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Question Number 18396    Answers: 1   Comments: 0

∫ (√(1 + (1/x^2 ) + (1/((x + 1)^2 )))) dx

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

Question Number 18394    Answers: 0   Comments: 0

Question Number 18392    Answers: 1   Comments: 1

Question Number 18388    Answers: 1   Comments: 0

Solve simultaneously. x + y = 5 ....... (i) 5^x + y = 15 ...... (ii)

$$\mathrm{Solve}\:\mathrm{simultaneously}.\: \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{5}\:\:\:\:\:.......\:\left(\mathrm{i}\right) \\ $$$$\mathrm{5}^{\mathrm{x}} \:+\:\mathrm{y}\:=\:\mathrm{15}\:\:\:\:......\:\left(\mathrm{ii}\right) \\ $$

Question Number 18386    Answers: 1   Comments: 1

Prove that ((2 + (√5)))^(1/3) + ((2 − (√5)))^(1/3) is a rational number.

$$\mathrm{Prove}\:\mathrm{that}\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:+\:\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:−\:\sqrt{\mathrm{5}}}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{rational}\:\mathrm{number}. \\ $$

Question Number 18385    Answers: 1   Comments: 0

Find all the integers which are equal to 11 times the sum of their digits.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{integers}\:\mathrm{which}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{11}\:\mathrm{times}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{digits}. \\ $$

Question Number 18384    Answers: 1   Comments: 0

Let a, b, c ∈ R, a ≠ 0, such that a and 4a + 3b + 2c have the same sign. Show that the equation ax^2 + bx + c = 0 can not have both roots in the interval (1, 2).

$$\mathrm{Let}\:{a},\:{b},\:{c}\:\in\:{R},\:{a}\:\neq\:\mathrm{0},\:\mathrm{such}\:\mathrm{that}\:{a}\:\mathrm{and} \\ $$$$\mathrm{4}{a}\:+\:\mathrm{3}{b}\:+\:\mathrm{2}{c}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{sign}.\:\mathrm{Show} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{can} \\ $$$$\mathrm{not}\:\mathrm{have}\:\mathrm{both}\:\mathrm{roots}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\left(\mathrm{1},\:\mathrm{2}\right). \\ $$

Question Number 19200    Answers: 0   Comments: 5

A river of width d is flowing with speed u as shown in the figure. John can swim with maximum speed v relative to the river and can cross it in shortest time T. John starts at A. B is the point directly opposite to A on the other bank of the river. If t be the time John takes to reach the opposite bank, match the situation in the column I to the possibilities in column II. Column I (A) John reaches to the left of B (B) John reaches to the right of B (C) John reaches the point B (D) John drifts along the bank while minimizing the time Column II (p) t = T (q) t > T (r) u < v (s) u > v

$$\mathrm{A}\:\mathrm{river}\:\mathrm{of}\:\mathrm{width}\:{d}\:\mathrm{is}\:\mathrm{flowing}\:\mathrm{with}\:\mathrm{speed} \\ $$$${u}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{figure}.\:\mathrm{John}\:\mathrm{can}\:\mathrm{swim} \\ $$$$\mathrm{with}\:\mathrm{maximum}\:\mathrm{speed}\:{v}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{river}\:\mathrm{and}\:\mathrm{can}\:\mathrm{cross}\:\mathrm{it}\:\mathrm{in}\:\mathrm{shortest}\:\mathrm{time} \\ $$$${T}.\:\mathrm{John}\:\mathrm{starts}\:\mathrm{at}\:{A}.\:{B}\:\mathrm{is}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{directly}\:\mathrm{opposite}\:\mathrm{to}\:{A}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{bank}\:\mathrm{of}\:\mathrm{the}\:\mathrm{river}.\:\mathrm{If}\:{t}\:\mathrm{be}\:\mathrm{the}\:\mathrm{time}\:\mathrm{John} \\ $$$$\mathrm{takes}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{bank},\:\mathrm{match} \\ $$$$\mathrm{the}\:\mathrm{situation}\:\mathrm{in}\:\mathrm{the}\:\mathrm{column}\:\mathrm{I}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{possibilities}\:\mathrm{in}\:\mathrm{column}\:\mathrm{II}. \\ $$$$\boldsymbol{\mathrm{Column}}\:\boldsymbol{\mathrm{I}} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{John}\:\mathrm{reaches}\:\mathrm{to}\:\mathrm{the}\:\mathrm{left}\:\mathrm{of}\:{B} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{John}\:\mathrm{reaches}\:\mathrm{to}\:\mathrm{the}\:\mathrm{right}\:\mathrm{of}\:{B} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{John}\:\mathrm{reaches}\:\mathrm{the}\:\mathrm{point}\:{B} \\ $$$$\left(\mathrm{D}\right)\:\mathrm{John}\:\mathrm{drifts}\:\mathrm{along}\:\mathrm{the}\:\mathrm{bank}\:\mathrm{while} \\ $$$$\mathrm{minimizing}\:\mathrm{the}\:\mathrm{time} \\ $$$$\boldsymbol{\mathrm{Column}}\:\boldsymbol{\mathrm{II}} \\ $$$$\left(\mathrm{p}\right)\:{t}\:=\:{T} \\ $$$$\left(\mathrm{q}\right)\:{t}\:>\:{T} \\ $$$$\left(\mathrm{r}\right)\:{u}\:<\:{v} \\ $$$$\left(\mathrm{s}\right)\:{u}\:>\:{v} \\ $$

