞ (H_n /(n2^n )) solution recall thatΣ_(n=1) ^∞ H_n x^n =((ln(1−x))/(x−1)) ∵((H_n x^n )/n)=∫_0 ^x H_n t^(n−1) dt Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=∫_0 ^x ((ln(1−t))/(t(t−1)))dt Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=∫_0 ^x ((ln(1−t))/(t−1))dt−∫_0 ^x ((ln(1−t))/t)dt=A−B let A=∫_0 ^x ((ln(1−t))/(t−1))dt=−∫_0 ^x ((ln(1−t))/(1−t))dt A=−[−ln^2 (1−t)]_0 ^x +∫_0 ^x −((ln(1−t))/(1−t))dt 2A=ln^2 (1−x) A=(1/2)ln^2 (1−x)......(1) then B=∫_0 ^x ((ln(1−t))/t)dt=∫_0 ^1 ((ln(1−t))/t)dt+∫_1 ^x ((ln(1−t))/t)dt B=−Li_2 (1)−[Li_2 (x)−Li_2 (1)] B=−Li_2 (x)....(2) but I=A−B I=Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=(1/2)ln^2 (2)+Li_2 (x) ∵Σ_(n=1) ^∞ ((H_n x^n )/n)=(1/2)ln^2 (2)+Li_2 (x) but x=(1/2) Σ_(n=1) ^∞ (H_n /(n2^n ))=(1/2)ln^2 (2)+Li_2 ((1/2))=(1/2)ln^2 (2)+(π^2 /(12))−(1/2)ln^2 (2) ∵Σ_(n=1) ^∞ (H_n /(n2^n ))=(π^2 /(12)) by mathdave(06/09/2020)

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