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Question Number 19329 Answers: 0 Comments: 5

Solve the equation y^3 = x^3 + 8x^2 − 6x + 8 for positive integers x and y.

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:{y}^{\mathrm{3}} \:=\:{x}^{\mathrm{3}} \:+\:\mathrm{8}{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:\mathrm{8} \\ $$$$\mathrm{for}\:\mathrm{positive}\:\mathrm{integers}\:{x}\:\mathrm{and}\:{y}. \\ $$

Question Number 19324 Answers: 0 Comments: 0

y=tan^(−1) 3a^2 x−x^3 /a(a^2 −3x^2 )

$$\mathrm{y}=\mathrm{tan}^{−\mathrm{1}} \mathrm{3a}^{\mathrm{2}} \mathrm{x}−\mathrm{x}^{\mathrm{3}} /\mathrm{a}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{3x}^{\mathrm{2}} \right) \\ $$

Question Number 19323 Answers: 0 Comments: 0

y=sin (2tan^(−1) (√(1−x/1+x)))

$$\mathrm{y}=\mathrm{sin}\:\left(\mathrm{2tan}^{−\mathrm{1}} \sqrt{\left.\mathrm{1}−\mathrm{x}/\mathrm{1}+\mathrm{x}\right)}\right. \\ $$

Question Number 19313 Answers: 1 Comments: 0

Prove that ∣z_1 ± z_2 ∣^2 = ∣z_2 ∣^2 + ∣z_1 ∣^2 ± 2Re(z_1 z_2 ^ ) = ∣z_1 ∣^2 + ∣z_2 ∣^2 ± 2Re(z_1 ^ .z_2 )

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:\pm\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:\pm \\ $$$$\mathrm{2Re}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\pm\:\mathrm{2Re}\left(\bar {{z}}_{\mathrm{1}} .{z}_{\mathrm{2}} \right) \\ $$

Question Number 19312 Answers: 1 Comments: 0

Product of n, n^(th) roots of unity = 1.α.α^2 .α^3 ..... α^(n−1) = (−1)^(n−1) Why? How to get RHS?

$$\mathrm{Product}\:\mathrm{of}\:{n},\:{n}^{\mathrm{th}} \:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity} \\ $$$$=\:\mathrm{1}.\alpha.\alpha^{\mathrm{2}} .\alpha^{\mathrm{3}} \:.....\:\alpha^{{n}−\mathrm{1}} \:=\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \\ $$$$\mathrm{Why}?\:\mathrm{How}\:\mathrm{to}\:\mathrm{get}\:\mathrm{RHS}? \\ $$

Question Number 19321 Answers: 1 Comments: 1

Parallel tangents to a circle at A and B are cut in the points C and D by a tangent to the circle at E. Prove that AD, BC and the line joining the middle points of AE and BE are concurrent.

$$\mathrm{Parallel}\:\mathrm{tangents}\:\mathrm{to}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{at}\:\mathrm{A} \\ $$$$\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{cut}\:\mathrm{in}\:\mathrm{the}\:\mathrm{points}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D} \\ $$$$\mathrm{by}\:\mathrm{a}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{at}\:\mathrm{E}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{AD},\:\mathrm{BC}\:\mathrm{and}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{joining}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{points}\:\mathrm{of}\:\mathrm{AE} \\ $$$$\mathrm{and}\:\mathrm{BE}\:\mathrm{are}\:\mathrm{concurrent}. \\ $$

Question Number 19301 Answers: 1 Comments: 0

e^(iπ) +1=0

$${e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$

Question Number 19325 Answers: 0 Comments: 0

xcos^(−1) x/(√(1−x^2 ))

$$\mathrm{xcos}^{−\mathrm{1}} \mathrm{x}/\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$

Question Number 19326 Answers: 0 Comments: 1

tan^(−1) ((√x)−x/1+x^(3/2) )

$$\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{x}}−\mathrm{x}/\mathrm{1}+\mathrm{x}^{\mathrm{3}/\mathrm{2}} \right) \\ $$

Question Number 19294 Answers: 1 Comments: 0

Question Number 19293 Answers: 1 Comments: 3

Let AC be a line segment in the plane and B a point between A and C. Construct isosceles triangles PAB and QBC on one side of the segment AC such that ∠APB = ∠BQC = 120° and an isosceles triangle RAC on the other side of AC such that ∠ARC = 120°. Show that PQR is an equilateral triangle.

