2023 National Graduate Entrance Examination Mathematics I Past Paper

Approach:

This question tests the method of finding the asymptotes of a curve. The coefficients of the oblique asymptote $y=kx+b$ are determined by two limits: $k$ is the limit of $y/x$ and $b$ is the limit of $y-kx$.

Steps:

1. The slope: $k=\lim\limits_{x\to\infty}\dfrac{y}{x}=\lim\limits_{x\to\infty}\dfrac{x\ln\left(e+\dfrac{1}{x-1}\right)}{x}=\lim\limits_{x\to\infty}\ln\left(e+\dfrac{1}{x-1}\right)=e^0=1$.
2. The intercept: $b=\lim\limits_{x\to\infty}(y-kx)=\lim\limits_{x\to\infty}\left[x\ln\left(e+\dfrac{1}{x-1}\right)-x\right]=\lim\limits_{x\to\infty}x\left[\ln\left(e+\dfrac{1}{x-1}\right)-1\right]$.
3. Using $\ln(1+u)\sim u$ ($u\to0$) we get $\ln\left(e+\dfrac{1}{x-1}\right)-1=\ln\left(1+\dfrac{1}{e(x-1)}\right)\sim\dfrac{1}{e(x-1)}$, hence $b=\lim\limits_{x\to\infty}\dfrac{x}{e(x-1)}=\dfrac1e$.
4. Hence the equation of the oblique asymptote is $y=x+\dfrac{1}{e}$, so (B) is the correct choice.

Answer:

The equation of the oblique asymptote is $y=x+\dfrac{1}{e}$, so (B) is the correct choice.

Approach:

This question tests the relation between the boundedness of the solutions of a second-order linear differential equation with constant coefficients and the characteristic roots. That the solutions are bounded on $(-\infty,+\infty)$ is equivalent to all the roots of the characteristic equation being purely imaginary (with zero real part).

Steps:

1. The characteristic equation is $\lambda^2+a\lambda+b=0$; denote its discriminant by $\Delta=a^2-4b$.
2. When $\Delta>0$ there are two distinct real roots $\lambda_1,\lambda_2$, and the general solution is $y=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}$; as soon as one of the roots is nonzero, taking $C_1,C_2$ not both zero makes the solution unbounded.
3. When $\Delta=0$ there is a repeated root $\lambda=-\dfrac a2$, and the general solution is $y=(C_1+C_2x)e^{-\frac a2x}$; if $C_2\ne0$ the solution is unbounded as $x\to+\infty$.
4. When $\Delta<0$ there is a pair of conjugate complex roots $\lambda_{1,2}=-\dfrac a2\pm\dfrac{\sqrt{4b-a^2}}{2}i$, and the general solution is $y=e^{-\frac a2x}\left(C_1\cos\dfrac{\sqrt{4b-a^2}}{2}x+C_2\sin\dfrac{\sqrt{4b-a^2}}{2}x\right)$.
5. For the solutions to be bounded on the whole real axis, the factor $e^{-\frac a2x}$ must be identically 1, that is $a=0$; then $\Delta=a^2-4b<0$ gives $b>0$, so (C) is the correct choice.

Answer:

$a=0,\ b>0$, so (C) is the correct choice.

Approach:

This question tests the differentiability at the breakpoint and the continuity of the derivative of a function determined by parametric equations. First the function is written as a piecewise expression according to the sign of the parameter, and then the left and right derivatives and the second derivative are computed separately.

Steps:

