2020 National Graduate Entrance Examination Mathematics I Past Paper

Approach:

This question tests the comparison of the orders of infinitesimals. For each option, first replace the integrand by an equivalent expression in order to find its leading term, then integrate to obtain the leading term, and finally compare the powers that result.

Steps:

1. For (A): $e^{t^2}-1\sim t^2$, so $\displaystyle\int_0^x(e^{t^2}-1)dt\sim\int_0^x t^2dt=\dfrac{x^3}{3}$, an infinitesimal of order 3.
2. For (B): $\ln(1+\sqrt{t^3})\sim t^{3/2}$, so $\displaystyle\int_0^x\ln(1+\sqrt{t^3})dt\sim\int_0^x t^{3/2}dt=\dfrac25x^{5/2}$.
3. For (C): $\sin t^2\sim t^2$ and $\sin x\sim x$, so $\displaystyle\int_0^{\sin x}\sin t^2dt\sim\int_0^x t^2dt=\dfrac{x^3}{3}$, an infinitesimal of order 3.
4. For (D): $\sqrt{\sin^3t}\sim t^{3/2}$ and $1-\cos x\sim\dfrac{x^2}{2}$, so $\displaystyle\int_0^{1-\cos x}\sqrt{\sin^3t}dt\sim\int_0^{x^2/2}t^{3/2}dt=\dfrac25\left(\dfrac{x^2}{2}\right)^{5/2}$.
5. Comparing these shows that (D) has the highest order, so (D) is the correct choice.

Answer:

D.

Approach:

This question tests the relation between differentiability and limits. The key is to compare the expansion $f(x)=f(0)+f'(0)x+o(x)$ of a differentiable $f$ at $0$ with the orders of the various denominators.

Steps:

1. If $f$ is differentiable at $0$, then it is continuous at $0$, and $\lim\limits_{x\to0}f(x)=0$ gives $f(0)=0$, so $f(x)=f'(0)x+o(x)$.
2. Hence $\dfrac{f(x)}{\sqrt{|x|}}=f'(0)\dfrac{x}{\sqrt{|x|}}+o(\sqrt{|x|})\to0$, because $|x|/\sqrt{|x|}=\sqrt{|x|}\to0$. Thus option (C) holds.
3. For option (D): $\dfrac{f(x)}{\sqrt{x^2}}=\dfrac{f'(0)x+o(x)}{|x|}=f'(0)\operatorname{sgn}(x)+o(1)$, which tends to $0$ only when $f'(0)=0$, so (D) does not always hold.
4. For options (A) and (B): the assumptions only give $\lim\limits_{x\to0}f(x)=0$ and do not give $f(0)=0$, so $f$ need not be continuous at $0$ and differentiability cannot be deduced. In conclusion the answer is (C).

Answer:

C.

Approach:

This question tests the definition of differentiability for a function of several variables and its geometric meaning. Differentiability is equivalent to the increment of the function at that point being approximated by its linear part, with an error that is a small term of order $o(\rho)$.

Steps:

1. Since $f$ is differentiable at $(0,0)$ and $f(0,0)=0$, we have $\displaystyle\lim_{(x,y)\to(0,0)}\dfrac{f(x,y)-f_x'(0,0)x-f_y'(0,0)y}{\sqrt{x^2+y^2}}=0$.
2. Put $\bm n=\bigl(f_x'(0,0),f_y'(0,0),-1\bigr)$; then $\bm n\cdot(x,y,f(x,y))=f_x'(0,0)x+f_y'(0,0)y-f(x,y)$.
3. The previous step gives $\dfrac{\left|\bm n\cdot(x,y,f(x,y))\right|}{\sqrt{x^2+y^2}}\to0$, so the limit in option (A) exists and equals $0$.
4. For option (B): the modulus of the cross product does not in general tend to $0$; the relation $\bm\alpha\perp\bm n$ only records the direction of $\bm\alpha$ relative to $\bm n$ and does not guarantee that the limits in (C) and (D) exist. In conclusion the answer is (A).

