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Question-76586




Question Number 76586 by Master last updated on 28/Dec/19
Commented by Master last updated on 28/Dec/19
solve please
solveplease
Commented by Tony Lin last updated on 28/Dec/19
(5)  ∫_0 ^(π/2) sin^n xcos^n xdx  =(1/2^n )∫_0 ^(π/2) sin^n (2x)dx  let 2x=u, du=2dx  (1/2^(n−1) )∫_0 ^π sin^n udu  =(1/2^(n−2) )∫_0 ^(π/2) sin^n xdx  =(1/2^(n−2) )× { (((((n−1)(n−3)∙∙∙1)/(n(n−2)∙∙∙2))×(π/2), if n is even)),(((((n−1)(n−3)∙∙∙2)/(n(n−2)∙∙∙1)), if n is odd)) :}
(5)0π2sinnxcosnxdx=12n0π2sinn(2x)dxlet2x=u,du=2dx12n10πsinnudu=12n20π2sinnxdx=12n2×{(n1)(n3)1n(n2)2×π2,ifniseven(n1)(n3)2n(n2)1,ifnisodd
Commented by john santu last updated on 28/Dec/19
sorry sir missing in line 2  (1/2^n )(2sin xcos x)^n
sorrysirmissinginline212n(2sinxcosx)n
Commented by Tony Lin last updated on 28/Dec/19
(1)  a_(n+2) =(1/a_(n+1) )+a_n   ⇒a_(n+2) a_(n+1) =a_(n+1) a_n +1  ⇒a_(2019) a_(2018)   =a_(2018) a_(2017) +1  =a_(2017) a_(2016) +2  =a_1 a_2 +2017  =2018  ⇒a_(n+1) a_n =n  (a_(2019) /a_(2017) )=((2018)/(2017))  (a_(2017) /a_(2015) )=((2016)/(2015))  (a_5 /a_3 )=(4/3)  (a_3 /a_1 )=(2/1)  ⇒a_(2019) =a_1 ×((2×4×∙∙∙×2018)/(1×3×∙∙∙×2017))  =((2018!!)/(2017!!))
(1)an+2=1an+1+anan+2an+1=an+1an+1a2019a2018=a2018a2017+1=a2017a2016+2=a1a2+2017=2018an+1an=na2019a2017=20182017a2017a2015=20162015a5a3=43a3a1=21a2019=a1×2×4××20181×3××2017=2018!!2017!!
Commented by Master last updated on 28/Dec/19
2 and 3 , 4  solution plz
2and3,4solutionplz
Commented by Master last updated on 28/Dec/19
thanks
thanks
Commented by MJS last updated on 28/Dec/19
(3) all you need is calculate the determinant...  let p=2019  −2(x−p)(x−p+1)(x−p−1)(3px+3p^2 −1)=0  ⇒  x_1 =p−1=2018  x_2 =p=2019  x_3 =p+1=2020  x_4 =((1−3p^2 )/(3p))=−((12229082)/(6057))
(3)allyouneediscalculatethedeterminantletp=20192(xp)(xp+1)(xp1)(3px+3p21)=0x1=p1=2018x2=p=2019x3=p+1=2020x4=13p23p=122290826057
Answered by benjo 1/2 santuyy last updated on 29/Dec/19
(4) f(0)=4 →c=4  0≤f(x)≤4 for 0≤x≤3 →  ax^2  +bx+4≤4 ,   x(ax+b)≤0 , b must be negatif  so we has 0≤x≤−(b/a) .   we getting −(b/a)=3 → b =−3a , a >0 ■
(4)f(0)=4c=40f(x)4for0x3ax2+bx+44,x(ax+b)0,bmustbenegatifsowehas0x(b/a).wegetting(b/a)=3b=3a,a>0◼

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