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Question Number 19434    Answers: 1   Comments: 0

If (((1 + i(√3))/(1 − i(√3))))^n is an integer, then n is

$$\mathrm{If}\:\left(\frac{\mathrm{1}\:+\:{i}\sqrt{\mathrm{3}}}{\mathrm{1}\:−\:{i}\sqrt{\mathrm{3}}}\right)^{{n}} \:\mathrm{is}\:\mathrm{an}\:\mathrm{integer},\:\mathrm{then}\:{n}\:\mathrm{is} \\ $$

Question Number 19433    Answers: 1   Comments: 0

(((√(5 + 12i)) + (√(5 − 12i)))/((√(5 + 12i)) − (√(5 − 12i)))) =

$$\frac{\sqrt{\mathrm{5}\:+\:\mathrm{12}{i}}\:+\:\sqrt{\mathrm{5}\:−\:\mathrm{12}{i}}}{\sqrt{\mathrm{5}\:+\:\mathrm{12}{i}}\:−\:\sqrt{\mathrm{5}\:−\:\mathrm{12}{i}}}\:= \\ $$

Question Number 19419    Answers: 1   Comments: 1

sin z=200 find z

$$\mathrm{sin}\:\boldsymbol{{z}}=\mathrm{200} \\ $$$$ \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{z}} \\ $$

Question Number 19415    Answers: 1   Comments: 0

PS is a line segment of length 4 and O is the midpoint of PS. A semicircular arc is drawn with PS as diameter. Let X be the midpoint of this arc. Q and R are points on the arc PXS such that QR is parallel to PS and the semicircular arc drawn with QR as diameter is tangent to PS. What is the area of the region QXROQ bounded by the two semicircular arcs?

$${PS}\:\mathrm{is}\:\mathrm{a}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{of}\:\mathrm{length}\:\mathrm{4}\:\mathrm{and}\:{O} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:{PS}.\:\mathrm{A}\:\mathrm{semicircular} \\ $$$$\mathrm{arc}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{with}\:{PS}\:\mathrm{as}\:\mathrm{diameter}.\:\mathrm{Let} \\ $$$${X}\:\mathrm{be}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{this}\:\mathrm{arc}.\:{Q}\:\mathrm{and}\:{R} \\ $$$$\mathrm{are}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{arc}\:{PXS}\:\mathrm{such}\:\mathrm{that}\:{QR} \\ $$$$\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:{PS}\:\mathrm{and}\:\mathrm{the}\:\mathrm{semicircular} \\ $$$$\mathrm{arc}\:\mathrm{drawn}\:\mathrm{with}\:{QR}\:\mathrm{as}\:\mathrm{diameter}\:\mathrm{is} \\ $$$$\mathrm{tangent}\:\mathrm{to}\:{PS}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{region}\:{QXROQ}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{semicircular}\:\mathrm{arcs}? \\ $$

Question Number 19413    Answers: 1   Comments: 1

How many integer pairs (x, y) satisfy x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{integer}\:\mathrm{pairs}\:\left({x},\:{y}\right)\:\mathrm{satisfy} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{4}{y}^{\mathrm{2}} \:−\:\mathrm{2}{xy}\:−\:\mathrm{2}{x}\:−\:\mathrm{4}{y}\:−\:\mathrm{8}\:=\:\mathrm{0}? \\ $$

Question Number 19409    Answers: 2   Comments: 1

What is the sum of the squares of the roots of the equation x^2 − 7[x] + 5 = 0? (Here [x] denotes the greatest integer less than or equal to x. For example [3.4] = 3 and [−2.3] = −3.)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{squares}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:−\:\mathrm{7}\left[{x}\right]\:+\:\mathrm{5}\:=\:\mathrm{0}? \\ $$$$\left(\mathrm{Here}\:\left[{x}\right]\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\right. \\ $$$$\mathrm{less}\:\mathrm{than}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to}\:{x}.\:\mathrm{For}\:\mathrm{example} \\ $$$$\left.\left[\mathrm{3}.\mathrm{4}\right]\:=\:\mathrm{3}\:\mathrm{and}\:\left[−\mathrm{2}.\mathrm{3}\right]\:=\:−\mathrm{3}.\right) \\ $$