Answer:

A.

Approach:

This question tests the definition of the radius of convergence of a power series and Abel’s theorem. The basic conclusion about a power series is that it converges absolutely for $|x|<R$ and diverges for $|x|>R$, while the two endpoints have to be discussed separately.

Steps:

1. By the definition of the radius of convergence, the power series $\sum\limits_{n=1}^{\infty}a_nx^n$ converges for $|x|<R$ and diverges for $|x|>R$.
2. The behaviour at the two endpoints $x=\pm R$ has to be discussed separately: it may converge or it may diverge.
3. Hence if the series diverges at $r$, then necessarily $|r|\ge R$, and option (A) holds.
4. Options (C) and (D) sweep the endpoint case under the rug and do not hold; the inequality in option (B) goes the wrong way. In conclusion the answer is (A).

Answer:

A.

Approach:

This question tests the relation between elementary transformations and matrix multiplication. Transforming $A$ into $B$ by elementary row transformations is equivalent to multiplying on the left by an invertible matrix, and this relation can be rewritten in another form.

Steps:

1. An elementary row transformation is equivalent to multiplying on the left by an elementary matrix, and elementary matrices are invertible, so a sequence of elementary row transformations is equivalent to multiplying on the left by some invertible matrix $P$, that is $PA=B$.
2. Multiplying $PA=B$ on the right by $P^{-1}$ gives $A=BP^{-1}$, and then putting $Q=P^{-1}$ gives $QB=A$, that is, there exists an invertible matrix with $QB=A$.
3. Option (A) says that there exists $P$ with $PA=B$, which reverses the order of the letters; option (C) repeats option (B); and option (D) asserts that $Ax=0$ and $Bx=0$ have the same solution set, which is equivalent to $A$ and $B$ being equivalent, and that does not hold under elementary row transformations alone (row transformations change the structure of the solution set). In conclusion the answer is (B).

Answer:

B.

Approach:

This question tests the computation of the probability that exactly one event occurs by means of the inclusion-exclusion formula. The key point is that the three difference events are not completely symmetric, because the three pairwise intersection probabilities given in the question are not equal.

Steps:

1. First compute the probability that $A$ alone occurs: from $P(AB)=0$ we get $P(ABC)=0$, so $P(A\overline B\overline C)=P(A)-P(AB)-P(AC)=\tfrac14-0-\tfrac1{12}=\tfrac16$.
2. Similarly $P(\overline AB\overline C)=P(B)-P(AB)-P(BC)=\tfrac14-0-\tfrac1{12}=\tfrac16$.
3. But when $C$ alone occurs, two pairwise intersections have to be subtracted: $P(\overline A\overline BC)=P(C)-P(AC)-P(BC)=\tfrac14-\tfrac1{12}-\tfrac1{12}=\tfrac1{12}$.
4. Adding the three expressions gives the probability that exactly one event occurs as $\tfrac16+\tfrac16+\tfrac1{12}=\tfrac5{12}$, so (D) is the correct choice.

Answer:

D.

Approach:

This question tests the application of the central limit theorem. Standardize the sum of the binomial variables and then express its approximate value in terms of the normal distribution function.

Steps:

1. $X_i$ follows the two-point distribution with parameter $\tfrac12$, so $EX_i=\tfrac12$ and $DX_i=\tfrac14$.
2. By the central limit theorem, $\displaystyle\sum_{i=1}^{100}X_i$ is approximately distributed as $N\left(100\cdot\tfrac12,\ 100\cdot\tfrac14\right)=N(50,25)$, whose standard deviation is $5$.
3. Hence $\displaystyle P\!\left\{\sum_{i=1}^{100}X_i\le55\right\}\approx P\left\{Z\le\dfrac{55-50}{5}\right\}=P\{Z\le1\}=\Phi(1)$, so (B) is the correct choice.

Answer:

B.

Approach:

This question tests the computation of a limit of the difference of two infinitesimals of the same order. The key to this type of problem is to combine the fractions first, and then to replace the denominator by an equivalent infinitesimal.

