2022 National Graduate Entrance Examination Mathematics I Past Paper

Approach:

From $\lim\limits_{x\to1} \frac{f(x)}{\ln x}=1$ it follows that $f(x)\sim \ln x$ as $x\to1$.

Steps:

Because $\ln x\to0$ as $x\to1$, we get $\lim\limits_{x\to1} f(x)=0$.

Answer:

B

Approach:

Compute the partial derivatives and substitute them into the condition to solve the equation.

Steps:

$\frac{\partial z}{\partial x}=y f\left(\frac{y}{x}\right)-\frac{y^2}{x}f'\left(\frac{y}{x}\right)$, $\frac{\partial z}{\partial y}=x f\left(\frac{y}{x}\right)+y f'\left(\frac{y}{x}\right)$.

More directly, $z=xyf(y/x)$ is a homogeneous function of degree two in $x,y$, so by Euler’s formula

The right-hand side given in the problem is $y^2(\ln y-\ln x)=xy\cdot\frac{y}{x}\ln\frac{y}{x}$, so

Hence $f(1)=0$, $f'(u)=\frac{\ln u+1}{2}$, $f'(1)=\frac{1}{2}$.

Answer:

B

Approach:

Analyse the limit properties of composite functions and look for a counterexample.

Steps:

Put $y_n=\cos x_n$, then $y_n\in [0,1]$. If $\lim\limits_{n\to\infty} \sin(\cos x_n)$ and $\lim\limits_{n\to\infty} \cos x_n$ both exist, $x_n$ need not converge. For instance $x_n=\arccos a_n$, where $a_n$ converges but $x_n$ does not.

Answer:

D

Approach:

This question tests the comparison of definite integrals by the size of their integrands. The limits and the integrands of the three integrals are all defined on $[0,1]$, so comparing the integrands point by point gives the order of the three integrals.

Steps:

1. Denote the three integrands by $f_1(x)=\dfrac{x}{2(1+\cos x)}$, $f_2(x)=\dfrac{\ln(1+x)}{1+\cos x}$, $f_3(x)=\dfrac{2x}{1+\sin x}$.
2. Note that on $[0,1]$ both $1+\cos x$ and $1+\sin x$ are positive and neither exceeds $2$, so each integrand keeps the same sign as $\ln(1+x)$ and $x$ respectively, and after taking the differences the order can be decided point by point.
3. Comparing the three integrands point by point on $[0,1]$ (using the relative size of $\ln(1+x)$ and $x$ on this interval together with the values of $1+\cos x$ and $1+\sin x$) gives $f_2<f_1<f_3$.
4. By the order-preserving property of the definite integral (if $f<g$ then $\int f<\int g$), we get $I_2<I_1<I_3$, so (C) is the correct choice.

Answer:

C

Approach:

This question tests a sufficient condition for a matrix to be diagonalizable. A matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors; in particular, $n$ distinct eigenvalues guarantee diagonalizability.

Steps:

1. For option (A): by hypothesis the matrix $A$ has 3 distinct eigenvalues, so it has 3 linearly independent eigenvectors.
2. By the sufficient condition for diagonalizability (the number of linearly independent eigenvectors equals the order of the matrix), $A$ is diagonalizable, so option (A) fits the requirement of the question.
3. For each of the remaining options the number of eigenvectors is not guaranteed to reach the order of the matrix, so none of them can serve as a sufficient condition for diagonalizability.
4. Hence (A) is the correct choice.

Answer:

A

Approach:

Use elementary transformations of matrices together with the condition for equivalent systems.

Steps:

Put $y=\begin{pmatrix}x_1 \\ x_2\end{pmatrix}$, where $x_i(i=1,2)$ are $n$-dimensional column vectors.

If $\begin{pmatrix}A & B \\ O & A\end{pmatrix}y=0$ and $\begin{pmatrix}B & O \\ O & A\end{pmatrix}y=0$ are equivalent systems, that is,

$\begin{pmatrix}A & B \\ O & A\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=0$ and $\begin{pmatrix}B & O \\ O & A\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=0$,

i.e. $A x_1 + B x_2 = 0,\ A x_2 = 0$ and $B x_1 = 0,\ A x_2 = 0$.

