2023考研数一真题及答案

发布于:2025-06-27 ⋅ 阅读:(16) ⋅ 点赞:(0)

历年数一真题及答案下载直通车

  1. 曲线 y = x ln ⁡ ( e + 1 x − 1 ) y = x \ln(e + \frac{1}{x-1}) y=xln(e+x11)的渐近线方程为( )
    (A) y = x + e y = x + e y=x+e
    (B) y = x + 1 e y = x + \frac{1}{e} y=x+e1
    © y = x y = x y=x
    (D) y = x − 1 e y = x - \frac{1}{e} y=xe1

  2. 若微分方程 y ′ ′ + a y ′ + b y = 0 y'' + a y' + b y = 0 y′′+ay+by=0的解在 ( − ∞ , + ∞ ) (-\infty, +\infty) (,+)上有界,则( )
    (A) a < 0 , b > 0 a < 0, b > 0 a<0,b>0
    (B) a > 0 , b > 0 a > 0, b > 0 a>0,b>0
    © a = 0 , b > 0 a = 0, b > 0 a=0,b>0
    (D) a = 0 , b < 0 a = 0, b < 0 a=0,b<0

  3. 设函数 y = f ( x ) y = f(x) y=f(x) { x = 2 t + ∣ t ∣ y = ∣ t ∣ sin ⁡ t \begin{cases} x = 2 t + |t| \\ y = |t| \sin t \end{cases} {x=2t+ty=tsint确定,则( )
    (A) f ( x ) f(x) f(x)连续, f ′ ( 0 ) f'(0) f(0)不存在
    (B) f ′ ( 0 ) f'(0) f(0)存在, f ′ ( x ) f'(x) f(x) x = 0 x = 0 x=0处不连续
    © f ′ ( x ) f'(x) f(x)连续, f ′ ′ ( 0 ) f''(0) f′′(0)不存在
    (D) f ′ ′ ( 0 ) f''(0) f′′(0)存在, f ′ ′ ( x ) f''(x) f′′(x) x = 0 x = 0 x=0处不连续

  4. 已知 a n < b n ( n = 1 , 2 , … ) a_n < b_n (n = 1, 2, \ldots) an<bn(n=1,2,),若级数 ∑ n = 1 ∞ a n \sum_{n=1}^{\infty} a_n n=1an ∑ n = 1 ∞ b n \sum_{n=1}^{\infty} b_n n=1bn均收敛,则“ ∑ n = 1 ∞ a n \sum_{n=1}^{\infty} a_n n=1an绝对收敛”是“ ∑ n = 1 ∞ b n \sum_{n=1}^{\infty} b_n n=1bn绝对收敛的”( )
    (A) 充分必要条件
    (B) 充分不必要条件
    © 必要不充分条件
    (D) 既不充分也不必要条件

  5. 已知 n n n阶矩阵 A , B , C A, B, C A,B,C满足 A B C = 0 ABC = 0 ABC=0 E E E n n n阶单位矩阵,记矩阵 ( 0 A B C E ) \begin{pmatrix} 0 & A \\ BC & E \end{pmatrix} (0BCAE) ( A B C 0 E ) \begin{pmatrix} AB & C \\ 0 & E \end{pmatrix} (AB0CE) ( E A B A B 0 ) \begin{pmatrix} E & AB \\ AB & 0 \end{pmatrix} (EABAB0)的秩分别为 γ 1 , γ 2 , γ 3 \gamma_1, \gamma_2, \gamma_3 γ1,γ2,γ3,则( )
    (A) γ 1 ≤ γ 2 ≤ γ 3 \gamma_1 \leq \gamma_2 \leq \gamma_3 γ1γ2γ3
    (B) γ 1 ≤ γ 3 ≤ γ 2 \gamma_1 \leq \gamma_3 \leq \gamma_2 γ1γ3γ2
    © γ 3 ≤ γ 1 ≤ γ 2 \gamma_3 \leq \gamma_1 \leq \gamma_2 γ3γ1γ2
    (D) γ 2 ≤ γ 1 ≤ γ 3 \gamma_2 \leq \gamma_1 \leq \gamma_3 γ2γ1γ3

