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Question Number 35828    Answers: 1   Comments: 2

Question Number 35821    Answers: 0   Comments: 4

let f(t) = ∫_0 ^∞ ((e^(−ax) −e^(−bx) )/x^2 ) e^(−tx^2 ) dx with t>0 1) calculate f^′ (t) 2)find a simple form of f(t) 3) find the value of ∫_0 ^∞ ((e^(−2x) −e^(−x) )/x^2 ) e^(−3x^2 ) dx

$${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{ax}} \:−{e}^{−{bx}} }{{x}^{\mathrm{2}} }\:{e}^{−{tx}^{\mathrm{2}} } {dx}\:\:\:{with}\:{t}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{'} \left({t}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\mathrm{2}{x}} \:\:−{e}^{−{x}} }{{x}^{\mathrm{2}} }\:{e}^{−\mathrm{3}{x}^{\mathrm{2}} } {dx} \\ $$

Question Number 35811    Answers: 3   Comments: 6

Multiple Choice Questions Rio Mike. Q_1 . The value of 6 + 2 × 4 −15 ÷ 3 is; A) 3 B) 9 C) 4 D) 27 Q_2 . 80 as a product of its prime factor is. A) 2^3 × 5 B) 2 × 5^(3 ) C) 2^2 × 5^2 D) 2^4 × 5 Q_3 . The deteminant of (((6 4)),((3 2)) ) is A) 18 B) 0 C) 24 D) −24 Q_4 . the value of ((1/(16)))^(−(1/2)) is A) −(1/4) B) 4 C) −4 D) (1/2) Q_5 . if −8,m,n,19 are in AP then (m,n) is A) (1,10) B) (2,10) C) (3,13) D) (4,16)o Q_6 . if cos θ = ((12)/(13)) then cot^2 θ is; A) ((169)/(25)) B) ((25)/(169)) C)((169)/(144)) D) ((144)/(169)) Q_7 . the solution set of 3 ≤ 2x−1≤5 is; A) ]x≥−2,x≤(8/3)[ B) [x≥2,x≤−(8/3)[ C) ]x>−2,x≤(3/8)] D) [x≥−2,x≤(8/3)] Q_8 . Which is a factor of f(x)= x^2 −3x + 2 A) (x−1) B) (x−2) C) (x+1) D)(x+3) Q_(9 ) . The sum of the sum of roots and product of roots of the quadratic equation 3x^2 + 6x + 9=0 A) 1 B) −1 C) 32 D) 5 Q_(10) . 2log2−log2 = A) log 8 B) log 6 C) log 3 D) log 2. Q_(11) . Σ_(r=1) ^∞ 3^(2−r) = A) (9/4) B) (9/2) C) ((13)/3) D) (1/2) Q_(12) . The constand term in the binomial expansion of (x^2 + (1/x^2 ))^8 is A) 3^(rd) B) 4^(th) C) 5^(th) D) 6^(th) Q_(13) . cos(180−x) = A) −sinx B) sin x C) cos x D) −cos x Q_(14) . The value of p for which (2,1) , (6,3), and (4,p) are collinear is ; A) 2 B) 1 C)−1 D) −2 Q_(15) . ∫_1 ^2 x^3 dx = A) −sin 7x B) sin 7x C) −7sin7x D) 7sin7x Q_(16) . Given that g : → px −5, the value of p for which g^(−1) (3)=4 is A) (1/2) B) −(1/2) C) −2 D) 2 Q_(17) . The value of θ in the range 0°≤θ≤90° for which sinθ = cos θ is A) 90° B) 60° C) 30° D) 45° Questions