1. When $t\ge0$ we have $x=2t+|t|=3t,\ y=t\sin t$, hence $y=\dfrac{x}{3}\sin\dfrac{x}{3}$; when $t<0$ we have $x=t,\ y=-t\sin t$, hence $y=-x\sin x$.
2. Thus $y=\begin{cases}\dfrac{x}{3}\sin\dfrac{x}{3}, & x\ge0,\\-x\sin x, & x<0.\end{cases}$
3. Computing $f'(0)$: $f'(0)=\lim\limits_{x\to0^+}\dfrac{\sin\frac x3}{3}=0$, $f'(0)=\lim\limits_{x\to0^-}(-\sin x)=0$. The two sides are equal, so $f'(0)$ exists and equals 0.
4. Differentiating once more gives $y'(x)=\begin{cases}\dfrac13\sin\dfrac x3+\dfrac x9\cos\dfrac x3, & x\ge0,\\-\sin x-x\cos x, & x<0.\end{cases}$
5. Since the right-hand limit $\lim\limits_{x\to0^+}\dfrac13\cos\dfrac x3=\dfrac13$ and the left-hand limit $\lim\limits_{x\to0^-}(-\sin x-x\cos x)=0$ are not equal, $y''(0)$ does not exist, that is $f''(x)$ is not continuous at $x=0$, so (D) is the correct choice.

Answer:

$f'(0)$ exists ($f'(0)=0$), and $f''(x)$ is not continuous at $x=0$, so (D) is the correct choice.

Approach:

This question tests the comparison test for absolutely convergent series. Since $a_n<b_n$, the difference series $\sum(b_n-a_n)$ is a convergent series of positive terms, so the absolute convergence of the two series can be converted into each other.

Steps:

1. From $a_n<b_n$ we have $b_n-a_n>0$ and $\sum\limits_{n=1}^{\infty}(b_n-a_n)=\sum b_n-\sum a_n$ converges, so this series of positive terms converges absolutely.
2. If $\sum a_n$ converges absolutely, then from $|b_n|=|b_n-a_n+a_n|\le|b_n-a_n|+|a_n|$ and the comparison test, $\sum b_n$ converges absolutely.
3. If $\sum b_n$ converges absolutely, then from $|a_n|=|a_n-b_n+b_n|\le|a_n-b_n|+|b_n|$ and the comparison test, $\sum a_n$ converges absolutely.
4. The two implications go both ways, which shows that ”$\sum a_n$ converges absolutely” is neither a sufficient condition nor a necessary condition for $\sum a_n,\sum b_n$ both to converge, so (D) is the correct choice.

Answer:

Neither a sufficient nor a necessary condition, so (D) is the correct choice.

Approach:

This question tests the rank of block matrices and elementary transformations. Elementary transformations do not change the rank of a matrix, and the three block matrices can be reduced to block diagonal form, which reduces the problem to comparing $r(AB)$ with $r(ABAB)$.

Steps:

1. Apply an elementary transformation to $\begin{pmatrix}0&A\\BC&E\end{pmatrix}$: add $A$ times the second row to the first row, obtaining $\begin{pmatrix}-ABC&0\\BC&E\end{pmatrix}$; since $ABC=0$ this becomes $\begin{pmatrix}0&0\\BC&E\end{pmatrix}$, whose rank is $n$, that is $\gamma_1=n$.
2. An elementary transformation of $\begin{pmatrix}AB&C\\0&E\end{pmatrix}$ gives $\begin{pmatrix}AB&0\\0&E\end{pmatrix}$, so $\gamma_2=r(AB)+n$.
3. An elementary transformation of $\begin{pmatrix}E&AB\\AB&0\end{pmatrix}$ gives $\begin{pmatrix}E&0\\0&-ABAB\end{pmatrix}$, so $\gamma_3=n+r(ABAB)$.
4. Since $r(ABAB)\le r(AB)$, we get $\gamma_1\le\gamma_3\le\gamma_2$, so (B) is the correct choice.

Answer:

$\gamma_1\leq\gamma_3\leq\gamma_2$, so (B) is the correct choice.

Approach:

This question tests the necessary and sufficient condition for a matrix to be similar to a diagonal matrix. A matrix is diagonalizable if and only if the geometric multiplicity and the algebraic multiplicity of every eigenvalue are equal, that is, the defect of $r(A-\lambda E)$ equals the multiplicity.

