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If ∣z∣ = 3 , what is the maximum and minimum value of ∣z−1+i(√3) ∣ ? |
A rectangular cardboard is 8cm long and 6cm wide. What is the least number of beads you can arrange on the board such that there are at least two of the beads that are less than (√(10))cm apart. |
Let N be the greatest multiple of 36 all of whose digits are even and no two of whose digits are the same. Find the remainder when N is divided by 1000. |
Let K be the product of all factors (b−a) (not necessarily distinct) where a and b are integers satisfying 1≤a≤b≤10. Find the greatest integer n such that 2^n divides K. |
(√(bemath )) (1)∫ ((cos x)/(2−cos x)) dx (2) f(x) = ∣x^3 ∣ ⇒ f ′(x) ? |
if ψ(x)=(x/(1−x))+(x^3 /(1−x^2 ))+(x^5 /(1−x^5 ))+(x^7 /(1−x^7 )) show that ∫_0 ^1 [ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )]((ln((1/x)))/x)=((25832π^2 )/(1575)) soltion let the generating series form of ψ(x)=(x/(1−x))+(x^3 /(1−x^2 ))+(x^5 /(1−x^5 ))+(x^7 /(1−x^7 )) be ψ(x)=Σ_(i=0) ^∞ (x^(2i+1) /(1−x^(2i+1) )).........(1) from Ω=∫_0 ^1 ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )((ln((1/x)))/x)dx Ω=−∫_0 ^1 [ψ(x)+ψ(x^(1/3) )+ψ(x^(1/5) )+ψ(x^(1/7) )]((lnx)/x)dx there the series form be Ω=∫_0 ^1 [Σ_(n=0) ^∞ ψ(x^(1/(2n+1)) )((lnx)/x)]dx.........(2) putting equation (1) into (2) Ω=−∫_0 ^1 Σ_(n=0) ^∞ Σ_(i=0) ^∞ [(((x^(2i+1) .x^(1/(2n+1)) )/(1−x^(2i+1) .x^(1/(2n+1)) )))((lnx)/x)]dx Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 [((x^((2i+1)/(2n+1)) /(1−x^((2i+1)/(2n+1)) )))((lnx)/x)]dx let y=((2i+1)/(2n+1)) Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 [((x^y /(1−x^y )))((lnx)/x)]dx=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 ((x^(y−1) /(1−x^y )))lnxdx let the series form of (1/(1−x^y ))=Σ_(m=0) ^∞ x^(my) Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 (Σ_(m=0) ^∞ x^(my) .x^(y−1) lnx)dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ∫_0 ^1 (x^(my) .x^(y−1) .x^a )dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ∫_0 ^1 x^(my+y+a−1) dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ((1/(my+y+a))) Ω^′ (0)=Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ [(1/((my+y)^2 ))]=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )Σ_(m=0) ^∞ (1/((1+m)^2 )) let z=1+m at m=0 ,z=1 Ω(0)=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )Σ_(z=1) ^∞ (1/z^2 )=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )ζ(2)=(π^2 /6)(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 ) but y=((2i+1)/(2n+1)) Ω(0)=(π^2 /6)(Σ_(n=0) ^∞ Σ_(i=0) ^∞ (1/((((2i+1)/(2n+1)))^2 )))=(π^2 /6)Σ_(n=0) ^∞ (2n+1)^2 Σ_(i=0) ^∞ (1/((2i+1)^2 )) Ω=(π^2 /6)[1^2 +3^2 +5^2 +7^2 ]×[(1/1^2 )+(1/3^2 )+(1/5^2 )+(1/7^2 )] Ω=(π^2 /6)(1+9+25+49)×[(1/1)+(1/9)+(1/(25))+(1/(49))] Ω=(π^2 /6)•(84)•((12916)/(11025))=((25832π^2 )/(1575)) ∵∫_0 ^1 [ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )]((ln((1/x)))/x)=((25832π^2 )/(1575)) by mathew monday(05/09/2020) |
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calculate lim_(n→+∞) Π_(k=1) ^n (1+((√(k(n−k)))/n^2 )) |
calculate lim_(n→+∞) a_n ∫_0 ^1 x^(2n) sin(((πx)/2))dx with a_n =Σ_(k=1) ^n sin(((πk)/(2n))) |
f function continue on [0,1] find lim_(n→+∞) (1/n)Σ_(k=0) ^n (n−k)∫_(k/n) ^((k+1)/n) f(x)dx |
let u_n =Σ_(k=1) ^n (1/(√((k+n)(k+n+1)))) detremine lim_(n→+∞) u_n |
explicite f(x) =∫_0 ^(2π) ln(x^2 −2xcosθ +1)dθ (x≠+^− 1) |
calculate lim_(n→+∞) Σ_(k=1) ^n ln((n/(n+k)))^(1/n) |
caoculate ∫_0 ^(π/4) ln(1+2tanx)dx |
find lim_(x→1^+ ) ∫_x ^x^2 ((ln(t))/((t−1)^2 ))dx |
find I_n =∫_0 ^(π/4) (du/(cos^n u)) |
calculate lim_(n→+∞) Σ_(k=1) ^n (√((n−k)/(n^3 +n^2 k))) |
calculate lim_(n→+∞) Σ_(k=1) ^(2n) (k/(k^2 +n^2 )) |
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(√(bemath)) (1)lim_(x→∞) [(x^2 /(x+1)) − (x^2 /(x+3)) ] ? (2) prove that n^2 ≤ 2^n for ∀n∈N by mathematical induction |
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There are two cards; one is yellow on both sides and the other one is yellow in one side and red on the other. The cards have the same probability ((1/2)) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is yellow, then the probability that the under side is also yellow is. |
A chord which is a perpendicular bisector of radius of length 18cm in a circle, has length. |
Pg 998 Pg 999 Pg 1000 Pg 1001 Pg 1002 Pg 1003 Pg 1004 Pg 1005 Pg 1006 Pg 1007 |