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Question Number 120102 Answers: 2 Comments: 0

Θ = ∫ ((4x^(−1) +8x^(−3) )/(x^2 (√(x^4 +2x^2 +2)))) dx

$$\:\Theta\:=\:\int\:\frac{\mathrm{4}{x}^{−\mathrm{1}} +\mathrm{8}{x}^{−\mathrm{3}} }{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}}}\:{dx}\: \\ $$

Question Number 120092 Answers: 3 Comments: 1

Question Number 120091 Answers: 1 Comments: 0

When f(x) is divided by (x−1)(x+2), the remainder is (x+3), and when f(x) is divided by (x^2 +2x+5), the remainder is (2x+1). Find the remainder when f(x) is divided by (x−1)(x^2 +2x+5).

$$\mathrm{When}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right),\: \\ $$$$\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\left({x}+\mathrm{3}\right),\:\mathrm{and}\:\mathrm{when}\:{f}\left({x}\right) \\ $$$$\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\right),\:\mathrm{the}\:\mathrm{remainder} \\ $$$$\mathrm{is}\:\left(\mathrm{2}{x}+\mathrm{1}\right). \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{divided} \\ $$$$\mathrm{by}\:\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\right). \\ $$

Question Number 120090 Answers: 0 Comments: 0

Find the number of subsets of { 1,2,3,...,2000 } the sum of whose elements is divisible by 5

$${Find}\:{the}\:{number}\:{of}\:{subsets}\:{of} \\ $$$$\left\{\:\mathrm{1},\mathrm{2},\mathrm{3},...,\mathrm{2000}\:\right\}\:{the}\:{sum}\:{of}\: \\ $$$${whose}\:{elements}\:{is}\:{divisible}\:{by}\:\mathrm{5} \\ $$

Question Number 120089 Answers: 0 Comments: 0

If a continuous function f:R→R satisfies ∫_0 ^1 f(x)dx=∫_0 ^1 xf(x)dx=1 prove that ∫_0 ^1 (f(x))^2 dx≥4

$$\mathrm{If}\:\mathrm{a}\:\mathrm{continuous}\:\mathrm{function}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{satisfies} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {xf}\left({x}\right){dx}=\mathrm{1} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({f}\left({x}\right)\right)^{\mathrm{2}} {dx}\geqslant\mathrm{4} \\ $$

Question Number 120071 Answers: 1 Comments: 0

Question Number 120068 Answers: 2 Comments: 0

(i) lim_(x→0) ((cos x−1+(x^2 /2))/x^4 ) (ii) lim_(x→0) ((e^x −1−x−(x^2 /2)−(x^3 /6))/x^4 ) (iii) lim_(x→0) ((tan x−x)/(arc sin x−x))

$$\left({i}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}{{x}^{\mathrm{4}} }\: \\ $$$$\left({ii}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −\mathrm{1}−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{{x}^{\mathrm{4}} } \\ $$$$\left({iii}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}−{x}}{\mathrm{arc}\:\mathrm{sin}\:{x}−{x}} \\ $$

Question Number 120067 Answers: 2 Comments: 0

(i) lim_(x→0) ((x^2 sin (1/x))/(tan x)) (ii) Without L′Hopital rule lim_(x→0^+ ) ((1−cos x−xsin x)/(2−2cos x−sin^2 x))

$$\left({i}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \:\mathrm{sin}\:\left(\mathrm{1}/{x}\right)}{\mathrm{tan}\:{x}} \\ $$$$\left({ii}\right)\:{Without}\:{L}'{Hopital}\:{rule} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}−{x}\mathrm{sin}\:{x}}{\mathrm{2}−\mathrm{2cos}\:{x}−\mathrm{sin}\:^{\mathrm{2}} {x}} \\ $$

Question Number 120064 Answers: 1 Comments: 0

I = ∫ _0 ^(π/2) ((1/(ln (tan r))) + (1/(1−tan r)) ) dr

$${I}\:=\:\int\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\:}}\left(\frac{\mathrm{1}}{\mathrm{ln}\:\left(\mathrm{tan}\:{r}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{tan}\:{r}}\:\right)\:{dr} \\ $$

Question Number 120060 Answers: 2 Comments: 0

(i) ∫_(−2) ^0 (dx/(2x+3)) (ii)∫_3 ^5 (dx/( (((4−x)^2 ))^(1/3) ))

$$\left({i}\right)\:\underset{−\mathrm{2}} {\overset{\mathrm{0}} {\int}}\:\frac{{dx}}{\mathrm{2}{x}+\mathrm{3}} \\ $$$$\left({ii}\right)\underset{\mathrm{3}} {\overset{\mathrm{5}} {\int}}\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{4}−{x}\right)^{\mathrm{2}} }}\: \\ $$

Question Number 120059 Answers: 1 Comments: 0

Question Number 120058 Answers: 1 Comments: 0

(d^2 y/dx) +x (dy/dx) −y=0

$$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:+{x}\:\frac{{dy}}{{dx}}\:−{y}=\mathrm{0} \\ $$