$$\:{Multiple}\:{Choice}\:{Questions} \\ $$$$\:\:\:\boldsymbol{{Rio}}\:\boldsymbol{{M}}{ike}. \\ $$$$ \\ $$$${Q}_{\mathrm{1}} .\:{The}\:{value}\:{of}\:\:\mathrm{6}\:+\:\mathrm{2}\:×\:\mathrm{4}\:−\mathrm{15}\:\boldsymbol{\div}\:\mathrm{3} \\ $$$${is}; \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{3}\:\:{B}\right)\:\mathrm{9}\:{C}\right)\:\mathrm{4}\:{D}\right)\:\mathrm{27} \\ $$$$ \\ $$$${Q}_{\mathrm{2}} .\:\mathrm{80}\:{as}\:{a}\:{product}\:{of}\:{its}\:{prime} \\ $$$${factor}\:{is}. \\ $$$$\left.{A}\left.\right)\left.\:\mathrm{2}^{\mathrm{3}} ×\:\mathrm{5}\:{B}\right)\:\mathrm{2}\:×\:\mathrm{5}^{\mathrm{3}\:} \:{C}\right)\:\mathrm{2}^{\mathrm{2}} ×\:\mathrm{5}^{\mathrm{2}} \\ $$$$\left.{D}\right)\:\mathrm{2}^{\mathrm{4}} ×\:\mathrm{5} \\ $$$$ \\ $$$${Q}_{\mathrm{3}} .\:{The}\:{deteminant}\:{of}\:\:\begin{pmatrix}{\mathrm{6}\:\:\:\:\:\:\:\:\:\mathrm{4}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}\:{is} \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{18}\:\:{B}\right)\:\mathrm{0}\:\:{C}\right)\:\mathrm{24}\:{D}\right)\:−\mathrm{24} \\ $$$$ \\ $$$${Q}_{\mathrm{4}} .\:{the}\:{value}\:{of}\:\left(\frac{\mathrm{1}}{\mathrm{16}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {is} \\ $$$$\left.{A}\left.\right)\left.\:\left.−\frac{\mathrm{1}}{\mathrm{4}}\:\:\:{B}\right)\:\mathrm{4}\:\:\:{C}\right)\:−\mathrm{4}\:\:\:\:{D}\right)\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$${Q}_{\mathrm{5}} .\:{if}\:−\mathrm{8},{m},{n},\mathrm{19}\:{are}\:{in}\:{AP}\:{then}\: \\ $$$$\left({m},{n}\right)\:{is}\: \\ $$$$\left.{A}\left.\right)\left.\:\left.\left(\mathrm{1},\mathrm{10}\right)\:{B}\right)\:\left(\mathrm{2},\mathrm{10}\right)\:{C}\right)\:\left(\mathrm{3},\mathrm{13}\right)\:{D}\right)\:\left(\mathrm{4},\mathrm{16}\right){o} \\ $$$$ \\ $$$${Q}_{\mathrm{6}} .\:{if}\:{cos}\:\theta\:=\:\frac{\mathrm{12}}{\mathrm{13}}\:{then}\:{cot}^{\mathrm{2}} \theta\:{is}; \\ $$$$\left.{A}\left.\right)\left.\:\left.\frac{\mathrm{169}}{\mathrm{25}}\:\:{B}\right)\:\frac{\mathrm{25}}{\mathrm{169}}\:{C}\right)\frac{\mathrm{169}}{\mathrm{144}}\:{D}\right)\:\frac{\mathrm{144}}{\mathrm{169}} \\ $$$$ \\ $$$${Q}_{\mathrm{7}} .\:{the}\:{solution}\:{set}\:{of}\:\mathrm{3}\:\leqslant\:\mathrm{2}{x}−\mathrm{1}\leqslant\mathrm{5} \\ $$$${is};\: \\ $$$$\left.{A}\left.\right)\:\right]{x}\geqslant−\mathrm{2},{x}\leqslant\frac{\mathrm{8}}{\mathrm{3}}\left[\:\:\:{B}\right)\:\left[{x}\geqslant\mathrm{2},{x}\leqslant−\frac{\mathrm{8}}{\mathrm{3}}\left[\right.\right. \\ $$$$\left.{C}\left.\right)\left.\:\left.\right]{x}>−\mathrm{2},{x}\leqslant\frac{\mathrm{3}}{\mathrm{8}}\right]\:\:{D}\right)\:\left[{x}\geqslant−\mathrm{2},{x}\leqslant\frac{\mathrm{8}}{\mathrm{3}}\right] \\ $$$$ \\ $$$${Q}_{\mathrm{8}} .\:{Which}\:{is}\:{a}\:{factor}\:{of}\:{f}\left({x}\right)=\: \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}\:+\:\mathrm{2} \\ $$$$\left.{A}\left.\right)\left.\:\left.\left({x}−\mathrm{1}\right)\:{B}\right)\:\left({x}−\mathrm{2}\right)\:{C}\right)\:\left({x}+\mathrm{1}\right)\:{D}\right)\left({x}+\mathrm{3}\right) \\ $$$$ \\ $$$${Q}_{\mathrm{9}\:} .