Steps:

1. The matrix in option (A) is upper triangular, with the three distinct eigenvalues $1,2,3$, so it is certainly similar to a diagonal matrix.
2. The matrix in option (B) is a real symmetric matrix, and a real symmetric matrix is always similar to a diagonal matrix, so it too is diagonalizable.
3. The matrix in option (C) has the eigenvalues $1,2,2$, and for $\lambda=2$ we have $r(C-2E)=1$, so the multiplicity of the double eigenvalue is $2=3-r(C-2E)$; it is diagonalizable.
4. The matrix in option (D) also has the eigenvalues $1,2,2$, but $r(D-2E)\ne1$, that is the multiplicity of the double eigenvalue is $2\ne3-r(D-2E)$, so it cannot be similar to a diagonal matrix, and (D) is the correct choice.

Answer:

Option (D) cannot be similar to a diagonal matrix, so (D) is the correct choice.

Approach:

This question tests the linear representation of a vector group and the elementary row transformation of a matrix. Let the same vector be represented linearly by the $\alpha$ group and by the $\beta$ group; after the coefficients are combined one obtains a homogeneous system of linear equations, whose fundamental system of solutions is found by elementary row transformation.

Steps:

1. Put $r=x_1\alpha_1+x_2\alpha_2=y_1\beta_1+y_2\beta_2$, and after transposing the terms, $x_1\alpha_1+x_2\alpha_2-y_1\beta_1-y_2\beta_2=0$.
2. Writing this in matrix form and applying elementary row transformations: $\begin{pmatrix}1&2&-2&-1\\2&1&-5&0\\3&1&-9&-1\end{pmatrix}\to\begin{pmatrix}1&0&0&3\\0&1&0&-1\\0&0&1&1\end{pmatrix}$.
3. Hence the fundamental system of solutions is $(-3,1,-1,1)^T$, that is $(x_1,x_2,y_1,y_2)^T=c(-3,1,-1,1)^T,\ c\in\mathbb{R}$.
4. Then $r=-c\beta_1+c\beta_2=-c(2,5,9)^T+c(1,0,1)^T=(-c,-5c,-8c)^T=-c(1,5,8)^T$.
5. Put $k=-c$; then $r=k(1,5,8)^T,\ k\in\mathbb{R}$, so (D) is the correct choice.

Answer:

$r=k\begin{pmatrix}1\\5\\8\end{pmatrix},\ k\in\mathbb{R}$, so (D) is the correct choice.

Approach:

This question tests the computation of the expectation and the absolute expectation of a Poisson distribution. Write $|X-EX|$ piecewise according to whether $X$ is zero, and then sum with the probability formula of the Poisson distribution.

Steps:

1. A Poisson distribution with parameter 1 satisfies $EX=1$.
2. Thus $|X-EX|=\begin{cases}1, & X=0,\\X-1, & X=1,2,\cdots.\end{cases}$
3. Hence $E(|X-EX|)=1\cdot P\{X=0\}+\sum\limits_{k=1}^{\infty}(k-1)P\{X=k\}$.
4. From $P\{X=k\}=\dfrac{e^{-1}}{k!}$ we get $E(|X-EX|)=e^{-1}\sum\limits_{k}\dfrac{k-1}{k!}+e^{-1}=e^{-1}E(X-1)+e^{-1}=\dfrac{2}{e}$.
5. Hence (C) is the correct choice.

Answer:

$E(|X-EX|)=\dfrac{2}{e}$, so (C) is the correct choice.

Approach:

This question tests the distribution of the ratio of the sample variances of two independent normal samples. The key is to standardize the two sample variances into $\chi^2$ variables and then use the definition of the $F$ distribution to obtain the distribution of the ratio.

Steps:

1. $S_1^2=\dfrac{1}{n-1}\sum\limits_{i=1}^n(X_i-\overline{X})^2,\ S_2^2=\dfrac{1}{m-1}\sum\limits_{j=1}^m(Y_j-\overline{Y})^2$.
2. By the property of the sample variance of a normal population: $\dfrac{(n-1)S_1^2}{\sigma^2}\sim\chi^2(n-1)$, $\dfrac{(m-1)S_2^2}{2\sigma^2}\sim\chi^2(m-1)$.
3. The two samples are mutually independent, so the two $\chi^2$ variables above are mutually independent, and their ratio follows the $F$ distribution: $\dfrac{(n-1)S_1^2/\sigma^2}{(m-1)S_2^2/(2\sigma^2)}\sim F(n-1,m-1)$.
4. Rearranging gives $\dfrac{2S_1^2}{S_2^2}\sim F(n-1,m-1)$, so (D) is the correct choice.