Question Number 18379    Answers: 1   Comments: 0

Question Number 18361    Answers: 0   Comments: 0

N propositions are judged by 2k−1 people. Each person assigns “true” to exactly M propositions and “false” to the other N−M (M ≤ N). To say a proposition is “approved” means it is true according to at least k judges. Find the minimum and maximum numbers of approved propositions given N, M and k.

$${N}\:\mathrm{propositions}\:\mathrm{are}\:\mathrm{judged}\:\mathrm{by}\:\mathrm{2}{k}−\mathrm{1}\:\mathrm{people}. \\ $$$$\mathrm{Each}\:\mathrm{person}\:\mathrm{assigns}\:``\mathrm{true}''\:\mathrm{to} \\ $$$$\mathrm{exactly}\:{M}\:\mathrm{propositions}\:\mathrm{and}\:``\mathrm{false}'' \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:{N}−{M}\:\left({M}\:\leqslant\:{N}\right). \\ $$$$\mathrm{To}\:\mathrm{say}\:\mathrm{a}\:\mathrm{proposition}\:\mathrm{is}\:``\mathrm{approved}''\:\mathrm{means} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{true}\:\mathrm{according}\:\mathrm{to}\:\mathrm{at}\:\mathrm{least}\:{k}\:\mathrm{judges}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum}\:\mathrm{numbers} \\ $$$$\mathrm{of}\:\mathrm{approved}\:\mathrm{propositions}\:\mathrm{given}\:{N},\:{M}\:\mathrm{and}\:{k}. \\ $$

Question Number 18377    Answers: 0   Comments: 0

Suppose one is given two vector field A and B in region of space such that, A(x,y,z) = 4xi + zj + y^2 z^2 k B(x,y,z) = yi +3j − yzk Find: C(x,y,z) if C = A ∧ B Also prove that, C(x,y,z) is perpendicular to A(x,y,z)

$$\mathrm{Suppose}\:\mathrm{one}\:\mathrm{is}\:\mathrm{given}\:\mathrm{two}\:\mathrm{vector}\:\mathrm{field}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{in}\:\mathrm{region}\:\mathrm{of}\:\mathrm{space}\:\mathrm{such}\:\mathrm{that}, \\ $$$$\mathrm{A}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\:=\:\mathrm{4xi}\:+\:\mathrm{zj}\:+\:\mathrm{y}^{\mathrm{2}} \mathrm{z}^{\mathrm{2}} \mathrm{k} \\ $$$$\mathrm{B}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\:=\:\mathrm{yi}\:+\mathrm{3j}\:−\:\mathrm{yzk} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{C}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\:\mathrm{if}\:\mathrm{C}\:=\:\mathrm{A}\:\wedge\:\mathrm{B} \\ $$$$\mathrm{Also}\:\mathrm{prove}\:\mathrm{that},\:\:\mathrm{C}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{A}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right) \\ $$

Question Number 18721    Answers: 1   Comments: 0

Let ABC and ABC′ be two non- congruent triangles with sides AB = 4, AC = AC′ = 2(√2) and angle B = 30°. The absolute value of the difference between the areas of these triangles is

$$\mathrm{Let}\:{ABC}\:\mathrm{and}\:{ABC}'\:\mathrm{be}\:\mathrm{two}\:\mathrm{non}- \\ $$$$\mathrm{congruent}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sides}\:{AB}\:=\:\mathrm{4}, \\ $$$${AC}\:=\:{AC}'\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{and}\:\mathrm{angle}\:{B}\:=\:\mathrm{30}°. \\ $$$$\mathrm{The}\:\mathrm{absolute}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{difference} \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{areas}\:\mathrm{of}\:\mathrm{these}\:\mathrm{triangles}\:\mathrm{is} \\ $$