Since $A x = 0$ and $B x = 0$ have the same solutions, $A x = 0$ implies $B x = 0$, and conversely.

Hence $\begin{pmatrix}A & B \\ O & A\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=0$ is equivalent to $\begin{pmatrix}B & O \\ O & A\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=0$, so option (C) fits the requirement of the question.

Answer:

C

Approach:

Compute the determinants and analyse the linear dependence of the vector groups.

Steps:

$|\,\alpha_1,\alpha_2,\alpha_3\,|=\begin{vmatrix}\lambda & 1 & 1\\1 & \lambda & 1\\1 & 1 & \lambda\end{vmatrix}=\lambda^3-3\lambda+2=(\lambda-1)^2(\lambda+2)$

$|\,\alpha_1,\alpha_2,\alpha_4\,|=\begin{vmatrix}\lambda & 1 & 1\\1 & \lambda & \lambda\\1 & 1 & \lambda^2\end{vmatrix}=\lambda^4-2\lambda^2+1=(\lambda-1)^2(\lambda+1)^2$

When $\lambda=1$, $\alpha_1=\alpha_2=\alpha_3=\alpha_4=\begin{pmatrix}1\\1\\1\end{pmatrix}$, and at this point $\alpha_1,\alpha_2,\alpha_3$ and $\alpha_1,\alpha_2,\alpha_4$ are equivalent.

When $\lambda=-2$, $2=r(\alpha_1,\alpha_2,\alpha_3)<r(\alpha_1,\alpha_2,\alpha_4)=3$, so $\alpha_1,\alpha_2,\alpha_3$ and $\alpha_1,\alpha_2,\alpha_4$ are not equivalent.

When $\lambda=-1$, $3=r(\alpha_1,\alpha_2,\alpha_3)>r(\alpha_1,\alpha_2,\alpha_4)=1$, so $\alpha_1,\alpha_2,\alpha_3$ and $\alpha_1,\alpha_2,\alpha_4$ are not equivalent.

Hence $\lambda\in\{\lambda\mid \lambda\in\mathbb{R},\lambda\neq-1,\lambda\neq-2\}$.

Answer:

C

Approach:

Use the properties of the variance together with the covariance formula.

Steps:

From $X\sim U(0,3),\ Y\sim P(2)$ we know that $D(X)=\frac{3}{4},\ D(Y)=2$, hence

Answer:

B

Approach:

Apply Chebyshev’s inequality to estimate the probability.

Steps:

By Chebyshev’s inequality, $P\left\{|\overline{X}-\mu_1|\geq\varepsilon\right\}\leq \frac{D(X)}{n\varepsilon^2}$, where $D(X)=M_2-M_1^2$, so (C) is the correct choice.

Answer:

C

Approach:

Use the properties of the conditional distribution together with the correlation coefficient formula.

Steps:

$X\sim N(0,1)$, $Y|X=x\sim N(x,1)$, so $Y=X+Z$ with $Z\sim N(0,1)$ independent of $X$.

$\mathrm{Cov}(X,Y)=\mathrm{Cov}(X,X+Z)=\mathrm{Var}(X)=1$, $D(X)=1$, $D(Y)=D(X)+D(Z)=2$, so the correlation coefficient is $\rho=\frac{1}{\sqrt{2}}$, and (D) is the correct choice.

Answer:

D

Answer:

Approach:

Compute the gradient; the largest directional derivative equals the length of the gradient.

Steps:

From the given data $\frac{\partial f}{\partial x}=2x$, $\frac{\partial f}{\partial y}=4y$, so $\mathrm{grad}(0,1)=(0,4)$, and therefore

Answer:

Approach:

First write the integrand as $(\ln x)\,x^{-1/2}$ and then integrate by parts. Note that $d(\sqrt{x})=\frac{1}{2\sqrt{x}}dx$, so $\frac{dx}{\sqrt{x}}=2\,d(\sqrt{x})$, which turns $x^{-1/2}dx$ into “the differential of some function” and makes the change of variables / integration by parts easy to apply.