$$\mathrm{Let}\:{AC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{and}\:{B}\:\mathrm{a}\:\mathrm{point}\:\mathrm{between}\:{A}\:\mathrm{and}\:{C}. \\ $$$$\mathrm{Construct}\:\mathrm{isosceles}\:\mathrm{triangles}\:{PAB}\:\mathrm{and} \\ $$$${QBC}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{segment}\:{AC} \\ $$$$\mathrm{such}\:\mathrm{that}\:\angle{APB}\:=\:\angle{BQC}\:=\:\mathrm{120}°\:\mathrm{and} \\ $$$$\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle}\:{RAC}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{side}\:\mathrm{of}\:{AC}\:\mathrm{such}\:\mathrm{that}\:\angle{ARC}\:=\:\mathrm{120}°. \\ $$$$\mathrm{Show}\:\mathrm{that}\:{PQR}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}. \\ $$

Question Number 19292 Answers: 1 Comments: 1

Prove the equality sin (π/(2n)) sin ((2π)/(2n)) ... sin (((n − 1)π)/(2n)) = ((√n)/2^(n−1) ) .

$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{equality} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}}\:...\:\mathrm{sin}\:\frac{\left({n}\:−\:\mathrm{1}\right)\pi}{\mathrm{2}{n}}\:=\:\frac{\sqrt{{n}}}{\mathrm{2}^{{n}−\mathrm{1}} }\:. \\ $$

Question Number 19291 Answers: 1 Comments: 0

Prove that (1/(cos 6°)) + (1/(sin 24°)) + (1/(sin 48°)) = (1/(sin 12°)) .

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{24}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{48}°}\:=\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}\:. \\ $$

Question Number 19279 Answers: 0 Comments: 0

lim_(x→π) (((2x)/(cot(1/x))))

$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\left(\frac{\mathrm{2}{x}}{{cot}\frac{\mathrm{1}}{{x}}}\right) \\ $$

Question Number 19272 Answers: 0 Comments: 0

Question Number 19271 Answers: 1 Comments: 0

lim_(x→π) (2 − cos^2 x)^((2(√(2(1 + cos x))))/((x − π)^3 ))

$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\left(\mathrm{2}\:−\:\mathrm{cos}^{\mathrm{2}} \:{x}\right)^{\frac{\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{cos}\:{x}\right)}}{\left({x}\:−\:\pi\right)^{\mathrm{3}} }} \\ $$

Question Number 19268 Answers: 1 Comments: 0

Question Number 19264 Answers: 1 Comments: 2

Question Number 19250 Answers: 0 Comments: 2

Why arg(z) + arg(z^ ) = 2kπ, k ∈ Z? Shouldn′t it be always 0?

$$\mathrm{Why}\:\mathrm{arg}\left({z}\right)\:+\:\mathrm{arg}\left(\bar {{z}}\right)\:=\:\mathrm{2}{k}\pi,\:{k}\:\in\:{Z}? \\ $$$$\mathrm{Shouldn}'\mathrm{t}\:\mathrm{it}\:\mathrm{be}\:\boldsymbol{\mathrm{always}}\:\mathrm{0}? \\ $$

Question Number 19247 Answers: 1 Comments: 2

Question Number 20046 Answers: 0 Comments: 3

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Question Number 19245 Answers: 1 Comments: 1

Let f(x) be a quadratic polynomial with integer coefficients such that f(0) and f(1) are odd integers. Prove that the equation f(x) = 0 does not have an integer solution.

$$\mathrm{Let}\:{f}\left({x}\right)\:\mathrm{be}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{polynomial} \\ $$$$\mathrm{with}\:\mathrm{integer}\:\mathrm{coefficients}\:\mathrm{such}\:\mathrm{that}\:{f}\left(\mathrm{0}\right) \\ $$$$\mathrm{and}\:{f}\left(\mathrm{1}\right)\:\mathrm{are}\:\mathrm{odd}\:\mathrm{integers}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{equation}\:{f}\left({x}\right)\:=\:\mathrm{0}\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\:\mathrm{an} \\ $$$$\mathrm{integer}\:\mathrm{solution}. \\ $$

