2025 National Graduate Entrance Examination Mathematics I Past Paper

Approach:

Differentiate $f(x)$ and $g(x)$ separately and judge whether $x=0$ is an extremum point or an inflection point.

Steps:

1. $f'(x) = e^{x^2} \sin x$
2. $f''(x) = 2x e^{x^2} \sin x + e^{x^2} \cos x$
3. $f'(0) = 0,\ f''(0) = 1 > 0$, so $x=0$ is an extremum point of $f(x)$.
4. $g(x) = \int_0^x e^{t^2} dt \cdot \sin^2 x$
5. $g'(x) = e^{x^2} \sin^2 x + \sin 2x \int_0^x e^{t^2} dt$
6. $g''(x) = e^{x^2} \sin 2x + 2x e^{x^2} \sin^2 x + \sin 2x e^{x^2} + 2\cos 2x \int_0^x e^{t^2} dt$
7. $g'(0) = 0,\ g''(0) = 0,\ g'''(0) > 0$, so $(0,0)$ is an inflection point of $y=g(x)$.

Answer:

B

Approach:

Judge the convergence of the two series separately.

Steps:

1. $\left| \sin \dfrac{n^3\pi}{n^2+1} \right| = \left| \sin \left( \dfrac{n^3\pi}{n^2+1} - n\pi \right) \right| = \left| \sin \left( \dfrac{n}{n^2+1} \pi \right) \right| \sim \left| \dfrac{n}{n^2+1} \pi \right| \sim \dfrac{1}{n}$.
2. The series $\sum\limits_{n=1}^{\infty} \dfrac{1}{n}$ diverges, so it is not absolutely convergent.
3. $\sin \dfrac{n^3\pi}{n^2+1} = (-1)^n \sin \left( \dfrac{n^3\pi}{n^2+1} - n\pi \right) = (-1)^n \sin \dfrac{n}{n^2+1} \pi$, which is an alternating series.
4. The sequence $\sin \dfrac{n}{n^2+1} \pi$ is decreasing, so the series converges conditionally.
5. $\sum\limits_{n=1}^{\infty} (-1)^n \left( \dfrac{1}{\sqrt[3]{n^2}} - \tan \dfrac{1}{\sqrt[3]{n^2}} \right)$.
6. $\left| (-1)^n \left( \dfrac{1}{\sqrt[3]{n^2}} - \tan \dfrac{1}{\sqrt[3]{n^2}} \right) \right| = \left| \dfrac{1}{\sqrt[3]{n^2}} - \tan \dfrac{1}{\sqrt[3]{n^2}} \right| = \dfrac{1}{3n^2} + o\left( \dfrac{1}{n^2} \right)$.
7. The series $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^2}$ converges, so $\sum\limits_{n=1}^{\infty} (-1)^n \left( \dfrac{1}{\sqrt[3]{n^2}} - \tan \dfrac{1}{\sqrt[3]{n^2}} \right)$ converges absolutely.

Answer:

B

Approach:

This question tests the logical relations between the limit of a function, the limit of its derivative and the limit of its integral average. All four options are statements of the form “the existence of one limit implies the existence of another limit”; they can be judged one by one, and counterexamples are constructed where necessary.

Steps:

1. For option A: take $f(x)=\dfrac{\sin x^2}{x}$; then $\lim\limits_{x\to+\infty}f(x)=0$ exists, but $f'(x)=\dfrac{2x^2\cos x^2-\sin x^2}{x^2}$ has no limit as $x\to+\infty$, so A is wrong.
2. For option B: take $f(x)=\sqrt{x}$; then $f'(x)=\dfrac{1}{2\sqrt{x}}\to0$ exists, but $\lim\limits_{x\to+\infty}f(x)=+\infty$ does not exist, so B is wrong.
3. For option C: take $f(x)=\cos x$; then $\lim\limits_{x\to+\infty}\dfrac{\int_0^xf(t)dt}{x}=\lim\limits_{x\to+\infty}\dfrac{\sin x}{x}=0$ exists, but $\lim\limits_{x\to+\infty}f(x)$ does not exist, so C is wrong.
4. For option D: if $A=\lim\limits_{x\to+\infty}f(x)$ exists, then by L’Hopital’s rule $\displaystyle\lim_{x\to+\infty}\frac{\int_0^xf(t)dt}{x}=\lim_{x\to+\infty}\frac{f(x)}{1}=A$ exists, so D is correct, and D is the correct choice.