Steps:

1. Put

   $$u=\ln x,\qquad dv=\frac{dx}{\sqrt{x}}=x^{-1/2}dx.$$

   Then

   $$du=\frac{dx}{x}=d(\ln x),\qquad v=2\sqrt{x},$$

   because $d(\sqrt{x})=\frac{1}{2\sqrt{x}}dx \Rightarrow x^{-1/2}dx=2\,d(\sqrt{x})$.

2. Applying the formula for integration by parts

   $$\int u\,dv = uv - \int v\,du,$$

   we get

   $$\int_1^{e^2} \frac{\ln x}{\sqrt{x}}dx =\left.2\sqrt{x}\,\ln x\right|_{1}^{e^2} -2\int_1^{e^2} \sqrt{x}\,\frac{dx}{x}.$$

3. Write $\sqrt{x}\,\frac{dx}{x}$ once more as $x^{-1/2}dx$ and integrate the power function directly:

   $$\int_1^{e^2} \sqrt{x}\,\frac{dx}{x} =\int_1^{e^2} x^{-1/2}dx =\left.2\sqrt{x}\right|_{1}^{e^2}=2(e-1).$$

4. Substituting back gives

   $$\int_1^{e^2} \frac{\ln x}{\sqrt{x}}dx =2\left(e^2\ln e^2-1\cdot 0\right)-2\cdot 2(e-1) =4.$$

Answer:

$\frac{4}{e^2}$.

Approach:

Turn the problem into one of finding the maximum of a function.

Steps:

From the given data $k\geq\frac{x^2+y^2}{e^{x+y}}$, so the problem becomes computing the maximum of $f(x,y)=\frac{x^2+y^2}{e^{x+y}}=(x^2+y^2)e^{-(x+y)}$ on $x\geq0,y\geq0$.

Put $f(x,y)=(x^2+y^2)e^{-(x+y)}$. For $x>0,y>0$,

$\frac{\partial f}{\partial x}=e^{-(x+y)}(2x-x^2-y^2)$, $\frac{\partial f}{\partial y}=e^{-(x+y)}(2y-x^2-y^2)$.

Setting $\frac{\partial f}{\partial x}=0,\frac{\partial f}{\partial y}=0$ gives the stationary points $(0,0),(1,1)$.

At the stationary point $(0,0)$, $A=2,B=0,C=2,AC-B^2>0,A>0$, so $(0,0)$ is a local minimum point and $f_{\text{minimum}}=f(0,0)=0$.

At the stationary point $(1,1)$, $A=0,B=-2e^{-2},C=0,AC-B^2\lt0$, so $(1,1)$ is not an extremum.

When $x=0$, $f(0,y)=y^2e^{-y}(y>0)$, so $f'(0,y)=2ye^{-y}-y^2e^{-y}=y(2-y)e^{-y}$, which gives the stationary point $y=2$; then $f''(0,y)=(y^2-4y+2)e^{-y}$ and $f''(0,2)=-2e^{-2}\lt0$, so $f(0,2)=4e^{-2}$ is a maximum.

Similarly $f(2,0)=4e^{-2}$ is also a maximum.

Hence $k\geq f_{\max}(x,y)=\frac{4}{e^2}$.

Answer:

-1.

Approach:

Use the ratio test to judge the convergence of the series.

Steps:

Let $u_n(x)=\frac{n!}{n^n}e^{-nx}$, then

The solution is $x>-1$, so $a=-1$.

Answer:

$((E-(A-E)^{-1})^{-1}-E)A$

Approach:

Solve the matrix equation by using the inverse operation of matrices.

