Giả sử ∫ - 1 2 e x d x 2 + e x = ln a e + e 3 a e + b với a, b là các số nguyên dương. Tính giá trị của biểu thức P = sin πb a + 2017 π + cos πb a - sin 2018 π
A. 1
B. -1
C. 1 2
D. - 1 2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,y'=8x^3-9x^2+10x\\ \Rightarrow y''=24x^2-18x+10\\ b,y'=\dfrac{2}{\left(3-x\right)^2}\\ \Rightarrow y''=\dfrac{4}{\left(3-x\right)^3}\)
\(c,y'=2cos2xcosx-sin2xsinx\\ \Rightarrow y''=-5sin\left(2x\right)cos\left(x\right)-4cos\left(2x\right)sin\left(x\right)\\ d,y'=-2e^{-2x+3}\\ \Rightarrow y''=4e^{-2x+3}\)
a) \(\int\dfrac{2dx}{x^2-5x}=\int\left(\dfrac{-2}{5x}+\dfrac{2}{5\left(x-5\right)}\right)dx=-\dfrac{2}{5}ln\left|x\right|+\dfrac{2}{5}ln\left|x-5\right|+C\)
\(\Rightarrow A=-\dfrac{2}{5};B=\dfrac{2}{5}\Rightarrow2A-3B=-2\)
b) \(\int\dfrac{x^3-1}{x+1}dx=\int\dfrac{x^3+1-2}{x+1}dx=\int\left(x^2-x+1-\dfrac{2}{x+1}\right)dx=\dfrac{1}{3}x^3-\dfrac{1}{2}x^2+x-2ln\left|x+1\right|+C\)
\(\Rightarrow A=\dfrac{1}{3};B=\dfrac{1}{2};E=-2\Rightarrow A-B+E=-\dfrac{13}{6}\)
`a)TXĐ:R\\{1;1/3}`
`y'=[-4(6x-4)]/[(3x^2-4x+1)^5]`
`b)TXĐ:R`
`y'=2x. 3^[x^2-1] ln 3-e^[-x+1]`
`c)TXĐ: (4;+oo)`
`y'=[2x-4]/[x^2-4x]+2/[(2x-1).ln 3]`
`d)TXĐ:(0;+oo)`
`y'=ln x+2/[(x+1)^2].2^[[x-1]/[x+1]].ln 2`
`e)TXĐ:(-oo;-1)uu(1;+oo)`
`y'=-7x^[-8]-[2x]/[x^2-1]`
Lời giải:
a.
$y'=-4(3x^2-4x+1)^{-5}(3x^2-4x+1)'$
$=-4(3x^2-4x+1)^{-5}(6x-4)$
$=-8(3x-2)(3x^2-4x+1)^{-5}$
b.
$y'=(3^{x^2-1})'+(e^{-x+1})'$
$=(x^2-1)'3^{x^2-1}\ln 3 + (-x+1)'e^{-x+1}$
$=2x.3^{x^2-1}.\ln 3 -e^{-x+1}$
c.
$y'=\frac{(x^2-4x)'}{x^2-4x}+\frac{(2x-1)'}{(2x-1)\ln 3}$
$=\frac{2x-4}{x^2-4x}+\frac{2}{(2x-1)\ln 3}$
d.
\(y'=(x\ln x)'+(2^{\frac{x-1}{x+1}})'=x(\ln x)'+x'\ln x+(\frac{x-1}{x+1})'.2^{\frac{x-1}{x+1}}\ln 2\)
\(=x.\frac{1}{x}+\ln x+\frac{2}{(x+1)^2}.2^{\frac{x-1}{x+1}}\ln 2\\ =1+\ln x+\frac{2^{\frac{2x}{x+1}}\ln 2}{(x+1)^2}\)
e.