  6. 下列矩阵中不能相似于对角矩阵的是( )
    (A) ( 1 1 a 0 2 2 0 0 3 ) \begin{pmatrix} 1 & 1 & a \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix} 100120a23
    (B) ( 1 1 a 1 2 0 a 0 3 ) \begin{pmatrix} 1 & 1 & a \\ 1 & 2 & 0 \\ a & 0 & 3 \end{pmatrix} 11a120a03
    © ( 1 1 a 0 2 0 0 0 2 ) \begin{pmatrix} 1 & 1 & a \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} 100120a02
    (D) ( 1 1 a 0 2 2 0 0 2 ) \begin{pmatrix} 1 & 1 & a \\ 0 & 2 & 2 \\ 0 & 0 & 2 \end{pmatrix} 100120a22

  7. 已知向量 α 1 = ( 1 2 3 ) \alpha_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} α1= 123 α 2 = ( 2 1 1 ) \alpha_2 = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} α2= 211 β 1 = ( 2 5 9 ) \beta_1 = \begin{pmatrix} 2 \\ 5 \\ 9 \end{pmatrix} β1= 259 β 2 = ( 1 0 1 ) \beta_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} β2= 101 ,若 γ \gamma γ既可由 α 1 , α 2 \alpha_1, \alpha_2 α1,α2线性表示,也可由 β 1 , β 2 \beta_1, \beta_2 β1,β2线性表示,则 Y = Y = Y=( )
    (A) k ( 3 3 4 ) , k ∈ R k \begin{pmatrix} 3 \\3 \\ 4 \end{pmatrix}, k \in \mathbb{R} k 334 ,kR
    (B) k ( 3 5 10 ) , k ∈ R k \begin{pmatrix} 3 \\ 5 \\ 10 \end{pmatrix}, k \in \mathbb{R} k 3510 ,kR
    © k ( − 1 1 2 ) , k ∈ R k \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}, k \in \mathbb{R} k 112 ,kR
    (D) k ( 1 5 8 ) , k ∈ R k \begin{pmatrix} 1 \\ 5 \\ 8 \end{pmatrix}, k \in \mathbb{R} k 158 ,kR

  8. 设随机变量 X X X服从参数为 1 的泊松分布,则 E ( ∣ X − E X ∣ ) = E(|X - E X|) = E(XEX)=( )
    (A) 1 e \frac{1}{e} e1
    (B) 1 2 \frac{1}{2} 21
    © 2 e \frac{2}{e} e2
    (D) 1 1 1

  9. X 1 , X 2 , … , X n X_1, X_2, \ldots, X_n X1,X2,,Xn为来自总体 N ( μ 1 , σ 2 ) N(\mu_1, \sigma^2) N(μ1,σ2)的简单随机样本, Y 1 , Y 2 , … , Y m Y_1, Y_2, \ldots, Y_m Y1,Y2,,Ym为来自总体 N ( μ 2 , 2 σ 2 ) N(\mu_2, 2 \sigma^2) N(μ2,2σ2)的简单随机样本,且两样本相互独立,记 X ‾ = 1 n ∑ i = 1 n X i \overline{X} = \frac{1}{n} \sum_{i=1}^n X_i X=n1i=1nXi Y ‾ = 1 m ∑ i = 1 m Y i \overline{Y} = \frac{1}{m} \sum_{i=1}^m Y_i Y=m1i=1mYi S 1 2 = 1 n − 1 ∑ i = 1 n ( X i − X ‾ ) 2 S_1^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \overline{X})^2 S12=n11i=1n(XiX)2 S 2 2 = 1 m − 1 ∑ i = 1 m ( Y i − Y ‾ ) 2 S_2^2 = \frac{1}{m-1} \sum_{i=1}^m (Y_i - \overline{Y})^2 S22=m11i=1m(YiY)2,则( )
    (A) S 1 2 S 2 2 ∼ F ( n , m ) \frac{S_1^2}{S_2^2} \sim F(n, m) S22S12F(n,m)
    (B) S 1 2 S 2 2 ∼ F ( n − 1 , m − 1 ) \frac{S_1^2}{S_2^2} \sim F(n-1, m-1) S22S12F(n1,m1)
    © 2 S 1 2 S 2 2 ∼ F ( n , m ) \frac{2 S_1^2}{S_2^2} \sim F(n, m) S222S12F(n,m)
    (D) 2 S 1 2 S 2 2 ∼ F ( n − 1 , m − 1 ) \frac{2 S_1^2}{S_2^2} \sim F(n-1, m-1) S222S12F(n1,m1)