\:{The}\:{sum}\:{of}\:{the}\:{sum}\:{of}\: \\ $$$${roots}\:{and}\:{product}\:{of}\:{roots}\:{of}\:{the} \\ $$$${quadratic}\:{equation}\:\mathrm{3}{x}^{\mathrm{2}} +\:\mathrm{6}{x}\:+\:\mathrm{9}=\mathrm{0} \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{1}\:\:\:{B}\right)\:−\mathrm{1}\:\:{C}\right)\:\mathrm{32}\:\:{D}\right)\:\mathrm{5} \\ $$$$ \\ $$$${Q}_{\mathrm{10}} .\:\mathrm{2}{log}\mathrm{2}−{log}\mathrm{2}\:=\: \\ $$$$\left.{A}\left.\right)\left.\:{log}\:\mathrm{8}\:\:\:\:\:\:\:{B}\right)\:{log}\:\mathrm{6}\:\:\:\:{C}\right)\:{log}\:\mathrm{3} \\ $$$$\left.{D}\right)\:{log}\:\mathrm{2}. \\ $$$$ \\ $$$${Q}_{\mathrm{11}} .\:\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{3}^{\mathrm{2}−{r}} = \\ $$$$\left.{A}\left.\right)\left.\:\left.\frac{\mathrm{9}}{\mathrm{4}}\:\:{B}\right)\:\frac{\mathrm{9}}{\mathrm{2}}\:\:{C}\right)\:\frac{\mathrm{13}}{\mathrm{3}}\:{D}\right)\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$${Q}_{\mathrm{12}} .\:{The}\:{constand}\:{term}\:{in}\:{the}\: \\ $$$${binomial}\:{expansion}\:{of}\:\left({x}^{\mathrm{2}} +\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{8}} {is} \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{3}^{{rd}} \:\:{B}\right)\:\mathrm{4}^{{th}} \:{C}\right)\:\mathrm{5}^{{th}} \:{D}\right)\:\mathrm{6}^{{th}} \: \\ $$$$ \\ $$$${Q}_{\mathrm{13}} .\:{cos}\left(\mathrm{180}−{x}\right)\:=\: \\ $$$$\left.{A}\left.\right)\left.\:−{sinx}\:\:\:{B}\right)\:{sin}\:{x}\:\:{C}\right)\:{cos}\:{x}\: \\ $$$$\left.{D}\right)\:−{cos}\:{x} \\ $$$$ \\ $$$${Q}_{\mathrm{14}} .\:{The}\:{value}\:{of}\:{p}\:{for}\:{which}\: \\ $$$$\left(\mathrm{2},\mathrm{1}\right)\:,\:\left(\mathrm{6},\mathrm{3}\right),\:{and}\:\left(\mathrm{4},{p}\right)\:{are}\:{collinear} \\ $$$${is}\:; \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{2}\:\:{B}\right)\:\mathrm{1}\:{C}\right)−\mathrm{1}\:{D}\right)\:−\mathrm{2} \\ $$$$ \\ $$$${Q}_{\mathrm{15}} .\:\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{3}} {dx}\:=\: \\ $$$$\left.{A}\left.\right)\left.\:−{sin}\:\mathrm{7}{x}\:\:\:{B}\right)\:{sin}\:\mathrm{7}{x}\:\:{C}\right)\:−\mathrm{7}{sin}\mathrm{7}{x} \\ $$$$\left.{D}\right)\:\mathrm{7}{sin}\mathrm{7}{x} \\ $$$$ \\ $$$${Q}_{\mathrm{16}} .\:{Given}\:{that}\:{g}\::\:\rightarrow\:{px}\:−\mathrm{5},\:{the}\: \\ $$$${value}\:{of}\:{p}\:{for}\:{which}\:{g}^{−\mathrm{1}} \left(\mathrm{3}\right)=\mathrm{4}\:{is} \\ $$$$\left.{A}\left.\right)\left.\:\left.\frac{\mathrm{1}}{\mathrm{2}}\:{B}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\:{C}\right)\:−\mathrm{2}\:\:{D}\right)\:\mathrm{2} \\ $$$$ \\ $$$${Q}_{\mathrm{17}} .\:{The}\:{value}\:{of}\:\theta\:{in}\:{the}\:{range}\: \\ $$$$\mathrm{0}°\leqslant\theta\leqslant\mathrm{90}°\:{for}\:{which}\:{sin}\theta\:=\:{cos}\:\theta\:{is} \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{90}°\:\:{B}\right)\:\mathrm{60}°\:{C}\right)\:\mathrm{30}°\:{D}\right)\:\mathrm{45}°\: \\ $$$$ \\ $$$$\:{Questions}\: \\ $$