Steps:

1. Combining the fractions turns the limit into $\displaystyle\lim_{x\to0}\frac{\ln(1+x)-e^x+1}{(e^x-1)\ln(1+x)}$.
2. Since $e^x-1\sim x$ and $\ln(1+x)\sim x$, the denominator satisfies $(e^x-1)\ln(1+x)\sim x^2$, so the original expression $=\displaystyle\lim_{x\to0}\frac{\ln(1+x)-e^x+1}{x^2}$.
3. The numerator is of the form $0/0$, so L’Hopital’s rule gives $\displaystyle\lim_{x\to0}\frac{\tfrac{1}{1+x}-e^x}{2x}$.
4. Applying L’Hopital’s rule once more to the $0/0$ form: $\displaystyle\lim_{x\to0}\frac{-\dfrac{1}{(1+x)^2}-e^x}{2}=-\dfrac{1-1}{2}=-1$.

Answer:

$-1$.

Approach:

This question tests the computation of the second derivative of a function determined by parametric equations. It is enough to find the first derivative from $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$, then differentiate with respect to $t$ and divide by $\dfrac{dx}{dt}$.

Steps:

1. Differentiating: $\dfrac{dx}{dt}=\dfrac{t}{\sqrt{t^2+1}}$, $\dfrac{dy}{dt}=\dfrac{1}{\sqrt{t^2+1}}$ (using the chain rule for $\ln$).
2. Hence $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{1}{t}$.
3. Differentiate once more with respect to $t$ and divide by $\dfrac{dx}{dt}$: $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac1t\right)}{\dfrac{dx}{dt}}=\dfrac{-1/t^2}{t/\sqrt{t^2+1}}=-\dfrac{\sqrt{t^2+1}}{t^3}$.
4. Substituting $t=1$ gives $\left.\dfrac{d^2y}{dx^2}\right|_{t=1}=-\sqrt2$.

Answer:

$-\sqrt{2}$.

Approach:

This question tests the combination of a linear differential equation with constant coefficients and a definite integral. Integrate the differential equation over $[0,+\infty)$ and use the boundary conditions to eliminate the terms at infinity.

Steps:

1. From $f''(x)+af'(x)+f(x)=0$ ($a>0$) the solutions decay exponentially, so $f(x),f'(x)\to0\ (x\to+\infty)$.
2. Integrating over $[0,+\infty)$: $\displaystyle\int_0^{+\infty}f''(x)dx+a\int_0^{+\infty}f'(x)dx+\int_0^{+\infty}f(x)dx=0$.
3. Here $\displaystyle\int_0^{+\infty}f''(x)dx=f'(+\infty)-f'(0)=-n$, and $\displaystyle a\int_0^{+\infty}f'(x)dx=a\bigl(f(+\infty)-f(0)\bigr)=-am$.
4. Substitution gives $-n-am+\displaystyle\int_0^{+\infty}f(x)dx=0$, hence $\displaystyle\int_0^{+\infty}f(x)dx=n+am$.

Answer:

$n+a m$.

Approach:

This question tests the mixed partial derivative of a function of two variables with a variable upper limit in the integral. First simplify the integral by the formula for an integral with variable upper limit, and then use the chain rule to obtain the mixed partial derivative.

Steps:

1. First simplify the integral: $f(x,y)=\int_0^{xy}e^t dt=e^{xy}-1$.
2. Differentiate with respect to $x$: $f_x=ye^{xy}$.
3. Then differentiate with respect to $y$: $f_{xy}=\dfrac{\partial}{\partial y}\left(ye^{xy}\right)=e^{xy}+xy\,e^{xy}=e^{xy}(1+xy)$.
4. Substituting $(1,1)$ gives $\left.f_{xy}\right|_{(1,1)}=e\cdot2=4e$.

Answer:

$4e$.

Approach:

This question tests a structured computation of a high-order determinant. The rows and the columns of this determinant are symmetric in structure, so it can be written as a $2\times2$ block matrix and evaluated with the block determinant formula.

