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2024年3月17日 星期日

112年政大統計系碩士班-基礎數學詳解

國立政治大學112學年度碩士班招生考試

考試科目:基礎數學   系所別:統計學系

Part I: Calculus

解答:State:Let f be continuous on [a,b] and differentiable on (a,b). Then c(a,b) such that f(c)=f(b)f(a)baProve:Let {A(a,f(a))B(b,f(b)), then ¯AB:y=f(b)f(a)ba(xa)+f(a)And let g(x)=f(x)[f(b)f(a)ba(xa)+f(a)]By Rolle's theorem, c(a,b)g(c)=0. That is,g(x)=f(x)f(b)f(a)bag(c)=f(c)f(b)f(a)ba=0f(c)=f(b)f(a)ba,Q.E.D
解答:Let ϵ>0, we choose δ=1ϵ. Then, we have x>δ=1ϵ|1x20|=1x2<ϵ(x>1ϵ1x2<ϵ)limx1x2=0,Q.E.D.
解答:(a)2x2x+4x3+4xdx=(1x+x1x2+4)dx=1xdx+xx2+4dx141(x/2)2+1dx=lnx+12ln(x2+4)12tan1(x2)+C(b)f(x)1x4f(0)32f(x)dx(c)a0f(x)dx=a0f(ax)dxI=π/20sinnxsinnx+cosnxdx=π/20sinn(π/2x)sinn(π/2x)+cosn(π/2x)dx=π/20cosnxcosnx+sinnxdxI+I=2I=π/20sinnx+cosnxsinnx+cosnxdx=π/201dx=π2I=π4(d)I=0ex2dxI2=(0ex2dx)(0ey2dy)=00e(x2+y2)dxdyLet {x=rcosθy=rsinθ, then I2=π/200rer2drdθ=π/2012dθ=π4I=π4=π2(e)I=sin5xcos2xdx=sinx(1cos2x)2cos2xdxLet u=cosx, then du=sinxdxI=(1u2)2u2du=(u6+2u4u2)du=17u7+25u513u3+C=17cos7x+25cos5x13cos3x+C

解答:(tanxy)=(x+y)sec2xy(1yxyy2)=1+y1ysec2xy1=(1+xy2sec2xy)ydydx=y=1ysec2xy11+xy2sec2xy
解答:x2+y2=1{x=cosθy=sinθz=1x+y=1cosθ+sinθf(x,y,z)=x+2y+3z=cosθ+2sinθ+33cosθ+3sinθ=32cosθ+5sinθ=329(229cosθ529sinθ)=329sin(αθ){max(f)=3+29min(f)=329

Part II: Linear Algebra

解答:(a)λv=Av=A2v=A(Av)=A(λv)=λ(Av)=λ2vλ2vλv=0λ(λ1)v=0λ=0,1(b)By (a), the eigenvalues are 0 or 1, then rank(A)=number of 1rank(A)=tr(A)
解答:Q is orthogonal Q1=QTQQT=Idet(QQT)=det(Q)det(QT)=(det(Q))2=1det(Q)=±1,Q.E.D.
解答:N(T)={vVT(v)=0}(a)T(0)=00N(T)(b)Suppose that {v1N(T)v2N(T){T(v1)=0T(v2)=0T(v1+v2)=T(v1)+T(v2)=0v1+v2N(T)(c)Suppose that vN(T)T(v)=0T(cv)=cT(v)=0cvN(T)From (a),(b), and (c), we have N(T) is a subspace of VR(T)={wWw=T(v)vV}(a)T(0)=00R(T)(b)Suppose that {w1R(T)w2R(T)v1,v2V{T(v1)=w1T(v2)=w2w1+w2=T(v1)+T(v2)=T(v1+v2)w1+w2R(T)(c)Suppose that wR(T)vVT(v)=wT(cv)=cT(v)=cwcwR(T)From (a),(b), and (c), we have R(T) is a subspace of W.Q.E.D.
解答:A=(0213)=(2111)(1002)(1112)An=(0213)=(2111)(1n002n)(1112)=(2n+22n+1+22n12n+11)
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解題僅供參考,其他歷年試題及詳解

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