Answer:

D

Approach:

This question tests the interchange of the order of integration of a double integral. First determine the region of integration, then split the region into two parts according to the sign of $x$, and interchange the order of integration in one of them.

Steps:

1. From the original expression $\displaystyle\int_{-2}^2dx\int_{4-x^2}^4f(x,y)dy$ we see that the region of integration is $D=\{(x,y)\mid 4-x^2\le y\le4,\ -2\le x\le2\}$.
2. Denote $D_1=\{(x,y)\mid 4-x^2\le y\le4,\ -2\le x\le0\}$, $D_2=\{(x,y)\mid 4-x^2\le y\le4,\ 0\le x\le2\}$; then $I=\iint_{D_1}f\,d\sigma+\iint_{D_2}f\,d\sigma$.
3. Interchange the order of integration for $D_2$: with $y$ fixed, $-x^2+y-4\le0$ means $x^2\ge4-y$, and combined with $0\le x\le2$ this gives $\sqrt{4-y}\le x\le2$, and $0\le y\le4$, so $\displaystyle\iint_{D_2}f\,d\sigma=\int_0^4dy\int_{\sqrt{4-y}}^2f(x,y)dx$.
4. Meanwhile $\displaystyle\iint_{D_1}f\,d\sigma=\int_{-2}^0dx\int_{4-x^2}^4f(x,y)dy$ keeps the original order.
5. Adding the two gives the expression of option (A), so (A) is the correct choice.

Answer:

A

Approach:

This question tests the positive index of inertia of a quadratic form. The positive index of inertia equals the number of positive eigenvalues of the matrix of the quadratic form, so it is enough to find the eigenvalues.

Steps:

1. The matrix of the quadratic form is $A=\begin{pmatrix}1&1&1\\1&0&0\\1&0&0\end{pmatrix}$.
2. Compute the characteristic polynomial: $|\lambda E-A|=\begin{vmatrix}\lambda-1&-1&-1\\-1&\lambda&0\\-1&0&\lambda\end{vmatrix}=\lambda\bigl[\lambda(\lambda-1)-1-1\bigr]=\lambda(\lambda^2-\lambda-2)$.
3. Factoring gives $|\lambda E-A|=\lambda(\lambda-2)(\lambda+1)$, so the eigenvalues are $\lambda_1=0,\ \lambda_2=2,\ \lambda_3=-1$.
4. There is only one positive eigenvalue, namely $\lambda_2=2$, so the positive index of inertia is 1, and (B) is the correct choice.

Answer:

B

Approach:

This question tests the geometric meaning of the solution set of a system of linear equations and the rank of a matrix. First the ranks of the coefficient matrix and of the augmented matrix are obtained from the linear relations among the vectors, and then it is decided whether the solution set is a plane or a straight line.

Steps:

1. Denote $A=(\alpha_1,\alpha_2,\alpha_3)$. Since $\alpha_1,\alpha_2$ are linearly independent while $\alpha_1,\alpha_2,\alpha_3$ are linearly dependent, $r(A)=2$.
2. Denote $\overline A=(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$; from $\alpha_1+\alpha_2+\alpha_4=0$ we know that $\alpha_4$ can be represented as a linear combination of $\alpha_1,\alpha_2$, so $r(\overline A)=2$.
3. Hence $Ax=\alpha_4$ has infinitely many solutions, and the solution set is described by $x\alpha_1+y\alpha_2+z\alpha_3=\alpha_4$, which is a plane in space.
4. Moreover, if that plane passed through the origin, then $\alpha_4=0$, which contradicts the linear independence of $\alpha_1,\alpha_2$, so the plane does not pass through the origin.
5. Hence its geometric figure is a plane not passing through the origin, and (D) is the correct choice.

Answer:

D

Approach:

This question tests the operational rules for the rank of a matrix and the verification of the conclusions by counterexamples. As soon as one of the four conclusions fails to hold, a matrix satisfying the condition can be constructed to rule it out.