Answer:

$\dfrac{2S_1^2}{S_2^2}\sim F(n-1,m-1)$, so (D) is the correct choice.

Approach:

This question tests the use of a sample statistic as an unbiased estimator for a normal population. Unbiasedness requires $E(\hat\sigma)=\sigma$, so it is enough to compute $E|X_1-X_2|$ in order to determine the constant $a$.

Steps:

1. Since $X_1,X_2$ are independent and identically distributed, $Y=X_1-X_2\sim N(0,2\sigma^2)$, whose density is $f(y)=\dfrac{1}{\sqrt{2\pi}\sqrt2\,\sigma}e^{-\frac{y^2}{4\sigma^2}}$.
2. Computing the absolute expectation: $E|Y|=\int_{-\infty}^{+\infty}|y|f(y)dy=\dfrac{2}{\sqrt{2\pi}\sqrt2\,\sigma}\int_0^{+\infty}ye^{-\frac{y^2}{4\sigma^2}}dy$.
3. Here $\int_0^{+\infty}ye^{-\frac{y^2}{4\sigma^2}}dy=2\sigma^2$, so $E|Y|=\dfrac{4\sigma^2}{\sqrt{2\pi}\sqrt2\,\sigma}=\dfrac{2\sigma}{\sqrt\pi}$.
4. From unbiasedness $E(\hat\sigma)=aE|Y|=\sigma$ we get $a=\dfrac{\sqrt\pi}{2}$, so (A) is the correct choice.

Answer:

$a=\dfrac{\sqrt{\pi}}{2}$, so (A) is the correct choice.

Approach:

This question tests the definition of equivalent infinitesimals and Taylor expansions. That $f(x)$ and $g(x)$ are equivalent means that their ratio tends to 1, and comparing the coefficients of the linear and the quadratic terms determines $a,b$.

Steps:

1. From $\ln(1+x)=x-\dfrac{x^2}{2}+o(x^2)$ we get $f(x)=ax+bx^2+\ln(1+x)=(a+1)x+\left(b-\dfrac12\right)x^2+o(x^2)$.
2. From $e^{x^2}=1+x^2+o(x^2)$ and $\cos x=1-\dfrac{x^2}{2}+o(x^2)$ we get $g(x)=\dfrac32x^2+o(x^2)$.
3. Equivalent infinitesimals require $\lim\limits_{x\to0}\dfrac{f(x)}{g(x)}=1$: the coefficient of the linear term must vanish, that is $a+1=0$, which gives $a=-1$.
4. The coefficients of the quadratic terms are equal, that is $b-\dfrac12=\dfrac32$, which gives $b=2$.
5. Hence $ab=-2$.

Answer:

$ab=-2$.

Approach:

This question tests the method of finding the tangent plane to a surface. Write the surface in implicit form $F(x,y,z)=0$, find the normal vector, and then substitute the point to obtain the equation of the tangent plane.

Steps:

1. Put $F(x,y,z)=x+2y+\ln(1+x^2+y^2)-z$, then the surface is $F=0$.
2. The partial derivatives are $F_x=1+\dfrac{2x}{1+x^2+y^2}$, $F_y=2+\dfrac{2y}{1+x^2+y^2}$, $F_z=-1$.
3. The normal vector at the point $(0,0,0)$ is $(1,2,-1)$.
4. The equation of the tangent plane is $1\cdot(x-0)+2\cdot(y-0)-1\cdot(z-0)=0$, that is $x+2y-z=0$.

Answer:

The equation of the tangent plane is $x+2y-z=0$.

Approach:

This question tests the computation of the coefficients of a Fourier cosine series and the summation of an infinite series. First $a_n$ is obtained from the formula, and then the sum is evaluated at the odd indices.

