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Question Number 38373 Answers: 1 Comments: 7

Question Number 38367 Answers: 2 Comments: 2

If f(x)=x^3 +1 then f^(−1) (x)=?

$${If}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{1}\:{then}\:{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$

Question Number 38366 Answers: 1 Comments: 0

At time t,the force acting on a particle P of mass 2kg is (2ti + 4j)N.P is initially at rest at the point with position vector (i + 2j). Find: a) the velocity of P when t = 2. b) the position vector when t = 2.

$${At}\:{time}\:{t},{the}\:{force}\:{acting}\:{on}\:{a}\:{particle} \\ $$$${P}\:{of}\:{mass}\:\mathrm{2}{kg}\:{is}\:\left(\mathrm{2}\boldsymbol{{ti}}\:+\:\mathrm{4}\boldsymbol{{j}}\right){N}.{P} \\ $$$${is}\:{initially}\:{at}\:{rest}\:{at}\:{the}\:{point}\:{with} \\ $$$${position}\:{vector}\:\left(\boldsymbol{{i}}\:+\:\mathrm{2}\boldsymbol{{j}}\right). \\ $$$${Find}: \\ $$$$\left.{a}\right)\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}\:=\:\mathrm{2}. \\ $$$$\left.{b}\right)\:{the}\:{position}\:{vector}\:{when}\:{t}\:=\:\mathrm{2}. \\ $$

Question Number 38362 Answers: 0 Comments: 0

Question Number 38365 Answers: 1 Comments: 0

A particle P moves on a straightline from a fixed point O and the distance x from O after t seconds is given as x = (1/(4 )) t^4 − (3/2) t^2 + 2t. Find: a) the velocity of P when t = 2, b) the acceleration of P when t = 2, c) the time at which the speed P is Minimum.

$${A}\:{particle}\:{P}\:{moves}\:{on}\:{a}\:{straightline} \\ $$$${from}\:{a}\:{fixed}\:{point}\:{O}\:{and}\:{the}\:{distance} \\ $$$${x}\:{from}\:{O}\:{after}\:{t}\:{seconds}\:{is}\:{given}\:{as} \\ $$$$\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}\:}\:{t}^{\mathrm{4}} \:−\:\frac{\mathrm{3}}{\mathrm{2}}\:{t}^{\mathrm{2}} \:+\:\mathrm{2}{t}. \\ $$$${Find}: \\ $$$$\left.{a}\right)\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}\:=\:\mathrm{2}, \\ $$$$\left.{b}\right)\:{the}\:{acceleration}\:{of}\:{P}\:{when}\:{t}\:=\:\mathrm{2}, \\ $$$$\left.{c}\right)\:{the}\:{time}\:{at}\:{which}\:{the}\:{speed}\:{P}\: \\ $$$${is}\:{Minimum}. \\ $$

Question Number 38332 Answers: 1 Comments: 0

((3+(√5))/2) −(√((3+(√5))/2)) = ?!

$$\:\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:−\sqrt{\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\:=\:?! \\ $$

Question Number 38310 Answers: 1 Comments: 4

let f(x)=∫_0 ^(+∞) ((arctan(xt))/(1+t^2 ))dt with x≥0 1) calculate f^′ (x) then a simple form of f(x) 2) calculate ∫_0 ^(+∞) ((arctant)/(1+t^2 ))dt 3) calculate ∫_0 ^(+∞) ((arctan(2t))/(1+t^2 ))dt

$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{arctan}\left({xt}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{with}\:{x}\geqslant\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{'} \left({x}\right)\:{then}\:{a}\:{simple}\:{form}\:{of}\:\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{arctant}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{arctan}\left(\mathrm{2}{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$

Question Number 38323 Answers: 1 Comments: 1

find Σ_(n=1) ^(+∞) ((4n)/((2n−1)^2 (2n+1)^2 ))

$${find}\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:\:\:\frac{\mathrm{4}{n}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 38296 Answers: 0 Comments: 7

i have a suggestion pls comment...we are all virtual friends common bond is mathematics so may know each other by posting our self photo...if administator give permission..

$${i}\:{have}\:{a}\:{suggestion}\:{pls}\:{comment}...{we}\:{are}\:{all} \\ $$$${virtual}\:{friends}\:{common}\:{bond}\:{is}\:{mathematics} \\ $$$${so}\:{may}\:{know}\:{each}\:{other}\:{by}\:{posting}\:{our}\:{self} \\ $$$${photo}...{if}\:{administator}\:{give}\:{permission}.. \\ $$

