臺灣綜合大學系統113學年度學士班轉學生聯合招生考試
科目名稱:微積分B
解答:(a)f(x)=(x2−x+1)100⇒f′(x)=100(x2−x+1)99(2x−1)⇒f′(0)=100⋅1⋅(−1)=−100(b)f″(x)=9900(x2−x+1)98(2x−1)2+200(x2−x+1)99⇒f″(0)=9900+200=10100
解答:limx→0√3+cos3x−2x2=limx→0(√3+cos3x−2)′(x2)′=limx→0−3sin3x2√3+cos3x2x=limx→0−3sin3x4x√3+cos3x=limx→0(−3sin3x)′(4x√3+cos3x)′=limx→0−9cos3x4√3+cos3x+−6xsin3x√3+cos3x=−98解答:(a)f(x)=2x2=ex2ln2⇒f′(x)=2xln2ex2ln2⇒f′(2)=4ln2⋅16=64ln2(b)g(x)=x2x=e2xlnx⇒g′(x)=(ln2⋅2xlnx+2xx)e2xlnx⇒g′(2)=(4(ln2)2+2)24=32+64(ln2)2
解答:∫32xx2−5x+4dx=∫32x(x−4)(x−1)dx=∫32(4/3x−4−1/3x−1)dx=[43ln|x−4|−13ln|x−1|]|32=−13ln2−43ln2=−53ln2
解答:u=−√3x⇒du=−32√3xdx=32udx⇒dx=23udu⇒∫e−√3xdx=∫23ueudu=23(ueu−eu)=23(−√3xe−√3x−e−√3x)⇒∫∞3e−√3xdx=[23(−√3xe−√3x−e−√3x)]|∞3=23(3e−3+e−3)=83e−3
解答:f(x)=5√32+x⇒{f(0)=2f′(x)=15(32+x)4/5⇒{f′(0)=1/80f″(x)=−425(32+x)9/5⇒{f″(0)=−13200f‴(x)=36125(32+x)14/5⇒f‴(0)=9512000⇒f(x)=f(0)+f′(0)x+12f″(0)x2+16f‴(0)x3+⋯=2+180x−16400x2+31024000x3+⋯⇒{c1=1/80c2=−1/6400c3=3/1024000
解答:(a)f(x)=sin(3x)⇒{f′(x)=3cos(3x)f″(x)=−32sin(3x)f‴(x)=−33cos(3x)f[4](x)=34sin(3x)⇒f[n](x)={3ncos(3x)n≡1 mod 4−3nsin(3x)n≡2 mod 4−3ncos(3x)n≡3 mod 43nsin(3x)n≡0 mod 4111≡3 mod 4⇒f[111](0)=−3111cos0=−3111(b)sinx=∞∑n=0(−1)n1(2n+1)!x2n+1⇒g(x)=sin(x3)=∞∑n=0(−1)n1(2n+1)!x6n+3n=18⇒g(x)的x111係數為(−1)18⋅137!⇒g[111](0)=111!37!
解答:f(x,y)=(4x2+y2)e−2x⇒{fx=(8x−8x2−2y2)e−2xfy=2ye−2x⇒{fxx=(8−32x+16x2+4y2)e−2xfxy=−4ye−2xfyy=2e−2x⇒d(x,y)=fxxfyy−f2xy=(16−64x+32x2−8y2)e−4x{fx=0fy=0⇒(x,y)=(0,0),(1,0)⇒{d(0,0)=16>0⇒fxx(0,0)=8>0d(1,0)=−16e−4<0⇒critical points: (0,0,),(1,0), and local minimum: f(0,0)=0,saddle point: (0,0)
解答:{P(x,y)=x1/3y2/3g(x,y)=3x1/2+5y1/2−45⇒{Px=λgxPy=λgyg=0⇒{13x−2/3y2/3=λ⋅32x−1/2⋯(1)23x1/3y−1/3=λ⋅52y−1/2⋯(2)3x1/2+5y1/2=45⋯(3)⇒(1)(2)=y2x=3√y5√x⇒y1/2=65x1/2代入(3)⇒3x1/2+6x1/2=45⇒x1/2=5⇒y1/2=6⇒P(25,36)=251/3362/3
解答:改變積分順序:I=∫40∫2√x√x⋅sin(y2)y2dydx=∫20∫y20√x⋅sin(y2)y2dxdy=∫2023ysin(y2)dyu=y2⇒du=2ydy⇒I=∫4013sinudu=[−13cosu]|40=13(1−cos4)
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解題僅供參考,轉學考歷年試題及詳解
第7題的(b),sin(x^3)的泰勒級數不對,故答案也不對.
回覆刪除謝謝告知,已修訂
刪除提醒一下x^111項的係數沒錯,但題目是問g^(111)(0)答案應是(111!/37!)=111x110x...x38
刪除對,已修訂,謝謝!
刪除