Question Number 18357    Answers: 0   Comments: 0

From the topic transformer prove that: e = (√2) ε cos(ωt)

$$\mathrm{From}\:\mathrm{the}\:\mathrm{topic}\:\mathrm{transformer} \\ $$$$ \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\mathrm{e}\:=\:\sqrt{\mathrm{2}}\:\varepsilon\:\mathrm{cos}\left(\omega\mathrm{t}\right) \\ $$

Question Number 18355    Answers: 0   Comments: 0

Question Number 18349    Answers: 0   Comments: 0

Consider the iteration x_(k+1) =x_k −(([f(x)]^2 )/(f(x_k +f(x_k ))−f(x_k ))), k=0,1,2,... for the solution of f(x)=0. Explain the connection with Newton′s method, and show that (x_k ) converges quadratically if x_0 is sufficiently close to the solution.

$${Consider}\:{the}\:{iteration} \\ $$$${x}_{{k}+\mathrm{1}} ={x}_{{k}} −\frac{\left[{f}\left({x}\right)\right]^{\mathrm{2}} }{{f}\left({x}_{{k}} +{f}\left({x}_{{k}} \right)\right)−{f}\left({x}_{{k}} \right)},\:\:\:\:\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},... \\ $$$${for}\:{the}\:{solution}\:{of}\:{f}\left({x}\right)=\mathrm{0}.\:{Explain}\:{the} \\ $$$${connection}\:{with}\:{Newton}'{s}\:{method},\:{and}\:{show} \\ $$$${that}\:\left({x}_{{k}} \right)\:{converges}\:{quadratically}\:{if}\:{x}_{\mathrm{0}} \:{is} \\ $$$${sufficiently}\:{close}\:{to}\:{the}\:{solution}. \\ $$$$ \\ $$

Question Number 18342    Answers: 2   Comments: 0

Question Number 19199    Answers: 1   Comments: 0

y=tan x^(tan x^(tan x) )

$$\mathrm{y}=\mathrm{tan}\:\mathrm{x}^{\mathrm{tan}\:\mathrm{x}^{\mathrm{tan}\:\mathrm{x}} } \\ $$

Question Number 18369    Answers: 1   Comments: 0

Prove that a^4 + b^4 + c^4 ≥ abc(a + b + c)

$$\mathrm{Prove}\:\mathrm{that}\:{a}^{\mathrm{4}} \:+\:{b}^{\mathrm{4}} \:+\:{c}^{\mathrm{4}} \:\geqslant\:{abc}\left({a}\:+\:{b}\:+\:{c}\right) \\ $$

Question Number 18366    Answers: 1   Comments: 0

Question Number 18365    Answers: 1   Comments: 0

Question Number 18364    Answers: 0   Comments: 0

Question Number 18363    Answers: 0   Comments: 0

Question Number 18362    Answers: 0   Comments: 0

Question Number 18333    Answers: 0   Comments: 1

Question Number 18469    Answers: 1   Comments: 0

Find Z_x and Z_y for each of the functions below (a) Z = 8x^2 y + 14xy^2 + 5y^2 x^3 (b) Z = 4x^3 y^2 + 2x^2 y^3 − 7xy^5

$$\mathrm{Find}\:\mathrm{Z}_{\mathrm{x}} \:\mathrm{and}\:\mathrm{Z}_{\mathrm{y}} \:\mathrm{for}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{functions}\:\mathrm{below} \\ $$$$\left(\mathrm{a}\right)\:\:\mathrm{Z}\:=\:\mathrm{8x}^{\mathrm{2}} \mathrm{y}\:+\:\mathrm{14xy}^{\mathrm{2}} \:+\:\mathrm{5y}^{\mathrm{2}} \mathrm{x}^{\mathrm{3}} \\ $$$$\left(\mathrm{b}\right)\:\:\mathrm{Z}\:=\:\mathrm{4x}^{\mathrm{3}} \mathrm{y}^{\mathrm{2}} \:+\:\mathrm{2x}^{\mathrm{2}} \mathrm{y}^{\mathrm{3}} \:−\:\mathrm{7xy}^{\mathrm{5}} \\ $$

Question Number 18327    Answers: 1   Comments: 1

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