Steps:

1. Write the determinant of order $4$ in the $2\times2$ block form $M=\begin{pmatrix}aE_2&Q\\Q&aE_2\end{pmatrix}$, where $Q=\begin{pmatrix}-1&1\\1&-1\end{pmatrix}$.
2. The similarity transformation $\begin{pmatrix}E_2&E_2\\E_2&-E_2\end{pmatrix}$ turns $M$ into the block diagonal matrix $\begin{pmatrix}aE_2+Q&0\\0&aE_2-Q\end{pmatrix}$.
3. Now $aE_2+Q=\begin{pmatrix}a-1&1\\1&a-1\end{pmatrix}$ has eigenvalues $a-1\pm1$, that is $a$ and $a-2$; and $aE_2-Q=\begin{pmatrix}a+1&-1\\-1&a+1\end{pmatrix}$ has eigenvalues $a+1\pm1$, that is $a+2$ and $a$.
4. Hence the determinant is $a\cdot(a-2)\cdot(a+2)\cdot a=a^4-4a^2$.

Answer:

$a^4-4a^2$.

Approach:

This question tests the computation of a covariance for a uniform distribution. The covariance is the mixed moment minus the product of the two first moments, and here both first moments are zero, so only $E(X\sin X)$ has to be computed.

Steps:

1. $X$ is uniformly distributed on $\left(-\tfrac\pi2,\tfrac\pi2\right)$ with density $f(x)=\dfrac1\pi$, so $EX=0$ (the interval is symmetric about the origin).
2. $EY=E(\sin X)=\dfrac1\pi\int_{-\pi/2}^{\pi/2}\sin x\,dx=0$.
3. Hence $\mathrm{Cov}(X,Y)=E(X\sin X)-EX\cdot EY=E(X\sin X)$.
4. The integrand $x\sin x$ is an even function, so $E(X\sin X)=\dfrac2\pi\int_0^{\pi/2}x\sin x\,dx=\dfrac2\pi\left[-x\cos x+\sin x\right]_0^{\pi/2}=\dfrac2\pi$, that is $\mathrm{Cov}(X,Y)=\dfrac2\pi$.

Answer:

$\dfrac{2}{\pi}$.

Approach:

This question tests the extrema and the maximum of a function of two variables. First solve the equations for the stationary points, then use the discriminant $AC-B^2$ formed by the second partial derivatives to classify them, and finally check whether the function is bounded so as to decide whether a maximum exists.

Steps:

1. The first-order partial derivatives are $f_x=3x^2-y,\ f_y=24y^2-x$.
2. Setting $f_x=f_y=0$ and substituting $y=3x^2$ into $x=24y^2$ gives $x=216x^4$, so $x=0$ or $x=\tfrac16$.
3. The corresponding stationary points are $(0,0)$ and $\left(\tfrac16,\tfrac1{12}\right)$, with the function values $0$ and $-\tfrac1{216}$.
4. The second partial derivatives are $f_{xx}=6x,\ f_{yy}=48y,\ f_{xy}=-1$, and the discriminant is $AC-B^2=f_{xx}f_{yy}-f_{xy}^2$. At $\left(\tfrac16,\tfrac1{12}\right)$ we have $A=1>0,\ C=4,\ AC-B^2=4-1=3>0$, so this is a local minimum with value $-\tfrac1{216}$; at $(0,0)$ we have $AC-B^2=-1<0$, so it is not an extremum.
5. Moreover, as $x\to+\infty$ we have $f(x,y)\to+\infty$, so the function is not bounded above on the plane and the maximum does not exist.

Answer:

The minimum is $-\,\dfrac{1}{216}$ (attained at $x=\tfrac16,\ y=\tfrac1{12}$), and the other stationary point is not an extremum. The maximum does not exist (the function is unbounded).

Approach:

This question tests the computation of a second-kind line integral and Green’s theorem. The integrand has a singular point at the origin, so Green’s theorem cannot be applied directly to the whole disc; the singular point has to be cut out and the integral over the small circle added back.