Question Number 19403    Answers: 1   Comments: 1

Let P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9). What is the largest integer that is a divisor of P(n) for all positive even integers n?

$$\mathrm{Let}\:{P}\left({n}\right)\:=\:\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{3}\right)\left({n}\:+\:\mathrm{5}\right)\left({n}\:+\:\mathrm{7}\right)\left({n}\:+\:\mathrm{9}\right). \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{integer}\:\mathrm{that}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{divisor}\:\mathrm{of}\:{P}\left({n}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{even} \\ $$$$\mathrm{integers}\:{n}? \\ $$

Question Number 19457    Answers: 1   Comments: 0

prove that ∫tan^2 xdx =tan x−x

$${prove}\:{that}\:\int\mathrm{tan}\:^{\mathrm{2}} {xdx} \\ $$$$ \\ $$$$=\mathrm{tan}\:{x}−{x} \\ $$

Question Number 19394    Answers: 1   Comments: 0

Question Number 19393    Answers: 1   Comments: 0

Three tennis players A, B and C play each other only once. The probability that A will beat B is (3/5), that B will beat C is (2/3), and that A will beat C is (5/7). Find (1) the probability that A will not win both games (2) the probability that A will win not both games.

$$\mathrm{Three}\:\mathrm{tennis}\:\mathrm{players}\:\mathrm{A},\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{play}\:\mathrm{each} \\ $$$$\mathrm{other}\:\mathrm{only}\:\mathrm{once}.\:\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{A}\:\mathrm{will} \\ $$$$\mathrm{beat}\:\mathrm{B}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{5}},\:\mathrm{that}\:\mathrm{B}\:\mathrm{will}\:\mathrm{beat}\:\mathrm{C}\:\mathrm{is}\:\frac{\mathrm{2}}{\mathrm{3}},\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{A}\:\mathrm{will}\:\mathrm{beat}\:\mathrm{C}\:\mathrm{is}\:\frac{\mathrm{5}}{\mathrm{7}}.\:\mathrm{Find}\:\left(\mathrm{1}\right)\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{A}\:\mathrm{will}\:\mathrm{not}\:\mathrm{win}\:\mathrm{both}\:\mathrm{games} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{A}\:\mathrm{will}\:\mathrm{win}\:\mathrm{not}\:\mathrm{both}\:\mathrm{games}. \\ $$

Question Number 19555    Answers: 1   Comments: 0

Find center and radius of circle having equation zz^ + (1 − i)z + (1 + i)z^ − 1 = 0.

$$\mathrm{Find}\:\mathrm{center}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{having} \\ $$$$\mathrm{equation}\:{z}\bar {{z}}\:+\:\left(\mathrm{1}\:−\:{i}\right){z}\:+\:\left(\mathrm{1}\:+\:{i}\right)\bar {{z}}\:−\:\mathrm{1}\:=\:\mathrm{0}. \\ $$

Question Number 19391    Answers: 0   Comments: 1

Question Number 19389    Answers: 1   Comments: 0

What is the digital root of 3^(2017)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{digital}\:\mathrm{root}\:\mathrm{of}\:\:\mathrm{3}^{\mathrm{2017}} \\ $$

Question Number 19388    Answers: 0   Comments: 1

related to Q.19333 the side lengthes of a triangle are integer. if the perimeter of the triangle is 100, how many different triangles exist? what is the maximum area of them?

$$\mathrm{related}\:\mathrm{to}\:\mathrm{Q}.\mathrm{19333} \\ $$$$\mathrm{the}\:\mathrm{side}\:\mathrm{lengthes}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{are}\: \\ $$$$\mathrm{integer}.\:\mathrm{if}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\mathrm{is}\:\mathrm{100},\:\mathrm{how}\:\mathrm{many}\:\mathrm{different}\:\mathrm{triangles} \\ $$$$\mathrm{exist}?\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{of} \\ $$$$\mathrm{them}? \\ $$