Question Number 19243 Answers: 0 Comments: 3

Let a, b, c be the sides opposite the angles A, B and C respectively of a ΔABC. Find the value of k such that (a) a + b = kc (b) cot (A/2) + cot (B/2) = k cot (C/2).

$$\mathrm{Let}\:{a},\:{b},\:{c}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{opposite}\:\mathrm{the} \\ $$$$\mathrm{angles}\:\mathrm{A},\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{respectively}\:\mathrm{of}\:\mathrm{a} \\ $$$$\Delta\mathrm{ABC}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left(\mathrm{a}\right)\:{a}\:+\:{b}\:=\:{kc} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\:+\:\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:=\:{k}\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}}. \\ $$

Question Number 19239 Answers: 1 Comments: 0

Assume that a, b, c and d are positive integers such that a^5 = b^4 , c^3 = d^2 and c − a = 19. Determine d − b.

$$\mathrm{Assume}\:\mathrm{that}\:{a},\:{b},\:{c}\:\mathrm{and}\:{d}\:\mathrm{are}\:\mathrm{positive} \\ $$$$\mathrm{integers}\:\mathrm{such}\:\mathrm{that}\:{a}^{\mathrm{5}} \:=\:{b}^{\mathrm{4}} ,\:{c}^{\mathrm{3}} \:=\:{d}^{\mathrm{2}} \:\mathrm{and} \\ $$$${c}\:−\:{a}\:=\:\mathrm{19}.\:\mathrm{Determine}\:{d}\:−\:{b}. \\ $$

Question Number 19230 Answers: 1 Comments: 0

Question Number 19223 Answers: 1 Comments: 7

A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t = 0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t = 0 along the horizontal direction, with the speed v. Friction between the bead and the string may be neglected. Let t_P and t_Q be the respective times taken by P and Q to reach the point B. Then (a) t_P < t_Q (b) t_P = t_Q (c) t_P > t_Q (d) (t_P /t_Q ) = ((length of at arc ACB)/(length of chord AB))

$$\mathrm{A}\:\mathrm{particle}\:{P}\:\mathrm{is}\:\mathrm{sliding}\:\mathrm{down}\:\mathrm{a}\:\mathrm{frictionless} \\ $$$$\mathrm{hemispherical}\:\mathrm{bowl}.\:\mathrm{It}\:\mathrm{passes}\:\mathrm{the}\:\mathrm{point} \\ $$$${A}\:\mathrm{at}\:{t}\:=\:\mathrm{0}.\:\mathrm{At}\:\mathrm{this}\:\mathrm{instant}\:\mathrm{of}\:\mathrm{time},\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{component}\:\mathrm{of}\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{is} \\ $$$${v}.\:\mathrm{A}\:\mathrm{bead}\:{Q}\:\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{mass}\:\mathrm{as}\:{P}\:\mathrm{is} \\ $$$$\mathrm{ejected}\:\mathrm{from}\:{A}\:\mathrm{at}\:{t}\:=\:\mathrm{0}\:\mathrm{along}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{direction},\:\mathrm{with}\:\mathrm{the}\:\mathrm{speed}\:{v}. \\ $$$$\mathrm{Friction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{bead}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{string}\:\mathrm{may}\:\mathrm{be}\:\mathrm{neglected}.\:\mathrm{Let}\:{t}_{{P}} \:\mathrm{and}\:{t}_{{Q}} \\ $$$$\mathrm{be}\:\mathrm{the}\:\mathrm{respective}\:\mathrm{times}\:\mathrm{taken}\:\mathrm{by}\:{P}\:\mathrm{and} \\ $$$${Q}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{point}\:{B}.\:\mathrm{Then} \\ $$$$\left({a}\right)\:{t}_{{P}} \:<\:{t}_{{Q}} \\ $$$$\left({b}\right)\:{t}_{{P}} \:=\:{t}_{{Q}} \\ $$$$\left({c}\right)\:{t}_{{P}} \:>\:{t}_{{Q}} \\ $$$$\left({d}\right)\:\frac{{t}_{{P}} }{{t}_{{Q}} }\:=\:\frac{\mathrm{length}\:\mathrm{of}\:\mathrm{at}\:\mathrm{arc}\:{ACB}}{\mathrm{length}\:\mathrm{of}\:\mathrm{chord}\:{AB}} \\ $$

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