Steps:

1. Take $A=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}$, $B=\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}$, $C=E$; then $r(A)=1,\ r(B)=1,\ r(C)=3$, and $ABC=B$, $r(ABC)=1$.
2. Check the condition: $r(A)+r(B)+r(C)=1+1+3=5$, whereas $r(ABC)+2n=1+6=7$.
3. Conclusion (3) gives $r(A)=r(B)=r(C)=n$, that is all three equal 3, which contradicts $r(A)=1,\ r(B)=1$ in the example above, so (3) does not hold.
4. Conclusion (4) gives $r(AB)=r(BC)=n$, whereas in the example above $AB=0$ gives $r(AB)=0$ and $BC=B$ gives $r(BC)=1$, neither of which equals 3, so (4) does not hold.
5. Hence the correct conclusions can only be (1) and (2), that is option (A) is the correct choice.

Answer:

A

Approach:

This question tests the variance and the maximum of a linear combination of a two-dimensional normal distribution. Write $D(aX+bY)$ as a function of $a$, and then find its maximum under the constraint $a^2+b^2=1$.

Steps:

1. From $D(X)=D(Y)=1$ and $\mathrm{Cov}(X,Y)=\rho$, we get $D(aX+bY)=a^2DX+b^2DY+2ab\,\mathrm{Cov}(X,Y)=a^2+b^2+2ab\rho$.
2. From $a^2+b^2=1$, we get $D(aX+bY)=1+2ab\rho$.
3. From $a^2+b^2=1$ we know that $|ab|\le\dfrac12$, with equality when $a^2=b^2=\dfrac12$, so the maximum of $2ab\rho$ is $|\rho|$.
4. Hence the maximum of $D(aX+bY)$ is $1+|\rho|$, and (C) is the correct choice.

Answer:

C

Approach:

This question tests the approximation of a binomial distribution by a Poisson distribution. When $n$ is large and $p$ is small, $B(n,p)$ can be approximated by a Poisson distribution with parameter $np$.

Steps:

1. $T=\sum\limits_{i=1}^{20}X_i\sim B(20,0.1)$, with parameter $np=20\times0.1=2$.
2. Approximate by the Poisson distribution with parameter 2: $P\{T=k\}\approx\dfrac{2^k}{k!}e^{-2}$.
3. Hence $P\{T\le1\}=P\{T=0\}+P\{T=1\}\approx\dfrac{2^0}{0!}e^{-2}+\dfrac{2^1}{1!}e^{-2}=e^{-2}+2e^{-2}=3e^{-2}$.
4. Hence (C) is the correct choice.

Answer:

C

Approach:

This question tests the rejection region of a one-sided test for a normal population. The test statistic $\dfrac{\overline X-\mu}{\sigma/\sqrt n}$ follows the standard normal distribution, and substituting the boundary value of the null hypothesis gives the rejection region.

Steps:

1. Since $X_1,\dots,X_n$ come from $N(\mu,2)$, we know that $\sigma=\sqrt2$, and $\dfrac{\overline X-\mu}{\sigma/\sqrt n}\sim N(0,1)$.
2. For the one-sided test $H_0:\mu\le1,\ H_1:\mu>1$, the rejection region is $\dfrac{\overline X-1}{\sqrt2/\sqrt n}>Z_\alpha$.
3. Rearranging gives $\overline X>1+\dfrac{\sqrt2}{\sqrt n}Z_\alpha$.
4. Hence (D) is the correct choice.

Answer:

D

Approach:

This question tests the application of equivalent infinitesimals in the computation of a limit. Write $x^x$ as $e^{x\ln x}$ and then simplify with $e^u-1\sim u$.

Steps:

1. Since $x\to0^+$ implies $\ln x\to-\infty$ and $\ln(1-x)\sim-x$, we have $\ln x\cdot\ln(1-x)\sim-x\ln x$.
2. Also $x^x-1=e^{x\ln x}-1\sim x\ln x$.
3. Hence the original expression $=\lim\limits_{x\to0^+}\dfrac{x\ln x}{-x\ln x}=-1$.

Answer:

$-1$.

Approach:

This question tests the periodicity of the sum function of a Fourier series. The Fourier series of $f$, $\sum b_n\sin n\pi x$, has period 2, so first the point is moved back into $[0,1]$ by periodicity, and then the value is substituted.

Steps:

1. The least positive period of $\sin n\pi x$ is 2, so $S(x)$ has period 2.
2. Hence $S\left(-\dfrac72\right)=S\left(-\dfrac72+4\right)=S\left(\dfrac12\right)$.
3. Since $x=\dfrac12$ is a point of continuity of $f$, $S\left(\dfrac12\right)=f\left(\dfrac12\right)=\left(\dfrac12\right)^2=\dfrac18$.