Question Number 19385    Answers: 0   Comments: 5

tan^2 β=−1 find β.... lets solve for fun

$${tan}^{\mathrm{2}} \beta=−\mathrm{1} \\ $$$$ \\ $$$${find}\:\beta....\:{lets}\:{solve}\:{for}\:{fun} \\ $$

Question Number 19382    Answers: 2   Comments: 0

If log_2 (9^(x−1) +7)−log_2 (3^(x−1) +1)=2, then the values of x are

$$\mathrm{If}\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{9}^{{x}−\mathrm{1}} +\mathrm{7}\right)−\mathrm{log}_{\mathrm{2}} \left(\mathrm{3}^{{x}−\mathrm{1}} +\mathrm{1}\right)=\mathrm{2},\: \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{are} \\ $$

Question Number 19367    Answers: 1   Comments: 0

If α, β, γ and δ are four solutions of the equation tan (θ + ((5π)/4)) = 3 tan 3θ, then (1) Σtan α = 0 (2) Σtan α tan β = −2 (3) Σtan α tan β tan γ = −(8/3) (4) tan α tan β tan γ tan δ = −3

$$\mathrm{If}\:\alpha,\:\beta,\:\gamma\:\mathrm{and}\:\delta\:\mathrm{are}\:\mathrm{four}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{tan}\:\left(\theta\:+\:\frac{\mathrm{5}\pi}{\mathrm{4}}\right)\:=\:\mathrm{3}\:\mathrm{tan}\:\mathrm{3}\theta,\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\Sigma\mathrm{tan}\:\alpha\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\Sigma\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:=\:−\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\Sigma\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma\:=\:−\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma\:\mathrm{tan}\:\delta\:=\:−\mathrm{3} \\ $$

Question Number 19362    Answers: 0   Comments: 5

The block Q moves to the right with a constant velocity v_0 as shown in figure. The relative velocity of body P with respect to Q is (assume all pulleys and strings are ideal)

$$\mathrm{The}\:\mathrm{block}\:{Q}\:\mathrm{moves}\:\mathrm{to}\:\mathrm{the}\:\mathrm{right}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{constant}\:\mathrm{velocity}\:{v}_{\mathrm{0}} \:\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{figure}. \\ $$$$\mathrm{The}\:\mathrm{relative}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{body}\:{P}\:\mathrm{with} \\ $$$$\mathrm{respect}\:\mathrm{to}\:{Q}\:\mathrm{is}\:\left(\mathrm{assume}\:\mathrm{all}\:\mathrm{pulleys}\:\mathrm{and}\right. \\ $$$$\left.\mathrm{strings}\:\mathrm{are}\:\mathrm{ideal}\right) \\ $$

Question Number 19358    Answers: 1   Comments: 1

In the arrangement shown in figure two beads slide along a smooth horizontal rod. The relation between v and v_0 in given position will be

$$\mathrm{In}\:\mathrm{the}\:\mathrm{arrangement}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{figure} \\ $$$$\mathrm{two}\:\mathrm{beads}\:\mathrm{slide}\:\mathrm{along}\:\mathrm{a}\:\mathrm{smooth} \\ $$$$\mathrm{horizontal}\:\mathrm{rod}.\:\mathrm{The}\:\mathrm{relation}\:\mathrm{between} \\ $$$${v}\:\mathrm{and}\:{v}_{\mathrm{0}} \:\mathrm{in}\:\mathrm{given}\:\mathrm{position}\:\mathrm{will}\:\mathrm{be} \\ $$