Question Number 35798    Answers: 0   Comments: 3

Question Number 35789    Answers: 0   Comments: 1

Question Number 35791    Answers: 1   Comments: 0

A 5100 cm^2 trapezium has the perpendicular distance between the two parallel sides 60 m. If one of the parallel sides be 40 m then find the length of the other parallel side.

$$\mathrm{A}\:\mathrm{5100}\:\mathrm{cm}^{\mathrm{2}} \:\mathrm{trapezium}\:\mathrm{has}\:\mathrm{the}\:\mathrm{perpendicular} \\ $$$$\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{two}\:\mathrm{parallel}\:\mathrm{sides} \\ $$$$\mathrm{60}\:\mathrm{m}.\:\mathrm{If}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parallel}\:\mathrm{sides}\:\mathrm{be}\:\mathrm{40}\:\mathrm{m}\: \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{parallel}\:\mathrm{side}. \\ $$

Question Number 35771    Answers: 0   Comments: 2

Question Number 35769    Answers: 2   Comments: 0

Question Number 35768    Answers: 0   Comments: 0

Question Number 35766    Answers: 0   Comments: 1

let f(x)=arctan(1+x) and f_n (x)= e^(−nx) 1) define fof_n (x) and f_n of(x) 2) study the nature of the series Σ_(n=0) ^(+∞) fof_n (x) and Σ_(n=0) ^∞ f_n of(x) .

$${let}\:{f}\left({x}\right)={arctan}\left(\mathrm{1}+{x}\right)\:\:{and}\:\:{f}_{{n}} \left({x}\right)=\:{e}^{−{nx}} \\ $$$$\left.\mathrm{1}\right)\:{define}\:{fof}_{{n}} \left({x}\right)\:{and}\:{f}_{{n}} {of}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{nature}\:{of}\:{the}\:{series}\:\sum_{{n}=\mathrm{0}} ^{+\infty} {fof}_{{n}} \left({x}\right) \\ $$$${and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{f}_{{n}} {of}\left({x}\right)\:. \\ $$

Question Number 35763    Answers: 1   Comments: 0

Question Number 35743    Answers: 1   Comments: 1

find x log(log2+log_2 (x+1))=0

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{{x}} \\ $$$$\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{log}}\mathrm{2}+\boldsymbol{\mathrm{log}}_{\mathrm{2}} \left(\boldsymbol{{x}}+\mathrm{1}\right)\right)=\mathrm{0} \\ $$

Question Number 35739    Answers: 1   Comments: 2

Question Number 35738    Answers: 2   Comments: 0

Given that ∫_0 ^k x^2 = 16 find the value of k

$$\:{Given}\:{that}\:\:\int_{\mathrm{0}} ^{{k}} {x}^{\mathrm{2}} =\:\mathrm{16} \\ $$$${find}\:{the}\:{value}\:{of}\:{k} \\ $$

Question Number 35737    Answers: 0   Comments: 4

Question Number 35732    Answers: 1   Comments: 1

Question Number 35729    Answers: 1   Comments: 1

find the value of I =∫_0 ^π (dx/(2−cosx))

$${find}\:{the}\:{value}\:{of}\:\:\:{I}\:=\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{2}−{cosx}} \\ $$

Question Number 35833    Answers: 1   Comments: 1

fond lim_(n→+∞) n^a {ln(1+e^(−n) ) −e^(−n) } with a>0

$${fond}\:{lim}_{{n}\rightarrow+\infty} \:\:{n}^{{a}} \:\:\left\{{ln}\left(\mathrm{1}+{e}^{−{n}} \right)\:−{e}^{−{n}} \right\}\:{with}\:{a}>\mathrm{0} \\ $$

Question Number 35727    Answers: 0   Comments: 1

find lim_(x→0) ((sin(shx) −sh(sinx))/x)

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{sin}\left({shx}\right)\:−{sh}\left({sinx}\right)}{{x}} \\ $$

Question Number 35726    Answers: 1   Comments: 1

find lim_(x→0) ((1−cos(sinx))/x^2 )

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}−{cos}\left({sinx}\right)}{{x}^{\mathrm{2}} } \\ $$

Question Number 35723    Answers: 1   Comments: 0

Find the largest prime factor of the following: (1×2×3)+(2×3×4)+...+(2014×2015×2016)

$${Find}\:{the}\:{largest}\:{prime}\:{factor}\:{of}\:\:{the}\:{following}: \\ $$$$\left(\mathrm{1}×\mathrm{2}×\mathrm{3}\right)+\left(\mathrm{2}×\mathrm{3}×\mathrm{4}\right)+...+\left(\mathrm{2014}×\mathrm{2015}×\mathrm{2016}\right) \\ $$

Question Number 35715    Answers: 1   Comments: 1

Find lim_(n→∞) ((3^(n+1) +2^(n+1) )/(3^n +2^n ))

$${Find}\underset{{n}\rightarrow\infty} {\:{lim}}\frac{\mathrm{3}^{{n}+\mathrm{1}} +\mathrm{2}^{{n}+\mathrm{1}} }{\mathrm{3}^{{n}} +\mathrm{2}^{{n}} } \\ $$

Question Number 35750    Answers: 0   Comments: 4

Question Number 35736    Answers: 1   Comments: 0

lim_(x→0) ((ln (sec (ex)sec (e^2 x)......sec (e^(50) x)))/(e^2 −e^(2cos x) )) = ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{sec}\:\left({ex}\right)\mathrm{sec}\:\left({e}^{\mathrm{2}} {x}\right)......\mathrm{sec}\:\left({e}^{\mathrm{50}} {x}\right)\right)}{{e}^{\mathrm{2}} −{e}^{\mathrm{2cos}\:{x}} }\:=\:? \\ $$

Question Number 35703    Answers: 1   Comments: 0

Question Number 35696    Answers: 0   Comments: 7

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