\(y'=-7x^{-8}-\frac{(x^2-1)'}{x^2-1}=-7x^{-8}-\frac{2x}{x^2-1}\)
\(A=x^2-4x+1\)
\(A=\left(x^2-4x+4\right)-3\)
\(A=\left(x-2\right)^2-3\)
Vì \(\left(x-2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x-2\right)^2-3\ge-3\) với mọi x
\(\Rightarrow Amin=-3\Leftrightarrow x=2\)
\(B=4x^2+4x+11\)
\(B=\left(4x^2+4x+1\right)+10\)
\(B=\left(2x+1\right)^2+10\)
Vì \(\left(2x+1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(2x+1\right)^2+10\ge10\) với mọi x
\(\Rightarrow Bmin=10\Leftrightarrow x=-\dfrac{1}{2}\)
\(C=\left(x-1\right)\left(x+3\right)\left(x+2\right)\left(x+6\right)\)
\(C=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(C=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(C=\left(x^2+5x\right)^2-36\)
Vì \(\left(x^2+5x\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)
\(\Rightarrow Cmin=-36\Leftrightarrow x^2+5x=0\)
\(\Rightarrow x\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(D=5-8x-x^2\)
\(D=-\left(x^2+8x-5\right)\)
\(D=-\left(x^2+8x+16-16-5\right)\)
\(D=-\left(x^2+8x+16\right)+21\)
\(D=-\left(x+4\right)^2+21\)
Vì \(-\left(x+4\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x+4\right)^2+21\le21\) với mọi x
\(\Rightarrow Dmax=21\Leftrightarrow x=-4\)
\(E=4x-x^2+1\)
\(E=-\left(x^2-4x-1\right)\)
\(E=-\left(x^2-4x+4-4-1\right)\)
\(E=-\left(x^2-4x+4\right)+5\)
\(E=-\left(x-2\right)^2+5\)
Vì \(-\left(x-2\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x-2\right)^2+5\le5\) với mọi x
\(\Rightarrow Emax=5\Leftrightarrow x=2\)
a: \(y'=4\cdot3x^2-3\cdot2x+2=12x^2-6x+2\)
b: \(y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}=\dfrac{x-1-x-1}{\left(x-1\right)^2}=\dfrac{-2}{\left(x-1\right)^2}\)
c: \(y'=-2\cdot\left(\sqrt{x}\cdot x\right)'\)
\(=-2\cdot\left(\dfrac{x+x}{2\sqrt{x}}\right)=-2\cdot\dfrac{2x}{2\sqrt{x}}=-2\sqrt{x}\)
d: \(y'=\left(3sinx+4cosx-tanx\right)\)'
\(=3cosx-4sinx+\dfrac{1}{cos^2x}\)
e: \(y'=\left(4^x+2e^x\right)'\)
\(=4^x\cdot ln4+2\cdot e^x\)
f: \(y'=\left(x\cdot lnx\right)'=lnx+1\)
a: \(y'=\left(x^2+2x\right)'\left(x^3-3x\right)+\left(x^2+2x\right)\left(x^3-3x\right)'\)
\(=\left(2x+2\right)\left(x^3-3x\right)+\left(x^2+2x\right)\left(3x^2-3\right)\)
\(=2x^4-6x^2+2x^3-6x+3x^4-3x^2+6x^3-6x\)
\(=5x^4+8x^3-9x^2-12x\)
b: y=1/-2x+5
=>\(y'=\dfrac{2}{\left(2x+5\right)^2}\)
c: \(y'=\dfrac{\left(4x+5\right)'}{2\sqrt{4x+5}}=\dfrac{4}{2\sqrt{4x+5}}=\dfrac{2}{\sqrt{4x+5}}\)
d: \(y'=\left(sinx\right)'\cdot cosx+\left(sinx\right)\cdot\left(cosx\right)'\)
\(=cos^2x-sin^2x=cos2x\)
e: \(y=x\cdot e^x\)
=>\(y'=e^x+x\cdot e^x\)
f: \(y=ln^2x\)
=>\(y'=\dfrac{\left(-1\right)}{x^2}=-\dfrac{1}{x^2}\)
\(\int\left(3x^2-2x-4\right)dx=x^3-x^2-4x+C\)
\(\int\left(sin3x-cos4x\right)dx=-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+C\)
\(\int\left(e^{-3x}-4^x\right)dx=-\dfrac{1}{3}e^{-3x}-\dfrac{4^x}{ln4}+C\)
d. \(I=\int lnxdx\)
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x\end{matrix}\right.\)
\(\Rightarrow u=x.lnx-\int dx=x.lnx-x+C\)
e. Đặt \(\left\{{}\begin{matrix}u=x\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=x.e^x-\int e^xdx=x.e^x-e^x+C\)
f.
Đặt \(\left\{{}\begin{matrix}u=x+1\\dv=sinxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-cosx\end{matrix}\right.\)
\(\Rightarrow I=-\left(x+1\right)cosx+\int cosxdx=-\left(x+1\right)cosx+sinx+C\)
g.
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{2}x^2.lnx-\dfrac{1}{2}\int xdx=\dfrac{1}{2}x^2.lnx-\dfrac{1}{4}x^2+C\)
Ta có:
∫ - 1 2 e x d x 2 + e x = ln a e + e 3 a e + b = ln 2 + e x - 1 2 = ln 2 + e 2 - ln 2 + e - 1 ln 2 + e x 2 + e - 1 = ln 2 e + e 3 2 e + 1 = ln a e + e 3 a e + b
Suy ra a = 2; b = 1 hay
P = sin πb a + 2017 π + cos πb a - sin 2018 π = -1
Đáp án B