Question Number 120050 Answers: 3 Comments: 0

(i) y′′−4y′+5y=4sin^2 4x (ii) (x/2)+1 = (√(∣1−x^2 ∣))

$$\:\left({i}\right)\:{y}''−\mathrm{4}{y}'+\mathrm{5}{y}=\mathrm{4sin}\:^{\mathrm{2}} \mathrm{4}{x} \\ $$$$\:\left({ii}\right)\:\frac{{x}}{\mathrm{2}}+\mathrm{1}\:=\:\sqrt{\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}\: \\ $$

Question Number 120049 Answers: 2 Comments: 0

f(x+2)+f(x−2)=f(x) f(1)=1 ,f(2)=2,f(3)=3,f(4)=4 then f(100)=?

$$\:{f}\left({x}+\mathrm{2}\right)+{f}\left({x}−\mathrm{2}\right)={f}\left({x}\right) \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}\:,{f}\left(\mathrm{2}\right)=\mathrm{2},{f}\left(\mathrm{3}\right)=\mathrm{3},{f}\left(\mathrm{4}\right)=\mathrm{4} \\ $$$${then}\:{f}\left(\mathrm{100}\right)=? \\ $$

Question Number 120044 Answers: 2 Comments: 0

Given a_(n+1) = ((2a_n )/((2n+1)(2n+2))) find a_n .

$${Given}\:{a}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{2}{a}_{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)} \\ $$$${find}\:{a}_{{n}} . \\ $$

Question Number 120040 Answers: 2 Comments: 0

∫ (t^5 /( (√(2+t^2 )))) dt

$$\:\int\:\frac{{t}^{\mathrm{5}} }{\:\sqrt{\mathrm{2}+{t}^{\mathrm{2}} }}\:{dt}\: \\ $$

Question Number 120037 Answers: 1 Comments: 0

Suppose that R>0, x_0 >0, and x_(n+1) =(1/2)((R/x_n )+x_n ), n≥0 Prove: For n≥1, x_n >x_(n+1) >(√R) and x_n −(√R)≤(1/2^n ) (((x_0 −(√R))^2 )/x_0 )

$$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{R}>\mathrm{0},\:\mathrm{x}_{\mathrm{0}} >\mathrm{0},\:\mathrm{and} \\ $$$$\mathrm{x}_{\mathrm{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{R}}{\mathrm{x}_{\mathrm{n}} }+\mathrm{x}_{\mathrm{n}} \right),\:\mathrm{n}\geqslant\mathrm{0} \\ $$$$\mathrm{Prove}:\:\mathrm{For}\:\mathrm{n}\geqslant\mathrm{1},\:\mathrm{x}_{\mathrm{n}} >\mathrm{x}_{\mathrm{n}+\mathrm{1}} >\sqrt{\mathrm{R}}\:\mathrm{and} \\ $$$$\mathrm{x}_{\mathrm{n}} −\sqrt{\mathrm{R}}\leqslant\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:\frac{\left(\mathrm{x}_{\mathrm{0}} −\sqrt{\mathrm{R}}\right)^{\mathrm{2}} }{\mathrm{x}_{\mathrm{0}} } \\ $$

Question Number 120036 Answers: 1 Comments: 0

Question Number 120035 Answers: 1 Comments: 0

Question Number 120029 Answers: 0 Comments: 2

Montrer que ∀x∈R cos(sinx)>sin(cosx)

$$\mathrm{Montrer}\:\mathrm{que}\:\forall\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{cos}\left(\mathrm{sinx}\right)>\mathrm{sin}\left(\mathrm{cosx}\right) \\ $$

Question Number 120028 Answers: 2 Comments: 0

Question Number 120025 Answers: 2 Comments: 0

calculate f^′ (x) 1) f(x) =∫_0 ^∞ ((cos(xt))/(t^2 +x^2 ))dt 2)f(x)=∫_0 ^∞ ((sin(xt^2 +(√2)))/(t^2 +x^2 +3))dt

$$\mathrm{calculate}\:\mathrm{f}^{'} \left(\mathrm{x}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{xt}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\left.\mathrm{2}\right)\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\mathrm{xt}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \:+\mathrm{3}}\mathrm{dt} \\ $$

Question Number 120016 Answers: 2 Comments: 0

Question Number 120008 Answers: 2 Comments: 2

Question Number 120006 Answers: 0 Comments: 0

Question Number 119997 Answers: 4 Comments: 0

solve for x,a∈R. (√(x^2 +ax+a^2 ))+(√(x^2 −ax+a^2 ))=1

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x},\mathrm{a}\in\boldsymbol{\mathrm{R}}. \\ $$$$\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{ax}}+\boldsymbol{{a}}^{\mathrm{2}} }+\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{{ax}}+\boldsymbol{{a}}^{\mathrm{2}} }=\mathrm{1} \\ $$

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