2024 National Graduate Entrance Examination Mathematics I Past Paper

Approach:

Analyse the parity of the functions and use the properties of integral functions.

Steps:

Since $e^{\cos t}$ is an even function of $t$, $\int_0^x e^{\cos t}dt$ is an odd function of $x$, that is $f(x)$ is odd. Put $F(u)=\int_0^u e^{t^2}dt$; since $e^{t^2}$ is an even function, $F(u)$ is odd; and $\sin x$ is odd, so $g(x)=F(\sin x)$ is odd as well. Hence (C) is the correct choice.

Answer:

C

Approach:

Use the computation formula of the surface integral to turn the second-kind surface integral into a first-kind surface integral.

Steps:

Since $\frac{\partial z}{\partial x}=\frac{-x}{\sqrt{1-x^2-y^2}}=\frac{-x}{z},\ \frac{\partial z}{\partial y}=\frac{-y}{z}$, we have $\iint_{\Sigma} Pdy dz+Qdz dx=\iint_{\Sigma} P\cdot(-z_x)+Q\cdot(-z_y)dxdy=\iint_{\Sigma} \left(\frac{x}{z}P+\frac{y}{z}Q\right)dxdy$, so (A) is the correct choice.

Answer:

A

Approach:

Use the Taylor expansion of the logarithmic function to find the coefficients of the power series.

Steps:

From $\ln(1+x)=\sum\limits_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n},x\in(-1,1]$ we get

$\ln(2+x)=\ln2+\ln(1+\frac{x}{2})=\ln2+\sum\limits_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}\left(\frac{x}{2}\right)^n$

so $a_n=\begin{cases}\ln2, & n=0,\\ \frac{(-1)^{n-1}}{n\cdot2^n}, & n\geq1\end{cases}$, and hence $a_{2n}=\frac{(-1)^{2n-1}}{2n\cdot4^n}, n\geq1$.

$\sum\limits_{n=0}^{\infty} n a_{2n}=\sum\limits_{n=1}^{\infty} n a_{2n}=\sum\limits_{n=1}^{\infty}\frac{-1}{2\cdot4^n}=-\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{1-\frac{1}{4}}=-\frac{1}{6}$

Hence (A) is the correct choice.

Answer:

A

Approach:

Judge the correctness of each option by constructing counterexamples and by using the definition of the derivative.

Steps:

For option (A), take $f(x)=\begin{cases}mx, & x\neq0 \\ 1, & x=0\end{cases}$; then $\lim\limits_{x\to0} \dfrac{f(x)}{x}=m$, but $f(x)$ is not differentiable at $x=0$, so (A) is excluded.

For option (B), since $f'(0)=m$, the function $f(x)$ is continuous at $x=0$, so $f(0)=\lim\limits_{x\to0} f(x)=0$, and hence $\lim\limits_{x\to0} \dfrac{f(x)}{x}=\lim\limits_{x\to0} \dfrac{f(x)-f(0)}{x}=f'(0)=m$, so (B) is correct.

For option (C), take $f(x)=\begin{cases}mx, & x\neq0 \\ 1, & x=0\end{cases}$; then $\lim\limits_{x\to0} f'(x)=m$, but $f(x)$ is not differentiable at $x=0$, so (C) is excluded.

For option (D), take $f(x)=\begin{cases}mx+x^2\sin\dfrac{1}{x}, & x\neq0 \\ 0, & x=0\end{cases}$; then $f'(0)=m$, but $\lim\limits_{x\to0} f'(x)$ does not exist, so (D) is excluded. Hence (B) is the correct choice.

Answer:

B

Approach:

Analyse the relative position of the three planes and use the properties of the solutions of the system of linear equations.

Steps:

From the figure in the original paper, the three planes meet in a single straight line, so the system of equations

has infinitely many solutions, so $r\begin{pmatrix}\alpha_1\\\alpha_2\\\alpha_3\end{pmatrix}=r\begin{pmatrix}\beta_1\\\beta_2\\\beta_3\end{pmatrix}\lt3$, that is $m=n<3$. Moreover, the normal vectors of the three planes are coplanar, but those of $\pi_1$ and $\pi_2$ are not collinear, so $r=2$ follows, hence $m=n=2$, and (B) is the correct choice.

Answer:

B

Approach:

Use the condition of linear dependence of vectors and solve for the parameters by means of determinants.

Steps:

Since $\alpha_1,\alpha_2,\alpha_3$ are linearly dependent, $r(\alpha_1,\alpha_2,\alpha_3)\lt3$, so

which gives $a=1$ or $a=-2$. When $a=1$, $\alpha_1,\alpha_3$ are linearly dependent, contradicting the statement, so $a=-2$.

Also, from

we get $b=2$, so D is the correct choice.

Answer:

D

Approach:

Analyse the eigenvalues and eigenvectors of the matrix and compute the trace of a power of the matrix.