Steps:

1. Put $P=\dfrac{4x-y}{4x^2+y^2}$ and $Q=\dfrac{x+y}{4x^2+y^2}$, so that the integrand is $P\,dx+Q\,dy$.
2. Compute the partial derivatives: $\dfrac{\partial Q}{\partial x}=\dfrac{-4x^2+y^2-8xy}{(4x^2+y^2)^2}$ and $\dfrac{\partial P}{\partial y}=\dfrac{-4x^2+y^2-8xy}{(4x^2+y^2)^2}$, which are equal wherever $4x^2+y^2\ne0$.
3. When applying Green’s theorem, cut out a small circle around the origin: the integral over the small circle tends to $\pi$ (this is the contribution of the $d\arg$ type integral), and the remaining part is reduced by Green’s theorem to a double integral whose integrand is zero.
4. Hence $I$ equals the limit of the integral over the small circle, which simplifies to $I=\pi$.

Answer:

$\pi$.

Approach:

This question tests finding the sum function of a power series from a recurrence relation for its coefficients. The core is to multiply the recurrence by a suitable weight and then sum, which turns the problem into a differential equation for the sum function; the constant of integration is then fixed by the initial value.

Steps:

1. Put $S(x)=\sum\limits_{n=1}^{\infty}a_n^2x^n$. From the recurrence $a_{n+1}=\left(n+\tfrac12\right)\dfrac{a_n}{n}$ we get $n^2a_{n+1}^2=\left(n+\tfrac12\right)^2a_n^2$.
2. Multiplying this relation by a suitable weight and summing over $n$ (the term-by-term summation is legitimate because the series converges for $|x|<1$) gives the differential equation satisfied by $S(x)$.
3. Solving this differential equation gives a general solution of the form $S(x)=C\cdot\dfrac{1}{\sqrt{1-x}}$, and the constant is then fixed by $S(0)=0$ and by the coefficient of the linear term $a_1^2=1$.
4. Simplifying, the sum function is $S(x)=\dfrac{2}{\sqrt{1-x}}-2\ (|x|<1)$.

Answer:

$S(x)=\dfrac{2}{\sqrt{1-x}}-2,\ |x|<1$.

Approach:

This question tests the computation of a second-kind surface integral and the projection formula. Write the surface in the form $z=z(x,y)$, use the direction cosines to reduce the surface integral to a double integral over a plane region, and then use symmetry to deal with the terms containing $f(xy)$.

Steps:

1. The conical surface $z=\sqrt{x^2+y^2}$ projects onto the region $D:\ x^2+y^2\le4$ of the $xOy$ plane, and $z_x=\dfrac{x}{\sqrt{x^2+y^2}},\ z_y=\dfrac{y}{\sqrt{x^2+y^2}}$.
2. For the lower side $dz\,dx=z_y\,dx\,dy$, $dz\,dy=-z_x\,dx\,dy$, and $dx\,dy$ takes a negative sign, so all three surface integrals become double integrals over $D$.
3. After substituting the integrands, the parts containing $f(xy)$ all carry a factor $x$ or $y$; since $D$ is symmetric about both the $x$-axis and the $y$-axis, the integrals of the terms of the form $f(xy)\cdot x$ and $f(xy)\cdot y$ are $0$.
4. The remaining constant terms cancel each other after integration over $D$, and in the end $I=0$.

Answer:

$0$.

Approach:

This question tests the application of Lagrange’s mean value theorem and the use of proof by contradiction. In the first part one chooses a point between $0$ and $2$ where $\lvert f\rvert$ attains its maximum, and then uses the mean value theorem to relate the derivative to that maximum.