Question Number 38291 Answers: 1 Comments: 0

Question Number 38289 Answers: 1 Comments: 0

Question Number 38288 Answers: 1 Comments: 0

tan α−tan β = 2tan θ asin α−bsin β = lsin θ express sin α, sin β in terms of 𝛉.

$$\mathrm{tan}\:\alpha−\mathrm{tan}\:\beta\:=\:\mathrm{2tan}\:\theta \\ $$$${a}\mathrm{sin}\:\alpha−{b}\mathrm{sin}\:\beta\:=\:{l}\mathrm{sin}\:\theta \\ $$$${express}\:\mathrm{sin}\:\alpha,\:\mathrm{sin}\:\beta\:\:{in}\:{terms}\:{of}\:\boldsymbol{\theta}. \\ $$

Question Number 38286 Answers: 0 Comments: 1

(i) given the function f(t)=e^t and g(t)=lnt show that f○g(t)=g○f(t) (ii)if f(t)=at , g(t)=bt^2 +3 (fog)(2)=35 and (fog)(3)=75 find the value of a and b

$$\left(\boldsymbol{{i}}\right)\:\mathrm{given}\:\mathrm{the}\:\mathrm{function}\:\boldsymbol{{f}}\left(\boldsymbol{{t}}\right)=\boldsymbol{\mathrm{e}}^{\boldsymbol{{t}}} \:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{g}}\left(\boldsymbol{{t}}\right)=\boldsymbol{\mathrm{ln}{t}} \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{{f}}\circ\boldsymbol{{g}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{g}}\circ\boldsymbol{{f}}\left(\boldsymbol{{t}}\right) \\ $$$$\left(\boldsymbol{{ii}}\right)\mathrm{if}\:\boldsymbol{{f}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{at}}\:,\:\boldsymbol{{g}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{bt}}^{\mathrm{2}} +\mathrm{3} \\ $$$$\left(\boldsymbol{{fog}}\right)\left(\mathrm{2}\right)=\mathrm{35}\:\boldsymbol{\mathrm{and}}\:\left(\boldsymbol{{fog}}\right)\left(\mathrm{3}\right)=\mathrm{75} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\boldsymbol{{a}}\:\mathrm{and}\:\boldsymbol{{b}} \\ $$

Question Number 38284 Answers: 1 Comments: 2

Find all the complex number in the rectangular form such that (z−1)^4 =−1

$${Find}\:{all}\:{the}\:{complex}\:{number}\:{in}\:{the} \\ $$$${rectangular}\:{form}\:{such}\:{that} \\ $$$$\left({z}−\mathrm{1}\right)^{\mathrm{4}} =−\mathrm{1} \\ $$$$ \\ $$

Question Number 38282 Answers: 1 Comments: 0

Question Number 38281 Answers: 1 Comments: 0

Question Number 38262 Answers: 1 Comments: 0

Question Number 38252 Answers: 0 Comments: 2

if x^2 + 3xy − y^2 = 3 find (dy/dx) at point (1,1) hence differentiate ((sin x)/(1 + x)) with respect to x.

$${if}\:{x}^{\mathrm{2}} \:+\:\mathrm{3}{xy}\:−\:{y}^{\mathrm{2}} \:=\:\mathrm{3}\:{find}\: \\ $$$$\frac{{dy}}{{dx}}\:{at}\:{point}\:\left(\mathrm{1},\mathrm{1}\right)\:{hence} \\ $$$${differentiate}\:\frac{{sin}\:{x}}{\mathrm{1}\:+\:{x}}\:{with}\:{respect} \\ $$$${to}\:{x}. \\ $$

Question Number 38250 Answers: 0 Comments: 0

It is given that the first term of a GP is the last term of an AP. the second term of the AP is the third term of the GP..detemine the Geometric mean of the GP is the fourth term of the GP is 16.

$$\:\:{It}\:{is}\:{given}\:{that}\:{the}\:{first}\:{term}\:{of} \\ $$$${a}\:{GP}\:\:{is}\:{the}\:{last}\:{term}\:{of}\:{an}\:{AP}. \\ $$$${the}\:{second}\:{term}\:{of}\:{the}\:{AP}\:{is}\:{the} \\ $$$${third}\:{term}\:{of}\:{the}\:{GP}..{detemine} \\ $$$${the}\:{Geometric}\:{mean}\:{of}\:{the}\:{GP}\:{is}\: \\ $$$$\:{the}\:{fourth}\:{term}\:{of}\:{the}\:{GP}\:{is}\:\mathrm{16}. \\ $$