Steps:

From $A\alpha=0$ and $\alpha\neq0$ we get $\lambda_1=0$. Take nonzero vectors $\beta_1,\beta_2$ that are linearly independent (the plane perpendicular to $\alpha$ certainly contains two linearly independent vectors) and satisfy $\beta_1^T\alpha=\beta_2^T\alpha=0$; then $A\beta_1=\beta_1,A\beta_2=\beta_2$, and since $\beta_1,\beta_2$ are linearly independent, $\lambda=1$ is at least a double root, so $\lambda_1=0,\lambda_2=\lambda_3=1$. Hence the eigenvalues of $A^3$ are $0,1,1$, so $tr(A^3)=0+1+1=2$, and A is the correct choice.

Answer:

A

Approach:

Solve the problem using the properties of the normal distribution and the probability computation formula.

Steps:

$2X+Y\sim N(-2,10)$, $Y-X\sim N(-2,2^2)$,

so $P\{2X+Y<a\}=\Phi\left(\frac{a+2}{\sqrt{10}}\right)=P\{Y-X<0\}=\Phi\left(\frac{0+2}{2}\right)$.

Hence $\frac{a+2}{\sqrt{10}}=1\Rightarrow a=-2+\sqrt{10}$.

So B is the correct choice.

Answer:

B

Approach:

Compute the covariance by means of the conditional probability density and the joint probability density.

Steps:

When $0<x<1$, $f_{Y|X}(y|x)=\begin{cases}\dfrac{1}{1-x}, & x<y<1 \\ 0, & \text{other}\end{cases}$

$f(x,y)=\begin{cases}2, & x<y<1,0<x<1 \\ 0, & \text{other}\end{cases}$

$EXY=\iint xyf(x,y)dxdy=\int_0^1 dy\int_0^y 2xydx=\dfrac{1}{4}$

$EX=\int_0^1 x2(1-x)dx=\dfrac{1}{3}$

$EY=\iint yf(x,y)dxdy=\int_0^1 dy\int_0^y 2ydx=\dfrac{2}{3}$

$\mathrm{Cov}(X,Y)=EXY-EXEY=\dfrac{1}{36}$, so D is the correct choice.

Answer:

D

Approach:

Compute the distribution function of the random variable Z and judge the type of its distribution.

Steps:

The joint probability density of the random variables $X$ and $Y$ is $f(x,y)=\begin{cases}\lambda^2 e^{-\lambda(x+y)}, & x>0,y>0 \\ 0, & \text{other}\end{cases}$. The distribution function of the random variable $Z$ is $F_Z(Z)=P(Z\leq z)=P(|X-Y|\leq z)$,

When $z\leq0$, $F_Z(Z)=0$;

When $z>0$,

$=1-e^{-\lambda z}$

It is easy to see that the random variable $Z$ follows an exponential distribution with parameter $\lambda$, so (D) is the correct choice.

Answer:

D

Answer:

6

Approach:

Compute by means of equivalent infinitesimals and the properties of limits.

Steps:

From the given condition,

so $a=6$.

Answer:

5

Approach:

Compute the second derivative by means of the differentiation rule for a composite function of several variables.

Steps:

Put $u=\cos x,v=1+x^2$, then $y=f(u,v)$, and by the differentiation rule for a composite function of several variables,

Hence

Answer:

$-\frac{1}{\pi}$

Approach:

Compute the coefficients of the Fourier cosine series and then find the limit.

Steps:

First make the even extension of the function $f(x)$ and then its periodic extension,

Hence $\lim\limits_{n\to\infty} n^2\sin a_{2n-1}=\lim\limits_{n\to\infty} \frac{-4}{(2n-1)^2\pi}=-\frac{1}{\pi}$.

Answer:

$x=\tan\left(y+\frac{\pi}{4}\right)-y$

Approach:

Turn the differential equation into a separable one by a substitution.

Steps:

Put $u=x+y$, then $\frac{du}{dx}=1+y'$, and substituting into the equation gives,

After separating the variables, $u^2du=dx$, and integrating both sides gives $u-\arctan u=x+C$, so the general solution of the equation is $y-\arctan(x+y)=C$. Substituting the initial condition $y(1)=0$ gives $C=\frac{\pi}{4}$, so the particular solution required is $y-\arctan(x+y)=\frac{\pi}{4}$, that is $x=\tan\left(y+\frac{\pi}{4}\right)-y$.

Answer:

$a\geq0$

Approach:

Analyse the matrix inequality and determine the range of the parameter.

Steps:

By the associativity of matrix multiplication,

From

it follows that

Hence $\alpha^T A(\beta\alpha^T-\alpha\beta^T)A\beta=a(x_1y_2-x_2y_1)^2$.