Steps:

1. $f(x)$ is expanded as a cosine series, and by the coefficient formula $a_n=2\int_0^1(1-x)\cos n\pi x\,dx$.
2. Integrating by parts: $a_n=2\left(\int_0^1\cos n\pi x\,dx-\int_0^1x\cos n\pi x\,dx\right)=\dfrac{1}{n\pi}\sin n\pi x\Big|_0^1-\dfrac{1}{n\pi}\int_0^1x\,d\sin n\pi x$.
3. Continuing with integration by parts gives $a_n=-\dfrac{2}{n\pi}\left(\left[x\sin n\pi x\Big|_0^1-\int_0^1\sin n\pi x\,dx\right]\right)=-\dfrac{2}{n\pi}\cdot\dfrac{1-\cos n\pi}{n\pi}$.
4. That is $a_n=\dfrac{2(\cos n\pi-1)}{n^2\pi^2}$.
5. For $n=2k$ we have $\cos 2k\pi=1$, so $a_{2k}=0$, and hence $\sum\limits_{n=1}^{\infty}a_{2n}=0$.

Answer:

$\sum\limits_{n=1}^{\infty}a_{2n}=0$.

Approach:

This question tests the computation of a definite integral by means of the recurrence relation $f(x+2)-f(x)=x$ of the function. Turn $\int_2^3f(x)dx$ into $\int_0^1f(x+2)dx$ by a change of variables, and then use the recurrence relation to replace $f(x+2)$ by $f(x)+x$.

Steps:

1. $\displaystyle\int_0^3f(x)dx=\int_0^2f(x)dx+\int_2^3f(x)dx$, and the condition gives $\int_0^2f(x)dx=0$.
2. By the substitution $\displaystyle\int_2^3f(x)dx=\int_0^1f(x+2)dx=\int_0^1\left[f(x)+x\right]dx$.
3. Hence $\displaystyle\int_0^3f(x)dx=\int_0^2f(x)dx+\int_0^1f(x)dx+\int_0^1x\,dx=0+\int_0^1f(x)dx+\dfrac12$.
4. Integrating $f(x+2)-f(x)=x$ over $[0,1]$ gives $\displaystyle\int_2^3f(x)dx-\int_0^1f(x)dx=\int_0^1x\,dx=\dfrac12$, that is $\displaystyle\int_2^3f(x)dx=\int_0^1f(x)dx+\dfrac12$.
5. Eliminating $\displaystyle\int_0^1f(x)dx$ from the two equations gives $\displaystyle\int_0^3f(x)dx=\dfrac12$.

Answer:

$\displaystyle\int_0^3f(x)dx=\dfrac{1}{2}$.

Approach:

This question tests a system of linear equations built from inner products of vectors. After the condition $\gamma^T\alpha_i=\beta^T\alpha_i\ (i=1,2,3)$ is expanded it becomes a system of linear equations in the unknowns $k_1,k_2,k_3$, and after it is solved the sum of the squares is required.

Steps:

1. From $\gamma^T\alpha_i=\beta^T\alpha_i$ we get $k_1\alpha_1^T\alpha_i+k_2\alpha_2^T\alpha_i+k_3\alpha_3^T\alpha_i=\beta^T\alpha_i\ (i=1,2,3)$.
2. Computing the inner products: $\alpha_1^T\alpha_1=2,\ \alpha_2^T\alpha_1=0,\ \alpha_3^T\alpha_1=1$; $\alpha_2^T\alpha_2=2,\ \alpha_3^T\alpha_2=1$; $\alpha_3^T\alpha_3=2$; $\beta^T\alpha_1=0,\ \beta^T\alpha_2=-2,\ \beta^T\alpha_3=0$.
3. Hence the system of equations is $2k_1+k_3=0,\ 2k_2+k_3=-2,\ k_1+k_2+2k_3=0$.
4. Solving gives $k_1=-\dfrac12,\ k_2=-\dfrac32,\ k_3=1$.
5. Hence $k_1^2+k_2^2+k_3^2=\dfrac{1}{4}+\dfrac{9}{4}+1$.

Answer:

$k_1^2+k_2^2+k_3^2=\dfrac{11}{9}$.

Approach:

This question tests the computation of the probability that two independent binomial variables take the same value. First the possible values of $X,Y$ are written down, and then independence is used to split the joint probability into a product of the marginal probabilities.