Steps:

Since $(E-(A-E)^{-1})B=A$ and $A$ is invertible, we have

$A^{-1}(E-(A-E)^{-1})B=E$, so $B$ is invertible and $B=(E-(A-E)^{-1})^{-1}A$, hence

$B-A=(E-(A-E)^{-1})^{-1}A-A=((E-(A-E)^{-1})^{-1}-E)A$

Answer:

$\frac{5}{8}$.

Approach:

Compute with the conditional probability formula and the properties of independent events.

Steps:

From the statement $P(AB)=0$, $P(AC)=0$, $P(BC)=P(B)P(C)=\frac{1}{9}$, so the required probability is given by the conditional probability formula:

Substituting $P(A)=P(B)=P(C)=\frac{1}{3}$, $P(BC)=\frac{1}{9}$ gives:

Approach:

First solve the differential equation, then find the asymptote.

Steps:

From the statement we get

Since $y(1)=3$, we have $C=e$, hence

Let the equation of the asymptote of $y(x)$ be $y=kx+b$, then

Hence the oblique asymptote of $y(x)$ is $y=2x$.

Answer:

The oblique asymptote of $y(x)$ is $y=2x$.

Approach:

Compute the double integral by means of a polar coordinate transformation.

Steps:

The derivation of I is as follows.

Step 1: $I = \iint_D \frac{x^2-2 x y+y^2}{x^2+y^2}dxdy$

Step 2: $\iint_D \left[1-\frac{2 x y}{x^2+y^2}\right]dxdy$

Step 3: $\iint_D dxdy - \iint_D \frac{2 x y}{x^2+y^2}dxdy$

Step 4: $\pi + 2 - \int_0^{\frac{\pi}{2}} d\theta \int_0^2 \frac{2 \rho^2\cos\theta\sin\theta}{\rho^2} \rho d\rho$

Step 5: $\pi + 2 - 4 \int_0^{\frac{\pi}{2}} \cos\theta\sin\theta d\theta + 4 \int_0^{\frac{\pi}{2}} \frac{\cos\theta\sin\theta}{(\cos\theta+\sin\theta)^2} d\theta$

Step 6: $\pi + 4 \int_0^{\frac{\pi}{2}} \frac{\tan\theta}{\tan^2\theta+2 \tan\theta+1} d\theta$

Step 7: $\pi + 4 \int_0^{+\infty} \frac{t}{(t+1)^2(1+t^2)} dt$

Step 8: $\pi + 4 \cdot \frac{1}{2} \int_0^{+\infty} \left[\frac{1}{1+t^2} - \frac{1}{(t+1)^2}\right] dt$

Step 9: $\pi + 2 \arctan t \Big|_{0}^{+\infty} + \frac{1}{1+t} \Big|_{0}^{+\infty}$

Final result: $2\pi - 2$

Answer:

$I=2\pi-2$.

Approach:

This question tests a second-kind line integral together with Stokes’ formula. The line integral is written as the circulation of a vector field along a closed curve, Stokes’ formula turns it into a surface integral over a patch of the sphere, and the latter is then computed by parametrization.

Steps:

1. Collect the integrand by $dx,dy,dz$ into the form $P\,dx+Q\,dy+R\,dz$: $P=0$, $Q=2xy^2$, $R=yz^2-\cos z+2xyz+x\sin z$.
2. Thus $\displaystyle I=\oint_L P\,dx+Q\,dy+R\,dz$, where $L$ is the upper boundary of the spherical patch in the first octant, with its direction agreeing with the right-hand rule relative to the upper side.
3. By Stokes’ formula: $I=\iint_{\Sigma}(\nabla\times F)\cdot\bm{n}\,dS$. Computing the curl, $\nabla\times F=\left(R_y-Q_z,\ P_z-R_x,\ Q_x-P_y\right)=\left(z^2+2xz,\ -2yz-\sin z,\ 2y^2\right)$.
4. Take the unit outer normal on the upper side of $\Sigma$, $\bm{n}=(x,y,z)$, and parametrize $x=\sin\theta\cos\varphi,\ y=\sin\theta\sin\varphi,\ z=\cos\theta$, where $\theta,\varphi\in\left[0,\dfrac\pi2\right]$, $dS=\sin\theta\,d\theta\,d\varphi$.
5. Substituting and simplifying: $\displaystyle I=\iint_{\Sigma}\left[xz^2+2x^2z-y\sin z\right]dS$, and term by term $\iint_{\Sigma}xz^2dS=\dfrac{2}{15}$, $\iint_{\Sigma}x^2z\,dS=\dfrac{\pi}{16}$, $\iint_{\Sigma}y\sin z\,dS=3-2\sin 1-2\cos 1$.
6. Combining gives $I=\dfrac{2}{15}+\dfrac{\pi}{8}-3+2\sin 1+2\cos 1$.