If $\alpha^T A(\beta\alpha^T-\alpha\beta^T)A\beta\leq0$ always holds, that is if $-a(x_1y_2-x_2y_1)^2\leq0$ always holds, then $a\geq0$.

Answer:

$\dfrac{2}{3}$

Approach:

Solve for the probability parameter by means of the conditional probability formula.

Steps:

Let the event $A$ denote that the three experiments succeed at least once and the event $B$ denote that all three experiments succeed; then

By the conditional probability formula,

Solving gives $p=\dfrac{2}{3}$.

Answer:

$\sqrt{2}+\ln(1+\sqrt{2})-2$

Approach:

Use the symmetry of the region of integration and decompose the region of integration into simple subregions.

Steps:

Clearly the region of integration $D$ is symmetric about the $x$-axis, so denote the part of $D$ above the $x$-axis by $D_1$; then

Denote $D_2=\{(x,y)|0\leq x\leq1,0\leq y\leq1\},D_3=\{(x,y)|x^2+y^2\leq1,x\geq0,y\geq0\}$; then

Hence

Hence

Answer:

(1) $x+y+z=3$

(2) the maximum value is $21$, and the minimum value is $\frac{17}{27}$

Approach:

(1) Use the partial derivatives to find the equation of the tangent plane.

(2) Use the Lagrange multiplier method to find the extrema of a multivariable function.

Steps:

(1) From $\frac{\partial z}{\partial x}|_{(1,1,1)}=[3x^2-2(x+y)]|_{(1,1,1)}=1$, $\frac{\partial z}{\partial y}|_{(1,1,1)}=[3y^2-2(x+y)]|_{(1,1,1)}=-1$, the normal vector to the tangent plane $z=z(x,y)$ at $(1,1,1)$ is $(-1,-1,-1)$, so the equation of the tangent plane $T$ is

(2) From the statement, $D=\{(x,y)|x\geq0,y\geq0,x+y\leq3\}$. In the interior of the region $D$, from

the stationary points are $(0,0),(\frac{4}{3},\frac{4}{3})$, and it is easy to obtain $f(0,0)=3,f(\frac{4}{3},\frac{4}{3})=\frac{17}{27}$. In the region $D$, on the boundary $y=0,0\leq x\leq3$, we have $f(x,0)=x^3-x^2+3$, and from $\frac{df(x,0)}{dx}=3x^2-2x=0$ the stationary points are $x=0$ and $x=\frac{2}{3}$, at which $f(0,0)=3,f(\frac{2}{3},0)=\frac{77}{27}$. Similarly, in the region $D$, the stationary points on the boundary $x=0,0\leq y\leq3$ are $y=0,\frac{2}{3}$, and $f(0,0)=3,f(0,\frac{2}{3})=\frac{77}{27}$. In the region $D$, on the boundary $x+y=3,0\leq x\leq3$, we have $f(x,3-x)=x^3+(3-x)^3-6x+3$; put $\varphi(x)=x^3+(3-x)^3-6x+3$, then $\varphi'(x)=3x^2-3(3-x)^2-6=0$ gives the stationary point $x=\frac{3}{2}$, so the stationary point of the function $f(x,y)$ on $x+y=3$ is $(\frac{3}{2},\frac{3}{2})$, at which $f(\frac{3}{2},\frac{3}{2})=-\frac{3}{4}$. Moreover $f(3,0)=f(0,3)=21$, and comparison shows that $f(x,y)$ attains its maximum at the point $(3,0)$ or $(0,3)$, the maximum value being $21$, and attains its minimum at the point $(\frac{4}{3},\frac{4}{3})$, the minimum value being $\frac{17}{27}$.

Approach:

Prove the inequalities by means of the Taylor formula and the properties of integrals.

Steps:

(1) Expanding the function $f(x)$ by the Taylor formula at $x=0$ and at $x=1$ gives,

Noting that $f'(0)=f'(1)$, equations (1) and (2) give

Since $|f''(x)|\leq1$,

(2) From (1),

Integrating both sides of equation (3) over the interval $[0,1]$ gives

that is,

Hence

Answer:

(1) When $x\in(0,1)$, $|f(x)-f(0)(1-x)-f(1)x|\leq\dfrac{x(1-x)}{2}$;

(2) $\left|\int_0^1 f(x)dx-\dfrac{f(0)+f(1)}{2}\right|\leq\dfrac{1}{12}$.

Answer:

$\frac{4\pi}{25}\sqrt{5}\pi$

Approach:

Use Stokes’ formula to turn the line integral into a surface integral.

Steps:

Let the part of the plane $2x-z-1=0$ cut off by the sphere $x^2+y^2+z^2=2x$ be the surface $\Sigma$, with the direction taken to be the upper side; then the unit normal vector of the surface $\Sigma$ is $\frac{1}{\sqrt{5}}(-2,0,1)$.