Steps:

1. $X\sim B\left(1,\dfrac13\right)$, so $X$ takes only the values $0,1$, and $P(X=0)=\dfrac23,\ P(X=1)=\dfrac13$.
2. $Y\sim B\left(2,\dfrac12\right)$, so $Y$ takes the values $0,1,2$, and $P(Y=0)=\dfrac14,\ P(Y=1)=\dfrac12,\ P(Y=2)=\dfrac14$.
3. $X$ and $Y$ are mutually independent, so $P(X=Y)=P(X=0)P(Y=0)+P(X=1)P(Y=1)$.
4. Substituting gives $P(X=Y)=\dfrac23\cdot\dfrac14+\dfrac13\cdot\dfrac12=\dfrac16+\dfrac16=\dfrac13$.

Answer:

$P(X=Y)=\dfrac{1}{3}$.

Approach:

This question tests the establishment of a differential equation from a geometric condition and the extremum of an integral function. In the first part the statement “the intercept on the $y$-axis of the tangent equals the distance from the point to the $y$-axis” is turned into an equation; in the second part the integral function is differentiated to find its stationary points.

Steps:

1. (I) Let the equation of the tangent at the point $(x,y)$ be $y-y_0=y'(x_0)(x-x_0)$; putting $x=0$ shows that the intercept of the tangent on the $y$-axis is $y-xy'$.
2. The distance from the point to the $y$-axis is $x$, so $y-xy'=x$, that is $y=x(y-xy')$, which gives $y=x(2-\ln x)$.
3. The curve also passes through the point $(1,2)$, that is $y(1)=2$, so $y(x)=x(2-\ln x)$.
4. (II) $f(x)=\int_1^x(2-\ln t)t\,dt$, and by the fundamental theorem of calculus $f'(x)=x(2-\ln x)$; setting $f'(x)=0$ gives the stationary point $x=e^2$.
5. When $0<x<e^2$ we have $f'(x)>0$, and when $x>e^2$ we have $f'(x)<0$, so $f$ attains its maximum at $x=e^2$.
6. $f(e^2)=\int_1^{e^2}(2-\ln x)x\,dx=\dfrac14e^4-\dfrac54$.

Answer:

(I) $y(x)=x(2-\ln x)$;

(II) the maximum of $f(x)=\int_1^xy(t)dt$ is $\dfrac{1}{4}e^4-\dfrac{5}{4}$, attained at $x=e^2$.

Approach:

This question tests the criterion for the extrema of a function of two variables. First the stationary point equations are solved, and then the determinant $AC-B^2$ formed by the second partial derivatives is used to decide the case; when the determinant vanishes the criterion fails, and one has to approach along a curve to judge the sign of the function values.

Steps:

1. The partial derivatives are $f_x=-2x(y-x^2)-3x^2(y-x^3)$, $f_y=2y-x^2-3y^2+3x^3$; setting both equal to zero gives the stationary points $(0,0)$, $(1,1)$, $\left(\dfrac23,\dfrac{10}{27}\right)$.
2. The second partial derivatives are $f_{xx}=-(2y+3y-15x^2)$, $f_{xy}=-(2+3x)$, $f_{yy}=2$.
3. At $(0,0)$ we have $A=0,\ B=-2,\ C=2$, $AC-B^2=0$, so the sufficient condition fails; taking $y=x^2+kx^3\ (k>0)$ gives $f(x,y)=kx^5+o(x^5)$, and by the local sign preservation of limits the values of the function change sign near $(0,0)$, so it is not an extremum.
4. At $(1,1)$ we have $A=12,\ B=-5,\ C=2$, $AC-B^2=24-25<0$, so it is not an extremum.
5. At $\left(\dfrac23,\dfrac{10}{27}\right)$ we have $A=\dfrac{100}{27},\ B=\dfrac83,\ C=2$, $AC-B^2>0$ and $A>0$, so it is a local minimum point, and the minimum value is $\left(\dfrac{10}{27}-\dfrac49\right)\left(\dfrac{10}{27}-\dfrac8{27}\right)=-\dfrac4{729}$.