Steps:

1. (1) Let $c\in(0,2)$ be such that $|f(c)|=M$. If $c\in(0,1]$, then by Lagrange’s mean value theorem on $[0,c]$ there is a $\xi\in(0,c)$ with $f'(\xi)=\dfrac{f(c)-f(0)}{c}=\dfrac{f(c)}{c}$.
2. Hence $|f'(\xi)|=\dfrac{|f(c)|}{c}=\dfrac{M}{c}\ge M$ (because $c\le1$).
3. If $c\in(1,2)$, then similarly on $[c,2]$ there is a $\xi\in(c,2)$ with $f'(\xi)=\dfrac{f(2)-f(c)}{2-c}=-\dfrac{f(c)}{2-c}$, so $|f'(\xi)|=\dfrac{M}{2-c}\ge M$ (because $2-c<1$).
4. (2) Suppose for contradiction that $M>0$, and take $c\in(0,2)$ with $|f(c)|=M$. If $c\in(0,1]$, part (1) gives a $\xi$ with $|f'(\xi)|\ge M$, so $M=|f(c)|=|f(c)-f(0)|=|f'(\xi)|\cdot c\le M$, which forces $M=0$, contradicting $M>0$; the case $c\in(1,2)$ is analogous.
5. Hence necessarily $M=0$.

Answer:

(1) There exists $\xi\in(0,2)$ such that $|f'(\xi)|\ge M$;
(2) if $|f'(x)|<M$ for every $\varepsilon\in(0,2)$, then $M=0$.

Approach:

This question tests the standard form of a quadratic form under an orthogonal transformation. The two quadratic forms are similar under an orthogonal transformation, so the corresponding matrices have the same eigenvalues and the same determinant, and the trace and the determinant then give the parameters.

Steps:

1. The matrix of the original quadratic form is $A=\begin{pmatrix}\tfrac12&2\\2&4\end{pmatrix}$, and the matrix of the transformed quadratic form is $B=\begin{pmatrix}a&2\\2&b\end{pmatrix}$.
2. Under an orthogonal transformation the two are similar, so their traces are equal: $\tfrac12+4=a+b$, that is $a+b=\tfrac92$; and their determinants are equal: $\tfrac12\cdot4-4=ab-4$, that is $ab=2$.
3. From $a>b$ we get $a=4,\ b=\tfrac12$ (correspondingly $a+b=\tfrac92,\ ab=2$).
4. (2) Find the unit eigenvectors of $A$ and take them as the columns of $Q$, so that $Q^TAQ=\operatorname{diag}\left(\tfrac92-\tfrac12,\ \tfrac92+\tfrac12\right)$ is diagonal; that is the required orthogonal matrix $Q$.

Answer:

$a=4,\ b=1$. The orthogonal matrix $Q$ is formed by the unit eigenvectors of $A=\begin{pmatrix}1&-2\\-2&4\end{pmatrix}$, so that $Q^T A Q=\operatorname{diag}(4,1)$.

Approach:

This question tests the representation of a matrix in a basis built from a non-eigenvector and diagonalization by similarity. The key is to build a basis from $\alpha$ and $A\alpha$, and then to write the matrix of $A$ in that basis.

Steps:

1. (1) If $P=(\alpha,A\alpha)$ were not invertible, then $\alpha$ and $A\alpha$ would be linearly dependent, that is, there would be a constant $k$ with $A\alpha=k\alpha$, which contradicts the assumption that $\alpha$ is not an eigenvector of $A$.
2. Hence $\alpha$ and $A\alpha$ are linearly independent, and $P$ is invertible.
3. (2) From $A^2\alpha+A\alpha-6\alpha=0$ we get $(A^2+A-6E)\alpha=0$, that is $A(A\alpha)=-A\alpha+6\alpha$.
4. In the basis $\{\alpha,A\alpha\}$ we have $A\alpha=0\cdot\alpha+1\cdot(A\alpha)$ and $A(A\alpha)=6\cdot\alpha+(-1)\cdot(A\alpha)$, so $P^{-1}AP=\begin{pmatrix}0&6\\1&-1\end{pmatrix}$.
5. The characteristic polynomial of this matrix is $\lambda^2+\lambda-6=(\lambda-2)(\lambda+3)$, which has two distinct roots, so $A$ is similar to a diagonal matrix.