Answer:

$I=\dfrac{2}{15}+\dfrac{\pi}{8}-3+2\sin 1+2\cos 1$.

Approach:

Use the convexity of the function together with the mean value theorem for integrals to carry out the proof.

Steps:

Let $F(x)=(x-a)f\left(\frac{a+x}{2}\right)-\int_a^x f(t)dt$, then $F(a)=0$.

$F'(x)=f\left(\frac{a+x}{2}\right)+\frac{1}{2}(x-a)f'\left(\frac{a+x}{2}\right)-f(x)$

$=\frac{1}{2}(x-a)f'\left(\frac{a+x}{2}\right)+f\left(\frac{a+x}{2}\right)-f(x)$

By the Lagrange mean value theorem, $f(x)-f\left(\frac{a+x}{2}\right)=f'(\xi)\frac{1}{2}(x-a)$,

so $F'(x)=\frac{1}{2}(x-a)\left[f'\left(\frac{a+x}{2}\right)-f'(\xi)\right]$

Since $f''(x)\geq0$, $f'(x)$ is monotonically increasing, and therefore $f'\left(\frac{a+x}{2}\right)<f'(\xi)$, so $F'(x)\lt0$ and $F(x)$ is monotonically decreasing.

When $x>a$, $F(x)\lt0$, hence $F(b)\lt0$, that is $f\left(\frac{a+b}{2}\right)\leq \frac{1}{b-a}\int_a^b f(x)dx$.

Answer:

$f''(x)\geq0$ is the necessary and sufficient condition for $f\left(\frac{a+b}{2}\right)\leq\frac{1}{b-a}\int_a^b f(x)dx$ to hold for arbitrary real numbers $a,b$.

Approach:

First write down the matrix of the quadratic form, then find the eigenvalues and eigenvectors, and finally carry out the orthogonal transformation.

Steps:

(1) By definition, $f(x_1,x_2,x_3)=\sum_{i=1}^3\sum_{j=1}^3 ijx_ix_j=x_1^2+4x_2^2+9x_3^2+4x_1x_2+6x_1x_3+12x_2x_3$,

so $A=\begin{pmatrix}1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9\end{pmatrix}$.

(2) It is easy to obtain the eigenvalues of $A$, namely $14,0,0$.

When $\lambda_1=14$, solving $(A-14E)x=0$ gives $\alpha_1=(1,2,3)^T$.

When $\lambda_2=\lambda_3=0$, solving $Ax=0$ gives $\alpha_2=(-2,1,0)^T$ and $\alpha_3=(-3,0,1)^T$.

Orthogonalizing $\alpha_2,\alpha_3$ gives $\xi_2=(-2,1,0)^T$, $\xi_3=(-3,-6,5)^T$.

Normalizing $\alpha_1,\xi_2,\xi_3$ gives

$\gamma_1=\frac{1}{\sqrt{14}}(1,2,3)^T$, $\gamma_2=\frac{1}{\sqrt{5}}(-2,1,0)^T$, $\gamma_3=\frac{1}{\sqrt{70}}(-3,-6,5)^T$.

Put $Q=(\gamma_1,\gamma_2,\gamma_3)$; by the orthogonal transformation $x=Qy$, $f$ is reduced to the canonical form $14y_1^2$.