Question Number 19355    Answers: 0   Comments: 3

Two blocks are placed on a smooth horizontal surface and connected by a string pulley arrangement as shown. If a force F starts acting on block m_1 , then find the relation between acceleration of both masses and their values

$$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{are}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth} \\ $$$$\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{and}\:\mathrm{connected}\:\mathrm{by}\:\mathrm{a} \\ $$$$\mathrm{string}\:\mathrm{pulley}\:\mathrm{arrangement}\:\mathrm{as}\:\mathrm{shown}. \\ $$$$\mathrm{If}\:\mathrm{a}\:\mathrm{force}\:{F}\:\mathrm{starts}\:\mathrm{acting}\:\mathrm{on}\:\mathrm{block}\:{m}_{\mathrm{1}} , \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{both}\:\mathrm{masses}\:\mathrm{and}\:\mathrm{their}\:\mathrm{values} \\ $$

Question Number 19352    Answers: 1   Comments: 0

Prove that ∣z_1 − z_2 ∣^2 = ∣z_1 ∣^2 + ∣z_2 ∣^2 − 2∣z_1 ∣ ∣z_2 ∣ cos (θ_1 − θ_2 )

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:−\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \\ $$$$−\:\mathrm{2}\mid{z}_{\mathrm{1}} \mid\:\mid{z}_{\mathrm{2}} \mid\:\mathrm{cos}\:\left(\theta_{\mathrm{1}} \:−\:\theta_{\mathrm{2}} \right) \\ $$

Question Number 19351    Answers: 1   Comments: 2

Prove that ∣z_1 + z_2 ∣^2 = ∣z_1 ∣^2 + ∣z_2 ∣^2 ⇔ (z_1 /z_2 ) is purely imaginary number.

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\Leftrightarrow \\ $$$$\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }\:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{number}. \\ $$

Question Number 19350    Answers: 1   Comments: 0

Prove that ∣z_1 + z_2 ∣ = ∣z_1 − z_2 ∣ ⇔ arg(z_1 ) − arg(z_2 ) = (π/2)

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{1}} \:−\:{z}_{\mathrm{2}} \mid\:\Leftrightarrow \\ $$$$\mathrm{arg}\left({z}_{\mathrm{1}} \right)\:−\:\mathrm{arg}\left({z}_{\mathrm{2}} \right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$

Question Number 19349    Answers: 1   Comments: 2

Prove that ∣z_1 + z_2 + z_3 + .... + z_n ∣ ≤ ∣z_1 ∣ + ∣z_2 ∣ + ∣z_3 ∣ + .... + ∣z_n ∣

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:+\:....\:+\:{z}_{{n}} \mid\:\leqslant \\ $$$$\mid{z}_{\mathrm{1}} \mid\:+\:\mid{z}_{\mathrm{2}} \mid\:+\:\mid{z}_{\mathrm{3}} \mid\:+\:....\:+\:\mid{z}_{{n}} \mid \\ $$

Question Number 19452    Answers: 1   Comments: 0

Question Number 19345    Answers: 0   Comments: 0

A thin bi − convex lens rest on a plane mirror . it is found that a point objects placed 20cm above the object coincide with it own image. Determine the position and nature of the image when the object is placed (i) 8cm and (ii) 12 from the lens mirror combinatiom

$$\mathrm{A}\:\mathrm{thin}\:\mathrm{bi}\:−\:\mathrm{convex}\:\mathrm{lens}\:\mathrm{rest}\:\mathrm{on}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{mirror}\:.\:\:\mathrm{it}\:\mathrm{is}\:\mathrm{found}\:\mathrm{that}\:\mathrm{a}\:\mathrm{point} \\ $$$$\mathrm{objects}\:\mathrm{placed}\:\mathrm{20cm}\:\mathrm{above}\:\mathrm{the}\:\mathrm{object}\:\mathrm{coincide}\:\mathrm{with}\:\mathrm{it}\:\mathrm{own}\:\mathrm{image}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{position}\:\mathrm{and}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{image}\:\mathrm{when}\:\mathrm{the}\:\mathrm{object}\:\mathrm{is}\:\mathrm{placed} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{8cm}\:\:\mathrm{and}\:\:\left(\mathrm{ii}\right)\:\mathrm{12}\:\:\:\:\mathrm{from}\:\mathrm{the}\:\mathrm{lens}\:\mathrm{mirror}\:\mathrm{combinatiom} \\ $$

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