Answer:

$\dfrac{1}{8}$.

Approach:

This question tests the computation of a directional derivative. The directional derivative is the inner product of the gradient with the unit vector in that direction.

Steps:

1. The partial derivatives are $\dfrac{\partial u}{\partial x}=y^2z^3$, $\dfrac{\partial u}{\partial y}=2xyz^3$, $\dfrac{\partial u}{\partial z}=3xy^2z^2$.
2. The gradient at $(1,1,1)$ is $(1,2,3)$.
3. The length of the direction vector $\bm n=(2,2,-1)$ is $\sqrt{4+4+1}=3$, and its unit vector is $\bm n_0=\left(\dfrac23,\dfrac23,-\dfrac13\right)$.
4. The directional derivative $\left.\dfrac{\partial u}{\partial\bm n}\right|_{(1,1,1)}=(1,2,3)\cdot\left(\dfrac23,\dfrac23,-\dfrac13\right)=\dfrac23+\dfrac43-1=1$.

Answer:

$1$.

Approach:

This question tests a second-kind line integral together with Green’s formula. The curve $L$ is not closed, so a straight segment is added to close it, and then Green’s formula turns the integral into a double integral.

Steps:

1. Put $P=y+\cos x$, $Q=2x+\cos y$, then $\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}=2-1=1$.
2. Let $L_0$ be the straight segment from $(-1,0)$ to $(1,0)$; then $L$ and $L_0$ enclose the region $D=\{x^2+y^2\le1,\ y\ge0\}$.
3. By Green’s formula, the integral along the closed circuit formed by $L_0$ and $L$ is $I_1=\iint_D\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)d\sigma=\iint_D1\,d\sigma=\int_{-1}^1(1-x^2)dx=\dfrac43$.
4. The integral along $L_0$ is $I_2=\int_{-1}^1\cos x\,dx=2\sin1$ (here $y=0,\ dy=0$).
5. Since $L_0$ and $L$ have opposite directions, $I=I_1-I_2=\dfrac43-2\sin1$.

Answer:

$\dfrac{4}{3}-2\sin1$.

Approach:

This question tests the relation between a power of a matrix and the solutions of a system of equations. The fact that $A^2x=0$ and $Ax=0$ do not have the same solutions means that $A$ is not invertible, so the parameters can be determined from the vanishing of the determinant.

Steps:

1. If $A$ were invertible, then $A^2x=0\iff Ax=0$, and the two would have the same solutions, contradicting the statement; hence $A$ is not invertible, that is $|A|=0$.
2. Compute the determinant (expanding with respect to the doubling property of the second and the third column): $|A|=4\begin{vmatrix}0&-1\\1&-2\end{vmatrix}-(a-b)\begin{vmatrix}1&-1\\1&-2\end{vmatrix}$.
3. Here $\begin{vmatrix}0&-1\\1&-2\end{vmatrix}=1$ and $\begin{vmatrix}1&-1\\1&-2\end{vmatrix}=-1$, so $|A|=4-(a-b)\cdot(-1)=4+a-b$.
4. Setting $|A|=0$ gives $a-b=-4$.

Answer:

$-4$.

Approach:

This question tests the probability calculation for independent events and the conditional probability. First use independence to turn the probability of the union into an equation in $P(B)$, solve for $P(A),P(B)$, and then compute the conditional probability.

Steps:

1. Put $p=P(B)$, then $P(A)=2p$, and by independence $P(AB)=P(A)P(B)=2p^2$.
2. From $P(A\cup B)=P(A)+P(B)-P(AB)=\dfrac58$ we get $3p-2p^2=\dfrac58$, that is $16p^2-24p+5=0$, that is $(4p-1)(4p-5)=0$.
3. Take $p=\dfrac14$ (the value $p=\dfrac54>1$ is discarded), which gives $P(B)=\dfrac14,\ P(A)=\dfrac12$.
4. Hence the probability that “exactly one occurs” is $P(A\overline B)+P(\overline AB)=P(A)+P(B)-2P(A)P(B)=\dfrac12+\dfrac14-2\cdot\dfrac12\cdot\dfrac14=\dfrac12$.
5. Under the condition that $A\cup B$ occurs, the required conditional probability is $\dfrac{1/2}{5/8}=\dfrac45$.