Question Number 38247 Answers: 2 Comments: 2

Question Number 38235 Answers: 0 Comments: 4

A man 2m 50cm tall stands a distance of 3m in front of a large vertical plane mirror. i)what is the shortest length of the mirror that will enable the man see himself fully? ii)what is the answer of the above if the man were 5m away?

$${A}\:{man}\:\mathrm{2}{m}\:\mathrm{50}{cm}\:{tall}\:{stands}\:{a} \\ $$$${distance}\:{of}\:\mathrm{3}{m}\:{in}\:{front}\:{of}\:{a}\:{large} \\ $$$${vertical}\:{plane}\:{mirror}. \\ $$$$\left.{i}\right){what}\:{is}\:{the}\:{shortest}\:{length}\:{of}\:{the} \\ $$$${mirror}\:{that}\:{will}\:{enable}\:{the}\:{man}\:{see} \\ $$$${himself}\:{fully}? \\ $$$$\left.{ii}\right){what}\:{is}\:{the}\:{answer}\:{of}\:{the}\:{above} \\ $$$${if}\:{the}\:{man}\:{were}\:\mathrm{5}{m}\:{away}? \\ $$

Question Number 38261 Answers: 1 Comments: 0

Question Number 38232 Answers: 4 Comments: 1

Differentiate tan^(−1) ((((√(1+x^2 ))−1)/x)) without using any trigonometric substitution !

$$\mathrm{Differentiate}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}}{{x}}\right)\:\: \\ $$$${without}\:{using}\:{any}\:{trigonometric}\: \\ $$$${substitution}\:! \\ $$

Question Number 38222 Answers: 0 Comments: 1

If U={−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5} A={x/x^2 =25, x ∈ Z} B={x/x^2 +5=9, x ∈ Z} and C={x/−2≤ x ≤ 2, x ∈ Z} then (A ∩ B ∩ C)^c ∩ (A△B)^c =?

$$\mathrm{If}\:\mathbb{U}=\left\{−\mathrm{5},\:−\mathrm{4},\:−\mathrm{3},\:−\mathrm{2},\:−\mathrm{1},\:\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5}\right\} \\ $$$$\mathrm{A}=\left\{{x}/{x}^{\mathrm{2}} =\mathrm{25},\:{x}\:\in\:\mathrm{Z}\right\} \\ $$$$\mathrm{B}=\left\{{x}/{x}^{\mathrm{2}} +\mathrm{5}=\mathrm{9},\:{x}\:\in\:\mathrm{Z}\right\}\:\mathrm{and} \\ $$$$\mathrm{C}=\left\{{x}/−\mathrm{2}\leqslant\:{x}\:\leqslant\:\mathrm{2},\:{x}\:\in\:\mathrm{Z}\right\}\:\mathrm{then} \\ $$$$\left(\mathrm{A}\:\cap\:\mathrm{B}\:\cap\:\mathrm{C}\right)^{\mathrm{c}} \:\cap\:\left(\mathrm{A}\bigtriangleup\mathrm{B}\right)^{\mathrm{c}} =? \\ $$

Question Number 38211 Answers: 0 Comments: 2

let x>0 and F(x)= ∫_0 ^(+∞) ((arctan(xt^2 ))/(1+t^2 ))dt 1) find a simple form of F(x) 2)find the value of ∫_0 ^∞ ((arctan(2t^2 ))/(1+t^2 ))dt 3)find the value of ∫_0 ^∞ ((arctan(3t^2 ))/(1+t^2 ))dt.

$${let}\:{x}>\mathrm{0}\:{and}\:{F}\left({x}\right)=\:\int_{\mathrm{0}} ^{+\infty} \:\frac{{arctan}\left({xt}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{F}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\mathrm{2}{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left(\mathrm{3}{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}. \\ $$

Question Number 38210 Answers: 2 Comments: 4

let f(a)= ∫_0 ^π (dθ/(a +sin^2 θ)) (a from R) 1) find f(a) 2)calculate g(a)= ∫_0 ^π (dθ/((a+sin^2 θ)^2 )) 3)calculate ∫_0 ^π (dθ/(1+sin^2 θ)) and ∫_0 ^π (dθ/(2+sin^2 θ)) 4) calculate ∫_0 ^π (dθ/((3 +sin^2 θ)^2 )) .

$${let}\:{f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{{a}\:+{sin}^{\mathrm{2}} \theta}\:\:\:\left({a}\:{from}\:{R}\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:{g}\left({a}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{\left({a}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{d}\theta}{\mathrm{1}+{sin}^{\mathrm{2}} \theta}\:{and}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{d}\theta}{\mathrm{2}+{sin}^{\mathrm{2}} \theta} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{\left(\mathrm{3}\:+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:. \\ $$

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