  10. X 1 , X 2 X_1, X_2 X1,X2为来自总体 N ( μ , σ 2 ) N(\mu, \sigma^2) N(μ,σ2)的简单随机样本,其中 σ ( σ > 0 ) \sigma (\sigma > 0) σ(σ>0)是未知参数。若 σ ^ = a ∣ X 1 − X 2 ∣ \hat{\sigma} = a |X_1 - X_2| σ^=aX1X2 σ \sigma σ的无偏估计,则 a = a = a=( )
    (A) π 2 \frac{\sqrt{\pi}}{2} 2π
    (B) 2 π 2 \frac{\sqrt{2 \pi}}{2} 22π
    © π \sqrt{\pi} π
    (D) 2 π \sqrt{2 \pi} 2π

  11. x → 0 x \to 0 x0时,函数 f ( x ) = a x + b x 2 + ln ⁡ ( 1 + x ) f(x) = a x + b x^2 + \ln(1 + x) f(x)=ax+bx2+ln(1+x) g ( x ) = e x 2 − cos ⁡ x g(x) = e^{x^2} - \cos x g(x)=ex2cosx是等价无穷小,则 a b = a b = ab= ‾ \underline{\qquad}

  12. 曲面 z = x + 2 y + ln ⁡ ( 1 + x 2 + y 2 ) z = x + 2 y + \ln(1 + x^2 + y^2) z=x+2y+ln(1+x2+y2)在点 ( 0 , 0 , 0 ) (0, 0, 0) (0,0,0)处的切平面方程为 ‾ \underline{\qquad}

  13. f ( x ) f(x) f(x)为周期为 2 的周期函数,且 f ( x ) = 1 − x f(x) = 1 - x f(x)=1x x ∈ [ 0 , 1 ] x \in [0, 1] x[0,1],若 f ( x ) = a 0 2 + ∑ n = 1 ∞ a n cos ⁡ n π x f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos n \pi x f(x)=2a0+n=1ancosx,则 ∑ n = 1 ∞ a 2 n = \sum_{n=1}^{\infty} a_{2 n} = n=1a2n= ‾ \underline{\qquad}

  14. 设连续函数 f ( x ) f(x) f(x)满足 f ( x + 2 ) − f ( x ) = x f(x+2) - f(x) = x f(x+2)f(x)=x ∫ 0 2 f ( x ) d x = 0 \int_0^2 f(x) dx = 0 02f(x)dx=0,则 ∫ 1 3 f ( x ) d x = \int_1^3 f(x) dx = 13f(x)dx= ‾ \underline{\qquad}

  15. 已知向量 α 1 = ( 1 0 1 1 ) \alpha_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 1 \end{pmatrix} α1= 1011 α 2 = ( − 1 − 1 1 0 ) \alpha_2 = \begin{pmatrix} -1 \\ -1 \\ 1 \\ 0 \end{pmatrix} α2= 1110 α 3 = ( 0 1 − 1 1 ) \alpha_3 = \begin{pmatrix} 0 \\ 1 \\ -1\\ 1 \end{pmatrix} α3= 0111 β = ( 1 1 1 − 1 ) \beta = \begin{pmatrix} 1 \\ 1 \\ 1 \\ -1\end{pmatrix} β= 1111 γ = k 1 α 1 + k 2 α 2 + k 3 α 3 \gamma = k_1 \alpha_1 + k_2 \alpha_2 + k_3 \alpha_3 γ=k1α1+k2α2+k3α3,若 γ T α i = β T α i ( i = 1 , 2 , 3 ) \gamma^T \alpha_i = \beta^T \alpha_i (i = 1, 2, 3) γTαi=βTαi(i=1,2,3),则 k 1 2 + k 2 2 + k 3 2 = k_1^2 + k_2^2 + k_3^2 = k12+k22+k32= ‾ \underline{\qquad}

  16. 设随机变量 X X X Y Y Y相互独立,且 X ∼ B ( 1 , 1 3 ) X \sim B(1, \frac{1}{3}) XB(1,31) Y ∼ B ( 2 , 1 2 ) Y \sim B(2, \frac{1}{2}) YB(2,21),则 P { X = Y } = P\{X = Y\} = P{X=Y}= ‾ \underline{\qquad}