Answer:

$\dfrac{4}{5}$.

Approach:

This question tests the indefinite integral of a rational function. First make a partial fraction decomposition and then integrate term by term, where the quadratic factor is turned into an arctangent after completing the square.

Steps:

1. Make the partial fraction decomposition: $\dfrac{1}{(x+1)(x^2-2x+2)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2-2x+2}$; comparing coefficients gives $A=\dfrac15,\ B=-\dfrac13,\ C=\dfrac35$.
2. Hence the original integral $=\displaystyle\int_0^1\left(\dfrac{1}{5(x+1)}+\dfrac{-\frac13x+\frac35}{x^2-2x+2}\right)dx$.
3. Here $\dfrac{-\frac13x+\frac35}{x^2-2x+2}=-\dfrac16\cdot\dfrac{2x-2}{x^2-2x+2}+\dfrac{3}{5}\cdot\dfrac{1}{(x-1)^2+1}$.
4. Integrate term by term: $\dfrac15\ln2+\dfrac{1}{10}\ln\dfrac{2}{2}+\dfrac35\left[\arctan(x-1)\right]_0^1=\dfrac15\ln2+\dfrac35\cdot\dfrac{\pi}{4}$.
5. That is, the original integral $=\dfrac{3}{10}\ln2+\dfrac{1}{10}\pi$.

Answer:

$\dfrac{3}{10}\ln 2+\dfrac{1}{10}\pi$.

Approach:

This question tests the partial derivatives of a composite function of several variables and an ordinary differential equation with a parameter. First use the chain rule to express all the partial derivatives of $g$ in terms of $f'$ and $f''$, substitute them into the original equation to obtain an ordinary differential equation in $u$, and finally use the initial values to determine the constant.

Steps:

1. Put $u=\dfrac{x}{y}$, then $\dfrac{\partial g}{\partial x}=f'(u)\dfrac1y$ and $\dfrac{\partial g}{\partial y}=f'(u)\left(-\dfrac{x}{y^2}\right)$.
2. From $g(x,x)=f(1)=1$ and $\left.\dfrac{\partial g}{\partial x}\right|_{(x,x)}=f'(1)\dfrac1x=\dfrac2x$ we get $f'(1)=2$.
3. Compute the second partial derivatives: $\dfrac{\partial^2g}{\partial x^2}=f''(u)\dfrac1{y^2}$, $\dfrac{\partial^2g}{\partial x\partial y}=-\dfrac{x}{y^3}f''(u)-\dfrac1{y^2}f'(u)$, $\dfrac{\partial^2g}{\partial y^2}=\dfrac{x^2}{y^4}f''(u)+\dfrac{2x}{y^3}f'(u)$.
4. Substituting the above into the original equation and simplifying gives $u^2f''(u)+uf'(u)+\dfrac1{u^2}f'(u)=\dfrac1{u^2}$.
5. Put $p=f'(u)$, then $u^2p''+up'+\dfrac1{u^2}p'=\dfrac1{u^2}$, which gives $p=\dfrac{\ln u}{u}+\dfrac2u$.
6. Hence $f'(u)=\dfrac{\ln u}{u}+\dfrac2u$, and integrating gives $f(u)=\dfrac12\ln^2u+2\ln u+C$; since $f(1)=1$ we get $C=1$.

Answer:

$f(u)=\dfrac{1}{2}\ln^{2}u+2\ln u+1$.

Approach:

This question asks for the proof of a necessary and sufficient condition concerning the order of the slopes of secants. For sufficiency the monotonicity of $f'$ is squeezed out of the limit of the secant slopes (that is, the left and right derivatives); for necessity the Lagrange mean value theorem is used directly.

Steps:

Solution:

Sufficiency: suppose that in $(a,b)$, for arbitrary $x_1,x_2,x_3$, when $x_1<x_2<x_3$, we always have

then, taking in $(a,b)$ arbitrary $x_1<x_2<x_3<x_4<x_5$, we have

Letting both sides of $\frac{f(x_2)-f(x_1)}{x_2-x_1}<\frac{f(x_3)-f(x_2)}{x_3-x_2}$ tend to $x_2\to x_1^+$ gives