  17. 设曲线 y = y ( x ) ( x > 0 ) y = y(x) (x > 0) y=y(x)(x>0)经过点 ( 1 , 2 ) (1, 2) (1,2),该曲线上任一点 P ( x , y ) P(x, y) P(x,y) y y y轴的距离等于该点处的切线在 y y y轴上的截距。
    (i) 求 y ( x ) y(x) y(x)
    (ii) 求函数 f ( x ) = ∫ 1 x y ( t ) d t f(x) = \int_1^x y(t) dt f(x)=1xy(t)dt ( 0 , + ∞ ) (0, +\infty) (0,+)上的最大值。

  18. 求函数 f ( x , y ) = ( y − x 2 ) ( y − x 3 ) f(x, y) = (y - x^2)(y - x^3) f(x,y)=(yx2)(yx3)的极值。

  19. 设空间有界区域 Ω \Omega Ω由柱面 x 2 + y 2 = 1 x^2 + y^2 = 1 x2+y2=1与平面 z = 0 z = 0 z=0 x + z = 1 x + z = 1 x+z=1围成, Σ \Sigma Σ Ω \Omega Ω边界的外侧,计算曲面积分
    I = ∬ Σ 2 x z d y d z + x z cos ⁡ y d z d x + 3 y z sin ⁡ x d x d y . I = \iint_{\Sigma} 2 x z dy dz + x z \cos y dz dx + 3 y z \sin x dx dy. I=Σ2xzdydz+xzcosydzdx+3yzsinxdxdy.

  20. 设函数 f ( x ) f(x) f(x) [ − a , a ] [-a, a] [a,a]上具有二阶连续导数,证明:
    (i) 若 f ( 0 ) = 0 f(0) = 0 f(0)=0,则存在 ξ ∈ ( − a , a ) \xi \in (-a, a) ξ(a,a),使得 f ′ ′ ( ξ ) = 1 a 2 [ f ( a ) + f ( − a ) ] f''(\xi) = \frac{1}{a^2} [f(a) + f(-a)] f′′(ξ)=a21[f(a)+f(a)]
    (ii) 若 f ( x ) f(x) f(x) ( − a , a ) (-a, a) (a,a)内取得极值,则存在 η ∈ ( − a , a ) \eta \in (-a, a) η(a,a)使得 ∣ f ′ ′ ( η ) ∣ ≥ 1 2 a 2 ∣ f ( a ) − f ( − a ) ∣ . |f''(\eta)| \geq \frac{1}{2 a^2} |f(a) - f(-a)|. f′′(η)2a21f(a)f(a)∣.

  21. 已知二次型 f ( x 1 , x 2 , x 3 ) = x 1 2 + 2 x 2 2 + 2 x 3 2 + 2 x 1 x 2 − 2 x 1 x 3 f(x_1, x_2, x_3) = x_1^2 + 2 x_2^2 + 2 x_3^2 + 2 x_1 x_2 - 2 x_1 x_3 f(x1,x2,x3)=x12+2x22+2x32+2x1x22x1x3 g ( y 1 , y 2 , y 3 ) = y 1 2 + y 2 2 + y 3 2 + 2 y 2 y 3 g(y_1, y_2, y_3) = y_1^2 + y_2^2 + y_3^2 + 2 y_2 y_3 g(y1,y2,y3)=y12+y22+y32+2y2y3
    (i) 求可逆变换 x = P y x = P y x=Py,将 f ( x 1 , x 2 , x 3 ) f(x_1, x_2, x_3) f(x1,x2,x3)化为 g ( y 1 , y 2 , y 3 ) g(y_1, y_2, y_3) g(y1,y2,y3)
    (ii) 是否存在正交变换 x = Q y x = Q y x=Qy,将 f ( x 1 , x 2 , x 3 ) f(x_1, x_2, x_3) f(x1,x2,x3)化为 g ( y 1 , y 2 , y 3 ) g(y_1, y_2, y_3) g(y1,y2,y3)

  22. 设二维随机变量 ( X , Y ) (X, Y) (X,Y)的概率密度为 f ( x , y ) = { 2 π ( x 2 + y 2 ) , x 2 + y 2 ≤ 1 0 , 其他 f(x, y) = \begin{cases} \frac{2}{\pi} (x^2 + y^2), & x^2 + y^2 \leq 1 \\ 0, & \text{其他} \end{cases} f(x,y)={π2(x2+y2),0,x2+y21其他
    (i) 求 X X X Y Y Y的协方差;
    (ii) 判断 X X X Y Y Y是否相互独立;
    (iii) 求 Z = X 2 + Y 2 Z = X^2 + Y^2 Z=X2+Y2的概率密度。


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